1
$\begingroup$

The major product obtained in the following reaction is:

Bromination of aromayic allyl ketone

I wonder after bromonium ion $(\ce{Br+})$ is formed, the attack by MeOH is $\mathrm{S_N}$1 or $\mathrm{S_N}$2? If $\mathrm{S_N}$2, then product (b) should be formed and if $\mathrm{S_N}$1, then Product (a) should be formed.

I think here MeOH is bad nucleophile so $\mathrm{S_N}$1 should be preferred. But if we assume a case $\ce{Br2}$ with KOH then, after formation of bromonium ion product (a) Should be preferred. This because there is more deficiency of electrons on alpha position of ketonic group. Is this concept correct? I need some elaboration on this reaction.

New contributor
Alok Anand is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
4
  • 1
    $\begingroup$ This is clearly a one example of bad question. The original image hadonly mono-atomic $\ce{Br}$ to start with. $\endgroup$ Commented 5 hours ago
  • 1
    $\begingroup$ @MathewMahindaratne A bit of editing drama: not sure how you are viewing this post, but I would probably go with larger atom labels, use uniform font for the illustration, and get rid of colors as they don't convey any useful information. I can barely see the labels with 100% zoom on PC display, and it's a guesswork on a phone. The original illustration was unironically better to my eyes. $\endgroup$ Commented 4 hours ago
  • $\begingroup$ @andselisk: I change the image because I like to avoid overlapping groups. Also, want to include correct reagents for better understanding. I have only access to Marvin IS so cant get rid of colors :-(. Welcome to change as you wish! :-) $\endgroup$ Commented 4 hours ago
  • 1
    $\begingroup$ @MathewMahindaratne Marvin allows for B&W illustrations all right. It also supports ACS style out of the box by choosing it under File → Document Style → Settings… → Apply journal style preset: $\endgroup$ Commented 4 hours ago

1 Answer 1

Reset to default
1
$\begingroup$

After the bromonium ion is formed, there is a positive formal charge on the bromine atom. Consequently, the bromine atom essentially behaves as a protonated oxygen atom when we perform acid-catalysed ring opening.

So, the attack on bromonium ion ring would be like that of acid-catalysed epoxide opening, hence MeOH shall attack the site where the positive charge is stable. This site seems to be the one near the lone pair of oxygen atom rather than near the π-bond as that bond would be polarised to a high extent.

In such reactions of bromonium always the $\mathrm{S_N1}$ path which you say is preferred, even though it’s not strictly $\mathrm{S_N1}.$ Rather, it should be called $\mathrm{S_N1}$-like.

Hence the answer seems to be option (b).

$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.