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MCQMediumJEE 2025SN1 & SN2 Reactions

JEE Chemistry 2025 Question with Solution

Given below are two statements:

Statement I: CH3−O−CH2−Cl\mathrm{CH_3-O-CH_2-Cl} will undergo SN1S_N1 reaction though it is a primary halide.

Statement II: CH3−C(−CH3)(−CH3)−CH2−Cl\mathrm{CH_3-C(-CH_3)(-CH_3)-CH_2-Cl}

will not undergo SN2S_N2 reaction very easily though it is a primary halide.

In the light of the above statements, choose the most appropriate answer from the options given below:

  • A

    Statement I is incorrect but Statement II is correct.

  • B

    Both Statement I and Statement II are incorrect.

  • C

    Statement I is correct but Statement II is incorrect.

  • D

    Both Statement I and Statement II are correct.

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

  • Statement I concerns CH3−O−CH2−Cl\mathrm{CH_3-O-CH_2-Cl} and its tendency to undergo SN1S_N1 reaction though it is a primary halide.
  • Statement II concerns CH3−C(−CH3)(−CH3)−CH2−Cl\mathrm{CH_3-C(-CH_3)(-CH_3)-CH_2-Cl} and whether it undergoes SN2S_N2 reaction very easily though it is a primary halide.

Find: Which option correctly describes the truth of the two statements.

For Statement I, SN1S_N1 proceeds through carbocation formation. If Cl−\mathrm{Cl^-} leaves from CH3−O−CH2−Cl\mathrm{CH_3-O-CH_2-Cl}, the resulting cation is stabilized by the adjacent oxygen atom. The neighboring oxygen donates electron density and stabilizes the positive charge by resonance, so this primary halide can undergo an ionization-assisted SN1S_N1 pathway. Therefore, Statement I is correct.

For Statement II, the compound (CH3)3C−CH2−Cl\mathrm{(CH_3)_3C-CH_2-Cl} is neopentyl chloride. Although the carbon bearing chlorine is primary, SN2S_N2 requires backside attack at that carbon. The three methyl groups on the adjacent carbon create strong steric hindrance, making nucleophilic approach very difficult. Therefore, Statement II is correct.

Hence, both statements are correct.

Therefore, the correct option is D.

Mechanistic Explanation

Given:

  • CH3−O−CH2−Cl\mathrm{CH_3-O-CH_2-Cl}
  • CH3−C(−CH3)(−CH3)−CH2−Cl\mathrm{CH_3-C(-CH_3)(-CH_3)-CH_2-Cl}

Find: Evaluate the correctness of Statement I and Statement II.

Statement I: The compound is chloromethyl methyl ether, CH3−O−CH2−Cl\mathrm{CH_3-O-CH_2-Cl}. On loss of Cl−\mathrm{Cl^-}, the developing positive charge at the CH2\mathrm{CH_2} carbon is stabilized by the adjacent oxygen atom through resonance and electron donation. Thus, even though the substrate is formally primary, carbocation formation is comparatively facilitated.

So, Statement I is true.

Statement II: The second compound is neopentyl chloride, (CH3)3C−CH2−Cl\mathrm{(CH_3)_3C-CH_2-Cl}. In an SN2S_N2 reaction, the nucleophile must attack from the backside of the carbon attached to chlorine. Here, the adjacent quaternary carbon with three methyl groups causes severe steric crowding, which blocks effective backside attack.

Thus, SN2S_N2 reaction does not occur very easily.

So, Statement II is true.

Conclusion: Both statements are correct, so the correct option is D.

Common mistakes

  • Assuming that every primary halide always undergoes only SN2S_N2 reaction is incorrect. The nature of the neighboring group also matters; adjacent oxygen can stabilize the cationic intermediate and favor SN1S_N1-type behavior.

  • Ignoring steric hindrance in neopentyl halides is a common error. Even though the carbon attached to Cl\mathrm{Cl} is primary, the bulky neighboring tert-butyl group severely slows backside attack required for SN2S_N2.

  • Confusing carbocation stability with the degree of the carbon alone is incomplete. Resonance stabilization by a heteroatom such as oxygen can make an apparently unfavorable ionization pathway feasible.

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