Which of the following reactions can be used to prepare OH|CH3−C−C6H5|C2H5
A
O||CH3−C−C6H5+C2H5MgBr⟶[.]H3O+−−−−−−−−−−−−→
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B
O||C2H5−C−C6H5+CH3MgBr⟶[.]H3O+−−−−−−−−−−−−→
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C
O||CH3−C−C2H5+C6H5MgBr⟶[.]H3O+−−−−−−−−−−−−→
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D
Br|CH3−C−C6H5|C2H5Alc.KOH⟶Δ
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Solution
The correct options are AO||CH3−C−C6H5+C2H5MgBr⟶[.]H3O+−−−−−−−−−−−−→ BO||C2H5−C−C6H5+CH3MgBr⟶[.]H3O+−−−−−−−−−−−−→ CO||CH3−C−C2H5+C6H5MgBr⟶[.]H3O+−−−−−−−−−−−−→ The reactions mentioned in option A, B and C will be used to prepare the given tertiary alcohol.
In these reactions, a grignard reagent (alkyl magnesium bromide) attacks a ketone to form tertiary alcohol.
In the tertiary alcohol, 3 different alkyl groups are attached to C atom bearing -OH group. Hence, we can use three different combinations of Grignard reagents and ketones to prepare the product. In the reaction of option D, tertiary alkyl bromide will undergo dehydrobromination in the presence of alc. KOH to form an alkene.