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For nucleophilic substitution reactions two factors are crucial:
• Stability of the carbocation that would be produced in an $$\mathrm{S_N1}$$ mechanism.
• Ease of backside attack (absence of steric and electronic hindrance) for an $$\mathrm{S_N2}$$ mechanism.
List-I (type of halide) and List-II (behaviour) are compared one by one.
Case A: A is a normal primary alkyl halide, e.g. $$CH_3CH_2CH_2Cl$$.
• A primary carbocation is highly unstable, so $$\mathrm{S_N1}$$ is disfavoured.
• Back-side attack is unhindered, therefore the compound usually reacts by the bimolecular path $$\mathrm{S_N2}$$.
Hence A matches with “undergoes $$\mathrm{S_N2}$$ reaction” ⇒ Item III.
Case B: B is benzyl chloride, $$C_6H_5CH_2Cl$$.
• If chloride leaves, the benzyl carbocation $$C_6H_5CH_2^+$$ is stabilised by resonance with the aromatic ring.
• Such high stabilisation makes the unimolecular path very fast.
Therefore B matches with “undergoes $$\mathrm{S_N1}$$ reaction very readily” ⇒ Item I.
Case C: C is allyl chloride, $$CH_2=CHCH_2Cl$$.
• Loss of $$Cl^-$$ gives the allyl cation $$CH_2=CHCH_2^+$$ which is resonance-stabilised:
$$CH_2=CH-\overset{+}{C}H_2 \;\;\leftrightarrow\;\; ^{+}CH_2-CH=CH_2$$
• Resonance stabilisation again promotes the $$\mathrm{S_N1}$$ pathway.
Hence C is linked with “forms a resonance-stabilised carbocation” ⇒ Item IV.
Case D: D is vinyl chloride, $$CH_2=CHCl$$ (the halogen directly attached to a double-bonded carbon).
• The $$C-Cl$$ bond is polarised toward carbon because of $$sp^2$$-hybridisation and $$\pi$$ overlap; it acquires partial double-bond character.
• Nucleophilic substitution (both $$\mathrm{S_N1}$$ and $$\mathrm{S_N2}$$) becomes extremely slow; effectively the reaction does not occur under ordinary conditions.
Therefore D matches with “does not undergo normal nucleophilic substitution” ⇒ Item II.
Collecting all matches:
A → III B → I C → IV D → II
Among the given answer sets, this correspondence is given in Option B.
Hence the correct choice is:
Option B which is: A-III, B-I, C-IV, D-II.
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