Quick ether cleavage question

[Vapor1122]

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If we're cleaving a straight-chain ether with and acid (say, HBr), what determines which side becomes the alcohol and which becomes the alkyl halide? Example:

CCCOCC + HBr .→. CCCOH + BrCC ...or
............................CCCBr + HOCC

 

[Jaylee777]

Its an SN2 reaction so the BR- will attack preferentially in the order of methyl, primary, secondary, tert after the oxygen is protonated. If they are both the same types of carbons on both sides then I am not sure....I would say the side with less crowding/less substituents.

Also, if there is excess acid in the reaction then both sides of the ether become alkyl halides.

 

[Vapor1122]

If they are both the same types of carbons on both sides then I am not sure....I would say the side with less crowding/less substituents.

Yeah, that's what I meant. Anyone want to confirm this?

Also, if there is excess acid in the reaction then both sides of the ether become alkyl halides.

That I didn't know. But now that you mention it, in excess acid, why wouldn't you get an alcohol and alkyl halide, and then have the alcohol dehydrate to an alkene? Like:

 

[Jaylee777]

Because the strength of the nucleophiles produced from the strong acid used in this type of reaction (has to be strong acids like HBr and HI) are very weak bases.

 

[Vapor1122]

Okay, I see it now. So in excess acid, the weak base (Br-) undergoes SN2 with the alcohol, forming another alkyl halide in addition to the first one from the cleaved ether. Gotcha.

Anyone have an answer for the first question?