CBSE Class 11 Chemistry Notes For Chapter 12 Fundamental Concepts Of Organic Reaction Mechanism

Fundamental Concepts Of Organic Reaction Mechanism

In an organic reaction, the organic molecule (called the substrate) reacts with a suitable attacking species (called the reagent) to form products. The formation of produces) may take place either directly from the reactants (i.e., substrate and reagent) through a transition state or the formation of one or more intermediates. Some by-products may also be formed from the intermediate

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Intermediates

The reagents are mostly either positively or negatively charged. A positively charged reagent attacks the site which is rich in electrons while a negatively charged reagent attacks that site which is electron deficient. So, for the reaction to take place in the covalent bond of the substrate, the bond must have some degree of ionic character.

Although the bond of an organic compound is mainly covalent, that bond becomes partially ionic due to some permanent or temporary displacement of the bonding electrons.

The ionic nature of a bond may be attributed to the following reasons:

  1. Inductive effect
  2. Electromeric effect
  3. Resonance
  4. Hyperconjugation.

Besides these electronic effects, steric effect or steric hindrance plays a very important role in determining the reactivity of organic compounds

1. Inductive effect

Inductive effect Definition:

The permanent displacement of electrons along a carbon chain which occurs when some atom or group, either more or less electronegative than carbon is attached to the carbon chain is called the inductive effect

Inductive effect When a covalent bond is formed between two atoms having different electronegativities, the bonding electron pair is not shared equally by the two atoms. The electron pair being attracted by the more electronegative atom gets shifted more towards it. Consequently, the more electronegative atom acquires a partial negative charge (i.e., 6- ) and the less electronegative atom acquires a partial positive charge (Le., d+).

Example:

When an electronegative (electron-withdrawing) Clatom (or a group such as — NO2 ) is attached to the end of a carbon chain (whose carbon atoms are designated as 1, 2, 3,… etc.), the cr -electrons ofthe Cx —Cl bond are attracted by or displaced more towards the Cl-atom. As a result, the Cl-atom acquires a partial negative charge (δ-) and the carbon, C1, acquires a partial positive charge (δ+). As Cj is now somewhat positively charged, it in turn, attracts the cr -electrons of the C1-C2 bond towards it So, C2 acquires a partial positive charge (SS+) smaller in magnitude than that on C1. Similarly, C3 acquire a partial positive charge (δδδ+) even smaller than that on C2

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Positive Charge

Similarly, if an element less electronegative than carbon, such as lithium (Li) (or any other electron-releasing group or atom), is attached to the terminal C -atom bf a carbon chain, then a partial positive charge (<5+) is developed on the Li-atom and a partial negative charge (6-) is developed on the Cx -atom.

The small negative charge on C1, in turn, repels the cr -electrons of the C1—C2 bond towards C2. As a result, C2 acquires a partial negative charge (δδ-) smaller in magnitude than that of C1 Similarly, C3 acquires a partial negative charge (δδδ-) which is even smaller than that of C2

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Negative Charge

Inductive effect Characteristics:

  • This type of charge dispersal, which diminishes rapidly as the distance from the source increases, almost becomes negligible after the third carbon atom and is ignored.
  • It is to be noted that although the inductive effect causes some degree of polarity in covalent bonds, the bond is never cleaved due to the effect.
  • The inductive effect is represented by the symbol ( —<—). The arrow always points towards the more electronegative atom or group.

Measurement of inductive effect:

The inductive effect is always transmitted along a chain of carbon atoms. It cannot be expressed by any absolute value. The relative inductive effect of an atom or group is measured by taking H -atom of the R3C — H molecule as standard. When an atom or group- Z of the C — Z bond of the R3C —Z molecule attracts the bonding electrons more strongly than hydrogen of the C—H bond in the R3C— H molecule.

Then according to the definition introduced by Ingold, Z is said to have a negative inductive effect or electron-withdrawing inductive effect or -I effect. On the other hand, if the atom or group Z attracts the bonding electrons of the C — Z bond less strongly than the hydrogen atom of the C — H bond, then it is said to have a positive inductive effect or electron-releasing inductive effect or +1 effect

+I effect:  —NH > —O > — COO > (CH3)3C— > (CH3)2CH — > CH3CH2 — > CH3 — > D

– I effect:+NR3> — +SR2> — +NH3 > — NO2 > -SO2R > — CN > — COOH > — F > — Cl > — Br > — I > — OR > — OH

Impact of inductive effect and its explanation:

Some important properties of organic compounds such as acidic property, basic property, bond polarity, and chemical reactivity vary remarkably due to inductive effect. Some examples of the influence of + 1 and – I effects on the properties of organic compounds are discussed here

1. Strength of monocarboxylic acids

Rule 1:

The relative strengths of monocarboxylic acids can be explained by the inductive effect ofthe substituent present in the carbon chain. If an electron releasing group or atom is attached directly to the -COOH group or to the carbon chain close to the -COOH group, then the positive inductive effect (+1) of such group increases the electron density on the oxygen atom of the

O— H group and consequently, the shared pair of electrons of the O — H bond is less strongly attracted towards the oxygen atom. As a result, the dissociation of the O —H bond to give H+ ion is less favoured. Thus, a group having a + I effect, when present in a monocarboxylic acid molecule decreases the strength of that acid.

On the other hand, if an electron-withdrawing group or atom is attached to the carbon chain close to the —COOH group, the negative inductive effect (-1) of such group decreases the electron density on the oxygen atom of the O— H group and consequently, the shared pair of electrons of O —H bond are more strongly attracted towards the oxygen atom. As a result, the dissociation of the O—H bond to give H+ ion is facilitated. Thus, a group having – I effect, when present in a monocarboxylic acid molecule, increases the strength of that acid molecule.

Example: Chloroacetic acid is stronger than acetic acid

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Chloroacetic Acid

Rule 2:

The strength of carboxylic acid increases as the extent of the effect ofthe substituent increases.

Example:

-I effect of the halogens follows the order: fluorine > chlorine > bromine > iodine. So, among the halogen-substituted acetic acids, trifluoroacetic acid (FCH2COOH) is the strongest while iodoacetic acid (ICH2COOH) is the weakest

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Carboxylic Acid

Rule 3:

With the increase in the number of electron-attracting substituents, the strength of the acid increases.

Example:

Dichloroacetic acid is stronger than monochloroacetic acid while trichloroacetic acid is stronger than dichloroacetic acid.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Acid Increases

Rule 4:

As the distance ofthe electron attracting substituent from the carboxyl group increases, the strength of the acid decreases.

Example:

2-chlorobutanoic acid is a stronger acid than 3- chlorobutanoic acid which in turn is stronger than 4- chlorobutanoic acid

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Chlorobutanoic Acid

Acid strength can also be explained in terms of the relative stabilities of the acid and its conjugate base. Electron withdrawing groups (EWG) disperse the negative charge on the anion [i.e., conjugate base), thus stabilising it and hence increasing acidity. On the contrary, electron-donating groups EDG intensify the negative charge on the anion, thus destabilise it and hence decrease acidity

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques EWG Stabilise

2. The Basic Strength Of Amines

The increase in the strength of nitrogenous bases, e.g., amines, is related to the readiness with which they are prepared to take up protons and therefore, to the availability of the unshared pair ofelectrons on nitrogen.

Example:

We might expect the order of basic strength:

NH3<CH3NH2<(CH3)2NH<(CH3)3N> due to the increasing inductive effect (+1) of successive —CH3 groups making the N -atom more negatively charged, i.e., making the unshared pair of electrons more readily available. However, this sequence of basic strength of amines agrees with the results if measurements of basicity are made in the gas phase or in a solvent in which H -H-bonding does not take place

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Technique Amines

Basic strength increases (in the gas phase or in a solvent which does not form H-bond with amines). The introduction of electron-withdrawing groups close to the basic centre causes a decrease in F3Cthe basicity, due to their electron- Tri-trifluoromethylamine (virtually non-basic) withdrawing inductive effect.

An interesting example is the amine, (CF3)3N which is found to be virtually non-basic, due to the presence of three powerful electron-withdrawing — CF3 groups, each of which contains three highly electronegative F -atoms.

CBSE Class 11 Chemistry Notes For Chapter 12Organic Chemistry Basic Principles And Techniques Tri Trifluoromethylamine

The order of basic strength of amines in aqueous medium is:

⇒ \(\left(\mathrm{CH}_3\right)_2 \ddot{\mathrm{N}} \mathrm{H}\left(2^{\circ}\right)>\mathrm{CH}_3 \ddot{\mathrm{N}} \mathrm{H}_2\left(1^{\circ}\right)>\left(\mathrm{CH}_3\right)_3 \ddot{\mathrm{N}}\left(3^{\circ}\right)\)

Due to the combined effect of hydrogen bonding and +1 effect of— CH3, groups, the conjugate acid of (CH3)2 NH i.e.,   (CH3)2 +NH2 is the most stable while the conjugate acid of (CH3)3N, i.e., (CH3)3 NH is the least stable and for this reason, in the aqueous medium, the above order of basicity  observed

2.  Electromeric effect

The complete transfer of a pair of 7t -electrons of a multiple bond (double bonds such as C=C, C=0 and triple bonds such as C = C and G= N ) to one of the multiple bonded atoms (usually the more electronegative one) in the presence of an attacking reagent is called electromeric effect or E-effect.

The transfer of the electron pair is indicated by a curved arrow. As soon as the reagent is removed, this effect vanishes and the molecule reverts back to its original position. Since this effect occurs by the presence ofthe attacking reagent, it takes place in the direction which facilitates the reaction. The electromeric effect may be represented as follows

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Electromeric Effect

Types of electromeric effect:

+ E-effect:

If the electron pair of the -bond is transferred to that doubly bonded atom to which the attacking species gets finally attached, then the effect is called +E-effect

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Positive E Effect

– E-effect:

If the electron pair of the π-bond is transferred to that doubly bonded atom to which the attacking species Mous Lewis structures, which differ in the do not get finally attached, the effect is called -E-effect

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Negetive E Effect

The distinction between inductive and electromeric effect:

Organic Chemistry Basic Principles And Techniques The Distinction Between Inductive And Electromeric Effect

3. Resonance

Resonance Definition:

Various Lewis structures, which differ in the positions of non-bonding or ;r -electrons but not in the relative positions of atoms, are called resonance structures, contributing structures or canonical forms. This concept is known as resonance

There are some molecules or ions which cannot be represented adequately by a single electronic (Lewis) structure as all the properties of such molecules or ions do not correspond to a single Lewis structure. In such cases, it becomes necessary to represent the molecule or ion by writing two or more Lewis structures which differ in the arrangements of valence electrons but the basic structure involving cr -bonds remains the same.

It should be remembered that resonance is not a phenomenon, because there is no real existence of different resonance structures. These structures are all imaginary and are taken into consideration to explain the different physical and chemical properties of molecules or ions. The actual structure of the molecule or the ion lies in between these structures. We say that the actual molecule or ion is a resonance hybrid (weighted average) of all these resonance structures.

Resonance is also known as mesomerism. The various resonance structures are connected by double-headed arrows. They contribute to the actual structure in proportion to their stability. The magnitude of internal energy of the resonance hybrid of a molecule or ion is less than that of any resonating structure. Thus the molecule or ion gets stabilised by resonance

Examples:

1. Benzene molecule can be represented as a resonance ~ hybrid (III) of two Kekule structures:

I and II. Neither ofthe two structures can fully explain all the properties of benzene. For example, both structures I and II contain two types of carbon-carbon bonds such as C—C (1.54 Å) and C = C (1.34 Å). But actually, it has been found that all the 6 carbon-carbon bonds in benzene are of equal length (1.39 Å).

This suggests that the actual structure of benzene can neither be represented by I nor by II but by a resonance hybrid of these two structures in which all the six carbon-carbon bonds are of equal length and lie in between carbon-carbon single bond length of 1.54 Å and carbon-carbon double bond length of 1.34 Å. So, benzene is quite often represented by the non-Lewis structure III. The circle inside the ring indicates completely delocalised 6 n -electrons. Since I and II are exactly equivalent, They are of the same stability and make equal contributions to the hybrid

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Kekule Structures

2. Carbonate ion (CO32-) may be represented as a resonance hybrid ofthe following three structures:

IV, V and VI. None of the three structures can individually explain all the properties of carbonate ions. For example, in all three structures, the carbon-oxygen single bond (1.43Å) and carbon-oxygen double bond (1.20 Å) are present.

However, it has been found experimentally that all the carbon-oxygen bonds in carbonate ion are equal in length (1.28 A) and this bond length is slightly greater than that of the double bond but less than that of the single bond. All the three carbon-oxygen bonds are equivalent. So, the structural formula of the carbonate ion denotes a state equidistant from the three structures IV, V and VI and it is frequently expressed by the non-Lewis structure VII.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Lewis Structure

Rules for writing meaningful resonance structures:

The following rules are to be followed while writing realistic resonance structures:

  • The various resonance structures should differ only in the positions of electrons and not in the positions of atoms, Le., the basic structure involving cr -bonds should remain undisturbed.
  • The number of paired and unpaired electrons in each resonance structure must be the same.
  • All the atoms involved in the process of resonance must be coplanar (or nearly coplanar).
  • All the resonance structures should have nearly the same energy.
  • Each resonance structure must be a bona fide Lewis structure, i.e., all atoms in a resonance structure must exhibit proper valencies.
  • For example: There must not be any structure with pentavalent carbon, pentavalent nitrogen, bivalent hydrogen and so on.

Resonance energy:

The difference in internal energy between the actual molecule (observed value) and that of the resonance structure having the lowest internal energy or highest stability (obtained by calculation) is called resonance energy. The resonance energy is greater when

  • The contributing structures are all equivalent and
  • The number of contributing structures of roughly comparable energy is greater.

Calculation of resonance energy:

Resonance energy is not a measurable quantity. It can only be estimated from thermochemical data. If the theoretically calculated internal energy of a gram-mole of the most stable resonance structure is EC and the experimentally determined internal energy of the actual molecule (resonance hybrid) is EO, then the resonance energy ER = EC-EO. Resonance energy is expressed in kcal mol-1 or kj mol-1.

The greater the resonance energy, more the stability of the compound. The resonance energy becomes maximum when the contributing structures are equivalent, i.e., have equal energy content. Also, the more the number of resonance structures having a large contribution, the greater will be the resonance energy.

In determining the relative stabilities of similar molecules or ions, resonance energy is an important factor among various other factors like bond energy, internal strain etc.

Example:

The resonance energy of benzene can be calculated from the heat of hydrogenation values. The heat of hydrogenation is the quantity of heat evolved when the mol of an unsaturated compound is hydrogenated. Cyclohexene containing 1 double bond has a heat of hydrogenation of 28.6 kcal mol-1.

We might reasonably expect 1,3,5-cyclohexatriene to have a heat of hydrogenation of about three times as large as cyclohexene, i.e.,3 × 28.6 = 05.8 kcal .mol-1.

The value for benzene is 49.8 kcal .mol-1. It is 36 kcal .mol-1 less than the expected value. So, benzene evolves 36 kcal less energy per mole than predicted. This can only mean that benzene is more stable than hypothetical cyclohexadiene by 36 kcal .mol-1 energy. This 36 kcal. mol-1 energy is the resonance energy of benzene.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Resonance Of Benzene

Relative contributions of resonance structures towards resonance hybrid:

All resonance structures do not contribute equally towards resonance hybrid. Relative contributions of resonance structures towards resonance hybrid depend on their relative stabilities. The more stable the resonance structure, the more will be its contribution to resonance hybrid.

Factors that govern the stability of a resonance structure and its relative contribution towards hybrid are :

Rule 1:

Non-polar resonance structures, being more stable than the dipolar resonance structures, contribute more towards the resonance hybrid.

Example:

In the following alkadiene, the first resonance structure is more stable and thus contributes more than the second dipolar resonance structure

Organic Chemistry Basic Principles And Techniques Second Dipolar Reasonance Structure

Rule 2:

Resonance structures with a greater number of covalent bonds are more stable and contribute more towards the resonance hybrid.

Example:

In the following acyl cation, the second resonance structure having a greater number of covalent bonds is more stable and more contributing.

Organic Chemistry Basic Principles And Techniques Covalent Bonds Is More Stable

Rule 3:

In case of anions, the most stable structure is the one in which the negative charge resides on the most electronegative the one in which the positive charge resides on the least electronegative atom. So, these structures are more contributing.

Example:

Resonance structures II and IV of the following anion and cation are relatively more stable hence more contributing towards their respective resonance hybrids.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Reasonance Hybrids

Rule 4:

Canonical structures in which octets of all the atoms are fulfilled are relatively more stable and therefore, make a larger contribution towards the resonance hybrid.

Example:

The second resonance structure of the following acylium ion is more stable and more contributing because all the atoms have octets of electrons in their valence shells (except H which has a duplet).

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Second Reasonace

Rule 5:

Aromatic resonance structures are more stable and more contributing than the non-aromatic resonance structures having the same number of covalent bonds. Example: The aromatic resonance structure I of benzyl cation is more stable and more contributing than the non-aromatic resonance structure II.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Non Aromatic

Rule 6:

A resonance structure having two units of charge on the same atom is not stable and hence it has a very poor contribution. Again, structures having like charges on adjacent atoms are highly unstable and hence it has a negligible contribution. On the other hand, a resonance structure having two dissimilar charges close to each other is relatively more stable and more contributing than the structure in which the charges are further apart.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Contributing

Rule 7:

The resonance energy of a system involving monopolar resonance structures is greater than that involving dipolar resonance structures. So the former type of systems (i.e., molecules or ions) are more stable than the latter type.

Example:

Carboxylate ion is more stable (in fact more stabilised by resonance) than the corresponding carboxylic acid

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Polar Structures

Some facts about resonance structures:

  • Resonance structures are not real.
  • Resonance structures are not in equilibrium with each other.
  • Resonance structures are not isomers because the two isomers differ in the arrangement of both atoms and electrons, whereas resonance structures differ only in the arrangement ofelectrons

Effect of resonance on the properties of molecules:

The following properties of different molecules or ions can be explained by resonance

1. Bond length

Because of resonance, the single bond present in a molecule or ion may acquire a partial double bond character with a consequent decrease in bond length. Similarly, the double bond may acquire some single bond character with a consequent increase in bond length.

