Question

hexanamide + lialh4+3ch3i+ag2o+heat

Answer 1

Hex-1-ene

Answer 2

The given reaction sequence starts with hexanamide and involves reduction, exhaustive methylation, and Hofmann elimination.

Step 1: Reduction of hexanamide with LiAlH4.

Amides are reduced to primary amines by LiAlH4. Hexanamide is CH3(CH2)4CONH2. Reduction with LiAlH4 replaces the carbonyl group (C=O) with a methylene group (CH2). CH3(CH2)4CONH2 $\xrightarrow{\text{LiAlH}_4}$ CH3(CH2)4CH2NH2 The product is hexan-1-amine (or n-hexylamine), which is CH3(CH2)5NH2.

Step 2: Alkylation with 3 equivalents of CH3I.

Primary amines react with alkyl halides to form secondary amines, tertiary amines, and finally quaternary ammonium salts through a process called exhaustive alkylation. The use of excess CH3I (indicated by +3CH3I) ensures that the amine is fully methylated to the quaternary ammonium salt. CH3(CH2)5NH2 + CH3I $\rightarrow$ CH3(CH2)5NHCH3 + HI CH3(CH2)5NHCH3 + CH3I $\rightarrow$ CH3(CH2)5N(CH3)2 + HI CH3(CH2)5N(CH3)2 + CH3I $\rightarrow$ [CH3(CH2)5N(CH3)3]+ I- The product is N,N,N-trimethylhexan-1-aminium iodide, a quaternary ammonium salt.

Step 3: Reaction with Ag2O and heat (Hofmann elimination).

Quaternary ammonium halides react with silver oxide (Ag2O) in the presence of water to form the corresponding quaternary ammonium hydroxide. Ag2O reacts with water to form AgOH. [CH3(CH2)5N(CH3)3]+ I- + AgOH $\rightarrow$ [CH3(CH2)5N(CH3)3]+ OH- + AgI(s) Heating the quaternary ammonium hydroxide salt results in Hofmann elimination, which is an E2 elimination reaction. The hydroxide ion acts as a base, abstracting a beta-hydrogen, leading to the formation of an alkene, water, and a tertiary amine as the leaving group. Hofmann elimination follows the Hofmann rule, which states that the least substituted alkene is preferentially formed. The quaternary ammonium salt is [CH3(CH2)5N(CH3)3]+ OH-. Let's write the structure of the cation: CH3-CH2-CH2-CH2-CH2-CH2-N+(CH3)3. Let's label the carbons of the hexyl chain starting from the end attached to nitrogen: C6-C5-C4-C3-C2-C1-N+(CH3)3 The nitrogen is attached to C1 (alpha-carbon on the hexyl chain). The carbons attached to the nitrogen are C1 and the three methyl carbons. Beta-hydrogens are located on carbons adjacent to the alpha-carbon.

• On the hexyl chain, the carbon adjacent to C1 is C2. C2 has two beta-hydrogens.
• The methyl groups (CH3) are attached directly to the nitrogen. While they are alpha-carbons to the positive charge on N, there are no carbons adjacent to these methyl carbons that bear hydrogens. Thus, elimination cannot occur by abstracting a hydrogen from a carbon adjacent to the methyl carbon.

The only available beta-hydrogens for elimination are on C2 of the hexyl chain. The hydroxide base abstracts a beta-hydrogen from C2. The electrons from the C2-H bond form a double bond between C1 and C2, and the C1-N bond breaks, expelling trimethylamine N(CH3)3. CH3-CH2-CH2-CH2-CH2(H)-CH2-N+(CH3)3 OH- ^beta-H ^beta-C ^alpha-C

The reaction is: [CH3(CH2)5N(CH3)3]+ OH- $\xrightarrow{\text{heat}}$ CH3-CH2-CH2-CH2-CH=CH2 + N(CH3)3 + H2O The product formed is hex-1-ene.