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 (c) is the transition state of an $S_N2$ reaction, not a stable intermediate.

Reasoning:

1. In $S_N2$, the nucleophile attacks as the leaving group departs in a single step — no intermediate is formed.
2. Species (c) has a pentacoordinate carbon with partial bonds, which is a high-energy transition state, not a stable intermediate. …

1. Identify the reaction type.

   The species shown — $\mathrm{HO^-}$ attacking $\mathrm{CH_3Cl}$ with inversion of configuration — is the classic bimolecular nucleophilic substitution ($\mathrm{S_N2}$). In an $\mathrm{S_N2}$ reaction, bond formation and bond breaking occur simultaneously in a single step. There is no discrete intermediate that can be isolated; the species in square brackets (c) is the transition state, not a stable intermediate.

2. What is the geometry and hybridisation at carbon in (c)?

   In the transition state, the carbon is partially bonded to five atoms: the three hydrogens (still in their original positions), the incoming $\mathrm{HO}$ group, and the leaving $\mathrm{Cl}$ atom. This arrangement is trigonal bipyramidal.

   For a trigonal-bipyramidal geometry, the central atom uses $sp^2$ hybridisation for the three equatorial bonds (here, the C–H bonds) and the two axial positions (the incoming and leaving groups) involve $p$ orbitals. So carbon is $sp^2$ hybridised in this transition state.

3. Evaluate each statement.

   • (i) Intermediate (c) is unstable because in this carbon is attached to 5 atoms.

     This is correct. Carbon normally forms only four covalent bonds (octet rule). In the transition state, carbon has five partial bonds — it is pentacoordinate. This is a high-energy, unstable arrangement. The transition state cannot be isolated; it exists only fleetingly at the energy maximum of the reaction coordinate.

   • (ii) Intermediate (c) is unstable because carbon atom is $sp^2$ hybridised.

     This is incorrect in reasoning. $sp^2$ hybridisation itself is not inherently unstable (e.g., in ethene, carbon is $sp^2$ and perfectly stable). The instability here comes from pentacoordination, not from hybridisation. So the cause given is wrong.

   • (iii) Intermediate (c) is stable because carbon atom is $sp^2$ hybridised. …

Concept: SN2 Reaction Mechanism and Transition State Stability

The reaction shown is a classic SN2 (bimolecular nucleophilic substitution) reaction. The species labelled (c) is the transition state, not a stable intermediate.

Method: Analysis of SN2 Transition State Geometry and Hybridisation

Step 1 – Identify the species

• (a) $\mathrm{HO^-}$ = nucleophile
• (b) $\mathrm{CH_3Cl}$ = substrate (methyl chloride)
• (c) $[\mathrm{HO\cdots CH_3\cdots Cl}]^-$ = transition state
• (d) $\mathrm{CH_3OH}$ = product (inverted configuration)
• (e) $\mathrm{Cl^-}$ = leaving group

Step 2 – Understand the geometry of (c)

• The transition state has a trigonal bipyramidal shape.
• The central carbon is simultaneously bonded (partially) to five atoms: three H atoms (in a plane), the incoming $\mathrm{HO}$ (one axial position), and the leaving $\mathrm{Cl}$ (opposite axial position).

Step 3 – Determine hybridisation of carbon in (c)

• In the transition state, the carbon is $sp^2$ hybridised.
• The three H atoms lie in a plane (trigonal planar arrangement around carbon), while the incoming and leaving groups occupy the axial positions using p-orbitals.

Step 4 – Evaluate stability

• A carbon with five attached atoms is highly unstable — it violates the octet rule. …

Common Mistakes & How to Avoid Them

This question tests your understanding of SN2 reaction mechanism and transition state theory. Here are the most frequent errors students make:

Mistake 1: Calling (c) an "intermediate" instead of a "transition state"

The error: Students treat (c) as a stable, isolable species with a finite lifetime.

Why it's wrong: The species (c) is shown in square brackets with partial bonds — this is the transition state, not an intermediate. Intermediates have finite lifetimes and appear in energy wells; transition states are energy maxima.

How to avoid: Always check the notation:

• Square brackets [ ] with dashed/partial bonds → transition state
• No brackets, full bonds, often with a charge → intermediate

Mistake 2: Assuming "5 groups around carbon" automatically means $sp^3d$

The error: Students see the trigonal-bipyramidal-LOOKING geometry (5 groups around carbon) and reach for $sp^3d$ -- the hybridisation used for period-3+ elements like phosphorus or sulfur in a genuine trigonal bipyramidal molecule (e.g. $PCl_5$).

Why it's wrong: Carbon is a period-2 element and has no accessible d-orbitals -- $sp^3d$ is not a chemically meaningful hybridisation for carbon at all. The correct picture: the three H atoms stay on $sp^2$ orbitals (becoming coplanar as the "umbrella" flattens mid-inversion), while the incoming $\mathrm{HO}$ and leaving $\mathrm{Cl}$ share the two lobes of carbon's remaining, unhybridised p-orbital -- giving an electronically "5-coordinate-looking" transition state without invoking any d-orbital.

How to avoid: The $sp^3d$/trigonal-bipyramidal model is for elements that genuinely have accessible d-orbitals (period 3 and below). For carbon specifically, always default to $sp^2$ + p-orbital for this 5-group transition-state picture, never $sp^3d$.

Mistake 3: Thinking the transition state is "stable"

The error: Selecting option (iii) — "Intermediate (c) is stable because carbon atom is $sp^2$ hybridised."

Why it's wrong: Even if the hybridisation were $sp^2$ (which it isn't), a pentacoordinate carbon is always unstable due to:

• Steric crowding — 5 atoms around a small carbon
• Partial bonds — bonds are neither fully formed nor fully broken
• High energy — it's the peak of the reaction energy profile

How to avoid: Remember: Transition states are always unstable — they represent the highest energy point along the reaction coordinate. Stability is not about hybridisation alone; it's about bond count, geometry, and energy.

Mistake 4: Misreading the question — "reaction intermediate" …