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1-chloro butane on treatment with alcoholic potash forms
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Verified Answer
The correct answer is:
1-butene
Alc. \(\mathrm{KOH}\) is a dehydrohalogenation agent, it reacts with an alkyl halide and alkene forms as a product. Reaction of process can be given by
\(\underset{\text { 1-chlorobutane }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}} \xrightarrow{\mathrm{Alc}, \mathrm{KOH}} \underset{\text { 1-butene }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}}=\mathrm{CH}_2\)
Hence, option (b) is correct.
\(\underset{\text { 1-chlorobutane }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}} \xrightarrow{\mathrm{Alc}, \mathrm{KOH}} \underset{\text { 1-butene }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}}=\mathrm{CH}_2\)
Hence, option (b) is correct.
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