Hoffman elimination

[BrownieDDD]

Dat destroyer odess 2012 dition ch. 6 # 6

2-bromohexane -----> ? using tertbutyl oxide

a) hexene (Antizeitsev)
b) 2-hexene (Zeitsev product)

answer: A
explanation: Hoffman elim occurs bc base is strong big and builky therefore E2 rxn occurs whereby the least substituated alkene forms

why isn't it B?
I always thought HOffman was with with amine groups..this is an alkyl halide....
also, i was taught things are antizeitsev if BOTH the substrate and base are bulky..here only the base is bulky....so it should be zeitsev..



same thing with #12 of ch. 6

2-butanol --1)SOCL2---> ? ----2) Tertbutyloxide---> ? --3) HBr, ROOR ---> 1-bromobutane

....shouldn't the double bond after step 2 be a zeitsev product.............DAt destroyer has it as the antizeitsev product.

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[invictusx]

For the 2-bromohexane, you want to produce the least substituted product because terbutyloxide is a VERY bulky base and making 2-hexene is "slower" relative to hexene. The bulkiness of the nucleophile inhibits the rate.

For the 2nd question, HBr, ROOR is anti-markovnikov because it uses PEROXIDES (ROOR). This peroxide step is only seen with HBr.

 

[BrownieDDD]

but here only the base is bulky....

 

[xtra333]

look at Wade organic chemistry (in library of your school) it gives a good explanation of why hoffman elim produces anti-saytseff product. like E2, the hoffman elim needs to be angled the H and leaving group in a certain way.