1
$\begingroup$

Example 2: Identify all the products formed in the following reaction and indicate the major product.

Example 2

Solution: Three products will be formed in the given reaction as shown below:

Solution

In this reaction the primary alcohol undergoes SN1 mechanism by forming a carbocation and then undergoing ring expansion, but primary alcohols prefer SN2 mechanism. So why is this primary alcohol undergoing SN1 mechanism?

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ The Sn1 mechanism should be considered relatively unreasonable here. You might consider a Sn1-like mechanism instead where ring expansion is the thing that displaces water, i.e., the C-C bond is the nucleophile. $\endgroup$
    – Zhe
    Apr 16, 2019 at 0:49
  • $\begingroup$ But the carbocation is formed in Sn1 mechanism and ring expansion can only take place if carbocation is formed,also in this reaction the product formed by Sn1 mechamism is major product $\endgroup$
    – saurya yadav
    Apr 16, 2019 at 6:11
  • $\begingroup$ "ring expansion can only take place if carbocation is formed" That is incorrect. The ring expansion could displace the leaving group. $\endgroup$
    – Zhe
    Apr 16, 2019 at 13:24
  • 2
    $\begingroup$ The bottom line here is the the provide solution is wrong. $\endgroup$
    – Zhe
    Apr 16, 2019 at 13:25
  • $\begingroup$ Is the given answer to the question regarding major product 1-bromo cyclohexane? $\endgroup$
    – J_B892
    Sep 13, 2019 at 8:05

1 Answer 1

Reset to default
4
$\begingroup$

It does not. The proposed mechanism is wrong.

The conversion of cyclopentylmethanol to 1-bromo-1-methylcyclopentane is not reported anywhere in the literature. If you could somehow generate the primary carbocation, then one would expect a rapid 1,2-hydride shift to generate the more stable tertiary carbocation, giving rise to 1-bromo-1-methylcyclopentane.

The ring-expansion product could be rationalized by a concerted 1,2-alkyl shift displacing $\ce{OH2^+}$ without violating first-principles.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.