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What is the mechanism for double dehydrohalogenation reactions?
Why does a double-dehydrohalogenation reaction always result in a triple bond? I have the molecule 1,2-dibromooctane, and from the IR data, it appears to only have Sp and Sp3 carbons, but no Sp2, suggesting that only the triple bonded isomer is formed, but not the double bonded one.
I guess, first thing, do halides count when you are counting up the substitution of the ß-carbons? (What I mean is, would R–CH2X be considered mono- or di-substituted?)
If they do not count, why is the Sp configuration the major product?
I'd check my book, but it only talks about double-dehydrohalogenations for like, one paragraph, and I can't find a mechanism on the internet.
2 Answers
- Anonymous1 decade agoFavorite Answer
RCH2X is considered a monosubsituted alkyl halide.
Thermodynamically speaking, the use of a strong enough base to remove a C-H from a saturated hydrocarbon is also strong enough to remove a C-H from an alkene because these hydrogens are even more acidic than those of the dihalogenated species. Thus, it would be very hard to make the reaction stop midway. The only possible reaction you could do to generate an alkene and not an alkyne would be to use a metal that has stable M^n and M^(n+2) charges (oxidative addition):
RCXHCXH2 + M^n ---> RCH=CH2 + MX2
I hope this helped clarify your problem.