Which is most reactive towards soda-lime deca | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The decarboxylation of carboxylic acids with soda-lime $(NaOH + CaO)$ proceeds through the formation of a carbanion intermediate. The rate of deca

Vedclass · Accepted Answer

$p-O_2N-C_6H_4-CH_2-COOH$

Vedclass · Answer

(B) The decarboxylation of carboxylic acids with soda-lime $(NaOH + CaO)$ proceeds through the formation of a carbanion intermediate. The rate of decarboxylation is directly proportional to the stability of the carbanion formed after the loss of $CO_2$. Electron-withdrawing groups (EWGs) stabilize the carbanion,thereby increasing the rate of decarboxylation,while electron-donating groups (EDGs) destabilize it. Comparing the substituents: $1$. $-NO_2$ is a strong electron-withdrawing group $(EWG)$ due to both $-I$ and $-M$ effects. $2$. $-OCH_3$ is an electron-donating group $(EDG)$ due to its $+M$ effect. Between the nitro-substituted compounds,the $-NO_2$ group at the ortho or para position exerts a stronger electron-withdrawing effect via resonance ($-M$ effect) compared to the meta position. However,in the case of $p-O_2N-C_6H_4-CH_2-COOH$,the $-NO_2$ group is at the para position relative to the $-CH_2COOH$ group. The electron-withdrawing effect of the $-NO_2$ group stabilizes the negative charge on the benzylic carbon formed during the transition state. Thus,$p-O_2N-C_6H_4-CH_2-COOH$ is the most reactive.

Which is most reactive towards soda-lime decarboxylation?

Similar Questions

Consider the given reaction: $RCOOAg \xrightarrow{Br_2/\Delta} R-Br$. Which one of the following acids will give the maximum yield of $R-Br$ in the above reaction?

What product is obtained when two moles of acetic acid are heated with $P_2O_5$?

When formic acid is heated with conc. $H_{2}SO_{4}$,the gas evolved is

The role of conc. $H_2SO_4$ in the esterification process is

Identify the reagents $A$ and $B$ in the following reaction: $CH_3 CH_3 \leftrightarrow{(B)} CH_3 COOH \rightarrow{(A)} CH_3 CH_2 OH$

Vedclass Test Series

Exam Paper Generator

Online Exam Module