Select the reagent that will yield the greate | vedclass

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

Question

(A) The reaction of $CH_3CH_2Br$ (a primary alkyl halide) with a base/nucleophile can proceed via $S_N2$ (substitution) or $E2$ (elimination) pathways

Vedclass · Accepted Answer

$CH_3CH_2OK$ in dimethyl sulfoxide $(DMSO)$

Vedclass · Answer

(A) The reaction of $CH_3CH_2Br$ (a primary alkyl halide) with a base/nucleophile can proceed via $S_N2$ (substitution) or $E2$ (elimination) pathways.$CH_3CH_2OK$ is a strong,unhindered nucleophile,which favors the $S_N2$ substitution reaction.$(CH_3)_3COK$ is a strong,sterically hindered base (potassium tert-butoxide),which significantly favors the $E2$ elimination reaction over substitution due to steric hindrance.Therefore,$CH_3CH_2OK$ will yield a greater amount of substitution product compared to $(CH_3)_3COK$.

Select the reagent that will yield the greater amount of substitution on reaction with $CH_3CH_2Br$.

Similar Questions

Among the following chloro-compounds,the one having the lowest dipole moment is:

Consider the following bromides.
The correct order of $S_N1$ reactivity is

Chloroform,when kept open,is oxidized to

Arrange the following compounds in order of decreasing rate of hydrolysis for $S_{N}1$ reaction:
$(I) \ C_6H_5-CH_2-Br$
$(II) \ p-CH_3-C_6H_4-CH_2-Br$
$(III) \ p-CH_3CH_2-C_6H_4-CH_2-Br$
$(IV) \ p-(CH_3)_2CH-C_6H_4-CH_2-Br$

Observe the following reactions:
$1$. $C_6H_5-CH=CH_2 + HBr \rightarrow X$
$2$. $C_6H_5-C(CH_3)=CH_2 + HBr \rightarrow Y$
$3$. $C_6H_5-CH=CH_2 + HBr \xrightarrow{(C_6H_5COO)_2} Z$
The correct order of reactivity of $X, Y, Z$ towards $S_N1$ reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module