15
$\begingroup$

I have my concerns regarding option (a) in this question.

enter image description here

During the dehydration step, wouldn't the resulting carbocation intermediate formed in option (a) rearrange?
That is, the ring should contract from a $7$-carbon to a $6$-carbon one because the resulting carbocation would be more stable. But the resulting product would have to be an addition product since elimination now becomes impossible.

I know that this situation can arise because I have tackled it before and the answer was an addition product in that case, but that reasoning contradicts with the official answer key in this question. (The answers are a,b and d).

The shift I called for was the bond between carbon (endocylic with nitrogen) and the carbocation.

$\endgroup$
  • 1
    $\begingroup$ @Chemstack The issue here is that the carbocation after the shift is not stabilized. The nitrogen is fully protonated and is thus just an electron withdrawing group attached to a carbocation. $\endgroup$ – Zhe May 20 '17 at 13:24
  • $\begingroup$ @Zhe can you please explain it a bit more perhaps write an answer.Thanks. $\endgroup$ – Pink May 24 '17 at 19:29
  • $\begingroup$ @Pink I'm not sure there's anything else to say. If the nitrogen is fully methylated, it's just a de-stabilized group to have next to a cation. If you're confused, it's probably more educational for you to draw out the mechanism for the rearrangement and look at the intermediate instead of my doing it. $\endgroup$ – Zhe May 25 '17 at 1:18
3
$\begingroup$

Unfortunately, I cannot back this with references at the moment but in my opinion there should be no carbocation formed in the first place and therefore no tendency to undergo any Wagner-Meerwein rearrangements.

Secondary carbocations, while being less unstable than primary ones, are still rather unstable. There are no stabilisation mechanisms that suggest themselves in the reaction conditions (concentrated sulphuric acid) in which the nitrogen will be protonated and therefore unable to stabilise a carbocation in α position. I would assume that the elimination proceeds much more $\mathrm{E2}$-like: likely with no intermediate carbocation at all but if there is one it should be extremely short lived and the neighbouring proton should dissociate practically instantly.

$\endgroup$
2
$\begingroup$

There are multiple possibilities for this reaction, elimination may not come first, Wagner-Meerwein rearrangements may occur before this. For both A and B, rearrangement occurs before elimination. For D it is straightforward, the -OH will be eliminated and the corresponding alkene will be formed. For B, the -OH will first be protonated, the carbon nitrogen bond of the bridgehead carbon farther away will be broken. The nitrogen atom will then form a bond with the carbon having the -OH2 substituent, the loss of water will follow. This generates D, but a carbocation is in place of the -OH group. This compound can undergo elimination to generate the same alkene as D generated. Your question is pertaining to the lack of rearrangement of the carbocation. It cannot rearrange to one of the tertiary bridgehead carbons, because of Bredt's rule. So when heating in concentrated sulfuric acid, the major elimination product will be the alkene mentioned previously. While B and D proceed through this alkene, A proceeds through a Wagner-Meerwein rearrangement similar to B. This yields a carbocation. This carbocation will rearrange (but not in the context you had) because of angle strain on the adjacent carbon. The rearrangement increases stability and elimination then occurs to yield the same product as B and D. This then undergoes transformation to cycloheptatriene under the conditions provided.

$\endgroup$
  • $\begingroup$ This is not correct because compound (a) does not undergo rearrangement. It also does not really answer the OP's question... $\endgroup$ – Zhe Jun 30 '17 at 13:31
  • $\begingroup$ Straightforward elimination could occur, but that does not rule out the possibility of Wagner-Meerwein rearrangements. In more complex orgo problems, there are numerous possibilities. There is no way to prove either of us right or wrong without experimental evidence. $\endgroup$ – AS_1000 Jun 30 '17 at 13:45
  • $\begingroup$ It absolutely rules out rearrangement because it would put the carbocation next to a fully substituted nitrogen. $\endgroup$ – Zhe Jun 30 '17 at 13:46
  • $\begingroup$ You do not understand the mechanism of a Wagner-Meerwein rearrangement, the resulting carbocation is beta to the nitrogen, and the degree of substitution on the nitrogen wouldn't rule out the possibility of a rearrangement. $\endgroup$ – AS_1000 Jun 30 '17 at 13:53
  • $\begingroup$ Yes, that's the whole point. The nitrogen would ordinarily stabilize a cation to form an iminium but it can't if it's fully protonated from sulfuric acid. You do not place an fully substituted, electron withdrawing nitrogen next to carbocation for a rearrangement. That rearrangement will not occur. And if you disagree, then we're not talking about the same thing. I suggest reading the OP's question again. $\endgroup$ – Zhe Jun 30 '17 at 14:01

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.