Example: Due to resonance, the C — Cl bond (1.69 Å) of vinyl chloride (CH2 = CHCl) becomes shorter than C — Cl bond (1.76 Å) of ethyl chloride and its C=C bond (1.38 Å) becomes longer than the C=C bond (1.34 Å) of ethene

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Vinyl Chloride

2. Dipole moment

As a result of resonance, both the magnitude ofthe charge separated (e) and the distance between two charged centres (d) in any molecule may increase. So, the value of dipole moment (p = e × d) increases.

Example:

Due to resonance, the amount of charge separated and the distance between the centres of charges in nitroethane (CH2=CHNO2) is greater as compared to nitroethane (CH3CH2NO2). Consequently, the dipole moment <p) nitroethane is greater than that of nitroethane.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Dipole Moment.

3. Acidity And Basicity Of Organic Compounds

Acidic character of phenol:

The greater the ease with which a compound releases proton (H+) in its aqueous solution, the stronger it will be as an acid. That phenol is acidic and is a stronger acid than alcohol can be well explained in terms of resonance. Phenol in its aqueous solution ionises to produce phenoxide ions as follows

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Phenoxide Ion

Phenol can be represented as a resonance hybrid of the following resonance structures (I -V):

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Resonance Structure

Due to the contribution ofthe resonance structures n, in an IV the O -atom becomes positively charged. Because of this, the polarity of the O—H bond increases and hence the tendency of O—H bond fission (to release a proton) also increases. On the other hand, no such resonance is possible in a molecule of alcohol.

So, the alcoholic O — H bond is relatively less polar and the tendency of bond cleavage resulting in proton release is indeed very small. Hence, phenol is more acidic than alcohol.

Alternative explanation:

This relative acidity may also be explained by considering the phenol-phenoxide ion and Alcohol-alkoxide ion equilibria. Like phenol, phenoxide ion may also be represented as a resonance hybrid of the following (VI – X) resonance structures:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Technique Structures Of Phenoxide

Three (II, III and IV) out of five resonance structures of phenol involve charge separation, but the resonance structures of phenoxide ion involve no charge separation. The negative charge is only delocalised. Because of this, phenoxide ion is more resonance stabilised than phenol.

As a consequence, the equilibrium of phenol-phenoxide ion tends to shift towards the right, i.e., phenol exhibits acidic properties by releasing proton (H+) easily. On the other hand, both alcohol and alkoxide ions can be satisfactorily represented by single (localised) structures. Due to the absence of differential stabilisation caused by resonance, alcohol is very reluctant to produce alkoxide ions. So, phenol is a stronger acid than alcohol.

The basic character of aniline:

Aniline is a weak base and its basicity is much weaker than aliphatic amines (RNH2). This can be explained by resonance. It can be represented as a resonance hybrid ofthe following resonance structures:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Aniline

1. An unshared pair of electrons on the N-atom of aniline becomes involved in resonance interaction with the ring. As a result, N-atom acquires a partial positive charge. Consequently, aniline exhibits little tendency to take up a proton. So aniline is a weak bases

2. However, in the case of aliphatic amines, similar delocalisation of electrons by resonance is not possible, Naturally the electron density on N-atom is not reduced. In fact, due to the +1 effect of the alkyl group (R-), the electron density on Natom is somewhat increased.

As a consequence, nitrogen can easily donate its electron pair to a proton (H+) to combine with it. Thus, aniline (C6H5NH2) is a weaker base than aliphatic amines (RNH2). Apart from this, relative basicity can also be explained by considering aniline-anilinium ion and amine ammonium ion equilibrium systems.

Aniline is a resonance hybrid of five resonance structures (I-V).In the conjugate acid anilinium ion, the lone pair of electrons on the N atom is localised in the N —H bond and so, only two structures (VI and VII) can be drawn for its hybrid. Therefore, aniline is more resonance-stabilised concerning the anilinium ion. As a result of this, protonation of aniline is disfavoured.

None of the aliphatic amine and its conjugate acid can be stabilised by resonance. The conjugate acid is stabilised by the weak +1 effect ofthe -R group. Protonation ofthe aliphatic amine is, therefore, not disfavoured and is is somewhat favoured. Thus, aromatic amines are weaker bases than aliphatic amines.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Aliphatic Amines

Finally, in aromatic amine, the amino group is attached to sp² -carbon (more electronegative), whereas in aliphatic amine, it is attached to sp³ -carbon (less electronegative). This factor is also partly responsible for decreased basicity of aromatic amines than aliphatic amines.

Resonance effect or mesomeric effect

Resonance effect or mesomeric effect Definition:

The displacement of non-bonding or electrons from one part of a conjugated system (having alternate single and double bonds) to the other part causing permanent polarity in the system (creating centres of high and low electron density) is called resonance effect (R-effect) or mesomeric effect (M-effect)

There are two types of resonance or mesomeric effect: 

1. +R or +M-effect:

An atom or a group is said to have a +R or +M effect if it involves the transfer of electrons away from the atom or the substituent group attached to a double bond or a conjugated system

+ R or +M groups: —OH, —OR, —SH , — NH2, —Cl, — Br, —I etc.

Example: +M effect of Cl -atom in vinyl chloride may be shown as follows

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Vinyl Chloride

2. — R or —M- effect:

An atom or a group is said to have a — R or — M effect if it involves the transfer of electrons towards the atom or the substituent group attached to a double bond or a conjugated system.

R or -M groups: >C= O, —CHO, —COOR, — CN, —NO2 etc.

Example: -M -effect of —CHO group in acetaldehyde may be shown as follows:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Acraldehyde

4. Hyperconjugation 

Hyperconjugation Definition:

When a carbon containing at least one H -atom is attached to multiple bonds such as C=C, C ≡ C, C= O, C ≡ N etc., the cr -electrons of the C — H bond become involved in delocalisation with the π electrons of the unsaturated system, i.e., there occurs a σ-π conjugation.

Similarly, σ-p type of conjugation may also take place when a carbon-containing at least one H-atom is attached to a carbon-containing partially filled or vacant p -p-orbital. This special type of resonance or conjugation, giving stability to the species (molecule, free radical or carbocation) is called hyperconjugation.

Hyperconjugation causes a permanent polarity in the molecule and is known as the hyperconjugation effect

Examples:

1. Incaseofpropene (CH3CH=CH2)

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Hyperconjugative

Although one C—H bond ofthe methyl group is shown to be broken in each hyperconjugative structure, H+ is never free from the rest of the molecule nor does it change its position in the molecule. However, from the point of view of apparent fission of the C —H bond, hyperconjugation is also called no-bond resonance.

2. In the case of ethyl cation, hyperconjugation may be shown as:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Ethyl Cation

3. In Case of ethyl radical:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Ethyl Radical

Conditions for effective hyperconjugation:

For effective hyperconjugation, the p-orbital concerned and the a-C —H bond, i.e., the sp³-s orbital must remain in the same plane. Orbital representations of hyperconjugation in propene and ethyl cation are shown as follows:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Hyperconjugation

Although the stability of a molecule, ion or free radical increases due to hyperconjugation, this stability is less than that contributed by resonance. After the names of the scientists who proposed this theory, hyperconjugation is also called the BakerNathan effect.

Effects of hyperconjugation:

1. Relative stabilities of alkenes:

The stabilities of alkenes can easily be explained by hyperconjugation. The greater the number of a -hydrogen atoms (II-atom present on the carbon atom attached directly to a double bonded carbon), i.e., the greater the number of hyper conjugative structures, the higher the stability of the alkene due to hyperconjugation

Example:

2-Methylpropcne [(CH)2C=CH2 having 6 hyperconjugable or-H atoms gives 6 no-bond resonance structures while the isomeric compound, 1 -butene (CH3CH2CH = CH2) having only 2 hyperconjugable a-H atoms gives only 2 no-bond resonance structures.

It thus follows that 2-methyl propyne is thermodynamically more stable than its isomer

Directive influence of alkyl groups:

The ortho- and paradirective influence of alkyl groups can be explained by hyperconjugation.

Example:

Directive influence of the —CH3 group in toluene, can be explained based on hyperconjugation as follows:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Alkyl Groups

As a result of hyperconjugation, the electron density at ortho- and para-positions increases and as a consequence, the electrophilic substitution reactions in toluene occur mainly at these two positions. It thus follows that the alkyl groups are o, p-directing.

Relative stabilities of carbocations:

Due to hyperconjugation, the C — H bonding electron pair is attracted towards the positively charged C -atom of the carbocation.

This helps in dispersing the positive charge in different parts of the alkyl group, i.c., charge delocalisation resulting in instability of the carbocation.

As the number of a-H atoms increases, the number of no-bond resonance structures of carbocation increases which enhances the extent of charge delocalisadon and consequent stabilisation.

Hence, the order of stability of ethyl, isopropyl and left-butyl cation, due to hyperconjugation is:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Carbocations

Relative stabilities of free radicals:

Because of hyperconjugation, the odd electron of a free radical undergoes delocalisation for which it becomes stabilised.

With the increase in the number of α-H atoms, the number of no-bond resonance structures of a free radical increases and as a result, delocalization of the odd electron takes place to a greater extent and the stability of free radicals also increases.

Therefore, the stability of ethyl, isopropyl and for-butyl radicals follow the order:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Tert Butyl Radicals

Bond length:

Because of hyperconjugation, the carbon-carbon single bond in propene (CH3—CH=CH2) acquires some double bond character and the carbon-carbon double bond acquires some single bond character.

As a result, the C — C bond in propene is found to be a little shorter (1.488 Å) than the normal C —C bond (1.543 Å) in ethane and the C=C bond is found to be a little longer (1.353 Å) than the normal C=C bond (1.334 Å) in ethylene

Electron-releasing power of alkyl groups attached to unsaturated systems or electron-deficient carbon atoms:

This depends on the number of or-H atoms. The methyl ( —CH3) group having three α-H atoms has the highest hyperconjugative effect while this effect is non-existent with the t-butyl group (Me3C— ) having no a-H atom. So, electron releasing power of various alkyl groups when attached to a double bond (or an electron-deficient carbon) follows the order:

CH3 →CH3CH3→ (CH3)2CH→ (CH3)3C—

This order is exactly the reverse of the order of the +I-effect of these alkyl groups

5. Steric hindrance or steric strain

Steric hindrance  Definition:

Steric hindrance or steric strain refers to repulsive interactions between non-bonded atoms or groups which arise when the atoms or groups come very close to each other.

When two non-bonded atoms in a molecule Are closer to each other than the sum of their van der Waals radii, they repel each other due to spatial -crowding.

The repulsion arises primarily due to electron-electron repulsive forces involving the non-bonded atoms. Such repulsive interactions between non-bonded atoms is known as steric hindrance or steric strain. Steric strain is responsible for decreased stability or destabilisation of molecules.

When the sheer bulk of groups at or near a reacting site of a molecule hinders or retards a reaction, it is called steric hindrance. On the other hand, if the constituent atoms or groups of a molecule or ion owing to their bulky nature require more space than what is available for them, i.e., when they are forced too close to one another, then mechanical interference forced too close to one another, mechanical interference ion is then said to be under steric strain. Steric strain makes the species unstable, i.e., its energy increases

Example:

In cis. 2-butene, two -CH3 groups lying on the same side of the double bond are quite close to each other and so they get involved in steric interaction. In trans-2- butene, two —CH3 groups lying on the opposite sides of the double bond are far apart from each other so they are not involved in steric interaction. Thus, cis-2-butene is thermodynamically less stable than Frans-2-butene

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Steric Strain

The heat of hydrogenation of cis-isomer is 28.6 kcal mol-1 and for trans-isomer is 27.6 kcal .mol-1. This observation agrees with the relative stabilities of these isomeric alkenes. Because of steric hindrance, tertbutyl chloride does not hydrolysis by an SN2 mechanism

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Steric Hindrance

Effect on Stability or reactivity:

Steric hindrance is responsible for decreasing the stability and increasing the reactivity of many compounds. Due to steric strain, resonance or delocalisation of electrons may be inhibited (Steric inhibition of resonance or SIR). Again, steric hindrance created at the reaction centre decreases the rate of that reaction or does not even allow the reaction to occur. So, steric hindrance plays a vital role in determining the reactivity of a compound

The acidic character of substituted aromatic acids, phenols and basic character of substituted aromatic amines:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Substituted Aromatic Acids

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Substituted Aromatic Acids.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Substituted Aromatic Acids..

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Substituted Aromatic Acids...

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Substituted Aromatic Acids....

CBSE Class 11 Chemistry Notes For Chapter 12 Classification Of Organic Compounds On The Basis of Functional Group

Classification Of Organic Compounds On The Basis of Functional Group

Hydrocarbon

Hydrocarbons are the binary compounds of C and H. Based on their structural formulae, they are given types

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Hydrocarbons

1. Saturated hydrocarbons or alkanes or paraffins:

The open chain hydrocarbons in which the carbon atoms (except methane, CH4, in which the single carbon is bonded to four H -atoms) of each molecule are linked mutually by single covalent σ -bonds, and rest of the valencies of C atoms are satisfied by single covalent σ -bonds with H atoms, are known as saturated hydrocarbons or alkanes.

Since these saturated hydrocarbons are quite less reactive due to absence of any functional group, they are also called paraffins (Latin: parum =little and affins = affinity or reactivity). The compounds of this class are representedby the general formula: Cn2n+ 2.

Classification of carbon and hydrogen atoms present in saturated hydrocarbons or alkanes:

The C-atoms present in an alkane molecule may be classified into four types primary (1°), secondary (2°), tertiary (3°), and quaternary (4°), C-atom as follows:

  • A C-atom attached to only one (or no other) C-atom is called primary C-atom It is designated as 1° carbon.
  • A C-atom attached to two other C-atoms is called a secondary C-atom.It is designated as 2°.
  • A C-atom attached to three other C-atoms is called a tertiary C-atom.It is designated as 3° carbon.
  • A C-atom attached to four other C-atoms is called a quaternary C-atom. It is designated as 4° carbon.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Quaternary C Atom

The hydrogen atoms may similarly be classified.

The hydrogen atoms attached to 1°, 2°, and 3° carbon atoms are called primary (1°), secondary (2°) and tertiary (3°) hydrogen atoms respectively. It is to be noted that unlike quaternary carbon atoms quaternary H atom has no existence because a quaternary carbon does not carry any hydrogen atom. The former example clearly illustrates the various types of carbon and hydrogen atoms.

2. Unsaturated hydrocarbons:

The open-chain hydrocarbons which contain at least one carbon-carbon double bond > C=C<or carbon-carbon triple (—C= C— ) bond in their molecules are called unsaturated hydrocarbons. These unsaturated hydrocarbons are further classified into two types: alkenes and alkynes.

The unsaturated hydrocarbons containing carbon-carbon double bonds are called alkenes, for example, ethylene (CH2=CH2), propylene (CH3CH=CH2) etc. A double bond is made up of one cr -bond and one n -bond. These hydrocarbons are also called olefins (Greek: olefiant = oil forming) because the lower members of this class react with chlorine to form products. The general formula of alkenes is CnH2n where n = 1, 2, 3— etc.

The unsaturated hydrocarbons containing carbon-carbon triple bonds are called alkynes, for example, acetylene (HC = CH), methyl acetylene (CH3C= CH) etc. A triple bond is made up of one σ-bond and two π-bonds. The general formula of alkynes is CnH2n-2 where n= 1, 2,3 – etc

Classification of hydrocarbon derivatives based on the functional group:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Classification Of Hydrocarbon

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Classification Of Hydrocarbon.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Classification Of Hydrocarbon..

CBSE Class 11 Chemistry Notes For Qualitative and Quantitative Analysis Of Organic Compounds

Qualitative Analysis Of Organic Compounds

Since organic compounds are either hydrocarbons or their derivatives, the elements which occur in them are carbon (always present), hydrogen (nearly always present), oxygen (generally present), nitrogen, halogens, sulphur (less commonly present), phosphorus and metals (rarely present). All these elements can be detected by suitable methods.

1. Detection of carbon and hydrogen(C-H)

Detection of carbon and hydrogen Principle:

A small amount of a pure and dry organic compound is strongly heated with dry cupric oxide in a hard glass test tube when carbon present in it is oxidised to carbon dioxide and hydrogen is oxidised to water.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques C And H

Liberated CO2 turns lime water milky and the liberated water vapours (H2O) which get condensed in the bulb of the delivery tube, turn white anhydrous CuSO4into blue hydrated copper sulphate (CuSO4-5H2O)

Ca(OH)2 (Lime Water)+ CO2→CaCO3↓ (Milky)+ H2O

CuSO4 Anhydrous (white)+ 5H2O→CuSO4.5H2O(Blue)

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles Detection C And H

If the organic compound is a volatile liquid or gas, then the vapours of the compound are passed through heated copper oxide and the presence of C02 and H2O in the liberated gas can be proved in the same way.

If the organic compound contains sulphur in addition to carbon and hydrogen, then sulphur is oxidised to SO2 which also turns lime water milky due to the formation of calcium sulphite. In that case, the resulting gases are first passed through an acidified solution of potassium dichromate which absorbs SO2 and then through lime water.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Lime Water

2. Detection of nitrogen

1. Sodalime test

A very small amount of an organic compound is strongly heated with soda lime (NaOH + CaO) In a test tube. The evolution of ammonia having a typical smell Indicates the presence of nitrogen in the compound.

Example:

CH3CONH2 (Acetamide)+ [NaOH + CaO]→ CH3COONa (Sodium acetate)+ NH3

 Sodalime test Limitation:

Organic compounds containing nitrogen as nitro ( — NO2) or azo (— N =N— ) groups do not evolve NH3 upon heating with soda lime, i.e., do not give this test.

2. Lassaigne’s test

Preparation of Lassaigne’s extract or sodium extract:

A pea-sized freshly cut dry sodium metal is heated gently in a fusion tube till it melts to a shining globule. A small amount ofthe organic compound is added and the tube is heated strongly for 2-3 minutes till it becomes red hot. The hot tube is then plunged into 10-15 mL of distilled water taken in a mortar. The mixture is then ground thoroughly by a pestle and filtered. The filtrate is known as sodium extract or Lassaigne’s extract.

