It is quite obvious that carbonyl compound is going to convert into alkane, but what confuses me is whether Sn2 or E2 will occur at bromide position

Although there is heating( which I believe favour E2) and a strong base, the carbon is 1° which would favor SN2 more than E2. Please help.

  • $\begingroup$ I think you will get an SN2 at the alkyl bromide to give the substituted hydrazine, this may go on to do a second substitution with further alkyl bromide. You certainly will not get b. a could be the product if by manipulating the hydrazine addition conditions you can get selectivity for the reaction at the ketone. I consider this unlikely. Remember that the intermediate hydrazide can also react with the bromide. $\endgroup$
    – Waylander
    Apr 30, 2020 at 7:15
  • $\begingroup$ But the answer is given B? $\endgroup$ Apr 30, 2020 at 8:43

1 Answer 1


This is simply wrong. The person setting this question has assumed that the hydrazine will be completely selective for the ketone over the alkyl bromide and that the strong basic conditions will also cause elimination of the Br- to give the styrene. Selectivity seems most unlikely give the typical forcing conditions used in W-K reductions

Hydrazine is a powerful nucleophile and reacts with alkyl bromides under mild conditions often giving polyalkylation reference here.

I consider it most unlikely there will be any selectivity unless confronted by literature example.


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