At 30,^circ C,the rate of reaction between 2- | vedclass

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

Question

(A) The total rate of reaction is the sum of the rates of $S_N2$ and $S_N1$ mechanisms.Rate $= 	ext{Rate}_{S_N2} + 	ext{Rate}_{S_N1}$Given: $	ext{R

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(A) The total rate of reaction is the sum of the rates of $S_N2$ and $S_N1$ mechanisms.Rate $= 	ext{Rate}_{S_N2} + 	ext{Rate}_{S_N1}$Given: $	ext{Rate}_{S_N2} = 3.20 	imes 10^{-5} [2	ext{-bromobutane}][OH^-]$Given: $	ext{Rate}_{S_N1} = 1.5 	imes 10^{-6} [2	ext{-bromobutane}]$For $[OH^-] = 1.0\,M$:$	ext{Rate}_{S_N2} = 3.20 	imes 10^{-5} [2	ext{-bromobutane}] 	imes 1.0 = 3.20 	imes 10^{-5} [2	ext{-bromobutane}]$$	ext{Rate}_{S_N1} = 1.5 	imes 10^{-6} [2	ext{-bromobutane}]$Total Rate $= (3.20 	imes 10^{-5} + 0.15 	imes 10^{-5}) [2	ext{-bromobutane}] = 3.35 	imes 10^{-5} [2	ext{-bromobutane}]$Percentage of $S_N2 = (	ext{Rate}_{S_N2} / 	ext{Total Rate}) 	imes 100$Percentage of $S_N2 = (3.20 	imes 10^{-5} / 3.35 	imes 10^{-5}) 	imes 100 \approx 95.52\% \approx 96\%$.

Similar Questions

The back-side attack on $2$-bromobutane by methoxide $(CH_3O^{-})$ gives the product shown below. Which Fischer projection represents $2$-bromobutane used as the reactant in this reaction?

The major product of the following reaction is $CH_{2}=CH-CH_{2}-OH \xrightarrow[\text{Excess}]{HBr} \text{Product}$

Identify the products $A$ and $B$ in the reactions :
$R-X + AgCN \longrightarrow A + AgX$
$R-X + KCN \longrightarrow B + KX$

The correct reactivity order of the following compounds towards alkaline hydrolysis is:-
$I. \ p-CH_3O-C_6H_4-CH_2Cl$
$II. \ C_6H_5-CH_2Cl$
$III. \ p-CH_3-C_6H_4-CH_2Cl$
$IV. \ m-CH_3-C_6H_4-CH_2Cl$

Vedclass Test Series

Exam Paper Generator

Online Exam Module