Test for nitrogen:

The Lassaigne extract is usually alkaline in nature because the excess of metallic sodium reacts with water to form sodium hydroxide. A small amount of freshly prepared FeSO4 solution is added to a part of sodium extract and the contents are boiled when a light green precipitate of Fe(OH)2 is obtained.

The mixture is then cooled and acidified with dil.H2SO4. The immediate appearance of blue or green colouration or deep blue coloured precipitate of Prussian blue, Fe4[Fe(CN)6)3 indicates the presence of nitrogen.

Chemistry of Lassaigne’s test:

1. When organic compound is fused with metallic Na, carbon and nitrogen present in it combine with Na to form NaCN

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques S Cyanide

2. When the sodium extract is heated with FeSO4 solution, Fe(OH)2 is obtained which reacts with sodium cyanide to form sodium ferrocyanide. This is also obtained when FeSO4 reacts with NaCN.

FeSO4 + 2NaOH →Fe(OH)2 + Na2SO4

6NaCN + Fe(OH)2 →Na4 [Fe(CN)6 ] (Sodium ferrocyanide) + 2NaOH

FeSO4 + 6NaCN→Na2 SO4 + Na4 [Fe(CN)6 ]

Ferrous (Fe2+) ions present in hot alkaline solution undergo oxidation by O2 to give ferric (Fe3+) ions. These ferric ions then react with sodium ferrocyanide to produce ferric ferrocyanide (prussian blue)

4Fe(OH)2 + O2  + 2H2O→4 Fe(OH)3

2Fe(OH)3 + 3H2 SO4 →Fe2(SO4 )3 + 6H2O

3Na4 [Fe(CN)6] + 2Fe2(SO4)3→ Fe4[Fe(CN)6]3  (Prussian Blue)+ 6Na2SO4

1. In this test, it is desirable not to add FeCl3 solution because yellow FeCl3 causes Prussian blue to appear greenish. For the same reason, the alkaline extract should not be acidified with hydrochloric acid (which produces ferric chloride).

2. When the compound under investigation contains both nitrogen and sulphur, it may combine with sodium during fusion to form sodium thiocyanate (sulphocyanide). This gives blood red colouration with ferric chloride due to the formation of ferric thiocyanate. Thus, Prussian blue is not obtained.

 

⇒ \(\mathrm{Fe}^{3+}+3 \mathrm{NaSCN} \rightarrow \mathrm{Fe}(\mathrm{SCN})_3+3 \mathrm{Na}^{+}\)

3. If fusion is carried out with excess of sodium, the resulting thiocyanate decomposes to give sodium cyanide and sodium sulphide. So, in that case no blood red colouration is visualised.

NaSCN + 2Na→NaCN + Na2S

However, in that case, black precipitate of FeS is obtained when FeS04 solution is added to sodium extract. The black precipitate dissolves on the addition of dil. H2SO4 and the test of nitrogen is then performed with that clear solution.

FeSO4 + Na2S→FeS↓ (Black) + Na2SO4

FeS + H2SO4 →H2S↑+ FeSQ4

Lassaigne’s test Limitations:

  • Volatile organic compounds if present, escape before reacting when fused with metallic sodium.
  • Some nitro compounds may lead to explosion during the fusion process.

3. Detection of sulphur

1. Lassaigne’s test:

Lassaigne’s extract is prepared as described in the case of nitrogen Sulphur present in the compound (which does not contain nitrogen) reacts with sodium metal to form sodium sulphide:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques S Sulphide

The following tests are then performed with the extract to detect the presence of sulphur

Lead acetate test:

One part ofthe extract is acidified with acetic acid and then a lead acetate solution is added to it. The formation of a black precipitate of lead sulphide (PbS) confirms the presence of sulphur in the compound

Na2S + Pb(CH3COO)2→PbS↓(Black) + 2CH3COONa

Sodium nitroprusside test:

A few drops of sodium nitroprusside solution are added to another part of the extract. The appearance of a violet or purple colouration confirms the presence of sulphur in the compound.

Na2S + Na2[Fe(CN)5NO](Sodium nitroprusside)→ Sodium sulphonitroprusside Na4[Fe(CN)5NOS](Violet)

2. Oxidation test:

The organic compound is fused with a mixture of potassium nitrate and sodium carbonate and as a result, sulphur, if present, gets oxidised to sodium sulphate.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Oxidation State

The fused mass is extracted with distilled water and filtered. The filtrate is acidified with dil. HCl and then barium chloride solution is added to it. The formation of a white precipitate insoluble in hydrochloric acid indicates the presence of sulphur in the compound

Na2SO4 + BaCl2→2NaCl + BaSO4↓ (White)

4. Detection of halogens

1. Beilstein’s test:

A clean and stout Cu-wire flattened at one end is heated in the oxidising flame of a Bunsen burner until it imparts any green or bluish-green colour to the flame. The hot end of the Cu-wire is then touched with the organic compound under investigation and is once again introduced into the flame. The reappearance of green or bluish-green flame due to the formation of volatile copperhalide indicates the presence of halogens in the compound.

Beilstein’s test Limitations:

Many halogen-free compounds,

For example: 

Certain derivatives of pyridine and quinoline, purines, acid amides, urea, thiourea, cyano compounds etc. give this test presumably owing to the formation of volatile copper cyanides.

Therefore, this test is not always trustworthy. It does not indicate which halogen (Cl, Br or I) is present in the organic compound.

The test is not given by fluoro compounds since copper fluoride is non-volatile.

2.  Lassaigne’s test:

Lassaigne’s extract is prepared as described in the case of nitrogen. During fusion, Na combines with the halogen present in the compound to form sodium halide

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Sodium Halide

The extract is then boiled with dilute HNO3, cooled and a few drops of AgN03 solution are added to it White or yellow precipitation confirms the presence of halogen.

NaX + AgNO3 → AgX↓ + NaNO3 [X = Cl,Br,I]

1. Formation of a curdy white precipitate soluble in ammonium hydroxide solution indicates the presence of chlorine in the organic compound

NaCl + AgNO3→AgCU(White) + NaNO3

AgCl + 2NH4OH→[Ag(NH3)2]Cl(Soluble) + 2H2O

The formation of a pale yellow precipitate partially soluble in ammonium hydroxide solution indicates the presence of bromine in the organic compound.

NaBr + AgNO3→ AgBr4-(Pale yellow) + NaNO3

AgBr + 2NH4OH → [Ag(NH3)2]Br(Soluble) + 2H2O

The formation of a yellow precipitate insoluble in NH4OH solution indicates the presence of iodine in the organic compound.

Nal + AgNO3→AgI ↓ (Yellow) + NaNO3

Function Ol nitric acid:

If the organic compound contains nitrogen and sulphur along with halogens, Lassaigne’s extract contains sodium sulphide (Na2S) and sodium cyanide (NaCN) along with sodium halide (NaX). These will form a precipitate with silver nitrate solution and hence will interfere with the test.

Na2S + 2AgNO3→Ag2S↓  (Silver sulphide (Black) + 2NaNO3

NaCN + AgNO3→ AgCN↓   (Silver cyanide) (white)+ NaNO3

For this reason, before the addition of AgNO3 solution to the sodium extract for the detection of halogens, the extract is boiled with dilute HNO3 which decomposes sodium sulphide and sodium cyanide to vapours of H2S and HCN respectively

Alternatively, sulphide and cyanide ions can be removed by adding 5% nickel (II) nitrate solution which reÿct? with these ions forming precipitates of nickel (II) sulphide and nickel (II) cyanide

3. Chlorine water test for bromine and iodine:

  • A portion of Lassaigne’s extract is boiled with dilute nitric acid or dilute sulphuric acid to decompose NaCN and Na2S.
  • The solution is then cooled, and acidified with dil. H2SO4 and a few drops of carbon disulphide or carbon tetrachloride solution are added to it.
  • The resulting mixture is then shaken with a few drops of freshly prepared chlorine water and allowed to stand undisturbed for some time.

An orange or brown colouration in the carbon disulphide or carbon tetrachloride layer confirms the presence of bromine, whereas a violet colouration in the layer confirms the presence of iodine in the compound

2NaBr + Cl2→2NaCl + Br2 (turns CS2 or CCl4 layer orange)

2NaI + Cl2→2NaCl + I2 (turns CS2 or CCl4 layer violet)

5. Detection of phosphorus

1. The organic compound under investigation is fused with sodium peroxide (an oxidising agent) when phosphorus is oxidised to sodium phosphate

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Sodium Phosphate

2. The fused mass is then extracted with water, filtered and the filtrate is boiled with concentrated nitric acid. The mixture is then cooled and an excess of ammonium molybdate solution is added to it

3. The appearance of a yellow precipitate or colouration due to the formation of ammonium phosphomolybdate, (NH4 )3 PO4 -12MoO3, confirms the presence of phosphorus in the compound.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Phosphours

Quantitative Analysis Of Organic Compounds

After detecting the presence of various elements in a particular organic compound, the next step is to determine their respective percentages. This is known as quantitative analysis.

1. Estimation of carbon and hydrogen

Both carbon and hydrogen present in an organic compound are estimated by Liebig’s method.

Liebig’s method

Liebig’s method Principle:

A pure and dry organic compound of known mass is heated strongly with pure and dry copper oxide (CuO) in an atmosphere of air or oxygen-free from carbon dioxide. Both carbon and hydrogen present in the organic compound undergo complete oxidation and get converted into carbon dioxide and water respectively

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques CO2 And H2O

COthus produced is absorbed in a previously weighed potash bulb containing a strong KOH solution while water produced is absorbed in a previously weighed U-tube containing anhydrous CaCl2. The U-tube and the bulb are weighed again and from the difference between the two weights, the amount of CO2 and HaO are determined.

Liebig’s method Procedure:

  • The apparatus for the estimation of C and H is The apparatus consists of the following units:
    • Combustion tube,
    • U-tube containing anhydrous CaCl2 and
    • a bulb containing strong KOH solution.
  • The tube is heated l strongly for 2-3 hours till the whole of the organic compound is burnt up.
  • After combustion is over, the absorption units (the Utube and the potash bulb) are disconnected and weighed separately

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Tube

Results and calculations:

Let the mass of organic compound taken = w g. The increase in the mass of the potash bulb, i.e., the mass of CO2 formed = x g.

The increase in the mass of the U-tube, i.e., the mass of water formed =y g.

Percentage ot carbon: 1 mol of CO2 (44g) contains 1 gram atom of carbon (12 g).

∴ x g of CO2 contains = \(\frac{12x}{44}\) g of C

Now, \(\frac{12x}{44}\) g carbon is present in w g organic compound.

∴ The percentage of carbon in the compound

= \(\frac{2 y \times 100}{18 w}=\frac{2}{18} \times \frac{\text { Mass of } \mathrm{H}_2 \mathrm{O} \text { formed }}{\text { Mass of the compound taken }} \times 100\)

Percentage ot hydrogen:  1 mol of water (18g) contains 2 gram-atom hydrogen (2 g).

∴  y g of H2 O contains = \(\frac{2y}{18}\) g of H

Now, \(\frac{2y}{18}\) hydrogen is present in w g organic compound.

∴ The percentage of hydrogen in the compound

Estimation of C and H by Liebig’s method is suitable for organic compounds containing C, H and O only, but it requires some modifications for compounds containing nitrogen, halogens and sulphur.

1. Compounds containing nitrogen:

During combustion, N present in the organic compound undergoes oxidation to give its oxides (NO, NOz etc.) which are

Also absorbed in an alkali solution along with CO2. Oxides of nitrogen are decomposed back to nitrogen by placing a copper gauze roll near the exit of the tube. The N2 so produced is not absorbed by the alkali solution.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Alkali

2. Compounds containing halogens:

During combustion, halogens present in the organic compound get converted into volatile copper halides which partly decompose to give free halogens. These halogens and volatile copper halides get dissolved in an alkali solution. This can be prevented by placing a silver gauze roll near the exit of the combustion tube. Halogens combine with silver to give non-volatile silver halides.

2Ag + X2 →2AgX; 2Ag + CuX2 → 2AgX + Cu

3. Compounds containing only sulphur or sulphur and halogen:

During combustion, elemental sulphur present in the organic compound is oxidised to SO2 which is absorbed in the potash bulb.

This can be prevented by placing a layer of fused lead chromate near the exit of the tube. SO2 combines with lead chromate to produce non-volatile lead sulphate.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Sulphate

Lead chromate also reacts with copper halides and free halogens to form lead halides which remain in the combustion tube

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Halides

Liebig’s method Precautions:

  • All the joints ofthe combustion must be air-tight.
  • The combustion must be free from CO2 and water vapour.
  • The airflow is controlled in such a way that only 2-3 bubbles are generated per second. A faster flow of air may lead to the formation of carbon monoxide.
  • If carbon is deposited on the surface of the combustion mbe, then oxygen instead of air is to be passed at the final stage ofthe process

Example 1: 0.90g of an organic compound on complete combustion yields 2.20 g of carbon dioxide and 0.60 g of water. Calculate the percentages of carbon and hydrogen in the compound.
Answer:

Mass of the organic compound = 0.90 g

Mass of CO2 formed = 2.2 g & mass of H2O formed = 0.6 g

Percentage of carbon: 44 g CO2 contains = 12 g carbon

0.90 g organic compound contains = \(\frac{12 \times 2.20}{44}\) g of carbon

0.20 g organic compound contains = \(\frac{12 \times 2.20}{44}\) g carbon

Percentage of carbon = \(\frac{12 \times 2.20 \times 100}{44 \times 0.90}\)

= 66.67

Percentage of hydrogen:  18 g of H2O contains = 2g of hydrogen

0.60 g of H2O contains = \(\frac{2 \times 0.60}{18}\) of hydrogen

0.90 g compound contains = \(\frac{2 \times 0.60}{18}\) g hydrogen

Percentage of hydrogen = \(\frac{2 \times 0.60 \times 100}{18 \times 0.90}\)

= 7.41

2. Estimation ot nitrogen

The following two methods are commonly used for the estimation of nitrogen in an organic compound

  1. Duma’s method
  2. Kjeldahl’s method.

1. Dumas method

Dumas method Principle:

A weighed amount of the organic compound is heated with an excess of copper oxide in an atmosphere of CO2. Carbon and hydrogen present in the compound are oxidised to CO2 and H2O respectively while nitrogen is set free as nitrogen gas (N2). If any oxide of nitrogen is formed during this process, it is reduced back to nitrogen by passing over hot reduced copper gauze

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Dumas Method

When the gaseous mixture thus obtained is passed through a 40% KOH solution taken in a Schiff’s nitrometer, all gases except nitrogen are absorbed by KOH. The volume of nitrogen collected over the KOH solution is noted and from this, the percentage of nitrogen in the compound is calculated

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Copper Gauze

Dumas method Procedure:

  • The apparatus used for the estimation of nitrogen by the Dumas method ). It consists of:
    • CO2 generator combustion tube (a glass tube of diameter 15m/r and length 90 cm) and Schiff’s nitrometer.
    • The combustion tube is packed with [a] coarse CuO which prevents backward diffusion of gases produced,
    • an accurately weighed amount of organic compound of approximately 0.2g mixed with excess of CuO.
    • Coarse CuO and
    • A reduced copper gauze can reduce any oxides of nitrogen back to N2.
  • The Schiff’s nitrometer (connected to the combustion tube) consists of a graduated tube with a reservoir and a tap at the upper end. It has a mercury seal at the bottom to check the backward flow ofthe KOH solution.
  • CO2 generated by heating NaHCO3 or MgCO3 is dried by passing through concentrated HS04 and then passed through the combustion tube to displace the air present in the tube.
  • The combustion tube is then heated in the furnace. When the combustion is complete, a rapid stream of CO2 is passed through the tube to sweep away the last traces of N2.
  • The volume of nitrogen collected over the KOH solution in the nitrometer tube is recorded after careful levelling (by making the level of KOH solution both in the tube and reservoir the same).
  • The room temperature is recorded and the vapour pressure of the KOH solution at the experimental temperature is noted in the table.

Results and calculations:

Let the mass of the organic compound taken = w g, the volume of N2 gas collected = V mL, the atmospheric pressure = P mm, the room temperature = t°C and the vapour pressure of KOH solution at t°C =f mm. Hence, the pressure of dry N2 gas = (P- f) mm.

By applying the gas equation = \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

Volume of N2 at STP, = \(V_2=\frac{P_1 V_1 T_2}{P_2 T_1}=\frac{(P-f) \times V \times 273}{760(273+t)}\)

Percentage of nitrogen: 22400 mL of N2 gas at STP will weigh = 28 g.

2 mL of N2 gas at STP will weigh = \(\frac{28 \times V_2}{22400} \mathrm{~g}\)

Percentage of nitrogen:

= \(\frac{\text { Mass of nitrogen }}{\text { Mass of the compound }} \times 100\)

= \(\frac{28}{22400} \times \frac{V_2}{w} \times 100\)

The amount of nitrogen estimated by this method is 0.2% higher than the actual quantity. This is because a fraction of the small amount of air (nitrogen) entrapped in copper oxide gets deposited in the nitrometer

Example:

In Dumasg of an method organic for compound, the estimation gave 31.7 of nitrogen, mL of nitrogen at 14°C and 758 mm pressure. Calculate the percentage of nitrogen in the compound (vapour pressure of water at 14°C = 12 mm.
Answer:

If the volume ofnitrogen at STP is V mL, then

V = \(\frac{31.7(758-12)}{(273+14)} \times \frac{273}{760}\)

= 29. 6 mL

Now, at STP 22400 mL of nitrogen will weigh = 28 g

∴ 29.6 mL nitrogen at STP will weigh = \(\) = 0.037%

∴  Percentage of nitrogen = \(=\frac{0.037 \times 100}{0.1877}\)

= 19.71

2. Kjeldahl’s method

Kjeldahl’s method Principle:

A known mass of a nitrogenous organic compound is digested (heated strongly) with concentrated H2SO4 in the presence of a little potassium sulphate (to increase the boiling point of H2SO4) and a little CuSO4 (a catalyst) when the nitrogen present in the compound is quantitatively converted into ammonium sulphate. The resulting mixture is heated with excess of NaOH solution and the ammonia evolved is passed into a known but excess volume of a standard acid (HCl or H2SO4 ).

