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The Question

This was in a paper I gave. I want to know the mechanism by which both of these products will be formed and why is the 1st one the major product. My thinking is that the -OH group will be converted into a good leaving group by protonating it and then a 2-degree Carbocation will be formed which should give Product 2. But the answer given is product 1 (which will go through a 1-degree carbocation as an intermediate).
The 1st product:

The 1st Product

The 2nd product:

The 2nd product

Any input will be appreciated.

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    $\begingroup$ Consider the stability of the allylic cation and the size of the bromine anion. $\endgroup$
    – Waylander
    Jul 30 at 13:02
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    $\begingroup$ I think if the reaction is under kinetic control (low temperature) then it proceeds with the most stable intermediate, so product $2$. But if the reaction is under thermodynamic control (high temperature), then the more stable product, the more substituted alkene $1$ will be formed. They have not mentioned the temperature, so not sure which one to choose. $\endgroup$
    – TRC
    Jul 30 at 14:03
  • $\begingroup$ In such papers if there isn't a symbol of 'Delta" i.e. heat then it isn't high temperature. So, according to your theory product 2 must form as a major pdt, which is what I had marked in the exam, but its wrong according to the paper setters. Any other logic? @TRC $\endgroup$
    – GuyEternal
    Jul 30 at 14:53
  • $\begingroup$ Are you sure they mention heat every time? For instance, hydrolysis of the cyanide group to carboxylic acid makes heating essential - but I rarely see anything except $\ce{H3O^+}$ mentioned for the hydrolysis - in the same exam. $\endgroup$
    – TRC
    Jul 31 at 3:49
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    $\begingroup$ Besides, there is a famous question of addition of $\ce{HBr}$ to $1,3-$butadiene that most students come across while preparing for JEE. There the "high temperature" for thermodynamic control is $40^oC$. Do you think that temperature will be mentioned separately as heat? $\endgroup$
    – TRC
    Jul 31 at 3:52
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This reaction called Electrophilic addition to α,β-unsaturated carbonyl compounds.

Electrophilic addition to α,β-unsaturated carbonyl compounds is analogous to electrophilic addition to isolated double bonds, except that the electrophile adds to the carbonyl oxygen, the most basic atom in the molecule. After that, the nucleophile adds to the β carbon, and the resulting intermediate enol tautomerizes to the more stable carbonyl compound.

The 1st & 2nd products are incorrect.

What is the major product and why?


The major product reaction mechanism should be as follows enter image description here

and the reaction mechanism for the minor product is

enter image description here


Why?

As the carbonyl is protonated then the intermediate cation is a stronger electrophile and will react with the bromide ion.

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  • $\begingroup$ I believe this might be incorrect - as far as I know the protonation occurs fastest at that position where the most stable carbocation is first formed. Your carbocation forming during Michael addition is primary and unstable. The carbocation formed by protonating and removing the alcohol group is secondary and allyl - more stable. $\endgroup$
    – TRC
    Jul 31 at 3:42
  • $\begingroup$ I would expect acid-base reaction leading to removal of water a much faster reaction. We do have concentrated HBr here, since acid-base is way faster than electrophlilic addition, there may be a discrepancy with the source you have mentioned because it does not take into account the presence of alcohols. $\endgroup$ Jul 31 at 12:27
  • $\begingroup$ @ThermEster Thankyou for your answer but I think that the effect of alcoholic grp isn't taken into account in your answer (as pointed out by @napstablook). I understood the answer after discussing in the comments. $\endgroup$
    – GuyEternal
    Jul 31 at 17:17

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