Chemistry - Hydrocarbon Question with Solution | TestHub

${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}\frac{\text{ Conc }{\mathrm{H}}_2{\mathrm{SO}}_4}{160^{\circ }-180^{\circ }\mathrm{C}}\to {\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2$ $\stackrel{{\mathrm{Br}}_2}{⟶}\stackrel{\text{ alc. }\mathrm{KOH}}{⟶}{\mathrm{CH}}_3\mathrm{CHBr}-{\mathrm{CH}}_2\mathrm{Br}$ $\left[{\mathrm{CH}}_3\mathrm{C}(\mathrm{Br})={\mathrm{CH}}_2+{\mathrm{CH}}_3\mathrm{CH}=\mathrm{CHBr}\right]$ (A) (B) $\frac{{\mathrm{NaNH}}_2}{-\mathrm{HBr}}\to {\mathrm{CH}}_3\mathrm{C}\equiv \mathrm{CH}$ (Z) Alcoholic $\mathrm{KOH}$ brings about dehydrobromination of $Y$ and give a mixture of vinyl bromide (A and B) while ${\mathrm{NaNH}}_2$ being a strong base than alc. ${\mathrm{KOH}}^2$ readily brings about dehydrobromination of less reactive vinyl bromide to give propyne ${\mathrm{CH}}_3\mathrm{C}\equiv \mathrm{CH}$ i.e. (Z).