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Technique Acid

The excess acid left after the neutralisation of ammonia is estimated by titration with some standard alkali

2NaOH + H2SO4  →Na2SO4 + 2H2O

Kjeldahl’s Method Procedure

  •  The organic compound (0.5-5 g) is heated with a cone. H2SO4 in a Kjeldahl’s flask which is a long-necked round-bottomed flask with a loose stopper. A small amount of potassium sulphate and a few drops of mercury (or a little CuS ) are added.
  • The reaction mixture is heated for 2-3 hours when carbon and hydrogen present in the compound are oxidised to CO2 and H2O respectively while nitrogen is quantitatively converted to ammonium sulphate.
  • CO2 and water vapours escape through the loose stopper while ammonium sulphate remains in the flask.
  • The Kjeldahl’s flask is then cooled and the contents of the flask are transferred to a round-bottomed flask.
  • The mixture is diluted with water and excess caustic soda solution (40%) is added.
  • The flask is then connected to a Liebig condenser through Kjeldahl’s tap. The lower end of the condenser is dipped in a known volume of standard acid (N/10 HCl or H2SO4 ) taken in a conical flask.
  • The flask is then heated. The evolved ammonia gas is absorbed in the acid solution.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Kjeldahl Method

Results and calculations:

Let the mass of the organic compound = w g, volume of acid taken = V1 mL, normality of acid solution x(N) and volume of x(N) alkali required to neutralise unused acid = V2 mL.

Now, V2 mL  × (N) alkali = V2 mL × (N) acid.

∴ The quantity of acid used for neutralising ammonia = VjmL x(N) acid solution- V2 mL x(N) acid solution = (V1– V2) mL x(N) acid solution = (V1– V2)mL x(N) ammonia solution.

Now, 1000 mL (N) ammonia solution contains 17 g of NH3 or 14 g of nitrogen.

Therefore, (V1–  V2)mL × (N)

Ammonia solution contains = \(\frac{14 \times\left(V_1-V_2\right) x}{1000}\) g of nitrogen

This amount of nitrogen is present in w g compound

%  of nitrogen = \(\mathrm{n}=\frac{14 \times\left(V_1-V_2\right) x \times 100}{1000 \times w}=\frac{1.4\left(V_1-V_2\right) x}{w}\)

= \(\frac{1.4 \times \text { Vol. of acid used } \times \text { Normality of the acid used }}{\text { Mass of the compound taken }}\)

Kjeldahl’s method  Limitations:

  • Nitrogen present in pyridine, quinoline, diazo compound, azo compound, and nitro compound, cannot be converted into ammonium salts by this method.
  • So this method does not apply to such compounds. Thus although Dumas’s method applies to all compounds, Kjeldahl’s method has limited applications.

The percentage of nitrogen estimated in this method is not correct.

Kjeldahl’s method Utility:

This experiment can be performed quite easily and quickly. So in those cases where correct estimation of nitrogen content is not necessary

For example: Fertiliser, soil, food¬ stuff etc.), this method has its application.

This method is not troublesome and hence the possibility of error can be minimised by repeating the experiment several times.

Example:  0.257 g of a nitrogenous organic compound was analysed by Kjeldahl’s method and ammonia evolved was absorbed in 50 mL of (N/10) H2SO4. The unused acid required 23.2 mL of (N/10) NaOH solution for complete neutralisation. Determine the percentage of nitrogen in the compound.
Answer:

Acid taken = 50 mL (N/10) H2SO4 solution

= 5 mL(N) H2SO4 solution

Alkali required to neutralise the excess acid

= 23.2 mL (N/10) NaOH = 2.32 mL (N) NaOH

Now, 2.32 mL(N) NaOH = 2.32 mL(N) H2SO4 solution

The acid required to neutralise the ammonia evolved

= 5 mL (N) H2SO4– 2.32 mL (N) H2SO4 solution

= 2.68 mL (N) H2SO4 s 2.68 mL (N) NH3

[v KmL(N) H2SO4S VmL(N) ammonia]

1000 mL (N) ammonia solution contains 14g nitrogen

So, 2.68mL (N) ammonia solution contains \(\frac{14 \times 2.68}{1000}\) g nitrogen

So, 0.257g compound contains = \(\frac{14 \times 2.68}{1000}\) g nitrogen

Percentage of nitrogen = \(=\frac{14 \times 2.68 \times 100}{1000 \times 0.257}\)

= 14.6

3. Estimation of halogens: Carius method

Halogens Carius method Principle:

An organic compound containing halogen of known mass is heated with fuming nitric acid and a few crystals of silver nitrate. The halogen present in the compound is converted into the corresponding silver halide (AgX). From the mass of the compound taken and that of the silver halide formed, the percentage of halogen in the compound can be calculated.

Carius method Procedure:

  • About 5 mL of fuming nitric acid and 2 g of silver nitrate crystals are placed in a hard glass tube (Carius tube) of about 50 cm in length and 2 cm in diameter.
  • A small amount of accurately weighed organic compound is taken in a small tube and the tube is placed carefully into the Carius tube in such a way that nitric acid does not enter the tube.
  • The Carius tube is then sealed. Now, it is placed in an outer iron jacket and heated in a furnace at 550-560 K for about six hours
  • As a result, C, H and S present in the compound are oxidised to CO2, H2O and H2SO4 respectively. The halogen present in the compound gets converted into silver halide which is precipitated.
  • The tube is cooled and when the sealed capillary end is heated in a burner’s flame, a small hole is formed through which the gases escape.
  • The capillary end is now cut off and the precipitate of silver halide is filtered, washed, dried and weighed

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Carius

Results & calculations:

Let, the mass of organic compound taken = w g and the mass of silver halide (AgX) formed—x g. Now, 1 mol of AgX contains 1 gram-atom of X (X = Cl, Br or I ), i.e., (108 + atomic mass of X) g of AgX contain (atomic mass of X)g of X

∴ xg AgX contain = \(\frac{\text { atomic mass of } X}{(108+\text { atomic mass of } X)}\) X JCg of X

This amount of halogen (X) is present in w g compound.

Percentage of halogen in the compound

= \(\frac{\text { atomic mass of } \mathrm{X}}{(108+\text { atomic mass of } \mathrm{X})} \times \frac{x}{w} \times 100\)

1. Percentage of chlorine (atomic mass = 35.5 )

= \(\frac{35.5}{(108+35.5)} \times \frac{x}{w} \times 100\)

= \(\frac{35.5}{143.5} \times \frac{x}{w} \times 100\)

2. Percentage of bromine (atomic mass = 80 )

= \(\frac{80}{(108+80)} \times \frac{x}{w} \times 100\)

= \(\frac{80}{188} \times \frac{x}{w} \times 100\)

3. Percentage of iodine (atomic mass = 127 )

= \(\frac{127}{(108+127)} \times \frac{x}{w} \times 100\)

= \(\frac{127}{235} \times \frac{x}{w} \times 100\)

Example: In the Carius method, 0.256 g of an organic compound containing bromine gave 0.306 g of AgBr. Find out the percentage of bromine in the compound.
Answer:

The mass of the organic compound taken = 0.256 g and the mass of AgBr formed = 0.306 g.

Now, 1 mol of AgBr = 1 gram-atom of Br

or, (108 + 80) g or 188 g of AgBr = 80 gofBr

i.e., 188 g of AgBr contains = 80 g of bromine

0.306 g of Ag Br contain \(\frac{80}{188} \times 0.306\) g of bromine

This amount of bromine is present in 0.256g compound

Percentage6 of bromine = \(\frac{80}{188} \times \frac{0.306}{0.256} \times 100\)

= 50.9

4. Estimation of sulphur: Carius method

Sulphur Carius method Principle:

When an organic compound containing sulphur is heated with fuming nitric acid in a sealed tube (Carius tube), sulphur is quantitatively oxidised to sulphuric acid. It is then precipitated as barium sulphate by adding a barium chloride solution. The precipitate is then filtered, washed, dried and weighed. From the amount of BaSO4 formed, the percentage of sulphur is calculated.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Sulphur

Resultsandcalculations: Let the mass of organic compound

= w g and the mass of barium sulphate formed = xg

Now, 1 mol of BaSO4 contains 1 gram-atom of S, i.e.,

(137 + 32 + 64) g or, 233 g BaO4 contain 32 g sulphur.

∴ x g of BaSO4 contain \(\frac{32}{233}\) × x g of sulphur

So, this amount of sulphur is present in the w g compound.

∴ Percentage of sulphur = \(\frac{32}{233} \times \frac{x}{w} \times 100\)

Example: In sulphur estimation by the Carius method, 0.79 g of an organic compound gave 1.164 g of barium sulphate. Calculate the percentage of sulphur in the compound.
Answer:

The mass of the organic compound taken = 0.79 g and the mass of BaSO4 obtained = 1.164 g.

Now, 1 mol of BaSO4 = 1 gram-atom of S or,

233 g of BaSO4= 32 g of i.e.,

233 g of BaSO4 contain = 32 g ofsulphur.

1.164 g of BaSO4 contain \(\frac{32}{233} \times 1.164\) × 1.164 g of sulphur

So, 0.79 g compound contain \(\frac{32}{233} \times 1.164\) × 1.164 g of sulphur

Percentage of sulphur =  \(\frac{32}{233} \times \frac{1.164}{0.79} \times 100\)

= 20.24

5. Estimation of phosphorus: Carius method

Phosphorus Carius method principle:

An organic compound (containing P) of known mass is heated with fuming nitric acid in a sealed tube (Carius tube). Phosphorus present in the compound is oxidised to phosphoric acid which is precipitated as ammonium phosphomolybdate by heating it with concentrated nitric acid and then adding ammonium molybdate.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Nitric Acid

Ammonium phosphomolybdate (Yellow)

The precipitate of ammonium phosphomolybdate thus obtained is filtered, washed, dried and weighed.

Alternative method:

The phosphoric acid is precipitated as magnesium ammonium phosphate (MgNH4PO4) by the addition of magnesia mixture (a solution containing MgCl2, NH4Cl and NH4OH ).

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Ammonium

The precipitate is filtered, washed, dried and then ignited to give magnesium pyrophosphate (Mg2P2O)

CBSE Class 11 Chemistry Notes For ChapOrganic Chemistry Basic Principles And Techniques Megnesium

From the mass of magnesium pyrophosphate, the percentage of phosphorus in the compound can be easily calculated.

Results and calculations:

Let the mass of the organic compound taken = wg and the mass of ammonium phosphomolybdate formed = xg.

1 mol (NH4)3PO4-12MoO3 contains 1 gram-atom of P.

or, 3(14 + 4) + 31 + 4 × 16 + 12(96 + 3 × 16) = 1877g of

(NH4)3PO4. 12MOO3 = 31 g of P, i.e., 1877 g of

(NH4)3PO4 . 12MoO3 contain 31 g of phosphorus

xg of (NH4)3PO4 -12MoO3 contain \(\)

xg of phosphorus. This amount of phosphorus is present in w g of the compound.

Percentage of phosphorus = \(\frac{31}{1877} \times \frac{x}{w} \times 100\)

= \(\frac{31}{1877} \times \frac{\text { Mass of ammonium phosphomolybdate }}{\text { Mass of the compound taken }} \times 100\)

Alternative calculation: Let the mass of Mg2P20? formed

= xg. Now, 1 mol of (Mg2P2O) contains 2 gram-atom of P

or, (24 × 2 + 31 × 2 + 16 × 7) = 222 g of (Mg2P2O)

i.e., 222g of (Mg2P2O) contains 62 g of phosphorus.

x g of (Mg2P2O)contain \(\frac{62}{222} \times x\) g of phosphorus

w g organic compound contain \(\frac{62}{222} \times x\) xg phosphorus

Percentage of phosphorus = \(\frac{62}{222} \times \frac{x}{w} \times 100\)

= \(\frac{62}{222} \times \frac{\text { Mass of } \mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

Example: 0.35 g of an organic compound containing phosphorus gave 0.65 g magnesium pyrophosphate, Mg2P2O in the Carius method. Calculate the percentage of phosphorus in the given compound.
Answer:

The mass of the organic compound taken = 0.35 g and the mass of Mg2P2O formed = 0.65 g.

Now, 1 mol of Mg2P2O contains 2 gram-atom of P.

Or, (2 ×  24 + 2 × 31 + 16 × 7) g of Mg2P2O= 2 × 31 g of P

i.e., 222 g of Mg2P2O contains 62 g of phosphorus

0.65 g Mg2P2Ocontain \(\frac{62}{222} \times 0.65\) x 0.65 g of phosphorus

This amount of P is present in a 0.35 g compound.

Percentage of phosphorus = \(\frac{62}{222} \times \frac{0.65}{0.35} \times 100\)

6. Estimation of oxygen

Percentage of oxygen = 100 – (sum of the percentages of all other elements). However, the oxygen content of a compound can be estimated directly as follows:

Oxygen Principle:

An organic compound of known mass is decomposed by heating in a stream of nitrogen gas. The mixture of the gaseous products including oxygen is passed over red-hot coke when all the oxygen combines with carbon to form carbon monoxide. The mixture is then passed through warm iodine pentoxide (I2O5 ) when CO undergoes oxidation to CO2 and iodine is liberated.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Iodine

From the knowledge of the mass of iodine liberated or C02 produced, the percentage of oxygen in the compound can be calculated easily

Results and calculations: Let, the mass of the organic compound taken = wg and the mass of CO2 formed =xg. We find that each mole of oxygen liberated from the compound will produce 2 mol of CO2

x g. of CO2 is obtained from \(\frac{32}{88} \times x=\frac{16}{44} \times\) = x x g of O2

This amount of oxygen is present in w g ofthe compound.

Percentage of oxygen = \(\frac{16}{44} \times \frac{x}{w} \times 100\)

1. Percentage of carbon (Liebig’s method)

= \(\frac{12}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

2. Percentage of carbon (Liebig’s method)

= \(\frac{2}{18} \times \frac{\text { Mass of } \mathrm{H}_2 \mathrm{O} \text { formed }}{\text { Mass of the compound taken }} \times 100\)

3. Percentage of nitrogen (Dumas method)

= \(\frac{28}{22400} \times \frac{\text { Vol. of } \mathrm{N}_2 \text { gas at STP }}{\text { Mass of the compound taken }} \times 100\)

4. Percentage of nitrogen (KjeldahTs method)

= \(\frac{28}{22400} \times \frac{\text { Vol. of } \mathrm{N}_2 \text { gas at STP }}{\text { Mass of the compound taken }} \times 100\)

5. Percentage of halogen (Carius method)

= \(\frac{\text { At. mass of } X}{(108+\text { At. mass of } X)} \times \frac{\text { Mass of AgX }}{\text { Mass of the compound taken }} \times 100\)

X = Cl(35.5) , Br(80) or 1(127)

6. Percentage of sulphur (Carius method)

= \(\frac{32}{233} \times \frac{\text { Mass of } \mathrm{BaSO}_4 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

7. Percentage of phosphorus (Carius method)

\(\frac{62}{222} \times \frac{\text { Mass of } \mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

8. Percentage of oxygen

= \(\frac{16}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of compound taken }} \times 100\).

CBSE Class 11 Chemistry Notes For Purification And Analysis Of Organic Compounds

Class 11 Chemistry Organic Chemistry Basic Principles And Techniques Organic Chemistry Purification And Analysis Of Organic Compounds Introduction

The presence of impurities even in very small amounts may sometimes result in deviation of some properties of organic compounds to a marked degree. Therefore, to characterise an organic compound thoroughly, it is essential to obtain it in the purest form. Again, an organic compound must be in a certain state of purity before it can be analysed qualitatively and quantitatively to arrive at its correct molecular formula. The organic compounds whether isolated from a natural prepared in the laboratory are mostly impure. These are generally contaminated with some other substances. A large number of methods are available for the purification of organic compounds.

Different Methods For The Purification Of Organic Compounds

Some of the important methods which are commonly employed for the purification of organic compounds are as follows:

  1. Crystallisation
  2. Sublimation,
  3. Distillation,
  4. Extraction and
  5. Chromatography.

1. Crystallisation

Crystallisation Definition: 

Crystals are the purest form of a compound having definite geometrical shapes and the process by which an impure compound is converted into its crystals is known as crystallisation.

This is one of the most commonly used methods for the purification of solid organic compounds. It is based on the difference in solubilities ofthe compound and the impurities in a suitable solvent.

A solvent is said to be the most suitable one which fulfils the conditions such as: 

  • The organic solid must dissolve in the solvent on heating and must crystallise out on cooling
  • The solvent must not react chemically with the organic compound, and
  • The impurities should not be normally dissolved in the solvent or if they dissolve, they should be soluble to such an extent that they remain in the solution, i.e., in the mother liquor.

The various solvents which are commonly used for crystallisation are water, alcohol, ether, chloroform, carbon tetrachloride, benzene, acetone, petroleum ether etc.

Crystallisation Procedure:

  • A certain amount of an impure organic compound is added to a minimum amount of a suitable solvent and the mixture is then heated to get a hotsaturated solution ofthe compound. The hot solution is then filtered to remove the insoluble impurities, if present.
  • The clear solution is then allowed to cool down undisturbed when the solid organic compound separates in the form of fine crystals. The crystals do not separate even after a long time, the inner surface of the vessel is scratched with the round end of a glass rod to facilitate crystallisation.
  • The addition of a few crystals of the pure compound to the solution may also hasten the crystallisation process. The process of inducing crystallisation by adding a few crystals of the pure compound into its saturated solution is called seeding.
  • The crystallised compound is then filtered as usual. The crystals on the filter paper are washed with a small amount of solvent to remove the impurities.
  • The compound is then pressed in between the folds of filter paper to remove water as far as practicable. It is then dried in an esteem or air oven and finally in a vacuum desiccator.

Fractional crystallisation:

This method is used for the separation of a mixture of two (or more) compounds which have unequal solubilities in a particular solvent.

Fractional crystallisation Procedure:

  • A saturated solution of the mixture of compounds is prepared in a suitable solvent by applying heat and the hot solution is then allowed to cool when the less soluble component crystallises out earlier than the more soluble component.
  • The crystals are separated by filtration.
  • The mother liquor is then concentrated and the hot solution is allowed to cool when the crystals of the more soluble second component are obtained.
  • By repeating the process, all the components of the mixture are separated.
  • It thus follows that fractional crystallisation is the process of separation of different components of a mixture by repeated crystallisation.

