In the reaction CH_3CH_2CH_2Br xrightarrow{aq | vedclass

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

Question

(A) Step $1$: $CH_3CH_2CH_2Br + aq. NaOH ightarrow CH_3CH_2CH_2OH$ ($X$ is propan$-1-$ol).Step $2$: $CH_3CH_2CH_2OH \xrightarrow{Al_2O_3, 	ext{Heat

Vedclass · Accepted Answer

$CH_3-CH(OH)-CH_2Cl$

Vedclass · Answer

(A) Step $1$: $CH_3CH_2CH_2Br + aq. NaOH ightarrow CH_3CH_2CH_2OH$ ($X$ is propan$-1-$ol).Step $2$: $CH_3CH_2CH_2OH \xrightarrow{Al_2O_3, 	ext{Heat}} CH_3CH=CH_2$ ($Y$ is propene).Step $3$: $CH_3CH=CH_2 + HOCl ightarrow CH_3CH(OH)CH_2Cl$ (major product) and $CH_3CH(Cl)CH_2OH$ (minor product).According to Markovnikov's rule,the electrophile $Cl^+$ attacks the double bond to form the more stable carbocation,followed by the attack of $OH^-$. The major product is $CH_3CH(OH)CH_2Cl$.

In the reaction $CH_3CH_2CH_2Br$ $\xrightarrow{aq. NaOH} (X)$ $\xrightarrow{Al_2O_3, \text{Heat}} (Y)$ $\xrightarrow{HOCl} (Z)$,what is $Z$?

Similar Questions

In the reaction with $HCl$,an alkene reacts in accordance with the Markovnikov's rule to give a product $1-$chloro$-1-$methylcyclohexane. The possible alkene is

$CH_2=CHCl$ reacts with $HCl$ to form

$(A) (C_7H_{14}) \xrightarrow[Zn]{O_3} (B) + (C)$
Compound $(A)$ exists as geometrical isomers and $(B)$ gives Cannizzaro reaction. $(A)$ will be:

The major product of the reaction is:

Identify the $(C)$ product of the following reaction:

Vedclass Test Series

Exam Paper Generator

Online Exam Module