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I know that conc $\ce{H2SO4}$ is a dehydrating agent. And the most easiest mechanism for example would be this:

mechanism

But I want to know how to find the extent of dehydration of it.

I came across this question:

question

And the answer is:

d

My thinking:

(a) $\ce{O}$ is EN and will pull $\ce{-OH}$ towards itself inductively, thus decreasing the electron density on $\ce{-OH}$, and decreasing its power to pull $\ce{H}$ from concentrated $\ce{H2SO4}$.

Similarly for (b) and (c) and I think due to inductivity (c) will have more chances than (b).

As for the (d) I think $\ce{-CH3}$ group is an electron pusher so it should increase the electron density, and hence reducing the effect on $\ce{-OH}$ group .

1) But my doubt is: is this the only reason for the extent of dehydration by conc $\ce{H2SO4}$ on (d) to be more than (c)?

2) I don't think there would be much difference if it tautomerises to enol form.

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  • $\begingroup$ The extent of dehydration would relate to how much hydrate the starting material could potentialy form with the sulfuric acid? $\endgroup$
    – Technetium
    Oct 28, 2015 at 13:53
  • $\begingroup$ Could you please tell in more detail? $\endgroup$
    – shaistha
    Oct 28, 2015 at 14:40
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    $\begingroup$ You should think about the relative stabilities of the products. Which has the most stable dehydration product and why? $\endgroup$
    – bon
    Oct 28, 2015 at 15:13
  • $\begingroup$ I'll give you two more hints. (1) You must think in 3D. The ring on the molecule isn't flat. (2) a,b and c have OH moved relative to C=O group. The OH would have the most effect "meta" or "ortho" to the C=O, so b is very likely out off the bat. $\endgroup$
    – MaxW
    Oct 28, 2015 at 19:03
  • $\begingroup$ @MaxW thank you, it kind of helped. But I want to know why d over c? Is the reason what I have already written or something else. If I'm wrong, then please tell me where I'm wrong or I'm missing here. $\endgroup$
    – shaistha
    Oct 28, 2015 at 21:24

1 Answer 1

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The most stable Alkene is the most highly substituted Alkene. (d) gives a tri-substituted alkene, all others give the less disubstituted alkenes. Most stable Alkene (d) therefore more of it is formed! (Edited thanks to Bon for highlighting my error)

Also a factor to keep in mind is conjugation. (a), (b) and (d) will have conjugated double-bonds, while (c) will not. This conjugation also adds stability to the system. But the major stability contributor is the higher substituted alkene in (d).

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  • $\begingroup$ @bon only saw your comment after posting my answer...credit to you... $\endgroup$
    – Leeser
    Oct 29, 2015 at 8:50
  • $\begingroup$ No, credit goes to you because you posted an answer ;). However, you meant tri-substituted, didn’t you? $\endgroup$
    – Jan
    Oct 29, 2015 at 10:14
  • $\begingroup$ indeed I did..tri substituted it is, others are disubstituted.. $\endgroup$
    – Leeser
    Oct 29, 2015 at 11:24

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