A compound with molecular mass $180 \mathrm{u}$ is acylated with $\mathrm{CH}_{3} \mathrm{COCl}$ to get a…
- 2
- 5
- 4
- 6
Solution
$$
m_{1}\left(\mathrm{R}\left(\mathrm{NH}_{2}ight)_{x}ight)=m_{\mathrm{R}}+(14 \mathrm{u})_{x}
$$
$$
m_{2}\left(\mathrm{R}\left(\mathrm{CHCOCH}_{3}ight)_{x}ight)=m_{\mathrm{R}}+(56 \mathrm{u})_{x}
$$
Increase in molecular mass $=(56 \mathrm{u}-14 \mathrm{u})_{x}=(42 \mathrm{u})_{x}$
Given increase in the molecular mass $=210 \mathrm{u}$
Hence, $x=\frac{210 \mathrm{u}}{42 \mathrm{u}}=5$
Asked in: JEE-TOPICTESTS-CHEMISTRY