CBSE Class 12 Chemistry Amines Case Study Questions Set 02

Case Study 1

The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their very high reactivity. Substitution tends to occur at ortho and para-positions. To prepare monosubstituted aniline derivative, the activating affect of –NH₂ group be controlled. This can be done by protecting the –NH₂ group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.

Case Study 2

The conversion of primary amines into diazonium salts is known as diazotization. Arene diazonium salts are generally colourless crystalline solids highly soluble in water. These salts are more stable than aliphatic diazonium salts and undergo a number of substitution reactions due to excellent leaving ability of diazo group as N₂. Arene diazonium salts also couple with phenols and amines to form coloured azo dyes. Such type of reactions are known as coupling reactions.

Question 5. Why are arene diazonium salts more stable than aliphatic diazonium salts?

Answer:
Arene diazonium salts are more stable than aliphatic diazonium salts due to dispersal of positive charge on the benzene ring as shown below.

Teacher's Note:
1. The positive charge spreads to the two ortho and the para carbons – draw all five structures with correct arrows in the exam.

Question 6. How will you convert: Aniline to fluorobenzene?
Answer:
C₆H₅NH₂ (Aniline) →[NaNO₂/HCl, 273–278 K] C₆H₅N₂⁺Cl⁻ →[HBF₄] C₆H₅N₂⁺BF₄⁻ →[Δ] C₆H₅F (Fluorobenzene)

Teacher's Note:
1. Fluorobenzene cannot be made by Sandmeyer reaction; the diazonium fluoroborate route (Balz–Schiemann) is the one to write.

Question 7. Why are benzenediazonium salts soluble in water?
Answer:
Being ionic, they are soluble in water.

Teacher's Note:
1. Benzenediazonium chloride is a salt (C₆H₅N₂⁺ and Cl⁻), so the ions are hydrated by water.

Question 8. What is diazotisation?
Answer:
The reaction of converting aromatic primary amines into diazonium salts by treatment with a solution of nitrous acid at 273 K–278 K is called diazotisation.

Teacher's Note:
1. Always mention the low temperature (273–278 K) – above it the diazonium salt decomposes to phenol.

Question 9. What product is formed when aniline is first diazotised and then reacted with phenol in the alkaline medium?
Answer:
p-hydroxyazobenzene.

Teacher's Note:
1. Coupling with phenol occurs at the para position and gives an orange dye; with aniline it gives yellow p-aminoazobenzene.

Case Study 3

Amines have a lone pair of electrons on nitrogen atom due to which they behave as Lewis base. Larger the value of K_b or smaller the value of pK_b stronger is the base. Amines are more basic than alcohols, ethers, esters, etc. The basic character of aliphatic amines should increase with the increase of alkyl substitution. But it does not occur in a regular manner as a secondary aliphatic amine is unexpectedly more basic than a tertiary amine in solutions. Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron-donating groups such as —CH₃, —OCH₃, —NH₂, etc., increase the basicity while electron-withdrawing substituents such as —NO₂, —CN, halogens, etc. decrease the basicity of amines. The effect of these substituents is more at p- than at m-positions.

Question 10. Arrange the following in increasing order of their basic strength:
C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂, (C₂H₅)₂NH
Answer:
C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH

Teacher's Note:
1. Benzylamine is an aliphatic amine (N is not on the ring), so it is far more basic than aniline, though slightly less basic than ethylamine.

Question 11. Arrange the following compounds in increasing order of their acidic strength:
Methylamine, dimethylamine, aniline, N-methylaniline
Answer:
dimethylamine < methylamine < N-methylaniline < aniline.

Teacher's Note:
1. The acidic strength order of amines is simply the reverse of their basic strength order.

Question 12. Rearrange the following in increasing order of their basic strength:
p-toluidine, N, N-dimethyl-p-toluidine, p-nitroaniline, aniline
Answer:
p-nitroaniline < aniline < p-toluidine < N, N-dimethyl-p-toluidine.

Teacher's Note:
1. Each extra electron-donating CH₃ group (on the ring or on N) raises the electron density on nitrogen; –NO₂ lowers it.

Question 13. Which is more acidic, aniline or ammonia?
Answer:
Due to delocalization of the lone pair of electrons of the N-atom of aniline over the benzene ring, aniline is more acidic than ammonia.

