An optically active alkyl bromide C_4H_9Br re | vedclass

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

Question

(A) The optically active alkyl bromide $C_4H_9Br$ is $2$-bromobutane.Reaction of $2$-bromobutane with ethanolic $KOH$ (dehydrohalogenation) gives but-

Vedclass · Accepted Answer

Haloform test

Vedclass · Answer

(A) The optically active alkyl bromide $C_4H_9Br$ is $2$-bromobutane.Reaction of $2$-bromobutane with ethanolic $KOH$ (dehydrohalogenation) gives but-$2$-ene as the major product $[A]$.Reaction of but-$2$-ene $[A]$ with $Br_2$ yields $2,3$-dibromobutane $[B]$.Reaction of $2,3$-dibromobutane $[B]$ with alcoholic $KOH$ followed by $NaNH_2$ (dehydrohalogenation) yields but-$2$-yne $[C]$.Hydration of but-$2$-yne $[C]$ in the presence of $HgSO_4/H_2SO_4$ (Kucherov's reaction) yields butanone $(CH_3COCH_2CH_3)$,which is a methyl ketone.Methyl ketones are confirmed by the Haloform test (specifically the Iodoform test).

Similar Questions

$A$ compound $A$ has the molecular formula $C_2Cl_3OH$. It reduces Fehling's solution and on oxidation,it gives a monocarboxylic acid $B$. Compound $A$ is also obtained by the reaction of ethanol with chlorine. What is $A$?

The aldol condensation product of acetone on dehydration gives:

Which of the following does not form a stable hydrate on addition of $H_2O$?

When acetaldehyde reacts with $PCl_{5}$,the resulting compound is

In the given conversion,the best yield will be obtained with:

Vedclass Test Series

Exam Paper Generator

Online Exam Module