CH_3-CHCl-CHCl-CH_3 xrightarrow[dry ether]{N | vedclass

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

Question

(A) $1$. The reaction of $2,3-dichlorobutane$ with $Na$ in dry ether is a dehalogenation reaction (Wurtz-like elimination),which yields $X = CH_3-CH=C

Vedclass · Accepted Answer

$CH_3-CH_2-CH(OH)-CH_3$

Vedclass · Answer

(A) $1$. The reaction of $2,3-dichlorobutane$ with $Na$ in dry ether is a dehalogenation reaction (Wurtz-like elimination),which yields $X = CH_3-CH=CH-CH_3$ $(But-2-ene)$.$2$. The subsequent hydroboration-oxidation of $But-2-ene$ follows anti-Markovnikov addition of water across the double bond.$3$. Since $But-2-ene$ is a symmetric alkene,the addition of $OH$ group to either carbon $2$ or $3$ results in the same product,$Y = CH_3-CH_2-CH(OH)-CH_3$ $(Butan-2-ol)$.

$CH_3-CHCl-CHCl-CH_3$ $\xrightarrow[dry \ ether]{Na} X$ $\xrightarrow[(1) \ BH_3, THF; (2) \ H_2O_2, OH^-_{(aq)}] Y$; $Y$ is:

Similar Questions

The reaction,$H_2C=CH_2$ (Ethylene) $+ H_2O$ (water) $\xrightarrow{H_3PO_4, 300 \ ^\circ C/60 \ atm}$ $C_2H_5OH$ (Ethyl alcohol) is called:

$(CH_3)_3CH$ $\xrightarrow{KMnO_4} X$ $\xrightarrow[573 \ K]{Cu} Y$
The number of $sp^3$ and $sp^2$ carbons in $Y$ are respectively

Dilute aqueous $KMnO_4$ at room temperature reacts with $R-CH=CH-R$ to give:

Ethylene reacts with $Br_2$ to give $1, 2$-dibromoethane. Which intermediate is formed that leads to anti-addition?

The major product formed during the reaction of $1$-methylcyclopentene with $CH_3CO_3H$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module