Question

What is ${}^{\prime }{X}^{\prime }$ in the following sequence of reactions?

$\underset{\left(X\right)}{C_4{H}_{7}OCl}\stackrel{N{H}_3}{\to }{C}_4{H}_{9}ON\stackrel{B{r}_2/KOH}{\to }C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2$

• A. $(C{H}_3{)}_{2}CHCOCl$
• B. $C{H}_{3}C{H}_{2}CH\left(OH\right)C{H}_{2}Cl$
• C. $C{H}_{3}C{H}_{2}C{H}_{2}COCl$
• D. $ClC{H}_{2}C{H}_{2}C{H}_{2}CHO$

Answer 1

The correct option is C $C{H}_{3}C{H}_{2}C{H}_{2}COCl$

The reaction can be structurally written as,

$C{H}_{3}C{H}_{2}C{H}_{2}COCl\left(X\right)\stackrel{NH3}{\to }C{H}_{3}C{H}_{2}C{H}_{2}CON{H}_2\stackrel{Br2/KOH}{\to }C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2$

Here, X=$C{H}_{3}C{H}_{2}C{H}_{2}COCl$

X can be confirmed by its number of hydrogen atoms given in the molecular formula of X. And also, nitro n-propane can be obtained only from n-butanamide and this reaction is called a Hoffmann reaction. n-butanamide can be obtained only from n-butanyl chloride.

Thus, option -C is correct.