
The compound $ 2 - phenylbutan - 2 - ol $ can be prepared by which of the following combinations?
(A) $ {C_6}{H_5}COC{H_3} + {C_2}{H_5}MgBr $
(B) $ {C_2}{H_5}COC{H_3} + {C_6}{H_5}MgBr $
(C) $ {C_6}{H_5}CO{C_2}{H_5} + C{H_3}MgBr $
(D) All of these
Answer
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Hint: Carbonyl compounds are compounds consisting of $ C = O $ group. Aldehydes and ketones are the compounds that belong to carbonyl compounds. The Grignard reagent is an organometallic compound consisting of an alkyl part, metal, and non-metal-like halogens. When ketones are treated with Grignard reagent forms alcohols.
Complete answer:
Chemical compounds are classified into functional groups based on the groups present in them. Alcohols are compounds consisting of $ - OH $ groups. Ketones are the carbonyl compounds consisting of $ C = O $ attached to two alkyl groups. An organometallic compound is a compound consisting of alkyl part, metal, and non-metal-like halogens. The grignard reagent is the best example of an organometallic compound with the general formula of $ RMgX $ .
The desired compound is $ 2 - phenylbu\tan - 2 - ol $ , it has the structure of $ C{H_3}C{H_2}C(OH)({C_6}{H_5})C{H_3} $
When ketones are treated with Grignard reagent, the alkyl part in Grignard reagent attacks on the carbonyl carbon, and oxygen in the carbonyl group gets negative charge, upon hydrolysis forms alcohol.
The first preparation is $ {C_6}{H_5}COC{H_3} + {C_2}{H_5}MgBr \to {C_2}{H_5}C(OH)({C_6}{H_5})C{H_3} $
The other preparation is $ {C_2}{H_5}COC{H_3} + {C_6}{H_5}MgBr \to {C_2}{H_5}C(OH)({C_6}{H_5})C{H_3} $
Another preparation is $ {C_6}{H_5}CO{C_2}{H_5} + C{H_3}MgBr \to C{H_3}C(OH)\left( {{C_6}{H_5}} \right){C_2}{H_5} $
All the above-formed products are $ 2 - phenylbu\tan - 2 - ol $
Thus, from all the mentioned reagents, the desired product will be formed.
Option D is the correct one.
Note:
In general, alkyl groups get a positive charge as they are electron pushing or electron releasing groups. But in the case of organometallic compounds, the alkyl group gets a negative charge which is a different one. Thus, the alkyl part in the organometallic compound attacks on the carbonyl carbon as the carbonyl carbon has a positive charge.
Complete answer:
Chemical compounds are classified into functional groups based on the groups present in them. Alcohols are compounds consisting of $ - OH $ groups. Ketones are the carbonyl compounds consisting of $ C = O $ attached to two alkyl groups. An organometallic compound is a compound consisting of alkyl part, metal, and non-metal-like halogens. The grignard reagent is the best example of an organometallic compound with the general formula of $ RMgX $ .
The desired compound is $ 2 - phenylbu\tan - 2 - ol $ , it has the structure of $ C{H_3}C{H_2}C(OH)({C_6}{H_5})C{H_3} $
When ketones are treated with Grignard reagent, the alkyl part in Grignard reagent attacks on the carbonyl carbon, and oxygen in the carbonyl group gets negative charge, upon hydrolysis forms alcohol.
The first preparation is $ {C_6}{H_5}COC{H_3} + {C_2}{H_5}MgBr \to {C_2}{H_5}C(OH)({C_6}{H_5})C{H_3} $
The other preparation is $ {C_2}{H_5}COC{H_3} + {C_6}{H_5}MgBr \to {C_2}{H_5}C(OH)({C_6}{H_5})C{H_3} $
Another preparation is $ {C_6}{H_5}CO{C_2}{H_5} + C{H_3}MgBr \to C{H_3}C(OH)\left( {{C_6}{H_5}} \right){C_2}{H_5} $
All the above-formed products are $ 2 - phenylbu\tan - 2 - ol $
Thus, from all the mentioned reagents, the desired product will be formed.
Option D is the correct one.
Note:
In general, alkyl groups get a positive charge as they are electron pushing or electron releasing groups. But in the case of organometallic compounds, the alkyl group gets a negative charge which is a different one. Thus, the alkyl part in the organometallic compound attacks on the carbonyl carbon as the carbonyl carbon has a positive charge.
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