The reaction CH_3-CH(C_6H_5)-CH(Br)-C_6H_5 xr | vedclass

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

Question

(B) The reaction of $1-bromo-1,2-diphenylpropane$ with $alc.KOH$ is a dehydrohalogenation reaction ($E2$ mechanism).This reaction leads to the formati

Vedclass · Accepted Answer

$1,2-$diphenylprop$-1-$ene (trans-isomer)

Vedclass · Answer

(B) The reaction of $1-bromo-1,2-diphenylpropane$ with $alc.KOH$ is a dehydrohalogenation reaction ($E2$ mechanism).This reaction leads to the formation of an alkene by removing $H$ and $Br$ from adjacent carbons.The major product is the more stable alkene,which is the trans-isomer of $1,2-diphenylprop-1-ene$.In the trans-isomer,the bulky phenyl groups are on opposite sides of the double bond,minimizing steric hindrance.Thus,the product $A$ is $1,2-diphenylprop-1-ene$ (trans-isomer).

The reaction $CH_3-CH(C_6H_5)-CH(Br)-C_6H_5 \xrightarrow{alc.KOH} A$. Identify the product $A$.

Similar Questions

The correct order for increasing reactivity towards $S_N1$ substitution for the following compounds is:
$1. CH_3CH_2Br$
$2. CH_2=CHCH(Br)CH_3$
$3. CH_2=CHBr$
$4. CH_3CH(Br)CH_3$

Identify $A$ and $B$ in the following reactions:
$A \xrightarrow[\Delta]{aq. NaOH} C_2H_5OH \xleftarrow{AgOH} B$

Which of the following nucleophiles will show minimum reactivity towards $S_N2$ reaction:

What is the major product of the following reaction?
$CH_2=CH-CH_2-CH(Br)-CH_3 + KOH \xrightarrow[\text{heat}]{\text{alcohol}} \dots$

In which of the following reactions is an anti-Markovnikov product formed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module