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Helen [10]
3 years ago
5

1-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.

Chemistry
1 answer:
slava [35]3 years ago
8 0

Answer:

hex-1-ene with a hydroboration-oxidation reaction

Explanation:

In this case, we can start with the structure of 1-hexanol. In this alcohol, the <u>"OH" is on the first carbon</u>. Thus, the alkene that this alcohol can produce must have a double bond between carbons 1 and 2, that is, <u>hex-1-ene</u>.

Now, it is important to know which of the reactions we can use, <em>hydroboration-oxidation, </em>or <em>oxymercuration-reduction</em>. To decide which reaction to using, we must look again at the alcohol we want to produce. In this alcohol, the "OH" is attached to a primary carbon. Therefore, the addition of the "OH" must be made on the <u>least substituted carbon</u>, that is, carbon number 1.

Because of this, an <u>anti-Markovnikov reaction</u> (a reaction in which the "OH" is added on the less substituted carbon) must be used. The reaction, which serves us then is: <u>hydroboration-oxidation</u> since this reaction is anti-markovnikov.

See figure 1

I hope it helps!

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