[scientific]

Hello,

Please see the attached image.

I came across this question, and I was able to cross off 2/4 of the answers. I don't understand the final solution however. I believe this is an E2 reaction, and operated under the assumption that D was not ideal to remove.

(A) makes no sense because an H and the D get abstracted instead of the Br
(B) the configuration made no sense

I chose (C) because I thought isotopes made reactions occur more slowly. Or is deuterium more acidic than hydrogen?

[phth]

It will be C  if there is a syn-periplanar elimination deuterium. it will be D if there is an anti-periplanar elimination of deuterium.  Therefore, APP>SPP and the answer is D.  Syn-periplanar eliminations are possible but they are significantly higher in activation energy than the anti configuration.

a) not possible
b) SPP= high energy
c) SPP= high energy
d) answer

The H vs D isotope effect is the trick part of the question.  It is relatively insignificant.

[Babcock_Hall]

@OP, Do you know what an antiperiplanar arrangement is?  An oversimplified theory suggests that breaking a C-D bond will be up to sevenfold slower than breaking a C-H bond, but many factors contribute to the magnitude of the effect on the rate of the reaction.  I agree with pgk that this kinetic isotope effect (KIE) can be ignored in solving this problem.  I think that the deuterium is just functioning as a label here.