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

  • 2
    $\begingroup$ What is P.A.S? Also it should be $\ce{S_N2}$ and $\ce{S_N1}$ and not $\ce{S_{N^2}}$ and $\ce{S_{N^1}}$. $\endgroup$
    – ecneics
    Jul 18 at 18:15
  • $\begingroup$ 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 $\endgroup$
    – Azamat
    Jul 18 at 19:18
  • $\begingroup$ @Azamat No, I am not the author of the question. I've recently written a mock test for JEE conducted by a local institute. $\endgroup$ Jul 18 at 21:33
  • $\begingroup$ $S_n2$ 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. $\endgroup$ Jul 20 at 13:00
  • $\begingroup$ 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? $\endgroup$
    – user55119
    Sep 3 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 $\mathrm{4>1>2>3}$.

  • $\begingroup$ 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? $\endgroup$
    – Azamat
    Jul 20 at 6:09
  • $\begingroup$ @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/… $\endgroup$
    – blacknoir
    Jul 20 at 12:47

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