Teacher's Note:
1. The anilide ion C₆H₅NH⁻ is resonance-stabilised by the ring, but NH₂⁻ is not.

Question 14. (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution. Give reason.
Answer:
The basicity of amine in aqueous solution depends upon the stability of the substituted ammonium cation. Here the combination of three factors, +ve I effect of CH₃ groups, hydrogen bonding and steric hindrance favour greater stability for ammonium cation of dimethyl amine than ammonium cation of trimethyl amine. Hence dimethylamine is stronger base than trimethyl amine.

Teacher's Note:
1. In the gas phase, where there is no solvation, the order is simply 3° > 2° > 1° by +I effect.

Case Study 4

Ammolysis of alkyl halides gives a mixture of 1°, 2° and 3° amines. The actual composition of the mixture, however, depends upon the ratio of the alkyl halide and ammonia used. The boiling points of three isomeric 1°, 2° and 3° amines depend upon the extent of H-bonding. Further, due to the presence of a lone pair of electrons on the nitrogen atom, amines undergo acylation with acid chlorides and anhydrides.

Question 15. Aniline cannot be prepared by ammonolysis of chlorobenzene under normal conditions. Explain why ?
Answer:
In case of chlorobenzene, the C–Cl bond is quite difficult to break as it has acquired a partial double bond character due to resonance over the benzene ring. Therefore, under normal condition, ammonolysis of chlorobenzene does not yield aniline.

Teacher's Note:
1. Under drastic conditions (Cu₂O catalyst, about 475 K and high pressure) chlorobenzene does react with NH₃ to give aniline.

Question 16. N-Ethylethanamine boils at 329.3 K and butanamine boils at 350.8 K although both are isomeric in nature. Comment.
Answer:
(CH₃CH₂)₂N–H : N-Ethylethanamine (2°), M.F. C₄H₁₁N, b.p. 329.3 K ; CH₃CH₂CH₂CH₂NH₂ : Butanamine (1°), M.F. C₄H₁₁N, b.p. 350.8 K
2° Amines have only one while 1° amines have two hydrogen atoms on the N-atom. As a result, intermolecular association between H of one molecule and N of the other molecule is more in 1° amines than is 2° amines. Consequently, butanamine being a 1° amine has higher b.p. of 350.8 K than N-ethylethanamine being a 2° amine has lower b.p. of 329.3 K.

Teacher's Note:
1. Branching also lowers boiling point by reducing surface area, which adds to the effect of fewer N–H bonds in the 2° amine.

Question 17. Acylation of amines is carried out in the presence of pyridine. Give a plausible reason.
Answer:
During acylation of aniline, strong base pyridine is added. This is done in order to neutralize the HCl formed during the reaction and to shift the equilibrium to the right hand side.
C₆H₅NH₂ (Aniline) + Cl–CO–CH₃ (Acetyl chloride) \( \xrightarrow{\mathrm{Pyridine}} \) C₆H₅–NH–CO–CH₃ (Acetanilide) + HCl

Teacher's Note:
1. Without pyridine, the HCl formed would convert unreacted amine into its salt, which cannot be acylated.

Question 18. Why is CH₃NH₂ more basic than (CH₃)₃N while (CH₃CH₂)₃N is more basic than CH₃CH₂NH₂ ? Give reasons for your answer.
Answer:
The reversal of order of basicity of methyl and ethyl primary and tertiary amines in due to steric hindrance to H-bonding as explained below :
If the alkyl group is small, i.e., CH₃ then there is no steric hindrance to H-bonding of the conjugate acid formed by addition of a proton by H₂O molecules. In other words, the stability due to H-bonding predominates over the stability due to +I-effect of the CH₃ group and hence CH₃NH₂ is a stronger base than (CH₃)₃N.
However, if alkyl group is bigger than CH₃ group, i.e., C₂H₅ group, there will be some steric hindrance to H-bonding. As a result, stability due to +I-effect predominates over the stability due to H-bonding and hence (CH₃CH₂)₃N is more basic than CH₃CH₂NH₂.

