Question

Tertiary alcohols with two identical alkyl groups attached to the alcohol carbon can be made either from an ester and two moles of a Grignard reagent, or from a ketone and one mole of a Grignard reagent. Use retrosynthetic analysis to suggest one path of each type to synthesize 3‑methyl‑3‑pentanol (or 3‑methylpentan‑3‑ol). Reagent 1 + Reagent 2 ⟶⟶ −→−−H3O+ →H3O+ Route 1: Select an Ester, , and an appropriate Grignard, , to give 3‑methyl‑3‑pentanol (or 3‑methylpentan‑3‑ol) after reaction and an aqueous workup. Route 2: Select a ketone, , and an appropriate Grignard, , to give 3‑methyl‑3‑pentanol (or 3‑methylpentan‑3‑ol) after reaction and an aqueous workup

Answer 1

Answer:

Route 1: In order to obtain 3‑methyl‑3‑pentanol, the reagent 1 should be ethyl acetate (ester) and the reagent 2 should be ethyl magnesium bromide (Grignard).

Route 2: In order to obtain 3‑methyl‑3‑pentanol, the reagent 1 should be 3-pentanone (ketone) and the reagent 2 should be methyl magnesium bromide (Grignard).

Explanation:

In planning a Grignard synthesis it is important to notice the groups attached to the carbon atom bearing the alcohol group.

For route 1 the reagent 1 is an ester thus, the R group chousen will be added twice because the first step form a ketone and the second step form the tertiary alcohol. To obtain 3‑methyl‑3‑pentanol the reagent 1 should be ethyl acetate and the reagent 2 should be ethyl magnesium bromide because of that the only possibility.

For route 2 the reagent 1 is a ketone so, only one R group will be added. One possibility is 3-pentanone  as the ketone and methyl magnesium bromide as Grignard reagent.