# Comparing rates of substitution reactions

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?

• What is P.A.S? Also it should be $\ce{S_N2}$ and $\ce{S_N1}$ and not $\ce{S_{N^2}}$ and $\ce{S_{N^1}}$. Jul 18, 2021 at 18:15
• By "exam I gave recently" you mean that you are the author of this question? Or do you mean that you wrote this exam? I think it should depend on whether the solvent is protic or aprotic, but in neither case I see why this order is correct Jul 18, 2021 at 19:18
• @Azamat No, I am not the author of the question. I've recently written a mock test for JEE conducted by a local institute. Jul 18, 2021 at 21:33
• $S_n2$ in general is faster reaction than $S_n1$ and rate of $S_n2$ directly depends on stability of transition state (here bulky transition states are less stable). Also you can neglect $S_n2$ here because polar aprotic solvents favor $S_n2$ (since as transition state proceeds charge density reduces which is better stabilized by aprotic solvents). I would think (theoretically) the order is $S_n2$ based unless relevant data is provided. Jul 20, 2021 at 13:00
• How do you measure the rate of substitution based on RCl concentration when elimination is appreciable in the secondary and tertiary halides? Isn't the appearance of alcohol more relevant? Sep 3, 2021 at 18:16

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 $$\mathrm{4>1>2>3}$$.

• But can you say that $\ce{S_N1}$ is always faster than $\ce{S_N2}$? I don't understand what "Weak P. A. S." means but the rates of the two mechanisms and what mechanism is dominant depend on the solvent. Or am I missing something? Jul 20, 2021 at 6:09
• @Azamat No we can't say $S_N1$ is faster than $S_N2$ vice versa. without knowing the solvent .this may help you know about solvents also about "Weak P. A. S." [link]quora.com/… Jul 20, 2021 at 12:47
• The question is inane. The conditions are unknown the mechanisms are various and what is a weak PAS. Also the logic that SN1 is favored over SN2 makes the SN1 reaction faster than another SN2 reaction is fallacious. Sep 24 at 1:45