2. Sublimation

Sublimation Definition:

The Process of conversion of a solid into the gaseous state on heating without passing through the intervening liquid state and vice versa on cooling is called sublimation.

  • Only those substances, whose vapour pressures become equal to the atmospheric pressure much before their respective melting points, undergo ready sublimation when heated
  • This process is very useful for the separation of volatile solids which sublime on heating from the non-volatile impurities.

Sublimation Procedure:

  • The impure sample is taken in a china dish covered with perforated filter paper (or porcelain plate).
  • An inverted funnel is placed over the dish and its stem is plugged with cotton.
  • The dish is then heated gently when vapours of the volatile substance pass through perforations of the filter paper and condense on the cooler walls of the funnel leaving behind non-volatile impurities in the dish

Sublimation Applications:

Benzoic acid, camphor, naphthalene, anthracene, iodine etc. are purified by this method. In the case of other compounds like indigo which are very susceptible to thermal decomposition, sublimation is done under reduced pressure

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Sublimation

3. Distillation

Distillation Definition:

The process of conversion of a liquid into its vapours by heating followed by condensation of vapours thus produced by cooling is called distillation

  • The process of simple distillation is commonly used for the purification of liquids which do not undergo decomposition, on boiling Le., which are sufficiently stable at their boiling points and which contain non-volatile impurities.
  • Organic liquids such as benzene, ethanol, acetone, chloroform, carbon tetrachloride, toluene etc. can be purified by the process of simple distillation

Distillation Procedure:

  • The impure organic liquid is taken in a distillation flask which is fitted with a water condenser and a thermometer.
  • A receiver is attached to the lower end of the condenser.
  • One or two pieces of unglazed porcelain or glass beads are added to prevent bumping of the liquid during distillation.
  • The flack is then heated in a water bath or a sand bath (in the case of volatile and inflammable liquid) or directly (in the case of liquids having high boiling points) when the temperature rises gradually.
  • The liquid starts boiling when its vapour pressure becomes equal to the atmospheric pressure.
  • The vapours then pass through the water condenser and condense to form the liquid which is collected in the receiver
  • The non-volatile impurities are left behind in the distillation flask

Distillation:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Distillation

1. Fractional distillation:

Fractional distillation Definition:

The”distillation process in which a mixture of two or more miscible liquids having boiling points close to each other are separated is called fractional distillation

If the boiling points ofthe two liquids of a mixture are very close to each other, i.e., differ only by 10- 20K, their separation cannot be achieved by a simple distillation method. In such cases, the separation can be achieved by fractional distillation which involves repeated distillation and condensations by using a fractionating column

Fractional distillation Procedure:

  • The apparatus used is the same as in the simple distillation process except for a fractionating column.
  • When the mixture is heated, the temperature rises slowly and the mature starts boiling. The formed mainly consists of the more volatile liquid with a little of the less volatile liquid.
  • As these vapours travel up in the fractionating column, the vapours ofthe less volatile liquid condense more readily than those of the more volatile liquid.
  • Therefore, vapours rising become rich in the vapours of more volatile liquid and the liquid flowing down becomes rich in less volatile liquid.
  • This process is repeated throughout the length of the fractionating column.
  • As a consequence, the vapours which escape from the top of the column into the condenser consist of almost the more volatile liquid.
  • Thus, the distillate received contains the more volatile component almost in pure form whereas the liquid left behind in the flask is very rich in the less volatile component.
  • This process may sometimes be repeated to achieve complete separation of liquids.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Fractionating

Fractional distillation Applications:

One of the most important applications of fractional distillation is to separate crude petroleum into various fractions like gasoline, kerosene oil, diesel oil etc. 0 Fractional distillation is also used to separate methanol (b.p. 338 K) and acetone (b.p. 329 K) from pyroligeneous acid obtained by destructive distillation of wood.

2. Distillation under reduced pressure (Vacuum distillation):

The distillation process which involves the purification of high boiling liquids (which decompose at or below their boiling points) by reducing the pressure over the liquid surface is called vacuum distillation.

Some liquids which tend to decompose at or below their boiling points cannot be purified by ordinary distillation. Such liquids can be purified by distillation under reduced pressure. A liquid boils when its vapour pressure becomes equal to the external pressure.

Therefore if the pressure acting on It is reduced, the liquid boils at a lower temperature and so, its decomposition does not occur. Glycerol, for example, decomposes at its boiling point (563 K). However, if the external pressure is reduced to 12 mm, it boils at 453 K without decomposition. Some other compounds such as phenylhydrazine, diethyl malonate, ethyl acetoacetate etc. are also purified by this method.

Vacuum distillation Procedure:

  • The distillation is carried out in a two-necked flask called Claisen’s flask.”A capillary tube is fitted into one neck ofthe flask and is kept immersed in the liquid to be distilled.
  • The other neck of the flask is fitted with a thermometer. The side tube of this neck is connected to a condenser carrying a receiver at the other end.
  • The receiver is connected to a vacuum pump and a manometer.
  • The flask is usually heated in a sand or oil bath. To prevent bumping, a steady flow of air is maintained by the capillary tube with the help ofthe screw-type cock attached to it The desired pressure is maintained by using the vacuum pump

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Reduced Pressure

3. Steam distillation:

Steam distillation is applied for the separation and purification of those organic compounds which

  • Are insoluble in water
  • Are steam volatile,
  • Possess a vapour pressure of 10-15 mm Hg at 100°C and
  • Contain non-volatile organic or inorganic impurities.

Steam distillation Principle:

In steam distillation, the liquid boils when the sum of the vapour pressures of the organic liquid (pt) and that of water (p1) becomes equal to the atmospheric pressure (p), i.e.,

p = p1 + p2. Since pt is lower than p, the organic liquid boils at a temperature lower than its normal boiling point and hence its decomposition can be avoided. Thus, the principle of this method is similar to that of distillation under reduced pressure.

 Steam distillation Procedure:

  • The impure organic liquid is taken in a roundbottom flask. Steam from a steam generator is passed into the flask which is gently heated.
  • The mixture starts boiling when the combined vapour pressure becomes equal to the atmospheric pressure
  • At this temperature, the vapours of the liquid with the steam escape from the flask and after getting condensed it is collected in the receiver.
  • The distillate contains the desired organic liquid and water which can easily be separated by using a separating funnel

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Stem Distillation

4. Differential extraction

Differential extraction Method:

The method of separation of an organic compound from its aqueous solution by shaking with a suitable solvent is called differential extraction

A solid or liquid organic compound can be recovered from its aqueous solution by shaking the solution in a separating funnel with a suitable organic solvent which is insoluble in water but in which the organic compound is highly soluble. Some commonly employed solvents for extraction are ether, benzene, chloroform, carbon tetrachloride etc.

Differential extraction Procedure:

  • The aqueous solution ofthe organic compound is mixed with a small quantity of the organic solvent in a separating funnel
  • The funnel is stopped and its contents are shaken vigorously when the organic solvent dissolves out of the organic compound.
  • The separating funnel is then allowed to stand for some time when the solvent and water form two separate layers.
  • The lower aqueous layer (when the organic solvent used is ether or benzene) is run out by opening the tap of the funnel and the organic layer is collected.
  • The whole process is repeated to remove the organic compound completely from the aqueous solution.
  • The organic compound is finally recovered from the organic solvent by distilling off the latter.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Technique Extraction

5. Chromatography

Chromatography Definition:

The technique used for the separation of the components of a mixture in which the separation is achieved by the differential movement of individual components through a stationary phase under the influence of a mobile phase is called chromatography.

Chromatography is the most useful and modern technique extensively used for the separation of mixtures into their components, to purify the compounds and also to test the purity of compounds.

This method was first invented by M. Tswett, a Russian botanist in 1906 for the separation of coloured substances into individual components. The word ‘Chromatography’ was originally derived from the Greek word chroma means colour and graphy means writing.

Types of chromatography:

Depending upon the nature of the stationary phase (either a solid or a tightly held liquid on a solid support) and the mobile phase (either a liquid or a gas),

The various types of chromatographic techniques commonly used are:

  1. Column or Adsorption chromatography,
  2. Thin Layer Chromatography (TLC),
  3. High-performance liquid Chromatography (HPLC),
  4. Gas Liquid Chromatography (GLC),

Paper or partition Chromatography. In the first three cases, the mobile phases are liquid and the stationary phases are solid.

In the fourth case, the mobile phase is gas while the stationary phase is liquid and in the fifth case, both the mobile and stationary phases are liquid.

The chromatographic separation is based on the principle that the components of the mixture present in the moving phase move at different rates through the stationary phase and thus get separated. Now, depending on the basic principle, chromatography can be divided into two categories:

Absorption chromatography and Partition chromatography.

Adsorption chromatography Principle:

This category of chromatography is based upon the differential adsorption of the various components of a mixture on a suitable adsorbent such as alumina, cellulose, silica gel, magnesium oxide etc. Since some compounds undergo adsorption better than others, they travel through the column at different rates and thus get separated.

Adsorption chromatography is of two types:

  1. Column chromatography and
  2. Thin Layer Chromatography (TLC).

1. Column chromatography

It is the simplest of all the chromatographic techniques and is extensively used.

Chromatography Procedure:

  • A plug of cotton or glass wool is placed at the bottom of a clean and dry glass column and it is then covered with a layer of acid-washed sand.
  • A suitable adsorbent such as alumina, silica gel, magnesium oxide, starch etc. is made into a slurry with a non-polar solvent such as hexane or petroleum ether and the slurry is then added to the column gradually and carefully so that no air bubble is entraped in the column.
  • The excess of the solvent above the adsorbent is removed by opening the stop-cock. This constitutes the stationary phase.
  • The mixture of compounds (say A, B and C) to be separated is dissolved in a minimum volume of a suitable highly polar solvent in which it is readily soluble. It is then added to the top of the column with the help of a dropper or a microsyringe and allowed to pass slowly through it (if the mixture is liquid, it is added as such).
  • As the solution travels down, the different components of the mixture get adsorbed to different extents depending upon their polarity (say, A>B>C) and form a narrowband which is quite close to the top of the column. This band or zone is called a chromatogram.
  • A suitable solvent called eluent is then made to run through the column. The polarity of the solvent is gradually increased to elute the adsorbed materials. The eluent acts as the mobile phase.
  • As the solvent moves down the column, the components A, B and C present in the chromatogram begin to separate. The eluent dissolves out the different components selectively.
  • The component which is strongly adsorbed on the stationary phase moves slowly down the column, whereas the component that is weakly adsorbed moves at a faster rate. Therefore, the three components [A, B and C) form three bands at different places in the column.
  • As the addition of eluent is continued, the adsorbed components present in the bands are dissolved by the solvent and are then collected in the form of different fractions in separate conical flasks.
  • The eluent from each flask is then removed by evaporation or distillation to get the various components in pure form.
  • The process of separation of different components of the mixture from the adsorbent and their recovery with the help of a suitable solvent is called elution.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Chromatography

2. Thin Layer Chromatography (TLC):

It is another type of adsorption chromatography which involves the separation of the components of a mixture over a thin layer of adsorbent. This technique is particularly useful in rapid analysis ofthe purity of samples.

Chromatography Procedure:

  • A thin layer (0.2 mm thick) of adsorbent such as silica gel or alumina is spread over a plastic or glass plate of suitable size (5 cm  × 20 cm).
  • This thin layer of adsorbent acts as the stationary phase. This plate is called a thin-layer chromatography plate or TLC plate or chromatophore.
  • Two pencil lines are drawn across the width of the plate at distances about 1 cm from each end. The lower line is called the base line or starting line and the upper line is called the finish line or solvent front.
  • A drop of the solution of the mixture to be separated is placed on the starting line with the help of a capillary.
  • The plate is then dried and placed in a vertical position in a jar called a developing chamber containing a suitable solvent or a mixture of solvents. It acts as the mobile phase. The height of the solvent in the jar should be such that its upper surface does not touch the sample spot.
  • The chamber is then closed and kept undisturbed for half an hour
  • As the solvent slowly rises by the capillary action, the components of the mixture also move up along the plate to different distances depending upon their degree of adsorption and thus separation takes place.
  • When the solvent front reaches the finish line, the plate is removed from the jar and then dried. The spots of coloured components, due to their original colour, is visible on the TLC plate.

The spots of the colourless components which are invisible to the eye can be detected:

  • By placing the plate under a UV lamp because certain organic compounds produce a fluorescence effect in UV light
  • By placing the plate in a covered jar containing a few crystals of iodine because certain organic compounds which absorb iodine turn brown and
  • By spraying the plate with the solution of a suitable chemical reagent
  • For example – Ninhydrin in case of amino acids; 2,4-dinitrophenylhydrazine in case of carbonyl compounds
  • Different components developed on the TLC plate are identified through their retention factors (or retardation factor), i.e., R j values.

It may be defined as:

⇒ \(R_f=\frac{\text { Distance travelled by the compound from the baseline }}{\text { Distance travelled by the solvent from the baseline }}\)

  • If the two components of the mixture, for example, are A and B, then according to the given their  R2 values will be a/l and  b/l respectively.
  • The two compounds can be identified by comparing these Rf values with the Rf values of pure compounds. Since the solvent front on the TLC plate always moves faster than the compounds, Rf values are always less than 1.
  • Each spot is finally eluted separately with suitable solvents and collected

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Thin Layer

Partition chromatography

Unlike adsorption chromatography (column chromatography or TLC) which represents solid-liquid chromatography, partition chromatography is a liquid-liquid chromatography, i.e., both the stationary phase and mobile phase are liquids.

Partition chromatography Principle:

Partition chromatography is based on continuous differential partitioning (distribution) of components of a mixture between the stationary and the mobile phases Paper chromatography is a common example of partition chromatography.

In paper chromatography, a special type of paper known as chromatographic paper is used. Although the paper is made up of cellulose, the stationary phase in paper chromatography is not the cellulose but water which is adsorbed or chemically bound to cellulose. The mobile phase is usually a mixture of two or three liquids with water as one ofthe components.

Partition chromatography Procedure:

  • A suitable strip of chromatographic paper (20cm × 5cm, Whatman filler paper) is taken and a starting line (baseline) is drawn across the width of the paper at about 1 to 2 cm from the bottom.
  • The mixture to be separated is dissolved in a minimum amount of a suitable solvent and applied as a spot on the starting line with the help of a fine capillary or micro syringe.
  • The spotted chromatographic paper is then suspended in a suitable solvent or a mixture of solvents. The position of the paper should be such that the spot on the starting line remains above the surface of the solvent. Thus, the solvent acts as the mobile phase.
  • The solvent rises up the paper strip by capillary action and flows over the sample spot. The different components of the mixture depending upon their solubility (or partitioning between) in the stationary and mobile phases, travel through different distances.
  • When the solvent reaches the finish line the paper strip is taken out and dried in air. The paper strip so developed is called a chromatogram.
  • The spots of the separated coloured compounds are visible at different distances from their initial position on the starting line.
  • The spots of the separated colourless compounds are observed either by placing the paper strip under UV light or by using an appropriate spray reagent as discussed in TLC.
  • The components of the mixture are then identified by determining their R j values as discussed in TLC.
  • The process can also be performed by folding the chromatographic paper into a cylinder

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Paper

Criteria of purity of organic compounds

Several methods for the purification of organic compounds have already been discussed. The next important step is to test their purity, i.e., to know whether a particular compound has been purified or not. A pure organic solid has a definite and sharp melting point. An impure solid melts over a range of temperatures and even the presence of traces

Qualitative Analysis Of Organic Compounds

Since organic compounds are either hydrocarbons or their derivatives, the elements which occur in them are carbon (always present), hydrogen (nearly always present), oxygen (generally present), nitrogen, halogens, sulphur (less commonly present), phosphorus and metals (rarely present). All these elements can be detected by suitable methods.

1. Detection of carbon and hydrogen(C-H)

Detection of carbon and hydrogen Principle:

A small amount of a pure and dry organic compound is strongly heated with dry cupric oxide in a hard glass test tube when carbon present in it is oxidised to carbon dioxide and hydrogen is oxidised to water.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques C And H

Liberated CO2 turns lime water milky and the liberated water vapours (H2O) which get condensed in the bulb of the delivery tube, turn white anhydrous CuSO4into blue hydrated copper sulphate (CuSO4-5H2O)

Ca(OH)2 (Lime Water)+ CO2→CaCO3↓ (Milky)+ H2O

CuSO4 Anhydrous (white)+ 5H2O→CuSO4.5H2O(Blue)

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles Detection C And H

If the organic compound is a volatile liquid or gas, then the vapours of the compound are passed through heated copper oxide and the presence of C02 and H2O in the liberated gas can be proved in the same way.

If the organic compound contains sulphur in addition to carbon and hydrogen, then sulphur is oxidised to SO2 which also turns lime water milky due to the formation of calcium sulphite. In that case, the resulting gases are first passed through an acidified solution of potassium dichromate which absorbs SO2 and then through lime water.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Lime Water

2. Detection of nitrogen

1. Sodalime test

A very small amount of an organic compound is strongly heated with soda lime (NaOH + CaO) In a test tube. The evolution of ammonia having a typical smell Indicates the presence of nitrogen in the compound.

Example:

CH3CONH2 (Acetamide)+ [NaOH + CaO]→ CH3COONa (Sodium acetate)+ NH3

 Sodalime test Limitation:

Organic compounds containing nitrogen as nitro ( — NO2) or azo (— N =N— ) groups do not evolve NH3 upon heating with soda lime, i.e., do not give this test.

2. Lassaigne’s test

Preparation of Lassaigne’s extract or sodium extract:

A pea-sized freshly cut dry sodium metal is heated gently in a fusion tube till it melts to a shining globule. A small amount ofthe organic compound is added and the tube is heated strongly for 2-3 minutes till it becomes red hot. The hot tube is then plunged into 10-15 mL of distilled water taken in a mortar. The mixture is then ground thoroughly by a pestle and filtered. The filtrate is known as sodium extract or Lassaigne’s extract.

