Identify the products A and B in the followin | vedclass

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

Question

(B) $1$. The starting material is $1,2$-dibromo$-1-$phenylethane $(Ph-CH(Br)-CH_2-Br)$.$2$. Treatment with $(i) 	ext{ alc. KOH}$ followed by $(ii)

Vedclass · Accepted Answer

$A$ $\Rightarrow Ph-C \equiv CH, B$ $\Rightarrow Ph-C \equiv C-CH_2-CH_3$

Vedclass · Answer

(B) $1$. The starting material is $1,2$-dibromo$-1-$phenylethane $(Ph-CH(Br)-CH_2-Br)$.$2$. Treatment with $(i) 	ext{ alc. KOH}$ followed by $(ii) 	ext{ NaNH}_2$ causes dehydrohalogenation to form the terminal alkyne,phenylacetylene $(Ph-C \equiv CH)$. Thus,$A = Ph-C \equiv CH$.$3$. Phenylacetylene $(Ph-C \equiv CH)$ is then treated with $	ext{NaNH}_2$ to form the acetylide ion $(Ph-C \equiv C^-)$.$4$. This nucleophile reacts with ethyl chloride $(CH_3CH_2-Cl)$ via an $S_N2$ mechanism to form the product $B$,which is $1$-phenylbut-$1$-yne $(Ph-C \equiv C-CH_2-CH_3)$.

Identify the products $A$ and $B$ in the following reaction sequence:
$Ph-CH(Br)-CH_2-Br$ $\xrightarrow[(ii) NaNH_2]{(i) alc. KOH} A$ $\xrightarrow{NaNH_2, CH_3CH_2-Cl} B$

Similar Questions

Ozonolysis of acetylene $(HC \equiv CH)$ gives:

Chloroform,on warming with $Ag$ powder,gives:

Explain the acidic nature of ethyne.

When propyne is passed through a red hot iron tube at $873 \ K$,the reaction gives a product having the molecular formula of:

The trend in acidic behaviour of the following hydrocarbons is:
$a: HC \equiv CH$
$b: CH_2 = CH_2$
$c: CH_3 - CH_3$
$d: CH_3 - C \equiv CH$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Identify the products $A$ and $B$ in the following reaction sequence:$Ph-CH(Br)-CH_2-Br$ $\xrightarrow[(ii) NaNH_2]{(i) alc. KOH} A$ $\xrightarrow{NaNH_2, CH_3CH_2-Cl} B$