6
$\begingroup$

This is an objective type question I encountered in a objective test.

enter image description here

NBS (N-bromosuccinimide) is used for the allylic substitution of $\ce{Br}$ in a reaction. In this reaction, since the allylic carbon is the carbon adjacent to the carbon having double bonds, I thought that options B), C) and D) all can be the products but not A). However, the answer key gives the answer as D) and I can't understand why it should be?

|improve this question
$\endgroup$
  • 1
    $\begingroup$ This is weird. Unless there is some mechanism that I am overlooking, A1, B2, C2 and D2 should all not be reaction products. $\endgroup$ – Jan Apr 4 '17 at 13:17
  • $\begingroup$ Is there any other information? In particular, does it state that the starting material is racemic? $\endgroup$ – jerepierre Apr 22 '17 at 13:38
  • $\begingroup$ @jerepirrre no,no further information was provided to us. $\endgroup$ – Pink Apr 22 '17 at 15:14
2
+50
$\begingroup$

D is a product, but only one product. Not two products. Those two are the same molecule flipped on a mirror plane. Were the molecule chiral (having an non-superimposable asymmetric center), then it could be two products. Not as it sits now.

Notice D1 is the same as B1. D2 is also B1, but flipped. Six products.

D2 is an enantiomer of D1, but not identical to it. I cannot puzzle out how D is supposed to be the correct answer.

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ The structures in D are not the same compound, they are mirror images. They are chiral. You cannot "flip on a mirror plane". You can only rotate (flip) around an axis. $\endgroup$ – jerepierre Apr 22 '17 at 13:41
  • $\begingroup$ You are correct. The methyl group has shifted from the starting location. Fixing... $\endgroup$ – Pete Apr 24 '17 at 20:21
3
$\begingroup$

@Pete B1 is same as D1

But, D1 & D2 are not same.

@Aniline I see some problem in the options while it was framed/typed/published

I have shown the 6 possible stereoisomers, which have have labelled independently as A, B, C, D, E & F

For the sake of clarity have generated their names with some software [not from my own] to be assured that these are actually different or same compounds.

Showing the mechanism, how there is a possibility of getting SIX by a free radical mechanism:

enter image description here

This one a redraw of the options A, B, C & D with their names: enter image description here

The relationship of D1 & D2 as per question doesn't work to the the same:enter image description here

In my opinion, there are six stereoisomers possible, but the correct option does not exist, in the question[options given, in fact].

Hope it throws some more light on your query.

|improve this answer
$\endgroup$
0
$\begingroup$

The second product in Option D) is not a product of this reaction.

You cannot rotate the substrate molecule in such a way that the Me substituent ends up in the portraied position.

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ @SirJohnson So that means that the two products in option D are identical.we can rotate the first one 180 degree to get the second one. $\endgroup$ – Pink Mar 31 '17 at 1:30
  • $\begingroup$ To me, it looks like each group of two has an impossible product: the one where the methyl group’s carbon’s stereochemistry is inverted. Can you explain how the methyl group inversion would proceed? Especially since I am getting six products without methyl group inversion … $\endgroup$ – Jan Apr 4 '17 at 13:16
  • $\begingroup$ Absolute stereochemistry is irrelevant if the products in (B) are not wrong. $\endgroup$ – Zhe Apr 18 '17 at 19:42

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.