(a)(i) WHY $CH_3NH_2$ IS A STRONGER BASE THAN $(CH_3)_3N$ IN AQUEOUS SOLUTION

Step 1 - What decides basic strength in water.
In AQUEOUS solution the basic strength of an amine is governed by the COMBINED effect of
three factors:
(1) the $+I$ (electron-releasing) inductive effect of the alkyl groups,
(2) STERIC HINDRANCE around the nitrogen atom, and
(3) the extent to which the protonated cation (conjugate acid) is STABILISED BY
HYDROGEN BONDING with water molecules (solvation).

Step 2 - What the inductive effect alone would predict.
Each $-CH_3$ group pushes electron density towards nitrogen. On this basis alone,
$(CH_3)_3N$ (three methyl groups) should be the strongest base. But this is NOT what
is observed in water, so the other two factors must dominate.

Step 3 - Compare the solvation of the conjugate acids.
$CH_3\overset{+}{N}H_3$ has THREE N-H bonds $\Rightarrow$ it can form THREE hydrogen bonds
with water $\Rightarrow$ it is HIGHLY SOLVATED and therefore GREATLY STABILISED.
$(CH_3)_3\overset{+}{N}H$ has only ONE N-H bond $\Rightarrow$ it can form only ONE hydrogen
bond with water $\Rightarrow$ it is POORLY SOLVATED and much less stabilised.

Step 4 - Add the steric factor.
The three bulky methyl groups in $(CH_3)_3N$ crowd around the nitrogen atom and
STERICALLY HINDER the approach of a proton, making protonation difficult.

Step 5 - Conclusion.
Because the conjugate acid of $CH_3NH_2$ is far better stabilised by hydration, and
because its nitrogen is sterically free, the equilibrium
$$CH_3NH_2 + H_2O \rightleftharpoons CH_3\overset{+}{N}H_3 + OH^-$$
lies further to the right. Hence $CH_3NH_2$ IS A STRONGER BASE THAN $(CH_3)_3N$ IN
AQUEOUS SOLUTION.

(a)(ii) STRUCTURAL FORMULAE OF A AND B

Step 1 - Identify the first reaction.
$CH_3CONH_2$ (ethanamide, an amide) treated with $NaOBr$ (i.e. $Br_2$ + NaOH) undergoes
the HOFMANN BROMAMIDE DEGRADATION REACTION.
In this reaction an amide is converted into a PRIMARY AMINE containing ONE CARBON ATOM
LESS than the amide.

Step 2 - Write A.
$$CH_3CONH_2 \xrightarrow{NaOBr} CH_3-NH_2$$
$$\textbf{A} = CH_3NH_2 \quad \textbf{(Methanamine / Methylamine)}$$

Step 3 - Identify the second reaction.
A primary amine reacting with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base
undergoes ACYLATION (benzoylation), also called the SCHOTTEN-BAUMANN REACTION. The base
removes the HCl formed and drives the reaction forward. The product is a substituted
AMIDE.

Step 4 - Write B.
$$CH_3NH_2 + C_6H_5COCl \xrightarrow{\text{Base}} C_6H_5CO-NH-CH_3 + HCl$$
$$\textbf{B} = C_6H_5CONHCH_3 \quad \textbf{(N-Methylbenzamide)}$$

(b) IDENTIFICATION OF 'X'

Step 1 - Recall the Hinsberg test.
Hinsberg reagent is BENZENESULPHONYL CHLORIDE, $C_6H_5SO_2Cl$.
- A PRIMARY amine gives a sulphonamide that STILL HAS ONE HYDROGEN on nitrogen. This
hydrogen is acidic (because of the two strongly electron-withdrawing $S=O$ groups),
so the product DISSOLVES IN ALKALI, forming a salt.
- A SECONDARY amine gives a sulphonamide with NO HYDROGEN left on nitrogen. It cannot
form a salt, so the product is INSOLUBLE IN ALKALI.
- A TERTIARY amine has no N-H at all, so it DOES NOT REACT.

Step 2 - Apply the observation.
'X' REACTS with the reagent (so it is not tertiary) and the product is INSOLUBLE IN
ALKALI $\Rightarrow$ 'X' must be a SECONDARY AMINE.

Step 3 - Find the secondary amine of formula $C_3H_9N$.
The isomers of $C_3H_9N$ are:
$CH_3CH_2CH_2NH_2$ (propan-1-amine) - primary
$(CH_3)_2CHNH_2$ (propan-2-amine) - primary
$CH_3-NH-CH_2CH_3$ (N-methylethanamine) - SECONDARY
$(CH_3)_3N$ (trimethylamine) - tertiary

ANSWER: $\textbf{'X'} = CH_3-NH-C_2H_5$, N-METHYLETHANAMINE (ethylmethylamine)
$$C_6H_5SO_2Cl + CH_3NH C_2H_5 \rightarrow C_6H_5SO_2N(CH_3)(C_2H_5) + HCl$$
(the product has no N-H, hence insoluble in alkali)

(c) WHY THE $-NH_2$ GROUP OF ANILINE IS ACETYLATED BEFORE NITRATION

Step 1 - The problem with direct nitration.
The $-NH_2$ group is a VERY STRONG activator. Direct nitration of aniline creates three
difficulties:
(1) Nitrating mixture is strongly acidic, so aniline is protonated to the ANILINIUM ion,
$-\overset{+}{N}H_3$, which is meta-directing. A large amount (about 51%) of the
unwanted META product is formed.
(2) The ring is so highly activated that POLYSUBSTITUTION occurs.
(3) $HNO_3$ is an oxidising agent and aniline is easily oxidised, so it is partly
destroyed, giving dark tarry oxidation products and a poor yield.

Step 2 - How acetylation solves it.
Aniline is first treated with acetic anhydride to give ACETANILIDE:
$$C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_6H_5NHCOCH_3$$
In acetanilide, the lone pair on nitrogen is partly delocalised towards the electron-
withdrawing $C=O$ group of the acetyl group. Nitrogen therefore donates LESS electron
density into the ring.

Step 3 - The benefits.
(1) The ring is now only MODERATELY activated - it is CONTROLLED, so mono-substitution
occurs instead of polysubstitution.
(2) The amide nitrogen is far less basic, so protonation is avoided and the meta product
is not formed.
(3) The $-NHCOCH_3$ group protects the amine from OXIDATION by $HNO_3$.
(4) The bulky $-NHCOCH_3$ group blocks the ortho position sterically, so substitution
occurs predominantly at the PARA position.

Step 4 - Deprotection.
After nitration, the acetyl group is removed by acid or alkaline HYDROLYSIS to
regenerate the free amine:
$$p\text{-}O_2N\text{-}C_6H_4\text{-}NHCOCH_3 \xrightarrow[\text{or } OH^-/H_2O]{H_3O^+} p\text{-Nitroaniline}$$
Thus a good yield of pure p-nitroaniline is obtained.