Give a balanced chemical equation for each of the following –
(i) Preparation of ethane from sodium propionate.
(ii) Action of alcoholic KOH on bromoethane.
(i) ethane from sodium propionate : ethane is formed
C2H5COONasodium propionate+NaOHsodalime→CaOΔC2H6ethane+Na2CO3\underset{\text{sodium propionate}}{\text{C}_2\text{H}_5\text{COONa}} + \underset{\text{sodalime}}{\text{NaOH}} \underset{\Delta}{\xrightarrow{\text{CaO}}} \underset{\text{ethane}}{\text{C}_2\text{H}_6} + \text{Na}_2\text{CO}_3sodium propionateC2H5COONa+sodalimeNaOHΔCaOethaneC2H6+Na2CO3
(ii) ethyl alcohol is formed.
C2H5-Br Bromoethane [ethyl bromide]+KOH [aq.] alcoholic KOH→boilC2H5OH ethyl alcohol+KBr\underset{\text{ Bromoethane [ethyl bromide]}}{\text{C}_2\text{H}_5\text{-Br}} + \underset{\text{ alcoholic KOH}}{\text{KOH [aq.]}} \xrightarrow{\text{boil}} \underset{ \text{ ethyl alcohol}} {\text{C}_2\text{H}_5\text{OH}} + \text{KBr} Bromoethane [ethyl bromide]C2H5-Br+ alcoholic KOHKOH [aq.]boil ethyl alcoholC2H5OH+KBr
