$\ce{CCl4}$ does not react with hydroxide nucleophile whereas $\ce{CH3Cl}$ does.

I had initially thought that it was due to symmetrical geometry of $\ce{CCl4}$ and so the dipoles cancel out and the molecule is non-polar, which means there is no partially positively charged carbon atom in $\ce{CCl4}$, and so there is nothing for the hydroxide ion to attack (I figured inducing a dipole on one of the $\ce{C-Cl}$ bonds by the hydroxide ion upon collision would not be possible and out of the question due to the large difference in electronegativity between the $\ce{C}$ and $\ce{Cl}$ atom, so any possible partial positive charge on $\ce{Cl}$ atom would go away almost instantly) but then I saw the mark scheme to this question and it said that it was to do with the large size of the $\ce{Cl}$ atom - preventing the hydroxide ion from colliding with the carbon atom. Ehich makes sense I guess, but does the argument I made answer this question in any way? (I also have a feeling I may have a misconception on whether a partial charge exists on any of the atoms for a non-polar molecule, so please do clear that up if that error is included in my statement).

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    $\begingroup$ Your explanation is correct. Chlorine is a big atom, and Carbon is rather small. This prevents the hydroxide ion from colliding with the carbon atom. $\endgroup$
    – Maurice
    Commented Jan 18, 2022 at 20:43
  • $\begingroup$ How, exactly, does CHCl3 react with sodium hydroxide anyway? By nucleophilic substitution or by deprotonation? If the latter, then CCl4 does not have a dissociable proton. $\endgroup$ Commented Jan 18, 2022 at 21:03
  • $\begingroup$ Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. $\endgroup$
    – Buttonwood
    Commented Jan 18, 2022 at 23:10
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    $\begingroup$ @OscarLanzi Deprotonation is the first step of two (if you want it). $\endgroup$
    – Buttonwood
    Commented Jan 18, 2022 at 23:11
  • $\begingroup$ Related: Why are DCM and chloroform so resistant towards nucleophilic substitution? $\endgroup$ Commented Jan 26, 2022 at 22:58

1 Answer 1


In chloroform, there are three electron withdrawing groups ($\ce{Cl}$) which are able to stabilize a negative charge. Thus, the first step is the deprotonation of chloroform, typically with concentrated lye:

$$ \ce{CHCl3 + NaOH <=>[H2O] CCl^-_3 + H2O} $$

The second step is the loss of $\ce{Cl^-}$ to yield $\ce{NaCl}$ and dichlorocarben, $\ce{^{..}CCl2}$. This may be dangerous if performed neatly (one reason you don't dry dichloromethane over sodium). Under conditions of phase transfer catalysis ($\ce{NaCl}$ enters the aqueous phase, the carbene remains in the organic phase), it however may be tamed to be a useful procedure to install cyclopropyl groups:

enter image description here

An other base capable to deprotonate $\ce{CHCl3}$ is e.g., $\ce{KO^tBu}$ at $\pu{-30 ^\circ{}C}$ in diethyl ether (procedure in Org. Syn.).

This approach however has been superseded e.g., by the Simmons-Smith reaction (second line), attaching the cyclopropyl ring in one step with less safety concern, less drastic reaction conditions, and thus, wider scope of application.

Carbon tetrachloride, on the other hand, does not possess a proton $\ce{NaOH}$ could abstract.


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