2-methyl-2-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield. You do not have to consider stereochemistry.

See the explanation

Explanation:

In this question, we have to synthesize a molecule in which the "OH" group is placed in the most substituted carbon. Therefore we have to use a markovnikov reaction to the synthesis. The hydroboration is anti-markovnikov and the oxymercuration-reduction is markovnikov, therefore the oxymercuration-reduction is our best choice.

Now, the double bond can be placed in different places. Because the "OH" would be added in the tertiary carbon of the double bond. So, we can use 2-methylhex-1-ene or 2-methylhex-2-ene.

See figure 1.

I hope it helps

it's c. elements are placed in rows based on how may valencies electrons have.