The starting material is CH_3COCH_2CHO. The r | vedclass

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

Question

(B) The starting material is $CH_3COCH_2CHO$ ($3$-oxobutanal).Step $(i)$: Treatment with ethylene glycol and dry $HCl$ selectively protects the more r

Vedclass · Accepted Answer

but$-2-$en$-1-$al

Vedclass · Answer

(B) The starting material is $CH_3COCH_2CHO$ ($3$-oxobutanal).Step $(i)$: Treatment with ethylene glycol and dry $HCl$ selectively protects the more reactive carbonyl group (aldehyde) as a cyclic acetal. The ketone remains unaffected.Step $(ii)$: $LiAlH_4$ reduces the ketone group to a secondary alcohol. The product $A$ is $CH_3CH(OH)CH_2CH(OCH_2CH_2O)$.Step $(iii)$: Acidic hydrolysis $(H_3O^{+}/\Delta)$ deprotects the acetal back to an aldehyde and simultaneously causes dehydration of the $\beta$-hydroxy ketone formed,leading to an $\alpha,\beta$-unsaturated carbonyl compound.The final product $B$ is $CH_3CH=CHCHO$ (but$-2-$enal).

The starting material is $CH_3COCH_2CHO$. The reaction sequence is:
$CH_3COCH_2CHO$ $\xrightarrow[{(ii)LiAlH_4}]{{(i)ethylene\,glycol/dry\,HCl}} A$ $\xrightarrow[\Delta ]{H_3O^{+}} B$
What will be $B$?

Similar Questions

If formaldehyde and $KOH$ are heated,then we get

Why is there a large difference in the boiling points of butanal and butan$-1-$ol?

Identify the major product $A$ in the following reaction:
[Image of the reaction: $A$ substituted tetralin derivative with a ketone group,a chlorine atom,and a nitro group,reacting with $N_2H_4/KOH/H_2O$ (Wolff-Kishner reduction)]

Which of the following compounds will respond to the haloform reaction with $I_{2}$ and $KOH$?

In the given set of reactions,identify $D$:
$CH_3COOH$ $\xrightarrow{SOCl_2} A$ $\xrightarrow[Anhy. AlCl_3]{Benzene} B$ $\xrightarrow{HCN} C$ $\xrightarrow{H_2O} D$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The starting material is $CH_3COCH_2CHO$. The reaction sequence is:$CH_3COCH_2CHO$ $\xrightarrow[{(ii)LiAlH_4}]{{(i)ethylene\,glycol/dry\,HCl}} A$ $\xrightarrow[\Delta ]{H_3O^{+}} B$What will be $B$?