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When an optically active amine 'A' havin...

When an optically active amine 'A' having molecular formula `C_(4)H_(11)N` is subjected to Hoffmann's exhaustive methylation followed by hydrolysis, an alkene 'B' is produced which upon ozonolysis and subsequent hydrolysis yields formaldehyde and propanal. The amine 'A' is

A

`CH_(3)-underset(NH_(2))underset("| ")"CH"CH_(2)CH_(3)`

B

`CH_(3)NH-underset(CH_(3))underset("| ")"CH"-CH_(3)`

C

`CH_(3)-underset(CH_(3))underset("| ")"NC"H_(2)CH_(3)`

D

`CH_(3)CH_(2)CH_(2)CH_(2)NH_(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, let's break it down step by step: ### Step 1: Identify the structure of the optically active amine 'A' Given the molecular formula \(C_4H_{11}N\), we need to find an optically active amine. An optically active amine must have a chiral center. A suitable candidate is 2-aminobutane, which has the structure: ``` CH3 | CH3-CH-CH2-NH2 | CH3 ``` ### Step 2: Perform Hofmann's exhaustive methylation Hofmann's exhaustive methylation involves treating the amine with excess methyl iodide (\(CH_3I\)) and a strong base (like \(Ag_2O\)). This results in the formation of a quaternary ammonium salt. In our case, 2-aminobutane will be converted into a quaternary ammonium salt, which can be represented as: ``` CH3 | CH3-CH-CH2-N(CH3)3+ | CH3 ``` ### Step 3: Hydrolysis of the quaternary ammonium salt Upon hydrolysis, the quaternary ammonium salt will lose a methyl group, leading to the formation of an alkene. The hydrolysis will yield the alkene 'B', which is 2-butene: ``` CH3 | CH3-CH=CH2 ``` ### Step 4: Ozonolysis of alkene 'B' The ozonolysis of 2-butene will cleave the double bond, producing two carbonyl compounds. The ozonolysis reaction can be represented as follows: ``` CH3-CH=CH2 + O3 → CH3CHO + HCHO ``` ### Step 5: Hydrolysis of ozonolysis products After ozonolysis, the products are formaldehyde (HCHO) and propanal (CH3CHO). This confirms that the ozonolysis of alkene 'B' produces the required aldehydes. ### Conclusion Thus, the optically active amine 'A' is 2-aminobutane. ### Final Answer: The amine 'A' is **2-aminobutane**. ---
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