Which series of reagents can be used to conve | vedclass

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

Question

(C) Acetylene $(HC \equiv CH)$ is acidic in nature due to the $sp$ hybridized carbon atoms. $1$. Treatment with $NaNH_2$ (a strong base) removes the a

Vedclass · Accepted Answer

Both of these

Vedclass · Answer

(C) Acetylene $(HC \equiv CH)$ is acidic in nature due to the $sp$ hybridized carbon atoms. $1$. Treatment with $NaNH_2$ (a strong base) removes the acidic proton to form sodium acetylide $(HC \equiv C^- Na^+)$,which then reacts with an alkyl halide $(R - X)$ via an $S_N2$ mechanism to form a higher alkyne $(HC \equiv C - R)$. $2$. Alternatively,treatment with a Grignard reagent $(R - Mg - X)$ also removes the acidic proton to form the acetylide,which subsequently reacts with an alkyl halide $(R - X)$ to yield a higher alkyne. Therefore,both methods are effective for the synthesis of higher alkynes.

Which series of reagents can be used to convert acetylene into higher alkynes?

Similar Questions

The product$(s)$ obtained via oxymercuration $(HgSO_4 + H_2SO_4)$ of $1$-butyne would be:

Which of the following alkynes,upon treatment with $H_2$ $(2 \ mole)$ / $Pt$,gives an optically inactive compound?

When acetylene is passed through dilute $H_2SO_4$ in the presence of $HgSO_4$,the compound formed is .....

$CH_3-C \equiv CH \xrightarrow[dil. H_2SO_4]{HgSO_4} (A)$
$CH_3-C \equiv CH \xrightarrow[(1) BH_3.THF]{(2) H_2O_2/HO^{-}} (B)$
Products $(A)$ and $(B)$ can be differentiated by:

Which of the following hydrocarbons forms a salt?

Vedclass Test Series

Exam Paper Generator

Online Exam Module