This molecule can be synthesized from an alkyne anion and an alkyl bromide. However, there are two ways in which this molecule c
an 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.
In this case, we will have different options around the<u> triple bond</u>. On part 1 we have to use the <u>bigger anion</u>, so we have to use the anion in which the <u>benzene ring is included </u>therefore the <u>ethyl</u> would be the <u>halide</u> (see figure 1).
On part 2 we have to use the <u>smaller anion</u>, so the <u>ethyl group</u> would be the anion and the other part of the molecule would be the <u>alkyl halide</u> (see figure 2).