- ADehydration
- BDehydrogenation
- ✓Dehydrohalogenation
- DDehalogenation
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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Consider the following statements about borax :
$a.$ Each boron atom has four $B-O$ bonds
$b.$ Each boron atom has three $B-O$ bonds
$c.$ Two boron atoms have four $B-O$ bonds while other two have three $B-O$ bonds
$d.$ Each boron atom has one $-OH$ groups
Select correct statement(s) :
$ -\mathrm{OCH}_3,-\mathrm{NO}_2,-\mathrm{CN},-\mathrm{CH}_3-\mathrm{NHCOCH}_3, $
$ -\mathrm{COR},-\mathrm{OH},-\mathrm{COOH},-\mathrm{Cl}$