# What would be the major products of this reaction?

Obviously, A would be 3-Bromoo 3-methyl cyclo hexene:

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

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.)

• How are you introducing Chlorine with N-BROMO-succinimide? Are you aware that LiAlH4 reduces haloalkanes to alkanes? Mar 19, 2021 at 16:25
• @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 Mar 19, 2021 at 16:40
• There are 4 possible constitutional isomers possible from 2 allylic radicals. Why this one? Mar 19, 2021 at 17:30
• @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. Mar 21, 2021 at 5:23
• @MathewMahindaratne That rearrangement is possible. This is a bad question given that there is no definitive answer from the literature. Mar 21, 2021 at 8:01