Q.Answer on the basis of the following reaction (species labelled (a)–(d) as printed in the Exemplar):
$\mathrm{\underset{(a)}{HO^-} + \underset{(b)}{CH_3CH(Cl)CH_2CH_3} \rightarrow \underset{(c)}{CH_3CH(OH)CH_2CH_3} + \underset{(d)}{Cl^-}}$
(2-chlorobutane; in the printed diagram the central carbon of

The key idea is that the printed diagram itself tells you the mechanism: the product (c) is drawn with exactly the same spatial arrangement as the substrate (b) — $\mathrm{OH}$ simply occupies the in-plane position that $\mathrm{Cl}$ held, with $\mathrm{C_2H_5}$ still on the hashed bond and H still on the wedge. The configuration as drawn is not inverted.

Reasoning:

1. An $\mathrm{S_N2}$ reaction is a single concerted backside attack — it must invert the configuration at the carbon. A product drawn with its configuration unchanged therefore cannot have come from $\mathrm{S_N2}$, so statements (ii) and (iii) (the simultaneous attack/departure and its weakly-bonded transition state) do not describe this reaction. …

1. What the printed diagram actually shows

In the Exemplar's drawing, the substrate (b), 2-chlorobutane, has $\mathrm{CH_3}$ up, $\mathrm{C_2H_5}$ on a hashed bond, H on a wedge and $\mathrm{Cl}$ in the plane. The product (c), butan-2-ol, is drawn with the identical arrangement — $\mathrm{C_2H_5}$ still hashed, H still on the wedge, and $\mathrm{OH}$ sitting exactly where $\mathrm{Cl}$ was. In other words, the configuration at the stereocentre is drawn unchanged.

2. Why this rules out $\mathrm{S_N2}$

$\mathrm{S_N2}$ is a one-step, concerted process in which the nucleophile attacks from the side opposite the leaving group. That backside attack always turns the carbon inside-out (Walden inversion) — the product of an $\mathrm{S_N2}$ reaction must be drawn inverted. Since the printed product is not inverted, the reaction shown cannot be the concerted pathway. That eliminates:

• (ii) — "$\mathrm{OH^-}$ attaches from one side while $\mathrm{Cl^-}$ leaves simultaneously from the other" is precisely the concerted $\mathrm{S_N2}$ step; and
• (iii) — the "unstable intermediate in which $\mathrm{OH^-}$ and $\mathrm{Cl^-}$ are attached by weak bonds" is the partially-bonded $\mathrm{S_N2}$ transition state. On the $\mathrm{S_N1}$ path no such species arises: $\mathrm{Cl^-}$ has already left before $\mathrm{OH^-}$ arrives, so the two are never bonded to the carbon at the same time.

3. The mechanism the drawing depicts: $\mathrm{S_N1}$

2-Chlorobutane is a secondary alkyl halide — the borderline class that can react by either pathway depending on conditions, so the substrate class alone cannot decide the question. Here the stereochemical outcome decides it. In $\mathrm{S_N1}$:

1. The C–Cl bond ionises first (slow step), giving a planar, $sp^2$ carbocation and $\mathrm{Cl^-}$.
2. $\mathrm{OH^-}$ then attacks the flat carbocation (fast step) — from either face.

Because the intermediate is planar, attack on its two faces gives both possible configurations (which is why $\mathrm{S_N1}$ leads to racemisation overall). The molecule the Exemplar draws — the one that keeps the original arrangement — is a product only the carbocation pathway can deliver; a concerted backside attack could never give it. …

Method: Let the drawn stereochemistry decide the mechanism

Step 1 — Identify the substrate class

The substrate (b) is 2-chlorobutane, a secondary alkyl halide. Secondary substrates are the borderline case — they can react by $S_N1$ or $S_N2$ — so the substrate class alone cannot answer the question.

Step 2 — Read the printed drawing, group by group

• (b): $CH_3$ up, $C_2H_5$ on the hashed bond, H on the wedge, $Cl$ in the plane.
• (c): $CH_3$ up, $C_2H_5$ on the hashed bond, H on the wedge, $OH$ in the plane (where $Cl$ was).

The arrangement is unchanged — the drawn configuration is NOT inverted.

Step 3 — Apply the stereochemical rule

• $S_N2$: concerted backside attack ⇒ always inversion. A product drawn un-inverted cannot come from $S_N2$.
• $S_N1$: planar carbocation intermediate, attacked from either face ⇒ both configurations form; the un-inverted product the book draws is possible only via this pathway. …

Here's a breakdown of the common mistakes students make on this question, and how to avoid each.

Mistake 1: "Strong nucleophile, so it must be $\mathrm{S_N2}$"

Why it's wrong: $\mathrm{OH^-}$ is indeed a strong nucleophile, and as a general trend a secondary halide with a strong nucleophile often reacts by $\mathrm{S_N2}$. But this question does not ask about general trends — it says "on the basis of the following reaction", and the printed drawing shows the product with its configuration unchanged. A concerted $\mathrm{S_N2}$ attack must invert the carbon, so the outcome that is actually drawn cannot be $\mathrm{S_N2}$.

How to avoid: When a stereochemical drawing is given, it outranks a rule of thumb. Check whether the drawn product is inverted before reaching for the "strong nucleophile ⇒ $\mathrm{S_N2}$" shortcut.

Mistake 2: Misreading the drawing as an inversion

The error: Students glance at the wedge/hash marks, see stereo-bonds in both structures, and assume the familiar "inversion" story.

Why it's wrong: Compare the two drawn molecules group-by-group: in both, $\mathrm{C_2H_5}$ is on the hashed bond, H is on the wedge and $\mathrm{CH_3}$ points up; $\mathrm{OH}$ simply occupies the in-plane position $\mathrm{Cl}$ held. Nothing has flipped.

How to avoid: For every stereochemistry question, map each group's wedge/hash/in-plane role in the substrate and the product side by side before deciding anything.

Mistake 3: Picking (iii) because "substitution has a partially-bonded species"

The error: Option (iii)'s "unstable intermediate in which $\mathrm{OH^-}$ and $\mathrm{Cl^-}$ are attached by weak bonds" sounds familiar from the $\mathrm{S_N2}$ transition state, so students tick it.

Why it's wrong: That partially-bonded species exists only on the concerted ($\mathrm{S_N2}$) path. In $\mathrm{S_N1}$ — the mechanism this drawing depicts — the C–Cl bond breaks completely before the nucleophile arrives; at no point are $\mathrm{OH}$ and $\mathrm{Cl}$ both weakly bonded to the carbon.

How to avoid: First settle the mechanism, then judge every option against that one mechanism — never mix descriptions from the two pathways.

Mistake 4: Forgetting the multi-select instruction …