Question #78e48

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Dec 6, 2017

Answer:

Due to steric hindrance, When the base is bulky, it will preferentially remove the most accessible hydrogen to avoid steric repulsion.

Explanation:

Hofmann's Elimination reactions takes place:

• When the base is larger
• When alkyl halide contains one or more double bond at $\gamma$
carbon
• When the leaving group is poor such as F,NR3+,SR2+.

A bulky base removes the most accessible hydrogen and will be from the end of the substrate and hence least substitutedproduct.

{when alkyl halide contain one or more double bond then less substituted alkene may be major becase conjugated double bonds are more stable than isolated double bond.}

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