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Question

enter image description here

My Attempt

First a primary carbocation is formed and then methyl shift occurs to form a more stable secondary carbocation hence making (C) the major product, consequently I also arrived at answer (D) via hydride shift, and also option (A) would anyway be a possible product.

But the answer given is (B). Where am I going wrong?

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    $\begingroup$ Is the question wrong? Will (B) be the correct answer if it was asked much slower instead of faster? $\endgroup$ Commented Dec 12, 2015 at 17:52
  • $\begingroup$ SciFinder does not know any reactions starting from that reactant. So whatever the book says is based on theory, not experiment. Aka a bad question. $\endgroup$
    – Jan
    Commented Dec 21, 2015 at 20:32
  • $\begingroup$ @Jan though i couldn't agree with you more, aren't all questions in competitive exams based on theory, is my explanation below legitimate? $\endgroup$ Commented Dec 22, 2015 at 0:58
  • $\begingroup$ No, not necessarily. In fact, good exam questions should be based on experimental evidence not on theoretical thoughts unless the idea is explicitly consider how it may happen. Your explanation below doesn’t seem bad but I am hesitant to vote on it. $\endgroup$
    – Jan
    Commented Dec 22, 2015 at 1:03

1 Answer 1

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Apparently, (B) is wrong for obvious reasons.

  1. The product formed fastest will be kinetically determined. Shifting of bonds takes time so the fastest formed product will be the the one where minimum migrations occur hence (A).

  2. As @orthocresol stated below, a stable compound similar to cyclopropylmethyl carbocation is forming here. "The double bond helps to kick out the chloride ion to give a stabilised carbocation, which makes it react much faster than usual". Hence product (A) will be favored thermodynamically too (refer).
    enter image description here

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    $\begingroup$ I wouldn't say it's SN2 under those conditions. The double bond helps to kick out the chloride ion to give a stabilised carbocation, which makes it react much faster than usual. Go and read about the cyclopropylmethyl cation. $\endgroup$ Commented Dec 18, 2015 at 0:13

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