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In one of my textbooks, I came across a question as to why does $\ce{CCl4}$ not undergo hydrolysis. The answer given went along the lines of carbon lacking a vacant d-orbital to accept the lone pair of oxygen from water.

However I was curious to find out that other chlorine-derivatives of methane i.e. $\ce{CH3Cl}$, $\ce{CH2Cl2}$ and $\ce{CHCl3}$ (to some extent) do undergo hydrolysis, even though carbon lacks the vacant d-orbital.(Mechanism involved I figured would be Lone pair of oxygen attacking the vacant anti-bonding molecular orbital of carbon)

What is the actual reason that carbon tetra-chloride doesn't undergo hydrolysis while its other chlorine-derivatives seem to undergo hydrolysis effectively?

My only logical approach was the large size of chlorine provides stearic hindrance towards the attacking lone pair of water, But I didn't find any substantiation for this.

A detailed explanation would be appreciated! thanks in advance!

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  • $\begingroup$ "Does not undergo hydrolysis" is quite a bit of an overstatement. It does, albeit very slowly. Then again, other derivatives don't react fast either. $\endgroup$ – Ivan Neretin Jun 20 at 5:24
  • $\begingroup$ But the rate with CCl4 is negligible compared to the rate with other derivatives, As similar to the SF6 example, where no hydrolysis takes place and the reason is stearic hindrance contrary to SF4. $\endgroup$ – Aditya Iyer Jun 20 at 5:39
  • $\begingroup$ Plus in the case of carbontetrachloride , "for all purposes, we can assume no reaction takes place" right $\endgroup$ – Aditya Iyer Jun 20 at 6:35
  • $\begingroup$ Maybe this could help: Hydrolysis of $\ce{CCL4}$. $\endgroup$ – Firefox1921 Jun 20 at 6:46

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