Question

Given below are two statements: Statement I: CH\(_3\)-O-CH\(_2\)-Cl will undergo \( S_N1 \) reaction though it is a primary halide.
Statement II: {CH_3-C(-CH_3)(-CH_3)-CH_2-Cl}
will not undergo \( S_N2 \) 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.

Hint:

The \( S_N1 \) mechanism involves the formation of a carbocation and is favored by stable carbocations, while the \( S_N2 \) mechanism involves a backside attack and is hindered by steric crowding.

Answer 1

The Correct Option is D

To determine the correctness of the statements concerning the likelihood of certain chemical reactions involving primary halides, we must analyze the molecular structure and the mechanism involved.

Statement I: CH\(_3\)-O-CH\(_2\)-Cl will undergo \( S_N1 \) reaction though it is a primary halide. The \( S_N1 \) reaction typically involves formation of a carbocation intermediate, which is more stable in tertiary carbons. However, in this case, if the leaving group, Cl\(^-\), departs, the resulting carbocation is stabilized by the adjacent oxygen atom through resonance. Therefore, Statement I is correct because the resonance stabilization provided by the oxygen allows the primary alkyl halide to undergo a \( S_N1 \) reaction.

Statement II: CH_3-C(-CH_3)(-CH_3)-CH_2-Cl will not undergo \( S_N2 \) reaction very easily though it is a primary halide. The \( S_N2 \) mechanism involves a backside attack, which requires unobstructed access to the electrophilic carbon. Here, even though it is a primary halide, the three bulky methyl groups surrounding the reactive center hinder the approach of the nucleophile, thus impeding the \( S_N2 \) reaction. Therefore, Statement II is correct as the steric hindrance prevents the \( S_N2 \) mechanism.

Based on these analyses, the most appropriate answer is that both Statement I and Statement II are correct.

Answer 2

The compound is chloromethyl methyl ether, written as CH₃–O–CH₂–Cl (or Cl–CH₂–O–CH₃). Loss of Cl⁻ would give the cationic species Cl departure → CH₂⁺–O–CH₃, but the positive charge on CH₂ is stabilized by the adjacent oxygen by resonance/inductive effect (oxygen can donate electron density and form an oxonium-like resonance form). In other words the would-be carbocation is significantly stabilized by the neighbouring lone pairs on O, so the reaction can proceed by an S_N1 (or at least via an ionization-assisted pathway) despite the carbon being formally primary. 
In brief: \[ \text{Cl–CH}_2\text{–OCH}_3 \xrightarrow{\text{ionization}} \; [\text{CH}_2^{+}\!-\text{OCH}_3]\;\xrightarrow{\text{nucleophile}} \dots \] Stabilization by O makes ionization feasible → Statement I true. 
Why Statement II is correct
The given structure is neopentyl chloride, (CH₃)₃C–CH₂–Cl. Although the carbon bearing Cl is primary, the adjacent carbon is a quaternary center ((CH₃)₃C–). For S_N2 attack at the primary carbon the incoming nucleophile needs backside access to the C–Cl bond. Steric crowding from three methyl groups on the neighbouring carbon severely hinders this approach — so S_N2 is very slow. 
As for S_N1, formation of a stable carbocation would require generating the neopentyl carbocation (CH₃)₃C–CH₂⁺, which is not resonance-stabilized and is highly disfavoured. So neither pathway is favorable; in particular the S_N2 is strongly hindered. Hence the statement "will not undergo SN2 very easily" is correct. 
In short: steric hindrance prevents SN2 → Statement II true. 
Conclusion: Both statements are correct → Option 4.