Question

Complete the following chemical reaction : ${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{OH}\underset{{170}^{\mathrm{o}}\mathrm{C}}{\overset{\mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}\left(\mathrm{X}\right)\stackrel{{\mathrm{Br}}_2}{\to }\left(\mathrm{Y}\right)\stackrel{\mathrm{Excess}\mathrm{of}\mathrm{alcoholic}\mathrm{KOH}}{\to }\left(\mathrm{Z}\right)$

Answer 1

Completing the reaction:

Step 1: Dehydration reaction

• When alcohol reacts with concentrated sulphuric acid at high temperatures then dehydration occurs and alkene is formed.
• In a dehydration reaction, water is removed via the formation of a carbocation.
• The carbocation is stabilized by the formation of a double bond with the adjacent carbon atom.
• ${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{OH}\underset{{170}^{\mathrm{o}}\mathrm{C}}{\overset{\mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}C{H}_3—CH=C{H}_2$

Step 2: Reaction with bromine

• Alkene reacts with bromine to form alkyl halide.
• Bromine is added to the respective positions of the double bond.
• $C{H}_3—CH=C{H}_2+B{r}_2\to C{H}_3—CH\left(Br\right)—C{H}_2\left(Br\right)$

Step 3: Reaction with an excess of alcoholic KOH

• Alkyl halide reacts with alcoholic potassium hydroxide to form an alkyne.
• In an elimination reaction cations and anions are removed resulting in the formation of a pi-bond.
• This is an example of an elimination reaction.
• $C{H}_3—CH\left(Br\right)—C{H}_2\left(Br\right)\stackrel{ExcessofalcoholicKOH}{\to }C{H}_3—CH\equiv CH$

Therefore, the complete reaction is given by:

${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{OH}\underset{{170}^{\mathrm{o}}\mathrm{C}}{\overset{\mathrm{Conc}.{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}C{H}_{3}CH=C{H}_2\stackrel{{\mathrm{Br}}_2}{\to }C{H}_{3}CH\left(Br\right)—C{H}_2\left(Br\right)\stackrel{\mathrm{Excess}\mathrm{of}\mathrm{alcoholic}\mathrm{KOH}}{\to }C{H}_{3}C\equiv CH$