Identify the major product (P) in the given r | vedclass

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

Question

(A) Propanal $(CH_3CH_2CHO)$ has $\alpha$-hydrogens,so it undergoes aldol condensation in the presence of a dilute base $(NaOH)$.$1$. Two molecules of

Vedclass · Accepted Answer

$CH_3-CH_2-CH=C(CH_3)-CHO$

Vedclass · Answer

(A) Propanal $(CH_3CH_2CHO)$ has $\alpha$-hydrogens,so it undergoes aldol condensation in the presence of a dilute base $(NaOH)$.$1$. Two molecules of propanal react to form $3$-hydroxy-$2$-methylpentanal.$2$. Upon heating,this $\beta$-hydroxy aldehyde undergoes dehydration to form the $\alpha,\beta$-unsaturated aldehyde,which is $2$-methylpent-$2$-enal.Reaction: $2CH_3CH_2CHO \xrightarrow{dil. NaOH, \Delta} CH_3CH_2CH=C(CH_3)CHO + H_2O$.

Identify the major product $(P)$ in the given reaction of propanal with dilute $NaOH$ followed by heating.

Similar Questions

Acetone and acetaldehyde are differentiated by

$CaOCl_2 + H_2O \longrightarrow Ca(OH)_2 + X$
$X + CH_3CHO \longrightarrow Y$
$Y + Ca(OH)_2 \longrightarrow CHCl_3$
What is '$Y$'?

Product $(D)$ will be

$CH_3COOCH_2CH_3 + PhMgBr_{(excess)} \to P \xrightarrow{H_2O} Q$;
$Q$ is

What reagent and/or reaction conditions would you choose to bring about the conversion shown in the image?

Vedclass Test Series

Exam Paper Generator

Online Exam Module