Acidic cleavage of ethers (SN2)

by James

Description: When ethers are treated with strong acid in the presence of a nucleophile, they can be cleaved to give alcohols and alkyl halides. If the ether is on a primary carbon this may occur through an SN2 pathway.

Notes: Common acids for this purpose are HI and other hydrogen halides, as well as H2SO4 in the presence of H2O.

In the case where the ether being cleaved is secondary and has a stereocenter, there will be inversion of configuration.

Examples: 

Notes: The third example could also be written “H3O+”

Mechanism: Strong acid (HI) protonates the ether oxygen, which turns it into a better leaving group (Step 1, arrows A and B). Next, the iodide ion attacks the carbon in an SN2 reaction (Step 2, arrows C and D) to give the alcohol and methyl iodide.

Notes: In cases  where the ether being cleaved is secondary and has a stereocenter, there will be inversion of configuration.

{ 5 comments… read them below or add one }

Hao Luo March 12, 2014 at 2:00 am

Is HI a stronger nucleophile than HCl? If asked to list all the hydrogen halides in terms of their cleavage speed with an ether without side chains, what order should you arrange them? I always thought HF is the strongest and as you go down it gets weaker. But that wasn’t the case right? Could you please explain to me why.

Thanks,
Hao

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Jake the Dog July 26, 2014 at 11:39 am

HI is a better nucleophile than HCl because the Iodine atom is more polarizable and hence more able to take place in SN2 reaction. As you go down the halide group, the electron shells get larger and farther away from the nucleus and therefore more polarizable. Thus they get more nucleophilic down gorup as well

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Madhu88 May 10, 2014 at 1:44 am

what about unsymmetrical ether cleavage???????

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Petr Menzel May 22, 2014 at 5:20 am

Hi, in the mechanism is produced Ph-OH + CH3I. Why is not produced Ph-I + CH3-OH? Thx, P.

Reply

James May 22, 2014 at 2:52 pm

Because that would involve an SN2 of I- on an sp2 hybridized carbon, which is very slow – not to mention that “backside attack” would have to occur from the middle of the benzene ring!

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