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

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.

aliasger2709 aliasger2709 · Answer 2 · 2022-02-22T13:06:15+01:00

Ester for route 1 will be ethyl acetate and Grignard reagent will be ethyl magnesium bromide, and for route 2 will be 3-pentanone and methyl magnesium bromide.

What is ester, ketone and Grignard synthesis?

Ester and ketone are the functional groups derived from the acids and later one is of great importance in the industries and biology.

Organomagnesium halide that contains the halide group along with the alkyl and aryl groups is called Grignard reagent. For a Grignard synthesis, the alcohol attached to the carbon is taken into great consideration.

For route 1, the first reagent is the ester group and hence, R will be added twice to form tertiary alcohol. To get 3‑methyl‑3‑pentanol as the product of the reaction the reagent should be ethyl acetate and the second reagent will be ethyl magnesium bromide.

For route 2, the first reagent is the ketone group and only a single R group will be added, so reagent 1 will be 3-pentanone and reagent 2 will be methyl magnesium bromide.

Therefore, the ketone and ester groups along with the Grignard reagent are used for the preparation of the products.

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