Ph-CCl_2-CH_3 xrightarrow{3NaNH_2} text{Produ | vedclass

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

Question

(D) The reaction of $Ph-CCl_2-CH_3$ with $2$ equivalents of $NaNH_2$ undergoes dehydrohalogenation to form the alkyne $Ph-C \equiv CH$. Since $3$ equi

Vedclass · Accepted Answer

$Ph-C \equiv C^{\Theta} Na^{\oplus}$

Vedclass · Answer

(D) The reaction of $Ph-CCl_2-CH_3$ with $2$ equivalents of $NaNH_2$ undergoes dehydrohalogenation to form the alkyne $Ph-C \equiv CH$. Since $3$ equivalents of $NaNH_2$ are provided,the third equivalent acts as a strong base to abstract the acidic acetylenic proton from $Ph-C \equiv CH$. This results in the formation of the sodium acetylide salt: $Ph-C \equiv C^{\Theta} Na^{\oplus} (A)$.

$Ph-CCl_2-CH_3 \xrightarrow{3NaNH_2} \text{Product } (A)$; Product $(A)$ is

Similar Questions

$CH \equiv CH$ $\xrightarrow[H_2SO_4]{H_2O/Hg^{2+}} X$ $\xrightarrow{LiAlH_4} Y$ $\xrightarrow{P_4/Br_2} Z$. Here $Z$ is

$A$ compound is treated with $NaNH_2$ to give a sodium salt. Identify the compound.

When acetylene reacts with arsenic trichloride in the presence of anhydrous aluminium chloride,it produces

$CH_3C \equiv C^-Na^+ + (CH_3)_2CH-Cl \to$ What will be the product of this reaction?

Which of the following catalyst/reagent is used to convert $C \equiv C$ triple bond to $C = C$ double bond to form $cis$ isomer of alkene?

Vedclass Test Series

Exam Paper Generator

Online Exam Module