Question

Hydrolysis of an alkyl halide $(RX)$ by dilute alkali $[\overset{\ominus}{ O}H ]$ takes place simultaneously by $SN^2$ and $SN^1$ pathway. A plot of $-\frac{1}{[ RX ]} \frac{d[ R - x ]}{d t} VS [\overset{\ominus}{ O}H ]$ is a straight line of the slope equal to $2 \times 10^3\, mol ^{-1} L\, h ^{-1}$ and intercept equal to $1 \times 10^2\, h ^{-1}$. Calculate the initial rate $\left( mole \,L ^{-1} min ^{-1}\right)$ of comsumption of $RX$ when the

Answer 1

$\frac{-d[ RX ]}{d t}=k_2[ RX ][\overset{\ominus}{ O}H ] $ (by $SN^2 $ path way)
$k_2=$ rate constant of $ SN ^2 $ reaction
$\frac{-d[ RX ]}{d t}=k_1[ RX ] \left(b^2 SN ^1\right. $ path way)
$k_1=$ rate constant of $ SN ^1 $ reaction
$\frac{-d[ RX ]}{d t}=k_2[ RX ][\overset{\ominus}{ O}H ]+k_1[ RX ] $
$-\frac{1}{[ RX ]} \frac{d[ RX ]}{d t}=k_2[\overset{\ominus}{ O}H ]+k_1$
This is the equation of a straight line for $-\frac{1}{[ RX ]} \frac{d[ RX ]}{d t}$
vs $[\overset{\ominus}{ O}H ]$ plot with slope equal to $k_2$ and intercept equal to $k_1$
from question:
$k_2=2 \times 10^3\, mol ^{-1} L\,hr ^{-1}, k_1=1 \times 10^2 \,hr ^{-1} $
$[ RX ]=1.0 \,M $ and $[\overset{\ominus}{ OH }]=0.1\, M $
Hence,
$\frac{-d[ RX ]}{d t}=2 \times 10^3 \times 1 \times 0.1+1 \times 10^2 \times 1 $
$=300\, mol\, L ^{-1} hr ^{-1}$
$=5 \,mol \,L ^{-1} min ^{-1}$