Given below are two statements :
Statement I : CH₃-O-CH₂-Cl will undergo S_N 1 reaction though it is a primary halide.
Statement II : CH₃; |; CH₃-, C, -CH₂-Cl; |; CH₃ will not undergo S_N 2 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 :

Solution

Chemistry ma’am · 4 steps

Primary is only a label. Look at what sits next to the CH₂ carbon in each chloride.

1. Statement I. When Cl⁻ leaves CH₃-O-CH₂-Cl, the carbocation CH₃-O-CH₂⁺ forms. The oxygen next to the positive carbon donates a lone pair into it, so the positive charge is shared by resonance.

   CH₃-Ö-C(+)H₂ ↔ CH₃-O(+)=CH₂

2. Why SN1 works. In the second structure every atom has a complete octet, so this cation is much more stable than an ordinary primary carbocation. A stable carbocation is exactly what S_N1 needs. Statement I is correct.
3. Statement II. This is neopentyl chloride, (CH₃)₃C-CH₂-Cl. The bulky tert-butyl group sits right next to the CH₂ carbon and blocks the nucleophile from attacking from the back. S_N2 needs that backside attack, so it is very slow. Statement II is correct.
4. So. Both statements are correct, which is option (b).

The idea to keep. For S_N1, ask how stable the carbocation is, since an O or N lone pair next door helps a lot. For S_N2, ask how crowded the back side is. Primary or tertiary is only the first guess.

Answer from the NTA final answer key. Solution by PrepWiser.