Which of the following $α$ , $β$ -elimination?

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$H_{3} C - {CH}_{2} OH \to_{160^{o} - 180^{o} C}^{H_{2} {SO}_{4} conc.} {CH}_{2} = {CH}_{2} + H_{2} O$

$H_{2} CBr - {CH}_{2} Br \to_{or Nal/Acetone}^{Zn / {CH}_{3} OH or ether} {CH}_{2} = {CH}_{2} + {ZnBr}_{2}$

$H_{3} C - {CH}_{2} Cl \to_{60^{o} - 80^{o} C}^{KOH / alc} {CH}_{2} = {CH}_{2} + HCl$

All of the above

answer is D.

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Detailed Solution

In an elimination reaction we typically have a leaving group (denoted by “X”[Cl ,Br]in the figure above), The carbon that the leaving group is attached to is referred to as the "alpha" carbon. If a hydrogen is removed from the "beta" carbon, then we have formed an olefin and the reaction is called a beta elimination.

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