Question

In each reaction box, place the best reagent and conditions from the list below.

In each reaction box, place the best reagent and conditions from the list below.

Answer 1

The first thing to notice in this one is that you are extending the carbon chain. To do this, you will need an alkyne and an alkyl halide. Since the alkene can be converted to either, we'll hold that though for now.

The second things to notice is that the final product has a cis alkene. That's great because we know that cis alkenes can be formed from alkynes so this tells us where that triple bond was that we used to extend our chain. Now that we know the location of the bond, we can find out what we did with our starting material. By working backwards, we can see that the starting material was converted to the alkyne because it is the 4-carbon piece and our starting compound had 4 carbons.

So now I jump back to the original starting material and focus on making it in to the alkyne. To form an alkyne, you will need to do 2 elimination reactions, so we will need to form a dihalide. From here, we do our elimination to form the alkyne.

NaNH2 should be (in your mind) the official alkyne base. The reason why we need excess instead of 2 equivalents is because we not only want to form the pi bond, we want to have the anion as well. The 1st eq makes the double bond, the 2nd eq makes it a triple bond, and the 3rd eq removes the terminal hydrogen.

1. Br2

2. excess NaNH2

3. CH3CH2Br

4. H2, Lindlar's catalyst