Comparing rates of substitution reactions

Question

The following question was asked in an exam I gave recently

Compare the overall rates of the following substitution reactions $$\ce{CH3Cl ->[OH-][Weak P.A.S] CH3OH}$$

$$\ce{CH3CH2Cl ->[OH-][ Weak P.A.S] CH3CH2OH}$$ $$\ce{(CH3)2CHCl ->[OH-][ Weak P.A.S] (CH3)2CHOH}$$ $$\ce{(CH3)3CCl ->[OH-][ Weak P.A.S] (CH3)3COH}$$ (P.A.S means polar aprotic solvent). I know that $\ce{CH3Cl , CH3CH2Cl}$ react mainly via $S_{N^2}$ mechanism, $\ce{(CH3)2CHCl}$ react via both $S_{N^2} , S_{N^1}$ mechanisms considerably and $\ce{(CH3)3CCl}$ via $S_{N^1}$ mechanism.

The rates of the reactions can be given by the following rate equations

$r_1 = k_1[\ce{CH3Cl}][\ce{OH-}] + k_1^☆[\ce{CH3Cl}]$,

$k_1^☆$ being negligible

$r_2 = k_2[\ce{CH3CH2Cl}][\ce{OH-}] + k_2^☆[\ce{CH3CH2Cl}]$

,$k_2^☆$ being negligible

$r_3 = k_3[\ce{(CH3)2CHCl}][\ce{OH-}] + k_3^☆[\ce{(CH3)2CHCl}]$

can't neglect anything here,

$r_4 = k_4[\ce{(CH3)3CCl}][\ce{OH-}] + k_4^☆[\ce{(CH3)3CCl}]$ $k_4$ can be neglected here.

So, I would assume the reaction-3 occurs at least rate as ${k_3}$ and ${k_3^☆}$ are both low, but I don't know how to compare between $k_1 , k_2 , k_4^☆$.

Is there any theoretical way to compare them? If yes, then how?

Answer given:

The answer is 4>1>2>3

Answer 1

Actually one can't always find answers through logical formulations to these JEE problems on organic chemistry , but we can understand and compare logical constraints. I remember this question was asked in two different ways in my examinations - one considering only $\mathrm{S_N2}$ among the reactions and the other one is same as yours.

$$\ce{CH3Cl ->[OH-][Weak P.A.S] CH3OH}$$

$$\ce{CH3CH2Cl ->[OH-][ Weak P.A.S] CH3CH2OH}$$ $$\ce{(CH3)2CHCl ->[OH-][ Weak P.A.S] (CH3)2CHOH}$$ $$\ce{(CH3)3CCl ->[OH-][ Weak P.A.S] (CH3)3COH}$$

If we only consider $\mathrm{S_N2}$ among these reactions order will be $\mathrm{1>2>3>4}$.

But we have to consider how large the other atoms or functional groups bonded to the carbon are, compared to $\ce{OH^-}$ also known as steric inheritance or steric crowding, then

For this question we are considering a substitution reaction..

You said the answer yourself - $\ce{CH3Cl, CH3CH2Cl}$ react mainly via $\mathrm{S_N2}$ mechanism, $\ce{(CH3)2CHCl}$ react mainly via both $\mathrm{S_N2, S_N1}$ mechanisms considerably and $\ce{(CH3)3CCl}$ via $\mathrm{S_N1}$ mechanism.

Now you have to consider steric hindrance in $\ce{(CH3)3CCl}$ where the three $\ce{CH3}$ groups makes $\mathrm{S_N1}$ more favourable. Also the tert-butyl carbocation $\ce{(CH3)3C^+}$ is more stable beacause of 9 alpha hydrogens which involve in hyperconjugation. In most cases when steric factors and hyperconjugation both favour carbocation formation then $\mathrm{S_N1}$ is the major pathway preferred over $\mathrm{S_N2}$.

So for $\mathrm{4}$, strongly favoured $\mathrm{S_N1}$ pathway increases rate of reaction. For the other 3, $\mathrm{S_N2}$ is dominant which follows the order $\mathrm{1>2>3}$. Hence overall order is $\mathrm{4>1>2>3}$.