Compound X on heating with Zn dust gives comp | vedclass

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

Question

(C) The reaction sequence is as follows:$X$ $\xrightarrow{\Delta, Zn} Y$ $\xrightarrow{O_3, Zn/H_2O} 2 CH_3CH_2CHO$ (Propionaldehyde)Since the ozonoly

Vedclass · Accepted Answer

$3,4-$dibromohexane (vicinal)

Vedclass · Answer

(C) The reaction sequence is as follows:$X$ $\xrightarrow{\Delta, Zn} Y$ $\xrightarrow{O_3, Zn/H_2O} 2 CH_3CH_2CHO$ (Propionaldehyde)Since the ozonolysis product is two moles of propionaldehyde $(CH_3CH_2CHO)$,the alkene $Y$ must be hex$-3-$ene $(CH_3CH_2CH=CHCH_2CH_3)$.Heating a vicinal dibromide with $Zn$ dust causes debromination to form an alkene.Therefore,$X$ must be $3,4-$dibromohexane $(CH_3CH_2CH(Br)CH(Br)CH_2CH_3)$.Comparing this with the given options,the structure corresponds to $3,4-$dibromohexane.

Compound $X$ on heating with $Zn$ dust gives compound $Y$,which on treatment with $O_3$ followed by reaction with $Zn$ dust and $H_2O$ gives propionaldehyde. The structure of $X$ is:

Similar Questions

What is the major product obtained when $1$-butanol is heated with concentrated $H_2SO_4$?

Which of the following reagents can be used to convert propene into propan-$1$-ol?

Reaction $(1): CH_3-CH=CH-CH_3$ $\xrightarrow{\text{Cold } KMnO_4} (A)$ $\xrightarrow{NaIO_4} (B)$ ($2$ moles)
Reaction $(2): CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4/NaIO_4} (C)$ ($2$ moles)
Product $(B)$ and $(C)$ respectively are:

In the reaction sequence $CH_3CH = CHCH_3$ $\xrightarrow{O_3} A$ $\xrightarrow{H_2O/Zn} B$,the alkene forms product $B$. What is product $B$?

Identify the reagent$(s)$ $A$ and condition$(s)$ for the reaction shown in the image:

Vedclass Test Series

Exam Paper Generator

Online Exam Module