Which of the following compounds forms a Na-s | vedclass

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

Question

(A) The reaction with $NaNH_2$ (sodamide) is a characteristic test for terminal alkynes,which possess acidic hydrogen atoms.$NaNH_2$ is a strong base

Vedclass · Accepted Answer

$C_2H_2$

Vedclass · Answer

(A) The reaction with $NaNH_2$ (sodamide) is a characteristic test for terminal alkynes,which possess acidic hydrogen atoms.$NaNH_2$ is a strong base that can abstract the acidic proton from the terminal alkyne to form a sodium acetylide salt.Among the given options,$C_2H_2$ (acetylene or ethyne) is a terminal alkyne with the structure $HC \equiv CH$.The reaction is: $HC \equiv CH + NaNH_2 \rightarrow HC \equiv C^-Na^+ + NH_3$.Therefore,$C_2H_2$ forms a $Na$-salt.

Which of the following compounds forms a $Na$-salt upon reaction with $NaNH_2$?

Similar Questions

What is the major product of the following reaction?
$Cyclohexyl-C \equiv CH \xrightarrow{HOCl \text{ (excess)}} \text{major product}$

What is the major organic product of the following reaction?
$Ph-C \equiv CH \xrightarrow[H_2SO_4]{Hg^{2+}, H_2O} ?$

Identify the major product $A$ in the following reaction sequence:
$CH_3-CH=CH-CH_3$ $\xrightarrow[CCl_4]{Br_2}$ $\xrightarrow[(ii) NaNH_2]{(i) alc. KOH} (A)$

Predict the product $C$ obtained in the following reaction of but$-1-$yne: $CH_3-CH_2-C \equiv CH + HCl \to B \xrightarrow{HCl} C$

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