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What is the IUPAC name of the organic compound formed in the following chemical reaction? Acetone ā ( i i ) Ā H 2 O , Ā H + ( i ) Ā C 2 H 5 M g B r , d r y E t h e r \xrightarrow[(ii)\ H_2O,\ H^+]{(i)\ C_2H_5MgBr, dry Ether} ( i ) Ā C 2 ā H 5 ā M g B r , d ry Et h er ( ii ) Ā H 2 ā O , Ā H + ā Product
medium
Alcohols, Phenols and Ethers
2021
chemistry
Explanation To solve this problem, we need to understand the reaction of acetone with ethyl magnesium bromide ( C 2 H 5 M g B r C_2H_5MgBr C 2 ā H 5 ā M g B r ) followed by hydrolysis.
\newline This is a Grignard reaction, which involves the following steps:
\newline
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1. Reaction of acetone with
C 2 H 5 M g B r : C_2H_5MgBr:
\newline C 2 ā H 5 ā M g B r : ā¢ Acetone is a ketone with the structure
( C H 3 ) 2 C = O . (CH_3)_2C=O.
\newline ( C H 3 ā ) 2 ā C = O . ā¢ The Grignard reagent
C 2 H 5 M g B r C_2H_5MgBr C 2 ā H 5 ā M g B r acts as a nucleophile, attacking the carbonyl carbon of acetone.
ā¢ This results in the formation of an alkoxide intermediate:
( C H 3 ) 2 C ( O M g B r ) C 2 H 5 . (CH_3)_2C(OMgBr)C_2H_5.
\newline ( C H 3 ā ) 2 ā C ( OM g B r ) C 2 ā H 5 ā . 2. Hydrolysis of the alkoxide intermediate:
ā¢ The alkoxide is then treated with water and an acid (
), which protonates the alkoxide to form an alcohol.
ā¢ The final product is
( C H 3 ) 2 C ( O H ) C 2 H 5 , (CH_3)_2C(OH)C_2H_5, ( C H 3 ā ) 2 ā C ( O H ) C 2 ā H 5 ā , which is 2-methylbutan-2-ol.
Analyzing the options:
ā¢ Option 1: 2-methylpropan-2-ol is incorrect because it has one less carbon atom.
ā¢ Option 2: Pentan-2-ol is incorrect because the methyl group is not at the correct position.
ā¢ Option 3: Pentan-3-ol is incorrect because the hydroxyl group is not at the correct position.
ā¢ Option 4: 2-methylbutan-2-ol is correct as it matches the structure of the product.
Therefore, the correct IUPAC name of the product is 2-methylbutan-2-ol, corresponding to Option 4.