MCQ
Alcoholic solution of $KOH$ is used for
- ADehydration
- BDehydrogenation
- ✓Dehydrohalogenation
- DDehalogenation
✓
Answer
Correct option: C.
Dehydrohalogenation
c
(c) $C{H_3} - C{H_2} - Br + \mathop {KOH}\limits_{{\rm{alk}}}\,\, \xrightarrow{Dehydrohal ogenation} \,\,C{H_2} = C{H_2} + KBr + {H_2}O$
(c) $C{H_3} - C{H_2} - Br + \mathop {KOH}\limits_{{\rm{alk}}}\,\, \xrightarrow{Dehydrohal ogenation} \,\,C{H_2} = C{H_2} + KBr + {H_2}O$
In alcoholic $KOH$ alkoxide ions $(R{O^ - })$ are present which is a strong base. They abstract proton from ?-carbon of alkyl halide and favours elimination reaction
$\mathop {ROH}\limits_{{\rm{Alcohol}}} + KOH \to \mathop {ROK + {H_2}O}\limits_{{\rm{Potassium \,alkoxide}}} $
$ROK \to \mathop {R{O^ - }}\limits_{{\rm{Alkoxide \,ion}}} + {K^ + }$
$R{O^ - } + H - \mathop {C{H_2}}\limits^\beta - \mathop {C{H_2}}\limits^\alpha - Br \to ROH + C{H_2} = C{H_2} + Br$
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