Q.Arrange the compounds in increasing order of rate of reaction towards nucleophilic substitution. (Note printed in the Exemplar for questions 26 to 29.)

Concept: Nucleophilic Aromatic Substitution ($S_NAr$) reactivity for aryl halides — a nitro group at the ortho or para position strongly activates the ring toward nucleophilic substitution by stabilising the Meisenheimer complex through resonance. A meta nitro group cannot donate resonance, so its effect is much weaker.

Reasoning:

1. Chlorobenzene (a) has no activating group, so its rate is the lowest.
2. 1-chloro-3-nitrobenzene (c) has a meta nitro group — only inductive withdrawal, no resonance stabilisation of the intermediate. Rate is slightly higher than (a) but still low. …

Why this approach works

Nucleophilic aromatic substitution (SNAr) follows a two-step addition-elimination mechanism. The rate-determining step is the formation of a Meisenheimer complex (a negatively charged σ-adduct). Anything that stabilizes this negative charge speeds up the reaction.

Electron-withdrawing groups (EWGs) like $-\mathrm{NO_2}$ stabilize the negative charge by resonance and induction. But here's the key: resonance stabilization only works when the EWG is ortho or para to the leaving group. A meta-nitro group cannot delocalize the negative charge through resonance — it can only withdraw inductively, which is much weaker.

So the question reduces to: which positions of the nitro group actually help stabilize the Meisenheimer intermediate?

Step-by-step reasoning

1. Identify the reaction type and mechanism

All three compounds are aryl chlorides. In chlorobenzene, the chlorine is directly attached to an sp² carbon. Nucleophilic substitution here does not proceed via an SN1 or SN2 pathway (the C–Cl bond is too strong, and backside attack is blocked by the ring). Instead, it follows the SNAr (addition-elimination) mechanism.

The figure below shows the ground-state resonance of a halobenzene — the halogen's lone pair donating into the ring. This is background, not the reaction intermediate: it is what gives the aryl C–Cl bond its partial double-bond character, which is why the SN1/SN2 pathways are closed off. The negatively charged Meisenheimer complex discussed next is a different, separate species.

2. Analyze compound (a): chlorobenzene

No nitro group. No resonance or inductive stabilization of the intermediate. The negative charge is localized entirely on the ring, making the intermediate high in energy. This is the slowest of the three.

3. Analyze compound (c): 1-chloro-3-nitrobenzene (meta)

The $-\mathrm{NO_2}$ is meta to Cl (on C3, counting the C–Cl carbon as C1). Draw the resonance structures of the Meisenheimer complex formed by nucleophilic attack at the C–Cl (ipso) carbon: the negative charge delocalises only onto the carbons ortho and para to C1 — that is, C2, C4 and C6. It never reaches C3, the carbon bearing the nitro group, so a meta nitro group gets no opportunity to accept the charge through resonance. The only stabilization it provides comes from its inductive effect (withdrawal through σ-bonds), which is modest.

So (c) is faster than (a), but not by much. …

Concept: Effect of Substituents on Nucleophilic Aromatic Substitution (S$_N$Ar)

In nucleophilic aromatic substitution, the rate depends on how easily the nucleophile can attack the electron-deficient carbon bearing the leaving group. Electron-withdrawing groups (EWGs) like $-\mathrm{NO_2}$ increase the rate by stabilizing the negatively charged intermediate (Meisenheimer complex). The position of the EWG matters greatly.

Method: Activating Effect of Nitro Group in S$_N$Ar

Steps:

1. Identify the leaving group — Here, it's chlorine (Cl) on a benzene ring.

2. Analyze substituent effects:

   • Chlorobenzene (a): No activating group. Slowest rate.
   • 1-chloro-3-nitrobenzene (c): $-\mathrm{NO_2}$ is meta to Cl. Meta-EWGs cannot delocalize negative charge onto the nitro group during attack at the ipso carbon (where Cl is). So rate increases only slightly. …

Common Mistakes & How to Avoid Them

Mistake 1: Assuming all nitro groups activate the ring equally

The error: Students think that since $\mathrm{NO_2}$ is an electron-withdrawing group, it always increases the rate of nucleophilic substitution — regardless of position.

Why it's wrong: The $\mathrm{NO_2}$ group withdraws electrons by both inductive effect and resonance effect. The resonance effect is position-dependent:

• At ortho/para positions: $\mathrm{NO_2}$ can withdraw electron density directly from the carbon bearing Cl via resonance, stabilizing the Meisenheimer complex (the intermediate in $\mathrm{S_NAr}$).
• At meta position: Resonance withdrawal cannot reach the reaction site — only inductive withdrawal operates, which is weaker.

How to avoid: Always draw the resonance structures of the intermediate. For ortho/para nitro groups, you'll see negative charge delocalized onto the nitro oxygen — this stabilizes the intermediate and increases the rate.

Mistake 2: Confusing nucleophilic substitution with electrophilic substitution

The error: Students apply the same logic as for electrophilic substitution (where $\mathrm{NO_2}$ is meta-directing and deactivating) to nucleophilic substitution.

Why it's wrong: The two reactions have opposite electronic requirements:

• Electrophilic substitution: Ring must be electron-rich → $\mathrm{NO_2}$ deactivates (slows down).
• Nucleophilic substitution ($\mathrm{S_NAr}$): Ring must be electron-deficient → $\mathrm{NO_2}$ activates (speeds up), especially at ortho/para positions.

How to avoid: Memorise this key distinction:

For $\mathrm{S_NAr}$, electron-withdrawing groups (especially at ortho/para) increase the rate. For electrophilic substitution, they decrease the rate.

Mistake 3: Forgetting that chlorobenzene is the slowest

The error: Students rank chlorobenzene faster than the meta-nitro compound.

Why it's wrong: Chlorobenzene has no activating group — the Cl itself is weakly deactivating for $\mathrm{S_NAr}$ (it donates electrons by resonance but withdraws by induction). The meta-nitro compound has at least inductive withdrawal from $\mathrm{NO_2}$, making it faster than plain chlorobenzene.

How to avoid: Remember the order of activation for $\mathrm{S_NAr}$:

• No substituent (chlorobenzene) → slowest …