Question

This molecule can be synthesized from an alkyne anion and an alkyl bromide. However, there are two ways in which this molecule can be formed. One way uses a higher molecular weight alkyne anion (Part 1) and the other uses a lower molecular weight anion (Part 2). Draw the two versions in the boxes below. Omit spectator ions.

For Part 1: Draw the reactants (i.e., alkyne anion and alkyl bromide) needed for the pathway that uses a higher molecular weight alkyne anion:
For Part 2: Draw the reactants (i.e., alkyne anion and alkyl bromide) needed for the pathway that uses a lower molecular weight alkyne anion:

Answer 1

See explanation below

Step-by-step explanation:

First of all, we need to know the molecule that need to be synthesized. In the first picture attached, we have the molecule.

Now, according to that, the molecule can be formed in two ways. Both ways are rather different but we have one thing in common in both ways, and it's the fact that both ways requires the reaction with an alkyne. In one way, with the higher MW and the other with the lowest.

For the first way, when we need to use the higher MW, all we need to do is to put the triple bond in the highest molecule, in this case, a molecule that include the benzene ring, and the other reactant will be the alkyl bromide. You can see the picture of this product below, (Picture 2).

For the second way, the triple bond should be located in the lowest MW, in this case an ethyl group, instead of the benzene ring, and the molecule with the benzene ring would have the bromide. In both cases we will have the same product. See picture 2 for this.

Hope this helps