CH_3 - CH_2 - CH_2Br + KOH (alc.) to ProductT | vedclass

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

Question

(A) The reaction of $1$-bromopropane $(CH_3 - CH_2 - CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.This is an elimination reaction (

Vedclass · Accepted Answer

$CH_3 - CH = CH_2$

Vedclass · Answer

(A) The reaction of $1$-bromopropane $(CH_3 - CH_2 - CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.This is an elimination reaction ($E2$ mechanism) where a hydrogen atom is removed from the $\beta$-carbon and the bromine atom is removed from the $\alpha$-carbon.This results in the formation of a double bond between the $\alpha$ and $\beta$ carbons.The reaction is: $CH_3 - CH_2 - CH_2Br + KOH 	ext{ (alc.)} 	o CH_3 - CH = CH_2 + KBr + H_2O$.The product formed is propene $(CH_3 - CH = CH_2)$.

$CH_3 - CH_2 - CH_2Br + KOH$ (alc.) $\to$ Product
The product in the above reaction is:

Similar Questions

Determine the number of chiral products (including stereoisomers) formed upon $E_2$ elimination of the given substrate with $alc. KOH$.

The major product formed in the following reaction is
$1,2-dichloro-3-methyl-5-nitrocyclopentane$ $\xrightarrow[{Heat}]{{NaOCH_3(1 \ eq.)}}$ ?

The correct sequence of bond enthalpy of $C-X$ bond is:

The set of compounds in which the reactivity of the halogen atom increases in the order is ..........

Write the products of the following reactions:
$(i)$ $C_6H_5CH=CH_2 + HBr \rightarrow$
(ii) $CH_3-CH_2-CH=CH_2 + HCl \rightarrow$
(iii) $C_6H_5CH_2-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$CH_3 - CH_2 - CH_2Br + KOH$ (alc.) $\to$ ProductThe product in the above reaction is: