1.

Arrange the following alkyl halides in decreasing order of the rate of beta-elimination reaction with alcohlic KOH. (i) CH_(3) - underset(CH_(3))underset(|)overset(H)overset(|)(C) - CH_(2)Br (ii) CH_(3)-CH_(2)-Br (iii) CH_(3)-CH_(2)-CH_(2)-Br

Answer»

`i gt II gt iii`
`iii gt ii gt i`
`ii gt iii gt i`
`i gt iii gt ii`

Solution :The larger the number of alkyl groups attached to the DOUBLE bonded CARBON atoms, the more STABLE is the alkene and more reactive is the CORRESPONDING alkyl halide. So correct order is :
`i gt ii gt iii`.


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