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Answer :
The methyl ester can be represented as follows:
CH3COOCH3
The Grignard reagent used in this reaction is ethylmagnesium bromide (C2H5MgBr). The structure of the Grignard reagent can be represented as follows:
Br
|
CH3CH2Mg-Br
The Grignard reagent is formed by the reaction of magnesium (Mg) with ethyl bromide (C2H5Br). The bromine atom (Br) is attached to the carbon atom bonded to the magnesium atom (Mg), and the ethyl group (C2H5) is attached to the carbon atom bonded to the bromine atom.
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Final answer:
The methyl ester that reacts with two equivalents of a Grignard reagent to yield the alcohol (CH3CH2CH2CH2)2C(OH)CH3 is methyl acetate (CH3COOCH3). The Grignard reagent is ethyl magnesium bromide (CH3CH2MgBr). The process involves two nucleophilic additions to the ester's carbonyl carbon.
Explanation:
The student's question involves the reaction between a Grignard reagent and a methyl ester to produce a specific alcohol, (CH3CH2CH2CH2)2C(OH)CH3.
To determine the structure of the starting methyl ester and the Grignard reagent, we must consider the product of the reaction. The product alcohol suggests that the Grignard reagent used is ethyl magnesium bromide (CH3CH2MgBr), given that two ethyl groups are attached to the central carbon. With two equivalents of this Grignard reagent added, the starting methyl ester must have been acetone, which is a three-carbon ketone (CH3COCH3) converted to an ester form, so the correct ester is methyl acetate (CH3COOCH3).
The overall reaction would involve the methyl acetate reacting with two equivalents of ethyl magnesium bromide to yield the final alcohol after aqueous workup. The transformation involves nucleophilic attack by the Grignard reagent on the carbon of the ester's carbonyl group, followed by breakdown of the ester into an alcohol and a ketone, which subsequently reacts with the second equivalent of Grignard reagent to yield the tertiary alcohol after protonation.
