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An example of Wagner–Meerwein rearrangement given in my textbook is substitution by $\ce{AgNO2}$ in iodocyclohexane. One of the product, cyclopentylnitromethane is through ring contraction.

I don't understand how this took place. Can anyone explain the mechanism?

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    $\begingroup$ It would be a mistake to think that a five-member ring is very strained. $\endgroup$ – Zhe Jun 5 '20 at 15:14
  • $\begingroup$ I know it isn't 'very strained'. I am editing the question. The main question is how is the rearrangement happening. $\endgroup$ – B.Anshuman Jun 5 '20 at 16:01
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    $\begingroup$ This reference from 1952 claims that the product of the reaction is nitrocyclohexane. No rearrangement. doi.org/10.1021/ja01132a037. A previous study had claimed additionally the formation of 1-methylnitrocyclopentane, which is not cyclopentylnitromethane. What is the name of your textbook? $\endgroup$ – user55119 Jun 5 '20 at 20:33
  • $\begingroup$ @user55119, S.N.Sanyal - Reactions, Rearrangements and Reagents ; Publisher: Bharti Bhawan ; Page-185. You may find this book online. Also, in your link, it is clearly mentioned that Rosanow discovered rearranged products in 1915, i.e. Wagner-Meerwin Rearrangement is happening $\endgroup$ – B.Anshuman Jun 5 '20 at 20:58
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    $\begingroup$ I'll post a solution. $\endgroup$ – user55119 Jun 5 '20 at 23:42
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Kornblum1 demonstrated that nitrocyclohexane 3 was the major isolable product from the reaction of iodocyclohexane 1 and silver nitrite. Earlier work by Rosanow2 (1915) claimed the additional formation of tertiary nitro compound 7, a result that Kornblum did not confirm. If indeed nitrocyclopentane 7 did form, a ring contraction is required. Ring contraction of cyclohexyl carbocation 2 sacrifices the greater stability of a secondary carbocation for a primary carbocation (4) and presumably some loss in stability through ring contraction in forming cyclopentyl carbocation 6. There is no evidence that primary nitrocyclopentane 5 is formed. However, a hydride shift in 4 would afford the more stable tertiary carbocation 6 that would lead to 7.



1) N. Kornblum and C. Teitelbaum, J. Am. Chem. Soc., 1952, 71, 3076.
2) See reference in Kornblum paper.

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  • $\begingroup$ Structure 5 is given in my textbook as one of the products. $\endgroup$ – B.Anshuman Jun 6 '20 at 15:29
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    $\begingroup$ Structure 5 is not present in the Kornblum paper. There maybe a problem in interpreting the names of 5 and 7. OTOH, some information in textbooks is gleaned from memory. $\endgroup$ – user55119 Jun 6 '20 at 17:09

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