Question

Consider the following reaction:

$\mathrm{A}\underset{\left(\mathrm{ii}\right){\mathrm{H}}_3{\mathrm{O}}^{+}}{\overset{\left(\mathrm{i}\right){\mathrm{CH}}_3\mathrm{MgBr}}{\to }}\mathrm{B}\underset{573\mathrm{K}}{\overset{\mathrm{Cu}}{\to }}2-\mathrm{methyl}-2-\mathrm{butene}$

The mass percentage of carbon in $\mathrm{A}$ is:

Answer 1

The following observations can be deduced from the given reaction:

1. Product $\mathrm{B}$ react with $\mathrm{Cu}$ at $573K$ to form $2$-methyl-$2$-butene and we know that when the vapours of tertiary alcohol are passed over heated copper at $573K$ , the product formed is an alkene.
2. 
3. So product $\mathrm{B}$ is a tertiary alcohol.
4. Now it is also known that when ketones react with Grignard reagent $\left({\mathrm{CH}}_3\mathrm{MgBr}\right)$ in presence of acid $\left({\mathrm{H}}_3{\mathrm{O}}^{+}\right)$, tertiary alcohol is formed.
5. 
6. Therefore, product A is $1$- methyl ethanone.
7. Mass percentage of carbon in compound A is $\frac{W_{carbon}}{W_{compound}}$ where, $W_{carbon}$ is the weight of total carbon present in the compound i.e $12\times 4=48g$ and $W_{compound}$ is weight of compound i.e $72g$.

Mass percentage of carbon in compound $\mathrm{A}$ is $\frac{W_{carbon}}{W_{compound}}=\frac{48}{72}\times 100=66.67\%$

Therefore, the mass percentage of carbon in compound $\mathrm{A}$ is $66.67\%$.