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Consider the dehydration of the following molecule with conc. $\ce{H2SO4}$ at $443 \text K$:

The possible products are:

The first one is the Zaitsev product and the second one Hofmann product.

Which compound is the major product?

There are 2 factors to consider here:

  1. Zaitsev product is more stable than Hofmann product due to hyperconjugation
  2. Anti elimination is more favorable than syn elimination due to steric requirements

Obviously both the factors clash with each other, so I couldn't come to a definite conclusion. Can you help me out?

Nilay Ghosh
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newbie105
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1 Answers1

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The major product is the Zaitsev one i.e 1-methylcyclohexene. The OH needs to be protonated first in order for it to leave and moreover it is a very fast leaving group (also we have the case that $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ are very weak bases and considering that only elimination reactions occur tells us that it must follow an E1 elimination). When the OH is protonated and it leaves, we form a secondary carbocation, but that can be further more made stabilized via an hydride shift from the adjacent carbon producing a tertiary carbocation. Now either of the base $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ can capture the hydrogen from the adjacent carbon leading to the desired Zaitsev product. (Also the stereochemistry isn't necessary to worry about here, as the reaction proceeds through E1.)

mechanism

Floatoss
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  • I would humbly request the contributor to make an edit. The formation of carbocation by removal of H2O is actually a slow step. It is infact the RDS and should proceed slow as is characteristic of any. The addition of the H+ forming the protonated alcohol is the fast step that you've confused with. – Desai Jan 06 '21 at 02:22
  • but the water is such a nice leaving group that it leaves very fast and forms the cation. That is what I meant. – Floatoss Jan 06 '21 at 02:29
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    No, the slow step is carbocation formation. The rest are even faster. – orthocresol Jan 06 '21 at 11:16
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    @Desai, please stop using mod flags for everything. Those are meant for exceptional issues. If the answer is wrong, you can comment, or downvote. No need to flag for moderator intervention, or else we'd literally have to fact-check every answer on this site. – orthocresol Jan 06 '21 at 11:17
  • @orthocresol the carbocation is formed when the water leaves as a leaving group, but if water leaves fast (as its nature is), doesn't that imply that the carbocation formation is also faster? – Floatoss Jan 06 '21 at 11:20
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    OK, let's clarify a few things. (1) The "loss of water" refers to exactly the same step as the carbocation formation. What I'm saying is that that step, whatever you want to call it (water leaving or carbocation being formed), is the slow step. (2) You're still forming an unstable intermediate (a carbocation), which necessarily means that this step represents a large increase in energy, hence higher activation energy, hence slower. Water being the leaving group merely makes that slightly easier, but not to the point where it can be faster than the other steps in the mechanism. [...] – orthocresol Jan 06 '21 at 11:29
  • (3) You seem to be confusing "slow step" with a "slow" step. What I mean by this is that, the "slow step" need not even be slow by any means: it can be pretty fast in and of itself. It just needs to be the slowest of all the steps. That means that, even if you disagree with point (2) and say that carbocation formation is fast, it can still be the "slow step", as long as it's not as fast as everything else. – orthocresol Jan 06 '21 at 11:33
  • Aside from those two words which I nitpick on, however, your answer is perfectly correct. I think you could also mention that the syn vs anti requirement (as mentioned in the original question) is only relevant for E2 reactions. You implicitly say this by drawing out an E1 reaction, but it's better to leave no doubt. – orthocresol Jan 06 '21 at 11:36
  • @orthocresol Oh is it like even though water is a very nice/fast leaving group, the formation of carbocation is a very unstable step and so we say its "Rate Determining" as it generates an unstable product? – Floatoss Jan 06 '21 at 11:43
  • Well, sort of, but not directly: it is "rate determining" (or equivalently, the "slow step") simply because it is the slowest step in the entire mechanism. Typically, steps that form unstable intermediates are also the slowest, so there is a connection. – orthocresol Jan 06 '21 at 11:49
  • @orthocresol makes sense! I have one more doubt to clarify. When Desai said about this ** The addition of the H+ forming the protonated alcohol is the fast step that you've confused with** . Is this true? – Floatoss Jan 06 '21 at 11:53
  • Yes, I suspect that the protonation is probably the fastest step, although the rationale is a bit more involved. That's not really an issue worth worrying about in this mechanism, though. (In all fairness, knowing the slow step is not a particularly huge deal either. The important thing is knowing which mechanism it proceeds via, i.e. E1 or E2.) – orthocresol Jan 06 '21 at 12:00
  • Let us continue this discussion in chat. – Floatoss Jan 06 '21 at 12:15