Test for nitrogen:

The Lassaigne extract is usually alkaline in nature because the excess of metallic sodium reacts with water to form sodium hydroxide. A small amount of freshly prepared FeSO4 solution is added to a part of sodium extract and the contents are boiled when a light green precipitate of Fe(OH)2 is obtained.

The mixture is then cooled and acidified with dil.H2SO4. The immediate appearance of blue or green colouration or deep blue coloured precipitate of Prussian blue, Fe4[Fe(CN)6)3 indicates the presence of nitrogen.

Chemistry of Lassaigne’s test:

1. When organic compound is fused with metallic Na, carbon and nitrogen present in it combine with Na to form NaCN

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques S Cyanide

2. When the sodium extract is heated with FeSO4 solution, Fe(OH)2 is obtained which reacts with sodium cyanide to form sodium ferrocyanide. This is also obtained when FeSO4 reacts with NaCN.

FeSO4 + 2NaOH →Fe(OH)2 + Na2SO4

6NaCN + Fe(OH)2 →Na4 [Fe(CN)6 ] (Sodium ferrocyanide) + 2NaOH

FeSO4 + 6NaCN→Na2 SO4 + Na4 [Fe(CN)6 ]

Ferrous (Fe2+) ions present in hot alkaline solution undergo oxidation by O2 to give ferric (Fe3+) ions. These ferric ions then react with sodium ferrocyanide to produce ferric ferrocyanide (prussian blue)

4Fe(OH)2 + O2  + 2H2O→4 Fe(OH)3

2Fe(OH)3 + 3H2 SO4 →Fe2(SO4 )3 + 6H2O

3Na4 [Fe(CN)6] + 2Fe2(SO4)3→ Fe4[Fe(CN)6]3  (Prussian Blue)+ 6Na2SO4

1. In this test, it is desirable not to add FeCl3 solution because yellow FeCl3 causes Prussian blue to appear greenish. For the same reason, the alkaline extract should not be acidified with hydrochloric acid (which produces ferric chloride).

2. When the compound under investigation contains both nitrogen and sulphur, it may combine with sodium during fusion to form sodium thiocyanate (sulphocyanide). This gives blood red colouration with ferric chloride due to the formation of ferric thiocyanate. Thus, Prussian blue is not obtained.

 

⇒ \(\mathrm{Fe}^{3+}+3 \mathrm{NaSCN} \rightarrow \mathrm{Fe}(\mathrm{SCN})_3+3 \mathrm{Na}^{+}\)

3. If fusion is carried out with excess of sodium, the resulting thiocyanate decomposes to give sodium cyanide and sodium sulphide. So, in that case no blood red colouration is visualised.

NaSCN + 2Na→NaCN + Na2S

However, in that case, black precipitate of FeS is obtained when FeS04 solution is added to sodium extract. The black precipitate dissolves on the addition of dil. H2SO4 and the test of nitrogen is then performed with that clear solution.

FeSO4 + Na2S→FeS↓ (Black) + Na2SO4

FeS + H2SO4 →H2S↑+ FeSQ4

Lassaigne’s test Limitations:

  • Volatile organic compounds if present, escape before reacting when fused with metallic sodium.
  • Some nitro compounds may lead to explosion during the fusion process.

3. Detection of sulphur

1. Lassaigne’s test:

Lassaigne’s extract is prepared as described in the case of nitrogen Sulphur present in the compound (which does not contain nitrogen) reacts with sodium metal to form sodium sulphide:

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques S Sulphide

The following tests are then performed with the extract to detect the presence of sulphur

Lead acetate test:

One part ofthe extract is acidified with acetic acid and then a lead acetate solution is added to it. The formation of a black precipitate of lead sulphide (PbS) confirms the presence of sulphur in the compound

Na2S + Pb(CH3COO)2→PbS↓(Black) + 2CH3COONa

Sodium nitroprusside test:

A few drops of sodium nitroprusside solution are added to another part of the extract. The appearance of a violet or purple colouration confirms the presence of sulphur in the compound.

Na2S + Na2[Fe(CN)5NO](Sodium nitroprusside)→ Sodium sulphonitroprusside Na4[Fe(CN)5NOS](Violet)

2. Oxidation test:

The organic compound is fused with a mixture of potassium nitrate and sodium carbonate and as a result, sulphur, if present, gets oxidised to sodium sulphate.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Oxidation State

The fused mass is extracted with distilled water and filtered. The filtrate is acidified with dil. HCl and then barium chloride solution is added to it. The formation of a white precipitate insoluble in hydrochloric acid indicates the presence of sulphur in the compound

Na2SO4 + BaCl2→2NaCl + BaSO4↓ (White)

4. Detection of halogens

1. Beilstein’s test:

A clean and stout Cu-wire flattened at one end is heated in the oxidising flame of a Bunsen burner until it imparts any green or bluish-green colour to the flame. The hot end of the Cu-wire is then touched with the organic compound under investigation and is once again introduced into the flame. The reappearance of green or bluish-green flame due to the formation of volatile copperhalide indicates the presence of halogens in the compound.

Beilstein’s test Limitations:

Many halogen-free compounds,

For example: 

Certain derivatives of pyridine and quinoline, purines, acid amides, urea, thiourea, cyano compounds etc. give this test presumably owing to the formation of volatile copper cyanides.

Therefore, this test is not always trustworthy. It does not indicate which halogen (Cl, Br or I) is present in the organic compound.

The test is not given by fluoro compounds since copper fluoride is non-volatile.

2.  Lassaigne’s test:

Lassaigne’s extract is prepared as described in the case of nitrogen. During fusion, Na combines with the halogen present in the compound to form sodium halide

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Sodium Halide

The extract is then boiled with dilute HNO3, cooled and a few drops of AgN03 solution are added to it White or yellow precipitation confirms the presence of halogen.

NaX + AgNO3 → AgX↓ + NaNO3 [X = Cl,Br,I]

1. Formation of a curdy white precipitate soluble in ammonium hydroxide solution indicates the presence of chlorine in the organic compound

NaCl + AgNO3→AgCU(White) + NaNO3

AgCl + 2NH4OH→[Ag(NH3)2]Cl(Soluble) + 2H2O

The formation of a pale yellow precipitate partially soluble in ammonium hydroxide solution indicates the presence of bromine in the organic compound.

NaBr + AgNO3→ AgBr4-(Pale yellow) + NaNO3

AgBr + 2NH4OH → [Ag(NH3)2]Br(Soluble) + 2H2O

The formation of a yellow precipitate insoluble in NH4OH solution indicates the presence of iodine in the organic compound.

Nal + AgNO3→AgI ↓ (Yellow) + NaNO3

Function Ol nitric acid:

If the organic compound contains nitrogen and sulphur along with halogens, Lassaigne’s extract contains sodium sulphide (Na2S) and sodium cyanide (NaCN) along with sodium halide (NaX). These will form a precipitate with silver nitrate solution and hence will interfere with the test.

Na2S + 2AgNO3→Ag2S↓  (Silver sulphide (Black) + 2NaNO3

NaCN + AgNO3→ AgCN↓   (Silver cyanide) (white)+ NaNO3

For this reason, before the addition of AgNO3 solution to the sodium extract for the detection of halogens, the extract is boiled with dilute HNO3 which decomposes sodium sulphide and sodium cyanide to vapours of H2S and HCN respectively

Alternatively, sulphide and cyanide ions can be removed by adding 5% nickel (II) nitrate solution which reÿct? with these ions forming precipitates of nickel (II) sulphide and nickel (II) cyanide

3. Chlorine water test for bromine and iodine:

  • A portion of Lassaigne’s extract is boiled with dilute nitric acid or dilute sulphuric acid to decompose NaCN and Na2S.
  • The solution is then cooled, and acidified with dil. H2SO4 and a few drops of carbon disulphide or carbon tetrachloride solution are added to it.
  • The resulting mixture is then shaken with a few drops of freshly prepared chlorine water and allowed to stand undisturbed for some time.

An orange or brown colouration in the carbon disulphide or carbon tetrachloride layer confirms the presence of bromine, whereas a violet colouration in the layer confirms the presence of iodine in the compound

2NaBr + Cl2→2NaCl + Br2 (turns CS2 or CCl4 layer orange)

2NaI + Cl2→2NaCl + I2 (turns CS2 or CCl4 layer violet)

5. Detection of phosphorus

1. The organic compound under investigation is fused with sodium peroxide (an oxidising agent) when phosphorus is oxidised to sodium phosphate

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Sodium Phosphate

2. The fused mass is then extracted with water, filtered and the filtrate is boiled with concentrated nitric acid. The mixture is then cooled and an excess of ammonium molybdate solution is added to it

3. The appearance of a yellow precipitate or colouration due to the formation of ammonium phosphomolybdate, (NH4 )3 PO4 -12MoO3, confirms the presence of phosphorus in the compound.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Phosphours

Quantitative Analysis Of Organic Compounds

After detecting the presence of various elements in a particular organic compound, the next step is to determine their respective percentages. This is known as quantitative analysis.

1. Estimation of carbon and hydrogen

Both carbon and hydrogen present in an organic compound are estimated by Liebig’s method.

Liebig’s method

Liebig’s method Principle:

A pure and dry organic compound of known mass is heated strongly with pure and dry copper oxide (CuO) in an atmosphere of air or oxygen-free from carbon dioxide. Both carbon and hydrogen present in the organic compound undergo complete oxidation and get converted into carbon dioxide and water respectively

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques CO2 And H2O

COthus produced is absorbed in a previously weighed potash bulb containing a strong KOH solution while water produced is absorbed in a previously weighed U-tube containing anhydrous CaCl2. The U-tube and the bulb are weighed again and from the difference between the two weights, the amount of CO2 and HaO are determined.

Liebig’s method Procedure:

  • The apparatus for the estimation of C and H is The apparatus consists of the following units:
    • Combustion tube,
    • U-tube containing anhydrous CaCl2 and
    • a bulb containing strong KOH solution.
  • The tube is heated l strongly for 2-3 hours till the whole of the organic compound is burnt up.
  • After combustion is over, the absorption units (the Utube and the potash bulb) are disconnected and weighed separately

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Tube

Results and calculations:

Let the mass of organic compound taken = w g. The increase in the mass of the potash bulb, i.e., the mass of CO2 formed = x g.

The increase in the mass of the U-tube, i.e., the mass of water formed =y g.

Percentage ot carbon: 1 mol of CO2 (44g) contains 1 gram atom of carbon (12 g).

∴ x g of CO2 contains = \(\frac{12x}{44}\) g of C

Now, \(\frac{12x}{44}\) g carbon is present in w g organic compound.

∴ The percentage of carbon in the compound

= \(\frac{2 y \times 100}{18 w}=\frac{2}{18} \times \frac{\text { Mass of } \mathrm{H}_2 \mathrm{O} \text { formed }}{\text { Mass of the compound taken }} \times 100\)

Percentage ot hydrogen:  1 mol of water (18g) contains 2 gram-atom hydrogen (2 g).

∴  y g of H2 O contains = \(\frac{2y}{18}\) g of H

Now, \(\frac{2y}{18}\) hydrogen is present in w g organic compound.

∴ The percentage of hydrogen in the compound

Estimation of C and H by Liebig’s method is suitable for organic compounds containing C, H and O only, but it requires some modifications for compounds containing nitrogen, halogens and sulphur.

1. Compounds containing nitrogen:

During combustion, N present in the organic compound undergoes oxidation to give its oxides (NO, NOz etc.) which are

Also absorbed in an alkali solution along with CO2. Oxides of nitrogen are decomposed back to nitrogen by placing a copper gauze roll near the exit of the tube. The N2 so produced is not absorbed by the alkali solution.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Alkali

2. Compounds containing halogens:

During combustion, halogens present in the organic compound get converted into volatile copper halides which partly decompose to give free halogens. These halogens and volatile copper halides get dissolved in an alkali solution. This can be prevented by placing a silver gauze roll near the exit of the combustion tube. Halogens combine with silver to give non-volatile silver halides.

2Ag + X2 →2AgX; 2Ag + CuX2 → 2AgX + Cu

3. Compounds containing only sulphur or sulphur and halogen:

During combustion, elemental sulphur present in the organic compound is oxidised to SO2 which is absorbed in the potash bulb.

This can be prevented by placing a layer of fused lead chromate near the exit of the tube. SO2 combines with lead chromate to produce non-volatile lead sulphate.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Sulphate

Lead chromate also reacts with copper halides and free halogens to form lead halides which remain in the combustion tube

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Halides

Liebig’s method Precautions:

  • All the joints ofthe combustion must be air-tight.
  • The combustion must be free from CO2 and water vapour.
  • The airflow is controlled in such a way that only 2-3 bubbles are generated per second. A faster flow of air may lead to the formation of carbon monoxide.
  • If carbon is deposited on the surface of the combustion mbe, then oxygen instead of air is to be passed at the final stage ofthe process

Example 1: 0.90g of an organic compound on complete combustion yields 2.20 g of carbon dioxide and 0.60 g of water. Calculate the percentages of carbon and hydrogen in the compound.
Answer:

Mass of the organic compound = 0.90 g

Mass of CO2 formed = 2.2 g & mass of H2O formed = 0.6 g

Percentage of carbon: 44 g CO2 contains = 12 g carbon

0.90 g organic compound contains = \(\frac{12 \times 2.20}{44}\) g of carbon

0.20 g organic compound contains = \(\frac{12 \times 2.20}{44}\) g carbon

Percentage of carbon = \(\frac{12 \times 2.20 \times 100}{44 \times 0.90}\)

= 66.67

Percentage of hydrogen:  18 g of H2O contains = 2g of hydrogen

0.60 g of H2O contains = \(\frac{2 \times 0.60}{18}\) of hydrogen

0.90 g compound contains = \(\frac{2 \times 0.60}{18}\) g hydrogen

Percentage of hydrogen = \(\frac{2 \times 0.60 \times 100}{18 \times 0.90}\)

= 7.41

2. Estimation ot nitrogen

The following two methods are commonly used for the estimation of nitrogen in an organic compound

  1. Duma’s method
  2. Kjeldahl’s method.

1. Dumas method

Dumas method Principle:

A weighed amount of the organic compound is heated with an excess of copper oxide in an atmosphere of CO2. Carbon and hydrogen present in the compound are oxidised to CO2 and H2O respectively while nitrogen is set free as nitrogen gas (N2). If any oxide of nitrogen is formed during this process, it is reduced back to nitrogen by passing over hot reduced copper gauze

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Dumas Method

When the gaseous mixture thus obtained is passed through a 40% KOH solution taken in a Schiff’s nitrometer, all gases except nitrogen are absorbed by KOH. The volume of nitrogen collected over the KOH solution is noted and from this, the percentage of nitrogen in the compound is calculated

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Copper Gauze

Dumas method Procedure:

  • The apparatus used for the estimation of nitrogen by the Dumas method ). It consists of:
    • CO2 generator combustion tube (a glass tube of diameter 15m/r and length 90 cm) and Schiff’s nitrometer.
    • The combustion tube is packed with [a] coarse CuO which prevents backward diffusion of gases produced,
    • an accurately weighed amount of organic compound of approximately 0.2g mixed with excess of CuO.
    • Coarse CuO and
    • A reduced copper gauze can reduce any oxides of nitrogen back to N2.
  • The Schiff’s nitrometer (connected to the combustion tube) consists of a graduated tube with a reservoir and a tap at the upper end. It has a mercury seal at the bottom to check the backward flow ofthe KOH solution.
  • CO2 generated by heating NaHCO3 or MgCO3 is dried by passing through concentrated HS04 and then passed through the combustion tube to displace the air present in the tube.
  • The combustion tube is then heated in the furnace. When the combustion is complete, a rapid stream of CO2 is passed through the tube to sweep away the last traces of N2.
  • The volume of nitrogen collected over the KOH solution in the nitrometer tube is recorded after careful levelling (by making the level of KOH solution both in the tube and reservoir the same).
  • The room temperature is recorded and the vapour pressure of the KOH solution at the experimental temperature is noted in the table.

Results and calculations:

Let the mass of the organic compound taken = w g, the volume of N2 gas collected = V mL, the atmospheric pressure = P mm, the room temperature = t°C and the vapour pressure of KOH solution at t°C =f mm. Hence, the pressure of dry N2 gas = (P- f) mm.

By applying the gas equation = \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

Volume of N2 at STP, = \(V_2=\frac{P_1 V_1 T_2}{P_2 T_1}=\frac{(P-f) \times V \times 273}{760(273+t)}\)

Percentage of nitrogen: 22400 mL of N2 gas at STP will weigh = 28 g.

2 mL of N2 gas at STP will weigh = \(\frac{28 \times V_2}{22400} \mathrm{~g}\)

Percentage of nitrogen:

= \(\frac{\text { Mass of nitrogen }}{\text { Mass of the compound }} \times 100\)

= \(\frac{28}{22400} \times \frac{V_2}{w} \times 100\)

The amount of nitrogen estimated by this method is 0.2% higher than the actual quantity. This is because a fraction of the small amount of air (nitrogen) entrapped in copper oxide gets deposited in the nitrometer

Example:

In Dumasg of an method organic for compound, the estimation gave 31.7 of nitrogen, mL of nitrogen at 14°C and 758 mm pressure. Calculate the percentage of nitrogen in the compound (vapour pressure of water at 14°C = 12 mm.
Answer:

If the volume ofnitrogen at STP is V mL, then

V = \(\frac{31.7(758-12)}{(273+14)} \times \frac{273}{760}\)

= 29. 6 mL

Now, at STP 22400 mL of nitrogen will weigh = 28 g

∴ 29.6 mL nitrogen at STP will weigh = \(\) = 0.037%

∴  Percentage of nitrogen = \(=\frac{0.037 \times 100}{0.1877}\)

= 19.71

2. Kjeldahl’s method

Kjeldahl’s method Principle:

A known mass of a nitrogenous organic compound is digested (heated strongly) with concentrated H2SO4 in the presence of a little potassium sulphate (to increase the boiling point of H2SO4) and a little CuSO4 (a catalyst) when the nitrogen present in the compound is quantitatively converted into ammonium sulphate. The resulting mixture is heated with excess of NaOH solution and the ammonia evolved is passed into a known but excess volume of a standard acid (HCl or H2SO4 ).

