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Identify the major product C formed in the following reaction sequence: C H 3 − C H 2 − C H 2 − I → N a C N A → p a r t i a l h y d r o l y s i s O H − B → B r 2 N a O H C ( m a j o r ) \newline CH₃-CH₂-CH₂-I \xrightarrow{NaCN} A \xrightarrow[partial hydrolysis]{OH⁻} B \xrightarrow[Br₂]{NaOH} C \ (major) C H 3 − C H 2 − C H 2 − I N a CN A O H − p a r t ia l h y d ro l ys i s B N a O H B r 2 C ( maj or )
Explanation To solve this problem, we need to identify the major product C formed in the given reaction sequence.
\newline The sequence is: C H 3 − C H 2 − C H 2 − I → N a C N A → p a r t i a l h y d r o l y s i s O H − B → B r 2 N a O H C ( m a j o r )
\newline CH_3-CH_2-CH_2-I \xrightarrow{NaCN} A \xrightarrow[partial\ hydrolysis]{OH^-} B \xrightarrow[Br_2]{NaOH} C \ (major)
\newline C H 3 − C H 2 − C H 2 − I N a CN A O H − p a r t ia l h y d ro l ys i s B N a O H B r 2 C ( maj or )
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Let's analyze each step:
1. The first step involves the reaction of 1-iodopropane with sodium cyanide (NaCN):
• This is a nucleophilic substitution reaction where the iodide ion is replaced by the cyanide ion.
• The product A is propanenitrile (CH_3-CH_2-CH_2-CN).
2. The second step is the partial hydrolysis of propanenitrile:
• Under basic conditions (OH^-), the nitrile group (-CN) is partially hydrolyzed to form a carboxylate ion.
• The product B is propanoic acid (CH_3-CH_2-COOH).
3. The third step is the Hofmann bromamide reaction:
• Propanoic acid is treated with bromine and sodium hydroxide.
• This reaction converts the carboxylic acid into an amine with one less carbon atom.
• The product C is propylamine (CH_3-CH_2-CH_2-NH_2).
Analyzing the options:
• Option 1: Butanamide is incorrect as it has one more carbon than propylamine.
• Option 2: α-Bromobutanoic acid is incorrect as it is not an amine.
• Option 3: Propylamine is correct as it matches the product C.
• Option 4: Butylamine is incorrect as it has one more carbon than propylamine.
Therefore, the correct option is Option 3: Propylamine.