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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?

Text Solution

Verified by Experts

The rate expression involves `S_(N^(1))` and `S_(N^(2))` steps as:
`RX+OH^(-)rarrR^(+)+X^(-)overset(OH^(-))rarrROH+X^(-)S_(N^(1))`
`RX+OH^(-)rarrHO...R..X^(-)rarrHOR+X^(-)S_(N^(2))`
i.e., rate `= 4.7xxunderset(S_(N^(2)))(10^(-5))[RX][OH]+0.24xxunderset(S_(N^(2)))(10^(-5))[RX]`
`% rate of S_(N^(2))=[r_(S_(N^(2)))/(r_(S_(N^(1)))+r_(S_(N^(2))))]xx100`
`=(4.7xx10^(-5)[RX][OH^(-)])/(4.7xx10^(-5)[RX][OH^(-)]+0.24xx10^(-5)[RX])xx100`
`=(4.7xx10^(-5)[OH^(-)])/(4.7xx10^(-5)[OH^(-)]+0.24xx10^(-5))`
`=(4.7xx10^(-5)xx0.001)/(4.7xx10^(-5)xx0.001+0.24xx10^(-5))`
`=1.9%`
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