Sapling Homework Chapter 10 Problem 32

Problem 32 is a multistep synthesis problem that ties in concepts we've seen throughout this course. This problem starts with an 1-Butene and must end with 3-Hexene. The easiest way to do this problem is to do a reverse synthesis. We know that since the product has more carbons than the reactant, we must add carbons. The only way we have been taught to do that so far is by adding an alkyl halide to a terminal alkyne. So the first step is to make our product an alkyne. In the forward direction, to make an alkene from an alkyne, we know that we have to add H2. We would add H2 with Lindlar's catalyst because we want the cis alkene.  If we just added H2, the reaction would have proceeded straight to an alkane, which is not what we are looking for. Below, the reverse reaction is shown.

We now are left with an internal alkyne. To get this internal alkyne in the forwards direction, we would add an alkyl halide to a terminal alkyne. Since there are 6 carbons and our reactant only has 4, we would add CH3CH2Br to the terminal alkyne. The reverse reaction is shown below:

In the forward direction, we must now go from our reactant (1-Butene) to this terminal alkyne. Since we do not know a direct way of going from an alkene to an alkyne, we must first go the other way, from an alkene to an alkane. To do this, Br2 is added to 1-Butene and the bromines add on either side of the double bond as shown below. 

The next step is to perform a double elimination. By doing this double elimination, the alkane turns into an alkyne as shown below. An E2 elimination is performed because NaNH2 is a strong, aprotic base.

Finally, here is the whole mechanism all together. In the answer to this problem, NaNH2 is excess because it is used throughout.

Sources:

Sapling

All drawings done by me.