Question

Arrange the following in order of increasing reactivity (least $\to$ most) towards nucleophile.
$\underset{1}{C{H}_3\stackrel{O}{\stackrel{||}{C}})OCHC{H}_3}$ $\underset{2}{C{H}_3\stackrel{O}{\stackrel{||}{C}}Cl}$ $\underset{3}{C{H}_3\stackrel{O}{\stackrel{||}{C}}NHC{H}_3}$

• A. $1<2<3$
• B. $3<1<2$
• C. $1<3<2$
• D. $2<1<3$

Answer 1

The correct option is B $3<1<2$

$O$ $O$ $O$

$C{H}_3-\stackrel{||}{C}-OC{H}_{2}C{H}_3$ $C{H}_3-\stackrel{||}{C}-Cl$ $C{H}_3-\stackrel{||}{C}-NHC{H}_3$

$1$ $2$ $3$

The important features of carboxylic acid derivatives that influence their reactivity are governed by the substituents i.e. $-OC{H}_{2}C{H}_3,Cl,NHC{H}_3$. If the substituent is electron donating, then electrophilicity of carboxyl carbon is reduced, as a result it will be less reactive. Therefore $3$ is least reactive as it contains electron donating $NHC{H}_3$ group. If the substituent is electron-withdrawing, the electrophilicity is increased, therefore $2$ is most reactive. Reactivity of $1$ is in between $2$ and $3$. Therefore reactivity order: $3<1<2$ .