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 this is an S_N2 reaction, proceeding through a single transition state with inversion of configuration.

Step 1 – Identify the species. (a) $\mathrm{HO^-}$ is a nucleophile (donates a lone pair). (e) $\mathrm{Cl^-}$ is a leaving group (departs with the bonding pair). Both are electron-rich, so neither is an electrophile.

Step 2 – Analyse the transition state (c).

In the trigonal-bipyramidal transition state, the central carbon is simultaneously bonded (partially) to five atoms: the incoming $\mathrm{HO}$, the leaving $\mathrm{Cl}$, and the three H atoms. This geometry requires the carbon to be $sp^2$ hybridised — the three equatorial bonds (to H atoms) lie in a plane, while the axial partial bonds use the remaining p-orbital.

Step 3 – Evaluate each statement. …

1. Identify the reaction type

The species shown — $\mathrm{HO^-}$ attacking $\mathrm{CH_3Cl}$ through a single transition state with simultaneous bond-making and bond-breaking — is the classic $S_N2$ (bimolecular nucleophilic substitution) mechanism. The key signature: one step, no intermediates, and inversion of configuration at carbon (the "umbrella flip" mentioned in the diagram).

2. What is a nucleophile? What is an electrophile?

• Nucleophile: "nucleus-loving" — a species that donates a lone pair to form a new bond. It is electron-rich.
• Electrophile: "electron-loving" — a species that accepts a lone pair. It is electron-deficient.

In this reaction:

• (a) $\mathrm{HO^-}$ has a lone pair and a negative charge — it is the nucleophile that attacks carbon.
• (e) $\mathrm{Cl^-}$ is produced when the C–Cl bond breaks. It takes both electrons from that bond, so it is a leaving group. A leaving group is itself a nucleophile (it can donate a lone pair in other reactions), but in this specific step, it is not acting as a nucleophile — it is departing. However, the question asks about the species themselves, not their role in this single step. $\mathrm{Cl^-}$ is indeed a nucleophile in general (it can attack electrophiles). So statement (i) is correct: both (a) and (e) are nucleophiles.

Statement (iv) says both are electrophiles — false. $\mathrm{HO^-}$ is not electron-deficient; it is electron-rich. $\mathrm{Cl^-}$ is also not an electrophile (it has no empty orbital to accept electrons at low energy). So (iv) is wrong.

3. Hybridisation of carbon in the transition state (c)

In the reactant $\mathrm{CH_3Cl}$, carbon is $sp^3$ hybridised (four sigma bonds, tetrahedral geometry). In the product $\mathrm{CH_3OH}$, carbon is again $sp^3$ hybridised (tetrahedral). But what about the transition state (c)?

In the $S_N2$ transition state, the carbon is simultaneously bonded to:

• three H atoms (via sigma bonds),
• the incoming nucleophile $\mathrm{HO^-}$ (partial bond),
• the leaving group $\mathrm{Cl^-}$ (partial bond).

That makes five groups around carbon. This is a pentavalent carbon. For five electron pairs, the geometry is trigonal bipyramidal (as the problem states). The three H atoms lie in a plane (equatorial positions), and the incoming and leaving groups occupy the axial positions (above and below the plane). …

Concept: $S_N2$ Reaction Mechanism and Hybridisation in the Transition State

The reaction shown is a classic bimolecular nucleophilic substitution ($S_N2$). The key features are:

• One-step, concerted mechanism (bond breaking and bond forming happen simultaneously).
• The nucleophile attacks from the backside relative to the leaving group.
• This causes inversion of configuration at the carbon centre (the "umbrella" of H atoms flips).

Method: Identify the Role of Species and Hybridisation in the Transition State

Step 1: Identify nucleophiles and electrophiles

• (a) $\mathrm{HO^-}$: Has a lone pair and a negative charge → nucleophile.
• (e) $\mathrm{Cl^-}$: Also has lone pairs and a negative charge → by its inherent chemical character, also a nucleophile — even though its ROLE in this specific forward reaction is as the leaving group, that doesn't strip it of its underlying classification (chloride is a perfectly good nucleophile in plenty of other reactions).
• (b) $\mathrm{CH_3Cl}$: The carbon is partially positive (due to Cl's electronegativity) → electrophile.

Result: Both (a) and (e) are nucleophiles by character.

✓ Statement (i) is correct.

✗ Statement (iv) is incorrect (neither is electron-deficient).

Step 2: Determine hybridisation of carbon in the transition state (c)

• In the transition state, the carbon is simultaneously bonded to:
  • Three H atoms (from original $\mathrm{CH_3}$) — these become coplanar as the "umbrella" flattens mid-inversion. …

Let's break this down step by step.

What is the reaction?

This is a classic $S_N2$ reaction:

• Nucleophile: $\mathrm{HO^-}$ (a)
• Substrate: $\mathrm{CH_3Cl}$ (b)
• Transition state: $\mathrm{[HO\cdots CH_3\cdots Cl]^-}$ (c)
• Products: $\mathrm{CH_3OH}$ (d) and $\mathrm{Cl^-}$ (e)

The key features of $S_N2$:

• One step, no true intermediates (only a transition state).
• Backside attack by the nucleophile.
• Inversion of configuration at carbon (the "umbrella inversion" mentioned in the diagram).

Common mistakes and how to avoid them

✗ Mistake 1: Thinking (e) can't be a nucleophile because it's the leaving group here (rejecting Option (i))

The error: Students argue that since $\mathrm{Cl^-}$ is leaving in this specific reaction, it can't also be labelled a "nucleophile" -- so they reject option (i).

Why it's wrong: "Nucleophile" describes a species' inherent electronic character (electron-rich, has a lone pair to donate) -- not its role in one particular reaction. $\mathrm{Cl^-}$ genuinely is a nucleophile by nature (it's a perfectly good nucleophile in plenty of OTHER reactions); the fact that it happens to be leaving here, rather than attacking, doesn't strip it of that classification.

✓ How to avoid: Separate "what kind of species is this?" (a classification question) from "what role is it playing in this specific reaction?" (a mechanism question). Option (i) is asking the classification question, and by that measure, both (a) and (e) genuinely are nucleophiles.

✗ Mistake 2: Saying carbon in (c) is $sp^3$ hybridised (Option (ii))

Why it's wrong:

• In the transition state (c), the carbon is partially bonded to both the incoming $\mathrm{HO}$ and the leaving $\mathrm{Cl}$, on top of its three full bonds to H.
• $sp^3$ would mean four ordinary, full bonds in a tetrahedral arrangement -- that's not what's happening here (there are 3 full bonds + 2 partial bonds).

✓ How to avoid: Remember carbon has no accessible d-orbitals (it's period 2), so "trigonal bipyramidal, therefore $sp^3d$" -- the answer that works for phosphorus or sulfur -- doesn't apply to carbon. The three H atoms stay on $sp^2$ orbitals (becoming coplanar as the umbrella flattens), and the incoming/leaving groups share the remaining p-orbital's two lobes.

✗ Mistake 3: Saying carbon in (c) is $sp^2$ hybridised (Option (iii)) — this is correct

✓ How to confirm: As above -- the three H atoms are on $sp^2$ orbitals, and the partial bonds to the incoming/leaving groups use the unhybridised p-orbital. This is the standard, correct description of the $S_N2$ transition state's carbon (no d-orbitals needed or available).

✗ Mistake 4: Thinking both (a) and (e) are electrophiles (Option (iv))

Why it's wrong:

• (a) $\mathrm{HO^-}$ is electron-rich — it's a nucleophile, not an electrophile. …