(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) CONVERSION OF ANILINE TO BENZONITRILE

This is a two-step conversion through a diazonium salt.

Step 1 - DIAZOTISATION.
Aniline is treated with nitrous acid ($NaNO_2$ + dilute HCl, generated in situ) at a low
temperature of 273-278 K (0-5 $^\circ$C). Benzenediazonium chloride is formed.
$$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278\ K} C_6H_5\overset{+}{N_2}Cl^- + NaCl + 2H_2O$$
The low temperature is essential because the diazonium salt decomposes above 278 K,
giving phenol.

Step 2 - SANDMEYER REACTION.
The diazonium solution is warmed with cuprous cyanide (CuCN) dissolved in KCN. The
diazonium group is replaced by the cyanide group and nitrogen gas is evolved.
$$C_6H_5\overset{+}{N_2}Cl^- + CuCN/KCN \xrightarrow{\ \Delta\ } C_6H_5CN + N_2 \uparrow + CuCl$$
$$\textbf{Benzonitrile}$$

(An alternative for step 2 is the GATTERMANN REACTION, using copper powder with HCN/KCN,
which gives a slightly lower yield.)

Note: This diazonium route is the standard method, because direct replacement of the
$-NH_2$ group of aniline by $-CN$ is not possible.

(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) The 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 products and a poor yield.

Step 2 - How acetylation solves it.
Aniline is first converted to ACETANILIDE with acetic anhydride:
$$C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_6H_5NHCOCH_3$$
The nitrogen lone pair is now partly pulled towards the electron-withdrawing $C=O$
group, so less electron density is released into the ring.

Step 3 - The benefits.
(1) The ring is only MODERATELY activated, so substitution is CONTROLLED and stops at
mono-substitution.
(2) The amide nitrogen is much less basic, so protonation - and hence meta substitution -
is avoided.
(3) The $-NHCOCH_3$ group PROTECTS the amine from oxidation by $HNO_3$.
(4) The bulky acetyl group sterically blocks the ortho position, so substitution occurs
predominantly at the PARA position.

Step 4 - Deprotection.
After nitration, hydrolysis removes the acetyl group and regenerates the free amine,
giving a good yield of pure p-nitroaniline.