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    I need to find an alkyl halide that gives a mix of 3 alkenes on e2 elimination. I came up with ch3ch2ch(br)ch(br)ch3

    but this would have 4? 2 E/Z isomers that are ch3ch2c(br)=c(h)ch3 and ch3ch2c(h)=c(br)ch3
    then would have ch3ch2c(h)=c(h)ch3 and ch3ch2c(br)=c(br)ch3

    and an alkyl halide that gives a non zaitsev product on e2 - would an alkyl halide that gives a non zaitsev product on e2 elimination be something like ch3ch2ch(br)ch2ch3, as it has an equal number of carbons either side of the bromine atom that is going to be eliminated?
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    (Original post by ScreamYourHeartOut)
    I need to find an alkyl halide that gives a mix of 3 alkenes on e2 elimination. I came up with ch3ch2ch(br)ch(br)ch3

    but this would have 4? 2 E/Z isomers that are ch3ch2c(br)=c(h)ch3 and ch3ch2c(h)=c(br)ch3
    then would have ch3ch2c(h)=c(h)ch3 and ch3ch2c(br)=c(br)ch3

    and an alkyl halide that gives a non zaitsev product on e2 - would an alkyl halide that gives a non zaitsev product on e2 elimination be something like ch3ch2ch(br)ch2ch3, as it has an equal number of carbons either side of the bromine atom that is going to be eliminated?
    Why not CH3-CH2-CHX-CH3

    You get CH3CH=CHCH3 (two geometric isomers)

    and CH3CH2CH=CH2
 
 
 
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