Question

a. Starting with 3-methyl-1-butyne, how can you prepare the following alcohols?

1. 2-methyl-2-butanol

2. 3-methyl-1-butanol

b. In each case, a second alcohol would also be obtained. What alcohol would it be?

Short answer

3-methyl-1-butyne undergoes reaction with H₂in presence of Lindlar catalyst and H₂O and H₂SO₄to form 2-methyl-2-butanol.

Step-by-step solution

Step1:

3-methyl-1-butyne undergoes reaction with H₂in presence of Lindlar catalyst and H₂O and H₂SO₄to form 2-methyl-2-butanol.

Step 2:Both 1 and 2 will provide 3-methyl-2-butanol as a minor product. 

Because water will occasionally attack the secondary carbocation before it has a chance to rearrange to the tertiary carbocation, 3-methyl-2-butanol will be formed from 1.

Because boron can also add to the other sp² carbon in the second stage of synthesis, 3-methyl-2-butanol will be the minor product, and the transition state for its creation is less stable than the transition state leading to the major product. A carbocation rearrangement is not possible since the carbocation is not produced as an intermediate.

(The proton cannot join to the other sp² carbon in part 1 since this would necessitate the production of a primary carbocation in the second stage of the synthesis, which is extremely unstable.)