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For reaction RX + OH^(-) rarr ROH + X^(-...

For reaction `RX + OH^(-) rarr ROH + X^(-)`, rate expression is `R = 4.7 xx 10^(-5) [RX][OH^(-)] + 2.4 xx 10^(-5)[RX]`.
What `%` of reactant react by `S_(N)2` mechanism when `[OH^(-)] = 0.001` molar?

A

`1.9`

B

`66.2`

C

`95.1`

D

`16.4`

Text Solution

Verified by Experts

The correct Answer is:
a
The rate is made up of two parts `S_(N^(1))` and `S_(N^(2))`
`rate=4.7xx10^(-5)[RX][OH^(-)]+0.24xxunderset(S_(N^(1)))(10^(-5))[RX]`
Thus `% S_(N^(2))=[S_(N^(2))/(S_(N^(2))+S_(N^(1)))]xx100`
`=[(4.7xx10^(-5)[RX][OH^(-)])/(4.7xx10^(-5)[RX][OH^(-)]+0.24xx10^(-5)[RX])]xx100`
`=[(4.7[OH^(-)])/(4.7[OH^(-)]+0.24)]xx100`
`=(4.7xx0.001)/(4.7xx0.001+0.24)=1.9%`
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