Q.Consider the following reaction (species labelled (a)–(e) as printed in the Exemplar):
$\mathrm{\underset{(a)}{HO^-} + \underset{(b)}{CH_3Cl} \rightarrow \underset{(c)}{[HO\cdots CH_3\cdots Cl]^-} \rightarrow \underset{(d)}{CH_3OH} + \underset{(e)}{Cl^-}}$
In the printed diagram

The key idea is that the reaction shows a single-step, concerted substitution with inversion at carbon — the hallmark of $\mathrm{S_N2}$.

1. The diagram shows a trigonal-bipyramidal transition state (c) with partial bonds to both the incoming nucleophile ($\mathrm{HO^-}$) and the leaving group ($\mathrm{Cl^-}$). This is characteristic of an $\mathrm{S_N2}$ mechanism, not $\mathrm{S_N1}$ (which would involve a planar carbocation intermediate). …

Let’s understand why this reaction behaves the way it does. The key is the mechanism — the step-by-step path from reactants to products. The diagram shows a single step where the incoming $\mathrm{HO^-}$ approaches the carbon from the side opposite the leaving $\mathrm{Cl^-}$, passing through a high-energy transition state, and then forming $\mathrm{CH_3OH}$ with $\mathrm{Cl^-}$ expelled. This is the classic signature of an $\mathrm{S_N2}$ (substitution, nucleophilic, bimolecular) reaction.

In $\mathrm{S_N2}$, the nucleophile attacks the electrophilic carbon from the back side (the side opposite the leaving group). This forces the three other groups attached to that carbon to “umbrella” or invert — like an umbrella turning inside out in a strong wind. That inversion changes the spatial arrangement of the substituents, so if the starting carbon is chiral, the product has the opposite configuration.

Now, let’s examine each statement carefully.

1. Statement (i): The given reaction follows $\mathrm{S_N2}$ mechanism.

   Look at the species: (b) is $\mathrm{CH_3Cl}$ with tetrahedral geometry. (c) is the transition state $[\mathrm{HO\cdots CH_3\cdots Cl}]^-$ — a trigonal bipyramidal arrangement where both the incoming $\mathrm{HO}$ and the leaving $\mathrm{Cl}$ are partially bonded to the carbon. This is a single, concerted step with no intermediate. That is exactly what $\mathrm{S_N2}$ means: substitution, nucleophilic, bimolecular (both reactants are involved in the rate-determining step). So statement (i) is correct.

2. Statement (ii): (b) and (d) have opposite configuration.

   The diagram explicitly shows that the three H atoms in (b) are arranged tetrahedrally (one wedge, one dash, one in plane), and in the product (d) the “umbrella” of H atoms is drawn inverted. This inversion is a direct consequence of back-side attack. If the carbon were chiral (which it isn’t here because it has three identical H atoms, but the principle holds), the configuration would be reversed. For any $\mathrm{S_N2}$ reaction at a stereogenic centre, the product has the opposite configuration relative to the starting material. So statement (ii) is correct.

3. Statement (iii): (b) and (d) have same configuration.

   This directly contradicts the inversion shown in the diagram and the known mechanism of $\mathrm{S_N2}$. If the configuration were retained, that would suggest an $\mathrm{S_N1}$ mechanism (which proceeds through a planar carbocation intermediate and gives racemisation, not retention). But here, the product is clearly inverted. So statement (iii) is incorrect. …

Method: Mechanistic Analysis of Nucleophilic Substitution

This method uses the reaction profile (species, geometry, and bonding changes) to classify the mechanism and predict stereochemical outcome.

Steps

1. Identify the reaction type from the transition state geometry

• Species (c) is a trigonal-bipyramidal transition state with partial bonds to both the incoming nucleophile (HO⁻) and the leaving group (Cl⁻).
• This is the hallmark of an $\mathbf{S_N2}$ mechanism — a single, concerted step with a pentacoordinate transition state.
• In $\mathrm{S_N1}$, the intermediate would be a planar carbocation, not a trigonal-bipyramidal species.

2. Analyse the stereochemistry

• In $\mathrm{S_N2}$, the nucleophile attacks from the back side (opposite the leaving group).
• This causes inversion of configuration at the carbon centre — the “umbrella” of H atoms in (b) is inverted in (d). …

Let’s break this down step-by-step.

Step 1 – Identify the mechanism from the diagram

The diagram shows:

• A single step with a trigonal-bipyramidal transition state (c)
• The incoming HO⁻ and leaving Cl⁻ are both partially bonded in the transition state
• The product (d) has the umbrella of H atoms inverted compared to (b)

This is the classic picture of an Sₙ2 reaction — not Sₙ1.

Key fact:

• Sₙ2: one step, inversion of configuration, no carbocation intermediate.
• Sₙ1: two steps, racemization (or retention + inversion), carbocation intermediate.

So:

• (i) is correct — it follows Sₙ2.
• (iv) is incorrect — it does not follow Sₙ1.

Step 2 – Configuration of (b) and (d)

In Sₙ2, the nucleophile attacks from the back side, opposite the leaving group.

This causes inversion of configuration — like an umbrella turning inside out.

Thus:

• (b) and ** (d)** have opposite configuration.
• (ii) is correct, (iii) is incorrect.

Step 3 – Common mistakes students make