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Solomons and Fryhle has an example in which they have showed a very simple substitution reaction. The question was to synthesize methyl iodide. What they have done is taken methyl chloride and produced methyl iodide in the presence of $\ce{NaI}$ and ethanol through an $\mathrm{S_N2}$ mechanism.

Why did they use ethanol? I have read that $\mathrm{S_N2}$ reactions are favored under polar aprotic solvents and clearly ethanol is not that. Also, since it's a methyl and all substitution reactions on methyl go through $\mathrm{S_N2}$; so how would it affect my reaction if I would't use ethanol?

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Usually, nucleophilic substitution reactions of anionic nucleophiles occur more rapidly in polar aprotic solvents. In such solvents (e.g. DMSO) the typical relative reactivity order is
$$ \ce{SCN-} < \ce{I-} < \ce{Br-} < \ce{Cl-} < \ce{F-}$$

In protic hydrogen-bonding solvents (e.g. ethanol), anions are solvated. Therefore, the reactivity of nucleophiles is decreased. Since hard nucleophiles are more strongly solvated than soft nucleophiles, the relative reactivity of soft anions is increased in such solvents. The typical relative reactivity order is
$$ \ce{F-} < \ce{Cl-} < \ce{Br-} < \ce{I-} < \ce{SCN-}$$

In the given example, $\ce{-Cl}$ shall be replaced by $\ce{-I}$; and in alcohols, the nucleophilicity of $\ce{I-}$ is higher than the nucleophilicity of $\ce{Cl-}$.

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When compared to some polar, aprotic solvents, such as acetonitrile or dimethyl formamide, ethanol has a couple of advantages:

  • it is rather cheap
  • it is less toxic
  • the solubility of sodium iodide is rather high
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  • $\begingroup$ So you mean to say that it is suited only for this reaction of many other substitution reactions also? $\endgroup$ – Karan Singh Apr 16 '15 at 1:44
  • $\begingroup$ Please see my edit. $\endgroup$ – Karan Singh Apr 16 '15 at 1:49
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Where did you find the statement: "all substitution reactions on methyl go through $S_N1$"? It is obviously wrong. Methyl definitely favor $S_N2$.

Here is the answer. The conversion from $\ce{MeCl}$ to $\ce{MeI}$ go through $S_N2$ mechanism. The best solvent for $\ce{-Cl}$ to $\ce{-I}$ conversion is usually acetone. It is cheap, polar aprotic, good solubility for $\ce{NaI}$ and poor solubility for $\ce{NaCl}$. I think the reason to use ethanol is just because this reaction is too easy to be picky on solvent. It just works in ethanol.

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  • $\begingroup$ Also what do you mean by "too easy to be picky on solvent"? And what if I don't use ethanol or any other solvent(I mean just NaI and and MeCl). I am sorry if a reaction without that a solvent sounds weird -I have next to zero experience in a lab, so I don't know which one is present in a solid or liquid form! $\endgroup$ – Karan Singh Apr 16 '15 at 16:26
  • $\begingroup$ Neat (no solvent) condition is not possible because $MeCl$ is gas at room temperature. $\endgroup$ – Ian Fang Apr 17 '15 at 3:41

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