(This question is from a dubious source)

Friedel-Crafts Acylation of benzene with 1-chloro 2,2-dimethyl propanone

The reaction taking place here is Friedel-Crafts acylation of benzene with 1-chloro 2,2-dimethyl propanone (pivaloyl chloride?). To me, the options seem most peculiar. When the $\ce{C} - \ce{Cl}$ is broken, it gives rise to a carbocation at the carbon of $\ce{C}=\ce{O}$ bond.

According to my teacher, $\ce{C+}=\ce{O}$ now leaves with the electrons of the adjacent $\ce{C}-\ce{C}$ bond, giving rise to a stable tertiary butyl carbocation. This leads us to $\ce{PhCMe3}$.

Edit 2: this is the same reaction: Unusual Friedel–Crafts alkylation with pivaloyl chloride

Considering that the t-butyl group is para-directing, the options still don't seem right to me. If this conversion to a t-butyl carbocation happens once, can it not happen again? Would we simply not get two t-butyl groups para to each other?

(I still doubt multiple substitution is even possible)

Edit 3: as of 24/2/2024 this question has been confirmed to be of poor quality. See reasons in comments.

  • $\begingroup$ According to the person who gave me this question, the answer is B. $\endgroup$
    – zxen
    Feb 21 at 8:50
  • 2
    $\begingroup$ tBu group is similar to Me and electron-donating and so activating and p-directing $\endgroup$
    – Waylander
    Feb 21 at 8:57
  • $\begingroup$ See the answer to this SE question chemistry.stackexchange.com/questions/38454/… $\endgroup$
    – Waylander
    Feb 21 at 11:48
  • 2
    $\begingroup$ I strongly suspect that you would get at least some of the 1,4-di-tert-butyl benzene in the reaction mixture, possibly the major component. The acylium species is not going to stop being a source of tBu+ and the t-Bu-benzene is more activated than benzene itself. $\endgroup$
    – Waylander
    Feb 21 at 13:10
  • 2
    $\begingroup$ You'd get pretty much zero C, little D, but if you got B, then A would also be present and I second what Waylander said. $\endgroup$
    – Mithoron
    Feb 23 at 16:44


Your Answer

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