Teacher's Note:
1. Data check: pK_b of CH₃NH₂ = 3.38, (CH₃)₃N = 4.22, (C₂H₅)₃N = 3.25 and C₂H₅NH₂ = 3.29.

Case Study 5

Due to the presence of a lone pair of electrons on the nitrogen atom, amines act as Lewis bases. The basicity of aliphatic amines in aqueous solution, depends upon three factors : (i) +I-effect of the alkyl group, (ii) extent of H-bonding with water molecules and (iii) steric effect of the alkyl groups. Aromatic amines, on the other hand, are weaker bases than aliphatic amines due to delocalization of the lone pair of electrons of nitrogen over the benzene ring. Whereas electron-donating groups increase while electron-withdrawing groups decrease the basicity of amines. Further the base-strengthening effect of electron-donating groups and base-weakening effect of electron-withdrawing group is more marked at p-position than at m-position. However, due to ortho effect, o-substituted anilines are weaker bases than anilines regardless of the nature of substituent whether electron-donating or electron-withdrawing.

Question 19. Why is an alkylamine more basic than ammonia ? Give reason.
Answer:
Due to +I-effect of the alkyl group, the electron density on the N-atom of alkylamine (RNH₂) increases as compared to that on the N atom of ammonia (NH₃). As a result, alkylamine can more easily donate a pair electrons than NH₃ and hence alkylamine is more basic than NH₃.

Teacher's Note:
1. pK_b of methylamine (3.38) is lower than that of NH₃ (4.75), so it is the stronger base.

Question 20. Explain why aniline is more acidic than ammonia?
Answer:
Loss of a proton from aniline gives the anilide ion (I) while that from NH₃ gives the amide ion (II) :
C₆H₅NH₂ → C₆H₅N⁻H (I) + H⁺ ; NH₃ → NH₂⁻ (II) + H⁺
Since ion (I) is stabilized by resonance (the negative charge is delocalised over the benzene ring) while ion (II) is not, therefore, aniline is more acidic than ammonia.

Teacher's Note:
1. The same delocalisation of the N lone pair that makes aniline a weaker base makes it a (very weak) stronger acid than NH₃.

Question 21. Acetamide is a weaker base than ethylamine. Explain why ?
Answer:
Due to resonance, the lone pair of electrons on the nitrogen atom in CH₃CONH₂ is delocalized over the carbonyl group, C=O (CH₃–C(=O)–NH₂ ↔ CH₃–C(O⁻)=N⁺H₂). As a result, electron density on the N-atom in acetamide decreases. In contrast, in ethylamine, due to +I-effect of the ethyl group, electron density on the N-atom increases. Consequently, CH₃CONH₂ is a weaker base than CH₃CH₂NH₂.

Teacher's Note:
1. Amides are practically neutral; protonation, when it happens, occurs on the O atom, not on N.

Question 22. Arrange o-, m- and p-nitroanilines in increasing order of their basicity.
Answer:
The NO₂ group is electron-withdrawing and decreases the basicity of aniline. Its base-weakening effect is more marked at the p-position (−I and −R effects) than at the m-position (only −I effect), and o-nitroaniline is the weakest base due to ortho effect.
Increasing order of basicity : o-nitroaniline < p-nitroaniline < m-nitroaniline

Teacher's Note:
1. All three are weaker bases than aniline itself; pK_b values are about 14.3 (o), 13.0 (p) and 11.5 (m).

Case Study 6

Amines are vital organic compounds since they are widely distributed in nature in form of proteins, vitamins, hormones and alkaloids. In the laboratory, they can be prepared by a number of methods. These include reduction of cyanides, isocyanides, amides and nitro compounds. Besides these, two special methods, viz., Gabriel phthalimide reaction and Hofmann bromamide reaction are usually used for preparation of primary amines.

Case Study 7

Benzenediazonium salts are very versatile chemical intermediates which are used for preparation of a number of different classes of organic compounds. Since diazo group is a good leaving group, therefore, it can be easily substituted by a number of monovalent atoms or groups such as H, F, Cl, Br, I, OH, CN, NO₂, C₆H₅, etc. Besides these, diazo group can be reduced to hydrazines and diazonium salts undergo coupling reactions with amines in acidic medium and phenols in basic medium to form highly coloured azo compounds.