CH_3C equiv C^-Na^+ + (CH_3)_2CH-Cl to What w | vedclass

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

Question

(C) The reaction involves a strong nucleophile,the acetylide ion $(CH_3C \equiv C^-)$,and a secondary alkyl halide,isopropyl chloride $((CH_3)_2CH-Cl)

Vedclass · Accepted Answer

Propyne and propene

Vedclass · Answer

(C) The reaction involves a strong nucleophile,the acetylide ion $(CH_3C \equiv C^-)$,and a secondary alkyl halide,isopropyl chloride $((CH_3)_2CH-Cl)$.Since the nucleophile is a strong base and the substrate is a secondary alkyl halide,the reaction will primarily undergo an $E2$ elimination reaction rather than an $S_N2$ substitution.The acetylide ion acts as a base,abstracting a proton from the $\beta$-carbon of the isopropyl chloride,leading to the formation of propene $(CH_3-CH=CH_2)$,propyne $(CH_3-C \equiv CH)$,and sodium chloride $(NaCl)$.Therefore,the major products are propene and propyne.

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

Similar Questions

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?

The compound $X \ (C_5H_8)$ reacts with ammoniacal $AgNO_3$ to give a white precipitate and reacts with excess of $KMnO_4$ to give the acid,$(CH_3)_2CHCOOH$. Therefore,$X$ is:

Identify the product $D$ in the following series of reaction: $CH_3COOH$ $\xrightarrow{LiAlH_4} A$ $\xrightarrow{H^{+}, 443 \ K} B$ $\xrightarrow{Br_2} C$ $\xrightarrow{alc. KOH} D$

The major product in the following reaction is

The major product formed in the oxidation of acetylene by alkaline $KMnO_4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module