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Problem

Treating cyclohexanone with N-bromosuccinimide in the presence of carbon tetrachloride gives A. A on treatment with aqueous KOH gives B. Deduce the structures of A and B.

Solution

A is 2-bromocyclohexanone and B is potassium cyclopentane carboxylate.

Questions

  1. In the first step, why did the bromine atom get attached to the second position? The carbonyl group, being electron withdrawing, should direct the incoming electrophile away from it.
  2. In the second step, I am quite sure that the product mentioned is a result of Favorskii rearrangement. But there is also a possibility for the hydroxide ion to substitute bromine. Is this possible? If yes, why is the substituted product not a major product? If not, why?
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  • 2
    $\begingroup$ With regard to your first question - consider the role of keto/enol tautomerism. There is an error in the stated answer; B is not cyclohexane carboxylate potassium salt. $\endgroup$
    – Waylander
    Commented Apr 23, 2021 at 14:31
  • $\begingroup$ Count the number of carbons! $\endgroup$
    – user55119
    Commented Apr 23, 2021 at 15:06
  • $\begingroup$ @Waylander Could you elaborate your first statement? $\endgroup$ Commented Apr 23, 2021 at 15:21
  • 1
    $\begingroup$ I don't have a definitive answer for 2 but it is like a E2 vs SN2 for secondary haloalkanes. The solvent, temperature, and concentration will favour one or the other. See the last link below. However, it is also important to note that the hydrogen is rather acidic so it will likely favour deprotonation rather than substitution. chem.libretexts.org/Bookshelves/Organic_Chemistry/…. $\endgroup$
    – M.L
    Commented Apr 23, 2021 at 17:57
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    $\begingroup$ Also consider the steric factors; bromine is a large atom and the hydroxide ion must attack from the reverse face $\endgroup$
    – Waylander
    Commented Apr 23, 2021 at 18:09

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