(a) STRUCTURAL FORMULAE OF 'X' AND 'Y'

Step 1 - Deduce the mechanism for 'X'.
Rate depends ONLY on $[X]$:
$$\text{Rate} = k[X]$$
This is FIRST ORDER (unimolecular) $\Rightarrow$ the reaction follows the $S_N1$ mechanism.
$S_N1$ is favoured by TERTIARY halides, because the $3^\circ$ carbocation intermediate is
the most stable (stabilised by $+I$ and hyperconjugation of three alkyl groups).

Step 2 - Identify 'X'.
The only tertiary isomer of $C_4H_9Br$ is:
$$\textbf{'X'} = (CH_3)_3C-Br \qquad \textbf{2-Bromo-2-methylpropane (tert-butyl bromide)}$$

Step 3 - Deduce the mechanism for 'Y'.
Rate depends on BOTH concentrations:
$$\text{Rate} = k[Y][OH^-]$$
This is SECOND ORDER (bimolecular) $\Rightarrow$ the reaction follows the $S_N2$ mechanism.
$S_N2$ is favoured by primary and secondary halides (less steric crowding).

Step 4 - Identify 'Y'.
'Y' must also be $C_4H_9Br$ AND must be OPTICALLY ACTIVE, i.e. it must contain a CHIRAL
CARBON (a carbon with four different groups). The only such isomer is:
$$\textbf{'Y'} = CH_3-CH_2-\overset{*}{C}H(Br)-CH_3 \qquad \textbf{2-Bromobutane (sec-butyl bromide)}$$
The starred C-2 carries $-H$, $-Br$, $-CH_3$ and $-C_2H_5$, four different groups.

(b) WHICH ONE UNDERGOES RACEMISATION - AND WHY

ANSWER: 'X' (the $S_N1$ substrate).

Reason - step by step:
1. In the $S_N1$ mechanism the C-Br bond breaks FIRST (slow step), producing the
carbocation $(CH_3)_3C^+$.
2. This carbocation is $sp^2$ hybridised and therefore PLANAR (trigonal planar), with an
empty $p$ orbital perpendicular to the plane.
3. The nucleophile $OH^-$ can now attack this flat intermediate from EITHER FACE with
EQUAL PROBABILITY.
4. Attack from one face gives retention, attack from the other gives inversion, and the
two are formed in EQUAL AMOUNTS.
5. The result is a 50:50 mixture of the two enantiomers, i.e. a RACEMIC MIXTURE whose
net optical rotation is ZERO. This process is called RACEMISATION.

(c) WHICH ONE GIVES INVERSION OF CONFIGURATION - AND WHY

ANSWER: 'Y' (the $S_N2$ substrate).

Reason - step by step:
1. The $S_N2$ mechanism is a ONE-STEP (concerted) process: bond formation and bond
breaking occur simultaneously through a single transition state.
2. The nucleophile $OH^-$ must attack from the side EXACTLY OPPOSITE to the leaving
group $Br^-$ (BACK-SIDE ATTACK), because a front-side approach is blocked by the
electron cloud of the departing halide.
3. In the transition state the carbon is $sp^2$ hybridised, with $OH$ and $Br$ partially
bonded on opposite sides.
4. As the reaction completes, the other three bonds are pushed through to the other side,
exactly like an umbrella turning inside out in a strong wind.
5. The configuration at the chiral carbon is therefore INVERTED. This is called
WALDEN INVERSION, and the product is the enantiomer of the expected retained form.