What's the reaction of isopropyl chloride with sodium acetate? (The solvent is acetic acid)
Is it SN1 or SN2?

My thoughts:

  1. It's not that difficult to approach isopropyl chloride's carbon, for there isn't much steric crowding. It's just a 2 degree alkyl halide with two methyl groups. This favours SN2.

  2. The carbocation formed by the 2 degree alkyl halide, here in this case, is stabilised by the +I effect of two methyl groups and also six hyperconjugative structures. This should favour carbocation formation, hence SN1.

  3. Acetic acid is a polar protic solvent, and can solvate sodium acetate.

  4. Acetate ion isn't a great nucleophile (or is it?). I'm not sure, here. I know that acetic acid is a weak acid, so acetate ion is a strong base but I believe its nucleophilicity is suppressed by solvation.

  5. Chlorine isn't an amazing leaving group.

So, does SN1 take place or SN2? Could someone also give some experimental data to support their reasoning?

Here's something I found, it says that both SN1 and SN2 can occur. Now which one occurs here, is probably decided by the nucleophile? I'm not sure. Anyway, thought I'd put this table up here for reference: enter image description here

Source: Organic Chemistry - Clayden - Oxford University Press - 1st edition - Page 424

Edit: Someone in the comments section said that you can't predict which mechanism is major, and have to look for experimental facts. So, for the sake of the completeness, where are the facts? Could someone please post the relevant details as an answer?


2 Answers 2


Actually in these type of reactions, where the reactant can both possibly undergo $\mathrm{S_N1} $ and $\mathrm{S_N2} $, the reactions actually proceeds through a mixture of both type of reaction mechanism. In these type of reactions, rate is related as $$\text{Rate} = k_1 [\ce{R}] + k_2 [\ce{R}][\ce{X-}]$$ So, there is also the stereochemically inverted product and also racemised product.
Fraction of inverted product is $k_2/(k_1+k_2)$ and the fraction of racemised product is $k_1/(k_1 +k_2)$. But the actual values of $k_1$ and $k_2$ is experimentally determined and probably very difficult to determine by logical approach only.

  • $\begingroup$ Which mechanism is major? $\endgroup$ Commented Mar 18, 2018 at 14:32
  • 1
    $\begingroup$ I think you can't determine it logically, you have to look for experimental facts. $\endgroup$
    – Soumik Das
    Commented Mar 18, 2018 at 15:03
  • 1
    $\begingroup$ @schrodinger_16 "Which mechanism is major?" This question is ill-defined. Which mechanism dominates is highly dependent on reaction conditions, including concentration, solvent, and temperature. $\endgroup$
    – Zhe
    Commented May 17, 2018 at 17:47

In this case the transition state of the reaction is much polar than ground state if the reaction follows the SN1 Pathway. So in presence of polar medium like acetic acid the transition state will be more stabilised than ground state and hence activation energy of the reaction will decrease. And the rate of the reaction will be faster. But if the molecule follows SN2 pathway the transition state of the reaction will be less polar than ground state. As a result in presence of polar medium like acetic acid the ground state will be more stabilised than transition state.As a result the rate of reaction will decrease. So by analysing all aspects we can predict in acetic acid medium isopropyl chloride will wish to follow up SN1 reaction readily than SN2.


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