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Technique Acid

The excess acid left after the neutralisation of ammonia is estimated by titration with some standard alkali

2NaOH + H2SO4  →Na2SO4 + 2H2O

Kjeldahl’s Method Procedure

  •  The organic compound (0.5-5 g) is heated with a cone. H2SO4 in a Kjeldahl’s flask which is a long-necked round-bottomed flask with a loose stopper. A small amount of potassium sulphate and a few drops of mercury (or a little CuS ) are added.
  • The reaction mixture is heated for 2-3 hours when carbon and hydrogen present in the compound are oxidised to CO2 and H2O respectively while nitrogen is quantitatively converted to ammonium sulphate.
  • CO2 and water vapours escape through the loose stopper while ammonium sulphate remains in the flask.
  • The Kjeldahl’s flask is then cooled and the contents of the flask are transferred to a round-bottomed flask.
  • The mixture is diluted with water and excess caustic soda solution (40%) is added.
  • The flask is then connected to a Liebig condenser through Kjeldahl’s tap. The lower end of the condenser is dipped in a known volume of standard acid (N/10 HCl or H2SO4 ) taken in a conical flask.
  • The flask is then heated. The evolved ammonia gas is absorbed in the acid solution.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Kjeldahl Method

Results and calculations:

Let the mass of the organic compound = w g, volume of acid taken = V1 mL, normality of acid solution x(N) and volume of x(N) alkali required to neutralise unused acid = V2 mL.

Now, V2 mL  × (N) alkali = V2 mL × (N) acid.

∴ The quantity of acid used for neutralising ammonia = VjmL x(N) acid solution- V2 mL x(N) acid solution = (V1– V2) mL x(N) acid solution = (V1– V2)mL x(N) ammonia solution.

Now, 1000 mL (N) ammonia solution contains 17 g of NH3 or 14 g of nitrogen.

Therefore, (V1–  V2)mL × (N)

Ammonia solution contains = \(\frac{14 \times\left(V_1-V_2\right) x}{1000}\) g of nitrogen

This amount of nitrogen is present in w g compound

%  of nitrogen = \(\mathrm{n}=\frac{14 \times\left(V_1-V_2\right) x \times 100}{1000 \times w}=\frac{1.4\left(V_1-V_2\right) x}{w}\)

= \(\frac{1.4 \times \text { Vol. of acid used } \times \text { Normality of the acid used }}{\text { Mass of the compound taken }}\)

Kjeldahl’s method  Limitations:

  • Nitrogen present in pyridine, quinoline, diazo compound, azo compound, and nitro compound, cannot be converted into ammonium salts by this method.
  • So this method does not apply to such compounds. Thus although Dumas’s method applies to all compounds, Kjeldahl’s method has limited applications.

The percentage of nitrogen estimated in this method is not correct.

Kjeldahl’s method Utility:

This experiment can be performed quite easily and quickly. So in those cases where correct estimation of nitrogen content is not necessary

For example: Fertiliser, soil, food¬ stuff etc.), this method has its application.

This method is not troublesome and hence the possibility of error can be minimised by repeating the experiment several times.

Example:  0.257 g of a nitrogenous organic compound was analysed by Kjeldahl’s method and ammonia evolved was absorbed in 50 mL of (N/10) H2SO4. The unused acid required 23.2 mL of (N/10) NaOH solution for complete neutralisation. Determine the percentage of nitrogen in the compound.
Answer:

Acid taken = 50 mL (N/10) H2SO4 solution

= 5 mL(N) H2SO4 solution

Alkali required to neutralise the excess acid

= 23.2 mL (N/10) NaOH = 2.32 mL (N) NaOH

Now, 2.32 mL(N) NaOH = 2.32 mL(N) H2SO4 solution

The acid required to neutralise the ammonia evolved

= 5 mL (N) H2SO4– 2.32 mL (N) H2SO4 solution

= 2.68 mL (N) H2SO4 s 2.68 mL (N) NH3

[v KmL(N) H2SO4S VmL(N) ammonia]

1000 mL (N) ammonia solution contains 14g nitrogen

So, 2.68mL (N) ammonia solution contains \(\frac{14 \times 2.68}{1000}\) g nitrogen

So, 0.257g compound contains = \(\frac{14 \times 2.68}{1000}\) g nitrogen

Percentage of nitrogen = \(=\frac{14 \times 2.68 \times 100}{1000 \times 0.257}\)

= 14.6

3. Estimation of halogens: Carius method

Halogens Carius method Principle:

An organic compound containing halogen of known mass is heated with fuming nitric acid and a few crystals of silver nitrate. The halogen present in the compound is converted into the corresponding silver halide (AgX). From the mass of the compound taken and that of the silver halide formed, the percentage of halogen in the compound can be calculated.

Carius method Procedure:

  • About 5 mL of fuming nitric acid and 2 g of silver nitrate crystals are placed in a hard glass tube (Carius tube) of about 50 cm in length and 2 cm in diameter.
  • A small amount of accurately weighed organic compound is taken in a small tube and the tube is placed carefully into the Carius tube in such a way that nitric acid does not enter the tube.
  • The Carius tube is then sealed. Now, it is placed in an outer iron jacket and heated in a furnace at 550-560 K for about six hours
  • As a result, C, H and S present in the compound are oxidised to CO2, H2O and H2SO4 respectively. The halogen present in the compound gets converted into silver halide which is precipitated.
  • The tube is cooled and when the sealed capillary end is heated in a burner’s flame, a small hole is formed through which the gases escape.
  • The capillary end is now cut off and the precipitate of silver halide is filtered, washed, dried and weighed

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Carius

Results & calculations:

Let, the mass of organic compound taken = w g and the mass of silver halide (AgX) formed—x g. Now, 1 mol of AgX contains 1 gram-atom of X (X = Cl, Br or I ), i.e., (108 + atomic mass of X) g of AgX contain (atomic mass of X)g of X

∴ xg AgX contain = \(\frac{\text { atomic mass of } X}{(108+\text { atomic mass of } X)}\) X JCg of X

This amount of halogen (X) is present in w g compound.

Percentage of halogen in the compound

= \(\frac{\text { atomic mass of } \mathrm{X}}{(108+\text { atomic mass of } \mathrm{X})} \times \frac{x}{w} \times 100\)

1. Percentage of chlorine (atomic mass = 35.5 )

= \(\frac{35.5}{(108+35.5)} \times \frac{x}{w} \times 100\)

= \(\frac{35.5}{143.5} \times \frac{x}{w} \times 100\)

2. Percentage of bromine (atomic mass = 80 )

= \(\frac{80}{(108+80)} \times \frac{x}{w} \times 100\)

= \(\frac{80}{188} \times \frac{x}{w} \times 100\)

3. Percentage of iodine (atomic mass = 127 )

= \(\frac{127}{(108+127)} \times \frac{x}{w} \times 100\)

= \(\frac{127}{235} \times \frac{x}{w} \times 100\)

Example: In the Carius method, 0.256 g of an organic compound containing bromine gave 0.306 g of AgBr. Find out the percentage of bromine in the compound.
Answer:

The mass of the organic compound taken = 0.256 g and the mass of AgBr formed = 0.306 g.

Now, 1 mol of AgBr = 1 gram-atom of Br

or, (108 + 80) g or 188 g of AgBr = 80 gofBr

i.e., 188 g of AgBr contains = 80 g of bromine

0.306 g of Ag Br contain \(\frac{80}{188} \times 0.306\) g of bromine

This amount of bromine is present in 0.256g compound

Percentage6 of bromine = \(\frac{80}{188} \times \frac{0.306}{0.256} \times 100\)

= 50.9

4. Estimation of sulphur: Carius method

Sulphur Carius method Principle:

When an organic compound containing sulphur is heated with fuming nitric acid in a sealed tube (Carius tube), sulphur is quantitatively oxidised to sulphuric acid. It is then precipitated as barium sulphate by adding a barium chloride solution. The precipitate is then filtered, washed, dried and weighed. From the amount of BaSO4 formed, the percentage of sulphur is calculated.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Sulphur

Resultsandcalculations: Let the mass of organic compound

= w g and the mass of barium sulphate formed = xg

Now, 1 mol of BaSO4 contains 1 gram-atom of S, i.e.,

(137 + 32 + 64) g or, 233 g BaO4 contain 32 g sulphur.

∴ x g of BaSO4 contain \(\frac{32}{233}\) × x g of sulphur

So, this amount of sulphur is present in the w g compound.

∴ Percentage of sulphur = \(\frac{32}{233} \times \frac{x}{w} \times 100\)

Example: In sulphur estimation by the Carius method, 0.79 g of an organic compound gave 1.164 g of barium sulphate. Calculate the percentage of sulphur in the compound.
Answer:

The mass of the organic compound taken = 0.79 g and the mass of BaSO4 obtained = 1.164 g.

Now, 1 mol of BaSO4 = 1 gram-atom of S or,

233 g of BaSO4= 32 g of i.e.,

233 g of BaSO4 contain = 32 g ofsulphur.

1.164 g of BaSO4 contain \(\frac{32}{233} \times 1.164\) × 1.164 g of sulphur

So, 0.79 g compound contain \(\frac{32}{233} \times 1.164\) × 1.164 g of sulphur

Percentage of sulphur =  \(\frac{32}{233} \times \frac{1.164}{0.79} \times 100\)

= 20.24

5. Estimation of phosphorus: Carius method

Phosphorus Carius method principle:

An organic compound (containing P) of known mass is heated with fuming nitric acid in a sealed tube (Carius tube). Phosphorus present in the compound is oxidised to phosphoric acid which is precipitated as ammonium phosphomolybdate by heating it with concentrated nitric acid and then adding ammonium molybdate.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Nitric Acid

Ammonium phosphomolybdate (Yellow)

The precipitate of ammonium phosphomolybdate thus obtained is filtered, washed, dried and weighed.

Alternative method:

The phosphoric acid is precipitated as magnesium ammonium phosphate (MgNH4PO4) by the addition of magnesia mixture (a solution containing MgCl2, NH4Cl and NH4OH ).

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Ammonium

The precipitate is filtered, washed, dried and then ignited to give magnesium pyrophosphate (Mg2P2O)

CBSE Class 11 Chemistry Notes For ChapOrganic Chemistry Basic Principles And Techniques Megnesium

From the mass of magnesium pyrophosphate, the percentage of phosphorus in the compound can be easily calculated.

Results and calculations:

Let the mass of the organic compound taken = wg and the mass of ammonium phosphomolybdate formed = xg.

1 mol (NH4)3PO4-12MoO3 contains 1 gram-atom of P.

or, 3(14 + 4) + 31 + 4 × 16 + 12(96 + 3 × 16) = 1877g of

(NH4)3PO4. 12MOO3 = 31 g of P, i.e., 1877 g of

(NH4)3PO4 . 12MoO3 contain 31 g of phosphorus

xg of (NH4)3PO4 -12MoO3 contain \(\)

xg of phosphorus. This amount of phosphorus is present in w g of the compound.

Percentage of phosphorus = \(\frac{31}{1877} \times \frac{x}{w} \times 100\)

= \(\frac{31}{1877} \times \frac{\text { Mass of ammonium phosphomolybdate }}{\text { Mass of the compound taken }} \times 100\)

Alternative calculation: Let the mass of Mg2P20? formed

= xg. Now, 1 mol of (Mg2P2O) contains 2 gram-atom of P

or, (24 × 2 + 31 × 2 + 16 × 7) = 222 g of (Mg2P2O)

i.e., 222g of (Mg2P2O) contains 62 g of phosphorus.

x g of (Mg2P2O)contain \(\frac{62}{222} \times x\) g of phosphorus

w g organic compound contain \(\frac{62}{222} \times x\) xg phosphorus

Percentage of phosphorus = \(\frac{62}{222} \times \frac{x}{w} \times 100\)

= \(\frac{62}{222} \times \frac{\text { Mass of } \mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

Example: 0.35 g of an organic compound containing phosphorus gave 0.65 g magnesium pyrophosphate, Mg2P2O in the Carius method. Calculate the percentage of phosphorus in the given compound.
Answer:

The mass of the organic compound taken = 0.35 g and the mass of Mg2P2O formed = 0.65 g.

Now, 1 mol of Mg2P2O contains 2 gram-atom of P.

Or, (2 ×  24 + 2 × 31 + 16 × 7) g of Mg2P2O= 2 × 31 g of P

i.e., 222 g of Mg2P2O contains 62 g of phosphorus

0.65 g Mg2P2Ocontain \(\frac{62}{222} \times 0.65\) x 0.65 g of phosphorus

This amount of P is present in a 0.35 g compound.

Percentage of phosphorus = \(\frac{62}{222} \times \frac{0.65}{0.35} \times 100\)

6. Estimation of oxygen

Percentage of oxygen = 100 – (sum of the percentages of all other elements). However, the oxygen content of a compound can be estimated directly as follows:

Oxygen Principle:

An organic compound of known mass is decomposed by heating in a stream of nitrogen gas. The mixture of the gaseous products including oxygen is passed over red-hot coke when all the oxygen combines with carbon to form carbon monoxide. The mixture is then passed through warm iodine pentoxide (I2O5 ) when CO undergoes oxidation to CO2 and iodine is liberated.

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Iodine

From the knowledge of the mass of iodine liberated or C02 produced, the percentage of oxygen in the compound can be calculated easily

Results and calculations: Let, the mass of the organic compound taken = wg and the mass of CO2 formed =xg. We find that each mole of oxygen liberated from the compound will produce 2 mol of CO2

x g. of CO2 is obtained from \(\frac{32}{88} \times x=\frac{16}{44} \times\) = x x g of O2

This amount of oxygen is present in w g ofthe compound.

Percentage of oxygen = \(\frac{16}{44} \times \frac{x}{w} \times 100\)

1. Percentage of carbon (Liebig’s method)

= \(\frac{12}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

2. Percentage of carbon (Liebig’s method)

= \(\frac{2}{18} \times \frac{\text { Mass of } \mathrm{H}_2 \mathrm{O} \text { formed }}{\text { Mass of the compound taken }} \times 100\)

3. Percentage of nitrogen (Dumas method)

= \(\frac{28}{22400} \times \frac{\text { Vol. of } \mathrm{N}_2 \text { gas at STP }}{\text { Mass of the compound taken }} \times 100\)

4. Percentage of nitrogen (KjeldahTs method)

= \(\frac{28}{22400} \times \frac{\text { Vol. of } \mathrm{N}_2 \text { gas at STP }}{\text { Mass of the compound taken }} \times 100\)

5. Percentage of halogen (Carius method)

= \(\frac{\text { At. mass of } X}{(108+\text { At. mass of } X)} \times \frac{\text { Mass of AgX }}{\text { Mass of the compound taken }} \times 100\)

X = Cl(35.5) , Br(80) or 1(127)

6. Percentage of sulphur (Carius method)

= \(\frac{32}{233} \times \frac{\text { Mass of } \mathrm{BaSO}_4 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

7. Percentage of phosphorus (Carius method)

\(\frac{62}{222} \times \frac{\text { Mass of } \mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \text { formed }}{\text { Mass of the compound taken }} \times 100\)

8. Percentage of oxygen

= \(\frac{16}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of compound taken }} \times 100\).

General Anaesthesia Short Answer Questions

General Anaesthesia Short Answers

Question 1. Actions of nitrous oxide.

Answer:

Nitrous oxide is a colourless, odourless, non-inflammable gas.

Nitrous oxide Actions:

  • A moderate increase in pain threshold.
  • Slight amnesia effect
  • Euphoria is frequent
  • Decreased sense of Smell.
  • improved hearing.
  • Slight myocardial depression.
  • Minimal effect on respiration.

Pharmacology II Short Answer Questions

Question 2. Diazepam. Diazepam uses.

Answer:

  • To induce conscious sedation in small repeated doses or by slow infusion.
  • Sedative hypnotic.
  • Anxiety.
  • Centrally acting skeletal muscle relaxant
  • Combined with analgesic for the treatment of rheumatic disorders.
  • Antiepileptic drugs.

Question 3. Ketamine.

Answer:

Ketamine induces dissociative anaesthesia characterized by profound analgesia, immobility, amnesia, light sleep and a feeling of dissociation from one’s own body and surrounding.

Ketamine Site of action:

  • Cortex and subcortical areas.

Ketamine Dose:

  • 1-2 mg/kg slow IV or
  • 10 mg/kg IM
  • The onset of action -1 – 3 min.
  • Recovery – After 10-15 min.

Pharmacology II Short Answer Questions

Ketamine Indications:

  • Surgery of head and neck.
  • Children.
  • Asthmatic patients.
  • For short operations.
  • in patients who do not want to lose consciousness.

Ketamine Contra-indications:

  • Hypertensives.
  • ischaemic heart disease.
  • Pregnancy.
  • increased intracranial tension.
  • Psychiatric disorders.

Question 4. Methohexitone.

Answer:

Methohexitone is a more potent and shorter-acting thiopentone congener.

  • It is more rapidly metabolized.
  • It has quick and brief actions.
  • Induction – in 15 – 30 sec.
  • Recovery – within 2 – 3 min.

Methohexitone Actions:

  • Less hypotensive
  • Slight increase in heart rate
  • Moderate hypoventilation.

Methohexitone Contra/Indication:

  • Epileptic patients.

Question 5. Thiopentone sodium.

Answer:

Thiopentone sodium is ultra short-acting thiobarbiturate.

Thiopentone sodium The onset of action:

  • 15-20 sec when given intravenously.

Thiopentone sodium Duration of action:

  • 4 – 7 min.

Pharmacology II Short Answer Questions

Thiopentone sodium Properties:

  • Highly soluble in water.
  • Prepared freshly before injection.
  • Highly lipid soluble.
  • Poor analgesic.
  • Weak muscle relaxant
  • Extravasation of solution causes intense pain, necrosis and gangrene.

Thiopentone sodium Adverse effects:

  • Laryngospasm.
  • Shivering and delirium.

Question 6. Propofol.

