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I'm struggling to grasp how the reaction would turn out. Could you help?

My idea was that the bromide atom will get replaced by the amino group of $\ce{NaNH2}$, but when I google, I read that $\ce{NaNH2}$ is a strong base and so there might also be other possibilities like a deprotonation.

Disclaimer: also asked at https://www.reddit.com/r/chemhelp/comments/1bxea94/1bromo4tertbutylbenzene_reaction_with_nanh2/

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    $\begingroup$ Deprotonation to give benzyne seems most likely $\endgroup$
    – Waylander
    Commented Apr 6 at 16:17
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    $\begingroup$ A mixture of products seems even more likely. $\endgroup$
    – Oscar Lanzi
    Commented Apr 6 at 23:03
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    $\begingroup$ I’m voting to close this question because we discourage cross-posting within the network and outside it. $\endgroup$
    – Martin - マーチン ♦
    Commented Apr 7 at 11:35

1 Answer 1

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The reaction will proceed by deprotonation of the benzene ring to give the benzyne, an unstable intermediate, which then will react with ammonia/amide anion to give a mixture of m and p t-butyl aniline.

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  • $\begingroup$ How would $ortho$ product ever form? $\endgroup$
    – ron
    Commented Apr 7 at 18:30
  • $\begingroup$ By the addition of iso-butylene to aniline patents.google.com/patent/US4892974A/en $\endgroup$
    – Waylander
    Commented Apr 7 at 20:15
  • $\begingroup$ Sorry, I'm not following. The aryne intermediate can only produce products by addition to either end of the "triple" bond. This would yield products with the substituent oriented either $meta$ or $para$ to the $t$-butyl group, no $ortho$ product would be formed. Further, there is no isobutylene or silica catalyst, as claimed in your referenced patent, present in this reaction. $\endgroup$
    – ron
    Commented Apr 7 at 21:19
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    $\begingroup$ Sorry, my misunderstanding I thought you were asking how the o product is made. Have edited the answer $\endgroup$
    – Waylander
    Commented Apr 7 at 21:34

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