CH_3-CH_2-Br + Nu^{Theta} rightarrow CH_3-CH_ | vedclass

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

Question

(A) The rate of nucleophilic substitution reaction depends on the nucleophilicity of the nucleophile $Nu^{\Theta}$. Stronger nucleophiles react faster

Vedclass · Accepted Answer

$IV > III > I > II$

Vedclass · Answer

(A) The rate of nucleophilic substitution reaction depends on the nucleophilicity of the nucleophile $Nu^{\Theta}$. Stronger nucleophiles react faster. Comparing the basicity and nucleophilicity: $CH_3O^{\Theta}$ is a stronger base and nucleophile than $OH^{\Theta}$ due to the $+I$ effect of the $CH_3$ group. $OH^{\Theta}$ is a stronger nucleophile than $PhO^{\Theta}$ because the negative charge in $PhO^{\Theta}$ is delocalized over the benzene ring. $PhO^{\Theta}$ is a stronger nucleophile than $CH_3COO^{\Theta}$ because the negative charge in $CH_3COO^{\Theta}$ is delocalized over two oxygen atoms. Thus,the order of nucleophilicity is $(IV) CH_3O^{\Theta} > (III) OH^{\Theta} > (I) PhO^{\Theta} > (II) CH_3COO^{\Theta}$.

$CH_3-CH_2-Br + Nu^{\Theta} \rightarrow CH_3-CH_2-Nu + Br^{\Theta}$
The decreasing order of the reaction rate with nucleophile $Nu^{\Theta}$ is $Nu^{\Theta} = (I) PhO^{\Theta}; (II) CH_3COO^{\Theta}; (III) OH^{\Theta}; (IV) CH_3O^{\Theta}$

Similar Questions

Which of the following is the correct order of boiling points for alkyl halides?

The product $(C)$ in the following reaction sequence is:

An alkyl halide $X$ $(C_4H_9Br)$ undergoes nucleophilic substitution by $S_N2$ reaction. The product of $X$ on reaction with $Mg/\text{dry ether}$ followed by $D_2O$ is:

What is the product formed by the reaction of $C_6H_5CH = CHCH_3$ with $HBr$?

The optically inactive compound from the following is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$CH_3-CH_2-Br + Nu^{\Theta} \rightarrow CH_3-CH_2-Nu + Br^{\Theta}$ The decreasing order of the reaction rate with nucleophile $Nu^{\Theta}$ is $Nu^{\Theta} = (I) PhO^{\Theta}; (II) CH_3COO^{\Theta}; (III) OH^{\Theta}; (IV) CH_3O^{\Theta}$