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Question
Ethylene chloride and ethylidene chloride are isomers. Identify the correct statements.
(i) Both the compounds form same product on treatment with alcoholic KOH.
(ii) Both the compounds form same product on treatment with aq.NaOH.
(iii) Both the compounds form same product on reduction.
(iv) Both the compounds are optically active.
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Solution
(i) Both the compounds form same product on treatment with alcoholic KOH.
(iii) Both the compounds form same product on reduction.
Explanation:
\[\ce{\underset{(ethylidence chloride)}{H3C - CHCl2}}\] and \[\begin{array}{cc}
\phantom{}\ce{H2C - CH2}\phantom{}\\
\phantom{}|\phantom{....}|\\
\phantom{.}\ce{\underset{(ethylene dichloride)}{\phantom{}Cl\phantom{...}Cl}}\phantom{}
\end{array}\] are isomers.
(i) They give ethyne on treatment with alcoholic KOH.
\[\ce{CH3CHCl2 ->[alc][KOH] CH ≡ CH + 2KCl + 2H2O}\]
\[\ce{Cl - CH2 - CH2 - Cl ->[alc.KOH] CH ≡ CH + 2KCl + H2O}\]
(ii) On reduction with Zn dust in alcohol they give ethylene.
\[\ce{CH3CHCl2 + Zn ->[CH3OH] CH2 ≡ CH}\]
\[\ce{Cl - CH2 - CH2 - Cl + Zn ->[CH3OH] CH2 = CH2}\]
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