0
$\begingroup$

Reaction

In the book "Designing Organic Syntheses - A programmed approach to the Synthon Approach" pg 23, the author proposes that one route to 1-bromo-3-methylbut-2-ene would be by reacting 2-methylbut-3-en-2-ol with hydrobromic acid. I am confused as to why these two reactants wouldn't produce 3-bromo-3-methylbut-1-ene instead. I am guessing the first step in a reasonable reaction mechanism would be protonation of the alcohol to form H30, which then leaves via SN1 forming a tertiary carbocation with a trigonal planar geometry. I would have thought that the nucleophilic bromide would have preferentially attacked the tertiary carbon since over the primary carbon since:

  1. the tertiary carbocation is more stable than the primary carbocation
  2. the tertiary carbocation has a trigonal planar geometry so sterics wouldn't prevent it from being approached by the bromide

Reaxy's link to the reaction described- href https://www.reaxys.com/reaxys/secured/hopinto.do?context=R&query=RX.ID%3D175529&database=RX&origin=ReaxysOutput&ln=

Background - I am not a chemist but I have been reading about Organic chemistry, on and off, since I took it in college 12 years ago.

$\endgroup$
8
  • $\begingroup$ Check out this logic: cation formation and then resonance. I am not able to expand further, but my thought was originally formed because the cation is allylic in nature and is resonance stabilised. $\endgroup$ Dec 21, 2023 at 11:45
  • $\begingroup$ The sterics are still a large enough factor, bromine is a large atom $\endgroup$
    – Waylander
    Dec 21, 2023 at 13:22
  • $\begingroup$ Also the higher substituted double bond is the most stable and the thermodynamic more stable product is preferred. $\endgroup$
    – aventurin
    Dec 21, 2023 at 14:10
  • $\begingroup$ Hey, thanks for the responses so far and sorry for the delay - I was waiting for an email that someone replied to my post but I never got one. Anyway, @HarikrishnanM Allylic resonance is definitely at play, but what I thought was unusual in this case is that the electrons from the Pi bond are moving to form the less stable cation. I always read that primary cations basically dont form. $\endgroup$ Dec 22, 2023 at 20:20
  • $\begingroup$ @Waylander - I will try to see if I can find some material that discuss the effects of sterics on species with triagonal planar geometries. I assumed it wasn't a factor but you might be correct. $\endgroup$ Dec 22, 2023 at 20:21

2 Answers 2

0
$\begingroup$

Im a little hesitant to post an 'answer' to a question about the reaction mechanism for a particular reaction since there is no way to verify this information without performing specific tests or an ab initio calc, but someone posted an answer on Reddit that I thought made sense so I am repositing it here:

There are 2 possible react mechanisms here that give you this product:

SN1 where the driving force is greater stability of a tri-substituted alkene versus mono-substituted alkene. Both are allylic carbocations so the substitution of the alkene is more important.

SN2’ (read as SN2 prime) where HBr ptotonates the alcohol making it a good leaving group. Then Br- can nucleophilicly attack the terminal alkene that inturn displaces the water.

A key part for this reaction is temperature. The higher the heat, the more SN1 will be favored. I believe typical SN1 reactions need about 90C, but that is more a rule of thumb.

Since the reaction conditions on Reaxys are at 0 C I suspect that the mechanism for this reaction would be the SN2 route described above

$\endgroup$
1
  • 1
    $\begingroup$ Tertiary allylic carbonium ions form readily even at 0C. The protonating agent in hydrobromic acid is H3O+ not HBr making SN2 less likely and the stability of the alkene important. Higher temperatures in SN1[and SN2] reactions usually promote elimination over substitution, not SN2. The bromide ion is simply attracted to the entire delocalized cation and the more stable alkene is formed. $\endgroup$
    – jimchmst
    Feb 28 at 21:36
-1
$\begingroup$

First $HBr$ protonates the $-OH$ making it $-OH_2^+$, Now $-OH_2^+$ being a leaving group will leave the molecule creating a tertiary carbocation. But as you can observe there is p - $\pi$ Resonance Available. Hence we get 2 resonating structures. The above clarifies the presence of two Resonating Structures.

But deciding on the major product needs more clarification on the part of Reaction conditions. There are some interactions called ion pair interactions. The $Br^-$ present in the reaction medium mildly stabilizes the tertiary carbocation which dominate the stability from resonance. But we know that resonance doesn't change position of atoms rather changes only positions of bonds and electron pairs. But the caveat is that ion pair interactions are only dominant at around $-5^oC$. So if the reaction is done at around $-5^oC$ we can say the intermediate involving in the substitution will be the first i.e the tertiary carbocation and not primary one.

Now as we generally do organic reactions at room temperature or higher we can say that the significance of these ion pair interactions is gone and the stability of the product plays a bigger role in the formation of the final product and also if the reaction is performed at higher temperature, it tends to have a reversible nature in the product forming step, where by chance if any of the product is from the tertiary carbocation, it gets converted to the one with primary carbocation.

Hope this clarifies the doubt.

$\endgroup$
1
  • $\begingroup$ Ther are no primary and tertiary cations. There is one conjugated allylic cation that reacts with bromide to give the more stable alkene. $\endgroup$
    – jimchmst
    Feb 28 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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