1
$\begingroup$

Steric crowding is greater for the secondary halide, and this is important as the neucleophile will attack the back lobe of carbon atom. But the relative ability to leave is more for iodide than bromide.

So how can I compare the $\mathrm{S_N2}$ rates for sec-butyl iodide (2-iodobutane) and butyl bromide (1-bromobutane)?

$\endgroup$
  • 2
    $\begingroup$ It doesn’t matter. Steric hindrance is the more-to-worry about factor. See it this way; A group can leave only if the backside attack can happen. If that doesn’t even happen, no point in comparing the relative capability to leave. $\endgroup$ – MollyCooL Apr 4 '18 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.