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enter image description here

Obviously, A would be 3-Bromoo 3-methyl cyclo hexene:
enter image description here

as the stable free radical intermediate would be formed.
But B is what I am having problems with. I read that $\ce{LiAlH4}$ forms alkenes with three degree halides. So my guess was

2-Methyl-1,3-cyclohexadiene:
enter image description here

But when I checked my book, it said B would just be the Bromine getting substituted by Deuterium in (A)
(I'm sorry I don't know any site that allows making chemical structures with isotopes included so I had to describe it like this.)

Can someone please help me out with this?

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  • $\begingroup$ How are you introducing Chlorine with N-BROMO-succinimide? Are you aware that LiAlH4 reduces haloalkanes to alkanes? $\endgroup$
    – Waylander
    Mar 19, 2021 at 16:25
  • $\begingroup$ @Waylander Sorry, that was a mistake. I have made necessary corrections. And yes, I am aware that $LiAlH_4$ reduces haloalkanes to alkanes. But I also read that in case of three degree haloalkanes, $LiAlH_4$ gives alkenes instead of alkanes $\endgroup$
    – Prajwal Tiwari
    Mar 19, 2021 at 16:40
  • $\begingroup$ There are 4 possible constitutional isomers possible from 2 allylic radicals. Why this one? $\endgroup$
    – user55119
    Mar 19, 2021 at 17:30
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    $\begingroup$ @Prajwal Tiwari: A should be 3-bromo-1-methylcyclohexene (not 3-bromo-3-methylhexene; it is cyclohexene, any way). Thus, B should be 3-deutero-1-methylcyclohexene. $\endgroup$
    – Mathew Mahindaratne
    Mar 21, 2021 at 5:23
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    $\begingroup$ @MathewMahindaratne That rearrangement is possible. This is a bad question given that there is no definitive answer from the literature. $\endgroup$
    – Waylander
    Mar 21, 2021 at 8:01

1 Answer 1

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I will give this my best shot. There doesn't appear that there is any conclusive evidence for this reaction since some sources say elimination is favoured while others say substitution is the major product. However, for this reaction, the difference likely resides in the fact that it is an allylic bromide. This is extremely favourable for substitution reactions because the intermediate will be very stable. Just because of that, substitution may be favoured over elimination and result in the major product being Br substituted for a deuterium atom, just as your book mentioned.

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