Rate law for substitution reaction of 2-bromobutane with OHΘ in 75% ethanol and 25% water at 300C is rate =3.20×10−5 (2-bromobutane) [OH−]+1.5×10−6 (2-bromobutane) what percent of reaction take place by SN2 when [OH−]=1.00 M ? (Answer must be in nearest whole number)

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Rate law for substitution reaction of 2-bromobutane with $\overset{Θ}{OH}$ in 75% ethanol and 25% water at $30^{0} C$ is rate $= 3.20 \times 10^{- 5}$ (2-bromobutane) $[{OH}^{-}] + 1.5 \times 10^{- 6}$ (2-bromobutane) what percent of reaction take place by $S_{N} 2$ when $[{OH}^{-}] = 1.00 M$ ? (Answer must be in nearest whole number)

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answer is 95.

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Detailed Solution

$% by S_{N} 2 = \frac{S_{N} 2}{S_{N} 2 + S_{N} 1} \times 100 \begin{array}{l} = \frac{3.20 \times 10^{- 5} [2 - bromobutane] [1.00] \times 100}{3.20 \times 10^{- 5} [2 - bromobutane] [1.00] + 1.5 \times 10^{- 6} [2 - bromobutane]} \\ = \frac{3.20 \times 10^{- 5}}{3.20 \times 10^{- 5} + 0.15 \times 10^{- 5}} \times 100 = 95 to 96 % \end{array}$