Question

We see below that 3-methyl-3-hexanol can be synthesized from the reaction of 2-pentanone with ethylmagnesium bromide. There is a scheme of a chemical reaction, where CH3CCH2CH2CH3, with an O atom double-bonded to the second (from left to right) carbon (2-pentanone) reacts with CH3CH2MgBr (ethylmagnesium bromide). The arrow goes from the single bond between CH3CH2 and MgBr to the second C atom of 2-pentanone. The second arrow goes from the double bond between the second C atom and the O atom of 2-pentanone to the O atom. An intermediate is CH3CCH2CH2CH3, with a CH2CH3 group and an O minus MgBr plus group attached to the second carbon. In the presence of H3O plus the intermediate goes to CH3CCH2CH2CH3, with a CH2CH3 group and an OH group attached to the second carbon (3-methyl-3-hexanol, a tertiary alcohol). What other combinations of ketone and Grignard reagent could be used to prepare the same tertiary alcohol

Answer 1

Answer:

See explanation below

Explanation:

Ok, the first picture is your exercise. We have one type of reaction to form the 3-methyl-3-hexanol.

Now, according to the question, you want another form to get the 3-methyl-3-hexanol. In this case, the easiest way to do this, is using a ketone with the 6 carbons, and the grignard reagent to use should have the methyl group. Now, in which carbon should the carbonile group be? well, as the final product has the OH group in carbon 3, then the reaction is taking place there. So the reactants would be the hexan- 3 - one and the methyl magnesium bromide.

The second picture, you'll see the reactants and final product.

Hope this helps