Product-I \(\stackrel{\text { aq.KOH }}{\longleftarrow} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}…

Product-I \(\stackrel{\text { aq.KOH }}{\longleftarrow} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \stackrel{\text { alc.kOH }}{\longrightarrow}\) Product-II The correct statement is
  1. product-I is obtained by the elimination reaction
  2. product-II is obtained by the substitution reaction
  3. the molecular formula of product-I is \(\mathrm{C}_{2} \mathrm{H}_{4^{\prime}}\) while the molecular formula of product-II is \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\)
  4. product-I is the isomer of dimethyl ether, while product-II is the dehydrated compound of product-I.

Solution

$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \underset{ }{ } \stackrel{\mathrm{aq} \mathrm{KOH}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ (Product-I)
(Nucleophilic substitution reaction)
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \stackrel{\mathrm{de} \mathrm{KOH}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{4}$ (Product-II)
(elimination reaction)
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ (Product-I) isomer is $\mathrm{CH}_{3} \mathrm{OCH}_{3}$
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ denydration $\longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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