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.

{ 15 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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Gitesh January 27, 2015 at 8:03 am

You must be thinking that F being for electronegative stabilizes the negative charge on it the most. But remember thats not the only basis on which u decide the acidity. H-F bonds are hydrogen bonds(Very strong) so cleaving it would be the most difficult. While for the others hydration energy also plays a role. So the correct series is:
HI> HBr> HCl> HF

According to their acidic strength.

So I- acts as a very good nucleophile because it is a stable nucleophile.

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sotos February 15, 2015 at 7:41 am

But what about steric hindrance ? I would expect the chloride ion to be more reactive than the iodide due to its considerably smaller size. Also, that becomes even more important considering the fact that the reaction is Sn2.

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James February 18, 2015 at 9:23 pm

Steric hindrance isn’t really an issue with halides because the size is generally inversely proportional to the length of the bond they form with carbon [i.e. small Cl, shorter bond length, large I, longer bond length] and these two factors largely cancel out. More important is the type of solvent. In polar protic solvents (solvents that can hydrogen bond) iodide is a better nucleophile than chloride. See this posthttp://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/

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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.

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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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Yuvraj September 20, 2014 at 8:03 am

Suppose I have a polymeric structure, with multiple ether linkages, say for instance, an ether of polyethlene glycol. What is the probability of the cleavage affecting only the terminal linkage and not the other ether bonds??? Thanx.

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Carli November 6, 2014 at 10:33 pm

This was so helpful! Thank you so much! Do you have anything on epoxide ring opening?

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Zach W February 24, 2015 at 6:50 pm

In regards to reactivity, are more substituted carbon centers less reactive because during the backside attack there is more steric hinderance?

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James February 26, 2015 at 5:30 pm

Yes, exactly! (for SN2 reactions, anyway)

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Isaac March 22, 2015 at 4:12 am

Does this method work for complex molecules like LIGNIN which contain a lot of ether linkages?

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Amy Nguyen March 31, 2015 at 4:33 pm

For the first example with the clevage of C6H5OCH3, how come C6H5I and CH3OH don’t form ?

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James March 31, 2015 at 9:22 pm

Ah! This is an SN2 reaction…. and SN2 reactions require a backside attack. A backside attack would have to occur from inside the aromatic ring which is impossible (very crowded). Also, SN2 reactions don’t work on sp2 hybridized carbons (like alkenyl halides)

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