2

Compare the rates towards SN2 reaction:

a: 1‐bromopropan‐2‐one; b: 1-bromobutane; c: 1-bromo-2-methylpropane; d: 2-bromobutane

I got the point that 2-bromobutane d is a secondary halide, so the steric hinderance is highest making it least reactive among the four. Both b and c are primary halides, but c has branched isobutyl group, so because of steric effect the order should be b > c > d.

However, the answer claims that bromoacetone a is the most reactive towards SN2 despite also being primary halide. What is the exact reason for this?

My friend says if there is carbonyl on β-position, then it is highly reactive towards SN2. Is this even accurate? If yes, then what is the reason?

andselisk
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Arpit Raj Choudhary
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  • Does this question help? – Safdar Faisal Aug 12 '22 at 06:41
  • @SafdarFaisal that is a good question but how is it related with this? As in that question the reaction proceeds by elimination i guess, but here SN2 is asked! – Arpit Raj Choudhary Aug 12 '22 at 07:35
  • It was more the answer to the question. A positive charge on the alpha carbon is highly destabilized. Well here's a similar question – Safdar Faisal Aug 12 '22 at 07:56
  • I think this paper gives good details on this regard: https://pubs.acs.org/doi/10.1021/ja01075a025# – Reihani Aug 12 '22 at 13:49
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    As part of the paper: The favorable effect that the carbonyl and related have on sN2 reactivity probably stems in part from the absence of rate-retarding steric and field effects, coupled with a mildly rate-enhancing inductive effect. The very strong accelerating effect of these groups must, however, include some sort of an additional conjugating factor. The conjugation responsible for carbonyl activation is caused by an overlap of the p-orbital of the carbon atom of the carbonyl group with orbitals on the entering and leaving groups, rather than a π-bond type overlap. – Reihani Aug 12 '22 at 13:51

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