Question

n-Butane is produced by monobromination of ethane followed by Wurtz reaction. Calculate the volume of ethane at NTP required to produce $55g$ of n-butane, if the bromination takes place with $90$ percent yield and Wurtz reaction with $85$ percent yield.

Answer 1

Reactions involved here in are :

$C{H}_{3}C{H}_3+B{r}_2\underset{Monobromination}{\overset{h\nu }{\to }}C{H}_{3}C{H}_{2}Br+HBr$ .......(i)

$22.4$ litres at NTP Ethyl bromide

$109g$

$2C{H}_{3}C{H}_{2}Br+2Na\underset{Reaction}{\overset{Wurtz}{\to }}C{H}_{3}C{H}_2-C{H}_{2}C{H}_3+2NaBr$ .....................(ii)

Ethyl bromide n-butane

$2\times 109=218g$ $58g$

From Eqn. (ii) using weight-weight relationship we have,

$58g$ of n-butane are obtained from $C{H}_{3}C{H}_{2}Br=218g$

$\therefore 55g$ of n-butane will be obtained from $C{H}_{3}C{H}_{2}Br=\frac{218\times 55}{58}g$

But the yield of Wurtz reaction is only $85\%$

$\therefore$ Amount of $C{H}_{3}C{H}_{2}Br$ actually required

$=\frac{100}{85}\times \frac{218}{58}\times 55=243.2g$

From Eqn. (i) using weight volume relationship, we have,

$109g$ of $C{H}_{3}C{H}_{2}Br$ are obtained from ethane = $22.4$ litres at NTP

$\therefore$ $243.2g$ of $C{H}_{3}C{H}_{2}Br$ will be obtained from ethane

$=\frac{22.4\times 243.2}{109}$ litres at NTP

But the yield of bromination is only $90\%$

$\therefore$ Volume of ethane actually required

$=\frac{100}{90}\times \frac{22.4\times 243.2}{109}=55.33$ litres at NTP