# 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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