Complete The Following Chemical Reaction : C3H7OH → ConH2SO4 (X)→Br2 (Y)→ KOH (Z)

Given:

\(C_{3}H_{7}OH \xrightarrow[170^{\circ}C]{Conc.H_{2}SO4} (X) \overset{Br_{2}}{\rightarrow} (Y) \overset{Excess of alcoholic KOH}{\rightarrow} (Z)\)

According to the above given chemical reaction:

\(C_{3}H_{7}OH \) reacts with Conc.H_{2}SO4 at 170^{\circ}C\). Then \(CH_{3}- CH = CH_{2}\) is produce along with the elimination of water.

Later \(CH_{3}-CH=CH_{2}\) reacts in the presence of \(Br_{2}, CH_{3}-CH (Br)-CH-Br \)is produced.

Later on, reaction with the presence of excess alcoholic KOH will produce \(CH_{3}-C\equiv CH \)

Therefore,

X = \(CH_{3}- CH = CH_{2}\)

Y = \(CH_{3}-CH (Br)-CH-Br\)

Z = \(CH_{3}-C\equiv CH \)

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