Why is that when 3-methyl-2-butanol is treated with HBr, a single alkyl bromide is formed that...

Why is that when 3-methyl-2-butanol is treated with HBr, a single alkyl bromide is formed that results from a hydride shift and when 2-methyl-1-propanol is treated with HBr, no rearrangement occurs to form an alkyl bromide ??

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Expert Solution

Answer: According to the question , Here Two products will be possible. One more so than the other. 2-bromo-2-methyl butane and 2-bromo-3-methyl butane. The first one will be major. The reasoning behind this is that once you protonate the OH group to form a good leaving group, the H2O group can leave on its own, forming a secondary carbocation. A hydride shift will occur, quickly. Reason being that a shift will allow for the formation of a tertiary carbocation which is a more stable intermediate due to the inductive effect of the adjacent sigma bonds. Br- can then come in to add to the carbocation. The second product will be very minor, if not possible at all. I suppose wants you form the good leaving group, Br- could come in and do an Sn2 reaction. However, it is secondary and sterically crowded so that decreases the chance of Sn2 happening.

Hence thats why you can say that in when 2-methyl-1-propanol is treated with HBr, no rearrangement occurs to form an alkyl bromide.

So, it is all about the given question . Thank you :)

queen_honey_blossom answered 1 year ago

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