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

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(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

$96$

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(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 $S_N2 = (3.20 	imes 10^{-5} / 3.35 	imes 10^{-5}) 	imes 100 \approx 95.52\% \approx 96\%$.

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