Answer:

it is an oily liquid employed as a 1% emulsion

Propofol Pharmacokinetics:

  • Unconsciousness occurs in 15-45 sec and lasts approximately 10 minutes
  • Distributes rapidly
  • Plasma half-life – 2-4 min

Propofol Advantages:

  • Faster onset
  • Quick recovery
  • Antiemetic
  • Safer in pregnancy

Pharmacology II Short Answer Questions

Propofol Uses:

  • IV induction
  • Shorter duration procedures like endoscopies, bums dressings

Propofol Adverse Effects:

  • Pain at the injection site
  • Fall in BP
  • CVS and respiratory depression

NCERT Solutions For Class 8 Maths

CBSE Chemsitry Notes For Class 11 Aliphatic Hydrocarbons Alkanes

Class 11 Chemistry Hydrocarbons Aliphatic Hydrocarbons Alkanes

Open-chain saturated hydrocarbons are referred to as alkanes. At ordinary temperature and pressure, they generally do not show any affinity towards most of the reagents such as acids, bases, oxidizing, and reducing agents and because of this inertness, they are called paraffin (Latin: param = litde, affinis= affinity).

⇒ Each C-atom present in an alkane molecule is sp³ -hybridized. Four σ -bonds formed by each sp³ -hybridized carbon are directed towards the comers of a regular tetrahedron.

⇒  Thus, alkanes have a tetrahedral structure around each carbon atom. The molecular formula of alkanes is CnH2n + 2 [where n = 1, 2, ].

⇒  Their general formula is RH (R: alkyl group).

1. Nomenclature of alkanes

The nomenclature of alkanes according to the IUPAC system has been thoroughly Here, only the trivial names of the isomers of butane and pentane and the IUPAC names of some higher alkanes are mentioned

CBSE Chemsitry Notes For Class 11 Hydrocarbons IUPAC Names Of Some Higher Alkanes

2. Structure of alkanes

Alkanes contain only carbon-carbon and carbon-hydrogen single bonds. They have the following structural
characteristics:

  • Each C-atom is sp³ -hybridized. Four sp³ -hybrid orbitals are directed towards the comers of a regular tetrahedron. The carbon atom lies at the center of the tetrahedron.
  • All C—C and C—H bonds are strong sigma bonds. Each C —C cr -bond is formed as a result of the axial overlapping of two sp³ orbitals, one from each carbon atom, and each C—H bond is formed by the axial overlapping of one sp³ orbital of carbon with the s -orbital of hydrogen.
  • C—C and C—H bond lengths are 1.54A & 1.12A respectively. lv] All bond angles in alkanes (C —C —C, C —C —H, and H—C—H) have a value of 109°28′. Thus, alkanes possess tetrahedral structure

CBSE Chemsitry Notes For Class 11 Hydrocarbons Structure Of Ethane

  • Carbon atoms in an alkane molecule having three or more carbon atoms do not lie along a straight line. Instead, they form a zig-zag pattern. This is because each carbon atom is sp³ -hybridized and naturally the C—C— C bond angle is 109°28′ instead of 180°. It becomes clear from the structure of propane

CBSE Chemsitry Notes For Class 11 Hydrocarbons Structure Of Propane

  • C—C and C—H bond dissociation enthalpies are 83kcal -mol-1 and 99 kcal-mol-1 respectively

CBSE Chemsitry Notes For Class 11 Hydrocarbons Benzenoid Three And Two Dimensional Representation

3. Structural isomerism in alkanes

Alkanes (except methane, ethane, and propane) exhibit chain isomerism, a type of structural isomerism. This type of isomerism arises due to the difference like the carbon chain or the skeleton of the carbon atoms.

Example:

1. The two chain isomers having molecular formula (C4H10) are n-butane and isobutane. If a 1° or 2° H atom of a propane molecule is replaced by a methyl group, then these two isomers are formed.

CBSE Chemsitry Notes For Class 11 Hydrocarbons Two Chain Isomers of Molecular N butane And Isobutane

Example:

2. Three chain isomers of molecular formula C5H12 are n -pentane (CH3CH2CH2CH2CH3), isopentane [(CH3)2CHCH2CH3] and neopentane [(CH3)4C].

CBSE Chemsitry Notes For Class 11 Hydrocarbons Three Chain Form Of Isomers

3. These isomers are formed on the replacement of different H -atoms of n-butane and isobutane by methyl group

4 . Five chain isomers have molecular formula C6H14 and these by obtained by replacement of different types of H-atoms of n-pentane, isopentane, and neopentane by a methyl group.

These are as follows:

CBSE Chemsitry Notes For Class 11 Hydrocarbons Methyl Group

4. Conformational isomerism in alkanes

Conformational isomerism definition:

Electronic distribution of the sigma molecular orbital of a C—C bond is cylindrically symmetrical around the internuclear axis and as this is not disturbed due to rotation about its axis, free rotation about the C—C single bond is possible. An infinite number of spatial arrangements of atoms that result through rotation about a single bond are called conformations or conformational isomers or rotational isomers or simply conformers or rotamers and the phenomenon is called conformational isomerism.

The difference in potential energy between the most stable conformation and the conformation under consideration is called the conformational energy of the given conformation.It is to be noted that the rotation around a C—C single bond is not completely free.

It is hindered by a very small energy barrier of 1-20kl-mol-1due to very weak repulsive interaction between the electron clouds of different σ -bonds. Such repulsive interaction is called torsional strain.

Conformations are three-dimensional.

These are generally represented in paper by three projection formulae:

Flying wedge formula, sawhorse projection formula and Newman projection formula.

Conformations of ethane:

A molecule of ethane (CH3—CH3) contains a carbon-carbon single bond (σ -bond) and each carbon atom is attached to three hydrogen atoms. The two —CH3 groups can rotate freely around the C—C bond axis.

Rotation of one carbon atom keeping the other fixed results into an infinite number of spatial arrangements of hydrogen atoms attached to the rotating carbon atom concerning the hydrogen atoms attached to a fixed carbon atom.

  • These are called conformational isomers or conformations or conformers.
  • Thus, there are an infinite number of conformations of ethane. However, there are two extreme cases. The conformation in which the hydrogen atoms attached to two carbons are as close together as possible.
  • In which the dihedral angle between the two nearest C —H bonds of two — CH3 groups is zero, is called the eclipsed conformation.
  • The conformation in which the hydrogen atoms are as far apart as possible, i.e., the dihedral angle between two C —H bonds is 60° is called the staggered conformation.
  • The eclipsed conformation suffers from maximum torsional strain whereas in staggered conformation this strain is minimal.
  • So, the eclipsed conformation is much less stable than the staggered conformation.
  • Any other intermediate conformation i.e., the conformation in which the dihedral angle is between 0-60°, is called the skew conformation.

Its stability is in between the two extreme conformations. Therefore, the order of stability of these three conformations is:

Staggered > skew > eclipsed.

It is to be noted that in all these conformations, the bond angles and the bond lengths remain the same.

Saturated hydrocarbons containing more than two carbon atoms have different conformations. However, as there is only one carbon atom in methane, it does not exist in the above-mentioned conformations. The eclipsed and the staggered conformations of ethane can be represented by the flying wedge formula, sawhorse projection formula and

Newman projection formula is as follows:

1. Flying wedge formula:

In this representation, the two bonds attached to a carbon atom are shown in the plane of the paper and of the other two, one is shown above the plane and another below the plane. The bonds that are in the plane are shown by normal lines (—) but the bond

Above the plane is shown by a solid wedge ( —) bond below the plane is shown by a hashed wedge

CBSE Chemsitry Notes For Class 11 Aliphatic Hydrocarbons Alkanes

2. Sawhorse projection formula:

In this projection, the molecule is viewed along the molecular axis. It is then projected on paper by drawing the central C —C bond as a somewhat elongated line. The upper end of the line is slightly tilted towards the righthand side. The front carbon is shown at the lower end of the line, whereas the rear carbon is shown at the upper end.

Each carbon has three lines attached to it corresponding to three H -atoms. The lines are inclined at a 120° angle to each other.

CBSE Chemsitry Notes For Class 11 Hydrocarbons Sawhorse Conformation Formula

3. Newman projection formula:

In this projection, the molecule is viewed along the C —C bond. The C-atom nearer to the eye of the viewer (i.e., the front carbon) is represented by a point and the three H-atoms attached to the front C-atom are shown by the three lines drawn at an angle of 120° to each other. The C-atom situated farther from the eye of the viewer (i.e., the rear carbon) is represented by a circle, and the three hydrogen atoms attached to it are represented by three shorter lines drawn at an angle of 120° to each other.

Eclipsed and staggered conformations of ethane in I H H terms of the

Newman projection formula (along with dihedral angles, ) are shown below:

CBSE Chemsitry Notes For Class 11 Hydrocarbons Eclipsed And Staggered Conformation

The energy barrier between two extreme conformations is tiny and so, the rotation of two —CH3 groups takes place extremely rapidly. Due to this, it is not possible to separate the conformations of ethane. However, at any moment, the majority of ethane molecules exist in the staggered  conformation of minimum energy {i.e., maximum stability)

The eclipsed conformation is least stable because hydrogens and bonding pairs of electrons eclipsed C —H bonds involving adjacent C-atoms are very close to each other causing maximum repulsion. The staggered conformation is most stable because the hydrogens and bonding pairs of electrons of each pair of C —H bonds involving adjacent C-atoms are at a maximum distance. This causes minimum electronic as well as steric repulsion

The potential energy of the molecule is minimal for staggered conformation. It increases with rotation and reaches a maximum at eclipsed conformation. Experimentally, it has been found that the staggered conformation of ethane is 2.8 kcal-mol-1 more stable than the eclipsed conformation. (Eeclipsed  eclipsed –  Estaggered   = 2.8 kcal-mol-1 ).

Therefore, rotation about C—C bond is not completely free. However, this energy barrier is not large enough to prevent rotation at room temperature as collisions between the molecules supply sufficient kinetic energy to overcome this energy barrier

CBSE Chemsitry Notes For Class 11 Hydrocarbons Eclipsed And Staggered Conformation Potential Energy

Dihedral angle:

The dihedral angle (Φ) is the angle between the X—C—C and the C—C—Y plane of X—C—C—Y unit in a molecule. In ethane, it is the

CBSE Chemsitry Notes For Class 11 Hydrocarbons Dehydration Angle

The angle between the H—1C— 2C plane and 1C—2C—H plane, 1 2 i.e., it is the angle between the 2C—H bond and the C—H bond in the Newman projection formula. It is also called the angle of torsion.

Conformations of propane (1CH32CH23CH3):

In propane molecules, both C1—C2 & C2—C3 bonds are equivalent. An infinite number of conformations of propane can be obtained as a result of rotation about the C1—C2 (or C2—C3) bond. The two extreme conformations are the eclipsed conformation (I) and the staggered conformation

The staggered conformation is more stable than the eclipsed conformation by 3.4 kcal-mol-1

Conformations of n-butane (CH3-CH2-CH2-CH3):

n-butane contains two kinds of C —C bonds. So, conformations likely to be generated depend on that particular C —C bond around which C-atoms are made to rotate

CBSE Chemsitry Notes For Class 11 Hydrocarbons Conformations Of n Butane

1. Rotation about the C1—C2 bond:

Keeping C1 fixed, when C2 is rotated around the C1—C2 bond axis, infinite numbers of conformations are obtained. Among these, twoprincipal conformations are eclipsed (I) and staggered (II) conformations.

Their order of stability is:

Staggered > eclipsed, i.e., molecules of n-butane spend most of their time in staggered conformation (II).

2. Rotation about the C2-C3 bond:

An infinite number of conformations are possible if C3 is made to rotate around C2 — C3 bond axis, keeping C2 fixed.

Among these, the four chief conformations are:

  1. Anti-staggered (1)
  2. Gauchestaggered (3)
  3. Eclipsed (2)
  4. Fully eclipsed (4).

In anti-staggered conformation, the two —CH3 groups exist anti to each other, i.e., they are oriented at an angle of 180° (Φ = 180°). In the gauche-staggered conformation, the two —CH3 groups make an angle of 60° with each other (Φ = 60°). In the eclipsed conformation, the two pairs of —CH3 and H and one pair of H -atoms are in direct opposition, while in the fully eclipsed conformation, the two pairs of H-atoms and one pair of CH3 groups are in direct opposition.

The order of their stability:

1 >3 > 2 > 4, i.e., the molecules of n-butane pass most of their time in anti-staggered conformation (I).

Their Newman projection formulae are shown below:

CBSE Chemsitry Notes For Class 11 Hydrocarbons Newman Projection Formulae

The most stable and least stable conformations of n-butane are anti-staggered and fully eclipsed conformations respectively. The angular distance between two similar bonds in the anti-staggered conformation is maximum (180°). Thus, repulsion between electrons of such a bond pair is minimal.

Again, two —CH3 groups are located farthest from each other so, no steric hindrance or steric strain acts between them. On the other hand, the angular distance between two similar bonds in the fully eclipsed conformation is minimum (0°). Thus, the repulsion between electrons of each bond pair is maximum.

Again, two —CH3 groups are in direct opposition and hence there occurs severe steric strain involving these two CH3 groups. For this reason, anti-staggered conformation is the most stable while fully eclipsed conformation is the least stable conformation of  n-butane

The potential energy changes during rotation about the C2—C3 bond of n-butane is shown in the following diagram

CBSE Chemsitry Notes For Class 11 Hydrocarbons The Potential Energy Changes Rotation Of Bond Of n Butane

CBSE Class 12 Maths Important Question and Answers

NCERT Notes For Class 6 Maths

NCERT Notes For Class 6 Maths

NCERT Class 6 Science Chapter 4 Getting To Know Plants Multiple Choice Questions

NCERT Class 6 Science Chapter 4 Getting To Know Plants Multiple Choice Questions

Question 1. The plant which cannot be classified as a shrub is

  1. Jasmine
  2. Bougainvillea
  3. Carrot
  4. Lemon

Answer: 3. Carrot

Question 2. Which one of the following best describes the characteristics of a tree?

  1. Jasmine
  2. Bougainvillea
  3. Carrot
  4. Cotton

Answer: 2. Bougainvillea

Read and Learn More NCERT Class 6 Science MCQs

Question 3. Which of the following is the correct match between the characteristics of the stem and the category of the plant?

  1. Weak stem which cannot stand upright- creeper
  2. Green tender stem-shrub
  3. Thick, hard stem with branching near the base-tree
  4. Thick, hard stem with brandies high on the plant-herb

Answer: 1. Weak stem which cannot stand upright- creeper

Question 4. The picture shows a pumpkin. The spreading stems shown to indicate that a pumpkin should be classified as competency

Class 6 Science Chapter 4 Getting To Know Plants The spreding stems a pumpkin

  1. Herb
  2. Shru
  3. Creeper
  4. Tree

Answer: 3. Creeper

Question 5. The part of the plant which grows towards light and it also transports water, minerals and food.

  1. Stem
  2. Roots
  3. Leaf
  4. Flower

Answer: 1. Stem

Question 6. Which of the following is not the primary function of the stem?

  1. Conduction of water
  2. Photosynthesis
  3. Formation of branches
  4. Bears flowers and fruits

Answer: 2. Photosynthesis

Question 7. Which of the following is not a correct match?

  1. Petiole: attaches the leaf to the stem
  2. Lamina: green flat part of the leaf
  3. Margin: gives shape to the leaf
  4. Veins: transpiration

Answer: 4. Veins: transpiration

Question 8. Which of the following combinations of features would you observe in grass?

  1. Parallel venation and fibrous root
  2. Parallel venation and tap root
  3. Reticulate venation and fibrous root
  4. Reticulate venation and tap root

Answer: 1. Parallel venation and fibrous root

Question 9. Read the following sentences about photosynthesis.

  1. Sunlight, carbon dioxide, chlorophyll and water are necessary.
  2. Oxygen is absorbed.
  3. Leaves carry out photosynthesis.
  4. Proteins are made during photosynthesis.

Choose the correct pair of sentences that is true to photosynthesis

  1. 3 And 4
  2. 1 And 3
  3. 2 And 4
  4. 1 And 4

Answer: 2. 1 And 3

Question 10. Sarita pulled a herb out of the soil and observed that a plant part came out with it. Rohlni was watching the activity and saw some hair-like structures coming out from that part what could be the plant part?

  1. Flower
  2. Leaf
  3. Root
  4. Stem

Answer: 3. Root

Question 11. Which of the following plants does not have a tap root?

  1. Marigold
  2. Mango
  3. Maize
  4. Turnip

Answer: 3. Maize

NCERT Class 6 Science Chapter 4 Getting To Know Plants MCQs

Question 12. Which part of the plant grows in the soil?

  1. Stem
  2. Leaf
  3. Root
  4. Seed

Answer: 3. Root

Question 13. In the given diagram which part of the flower contains yellow powdery substances?

Class 6 Science Chapter 4 Getting To Know Plants Parts Of Flowers

  1. Ovary
  2. Filament
  3. Style
  4. Anther

Answer: 2. Filament

Question 14. Reproductive parts what characteristics would that plant portion have?

  1. Presence of stomata
  2. Presence of stamens and pistil
  3. Presence of midrib
  4. Presence of root hairs.

Answer: 2. Presence of stamens and pistil

Question 15. The male parts of the flower are called

  1. Pistil
  2. Carpel
  3. Stamen
  4. Style

Answer: 3. Stamen

Question 16. Which one of the following statements is incorrect?

  1. Leaves can only make food when there is light
  2. Male flowers will develop into fruits
  3. Roots hold the plant firmly to the ground
  4. Plants need air, water and sunlight to grow

Answer: 2. Male flowers will develop into fruits

Question 17. Which of the following terms constitutes the female part of the flower?

  1. Sepals, petals and stamen
  2. Stigma, style and ovary
  3. Ovary, stamen and stigma
  4. Ovary, style and stamen

Answer: 2. Stigma, style and ovary

Question 18. The underground plant part which anchors the plant to the soil is

  1. Stem
  2. Internode
  3. Leaves
  4. Roots

Answer: 2. Internode

Question 2. What part of the plant do we eat as food in turnip?

  1. Roots
  2. Stem
  3. Flower
  4. Trunk

Answer: 1. Roots

Question 3. Taproot is found in

  1. Onion
  2. Marigold
  3. Millet
  4. Wheat

Answer: 2. Marigold

Question 4. Which of the following plants does not have a tap root?

  1. Mustard
  2. Maize
  3. Tulsi
  4. Balsam

Answer: 2. Maize

Question 5. The lowermost and swollen part of the pistil is catted

  1. Ovule
  2. Ovary
  3. Style
  4. Filament

Answer: 2. Ovary.