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Why is RO (alkoxide) a better leaving group than OH, despite RO being more unstable due to the electron donating effect of the alkyl group on RO?

I read that the suitability of leaving groups is dependent also on their nucleophilicity, as good nucleophiles will tend to re-attack the molecule it left from. To what extent is this true?

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    $\begingroup$ Both aren't viable leaving groups at all. $\endgroup$ – Mithoron Apr 12 '17 at 16:28
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The alkoxide and the hydroxides aren't good leaving groups. Consider an alcohol, the $\ce{OH}$ group never leaves on its own. Oxygen donates a lone pair to the hydrogen of a hydronium ion (considering it to be in an aqueous solution). The water molecule now attached is a good leaving group (oxygen has a positive charge).

Nucleophilicity does not determine the suitability of a leaving group. A group is said to be a good leaving group when it can leave as a relatively stable ion. For example: iodine is a very good leaving group for an $\mathrm{S_N2}$ reaction.

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Both alkoxides and hydroxides are not great leaving groups, but since we are just comparing their leaving group character...

First of all, leaving group tendency does not depend on nucleophilicity. Rather, it depends on basicity. Both of these are often confused to be the same thing. Yes, it's true that coincidentally both of these, in most cases, suggest similar things, but they are not the same. Let us compare the two for OH and OR.

Nucleophilicity: OH is a better nucleophile than OR since it has more mobility(due to its small size)than OR. A good nucleophile should be small in order to attack sites easily.

Basicity: This does not depend on mobility. It only depends on electron donating tendency (inversely proportional). Here, OR is a weaker base, hence it is a better leaving group

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You are confusing hyperconjugation effect on a carbocation with basicity. Basicity mainly depends on the following factors in order of significance: Atom → Resonance → Inductance → Orbital. Atom is the same (oxygen), there is no resonance, so inductance is key. Carbon is more electronegative than H, so OR will better stabilize a negative charge, thus OR is a better leaving group stable weak base.

You can see hyperconjugation here

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  • $\begingroup$ Remember this is true for an aprotic solvent like DMSO.In acuose solution you can get confused because solvation takes huge effect in protic solvents. So in water -OH is better stabilized by solvation thus making it weakbase or better LG. But guess what most likely you are exploring the Trans-Esterification here either hidroxy or alcoxy can be expelled (even if -OH is a better acuose LG)but notice that when you expell hidroxy you are left in the same previous ester state. So no change. In contrast if you expell -OR you will form a carboxilic acid COOOH. Cheers! $\endgroup$ – Orniz J Quinones May 6 at 14:45

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