The major product 'P' formed in this reaction | vedclass

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

Question

(C) The given reactant is $2,2'$-diformylbiphenyl. Since both aldehyde groups are attached to the biphenyl system and lack $\alpha$-hydrogens,they und

Vedclass · Accepted Answer

$2-$(hydroxymethyl)biphenyl-$2$'-carboxylic acid

Vedclass · Answer

(C) The given reactant is $2,2'$-diformylbiphenyl. Since both aldehyde groups are attached to the biphenyl system and lack $\alpha$-hydrogens,they undergo an intramolecular Cannizzaro reaction in the presence of a base $(	ext{OH}^-)$. One aldehyde group is reduced to a primary alcohol $(-CH_2OH)$ and the other is oxidized to a carboxylate group $(-COO^-)$. Upon acidic workup $(	ext{H}^+)$,the carboxylate group is protonated to form a carboxylic acid $(-COOH)$. Thus,the product is $2$-(hydroxymethyl)biphenyl-$2'$-carboxylic acid.

The major product $'P'$ formed in this reaction is:

Similar Questions

The reaction is given as follows: $(CH_3)_2CHCHO + CH_3MgBr$ $\xrightarrow{Ether} A$ $\xrightarrow{H_3O^+} B$. The $IUPAC$ name of $B$ is:

The correct order of reactivity of $PhMgBr$ with:
$(I)$ $Ph-CO-Ph$
$(II)$ $CH_3-CHO$
$(III)$ $CH_3-CO-CH_3$
is:

Which substance will give a positive iodoform test?

What is the first intermediate formed during the reduction of carbonyl compounds by $H_2NNH_2$ and $KOH$?

Select the reaction with the correct major product.

Vedclass Test Series

Exam Paper Generator

Online Exam Module