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Question: The aldol condensation of acetaldehyde results in the formation of: (A) \(\text{C}{{\text{H}}_{3}}...

The aldol condensation of acetaldehyde results in the formation of:
(A) CH3CH2OH + CH3COOH\text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{2}}\text{OH + C}{{\text{H}}_{3}}\text{COOH}
(B) CH3CH2-CH(OH)-CHO\text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{2}}\text{-CH(OH)-CHO}
(C) CH3CO-CH(OH)-CH3-CHO\text{C}{{\text{H}}_{3}}\text{CO-CH(OH)-C}{{\text{H}}_{3}}\text{-CHO}
(D) CH3-CH(OH)-CH2-CHO\text{C}{{\text{H}}_{3}}\text{-CH(OH)-C}{{\text{H}}_{2}}\text{-CHO}

Explanation

Solution

For this problem first we have to study about the aldol condensation and its procedure and the condition on which molecule it is applicable. Then we have to write the complete reaction of the acetaldehyde undergoing aldol condensation.

Complete step by step solution:
-In the given question, we have to choose the correct product that is formed by the aldol condensation of the acetaldehyde.
-So, firstly we have to study about the aldol condensation that takes place in the aldehyde which consists of at least one carbon atom with alpha - hydrogen in the presence of dilute acid.
-Molecules such as aldehyde and ketone can undergo aldol condensation with alpha hydrogen.
-Firstly, we have to write the complete mechanism of the aldol condensation of the acetaldehyde.
-Firstly, the deprotonation of the acetaldehyde take place by the hydroxide ion i.e.
CH3CHO + OH  CH2CHO + H2O\text{C}{{\text{H}}_{3}}\text{CHO + O}{{\text{H}}^{-}}\text{ }\to \text{ }{}^{-}\text{C}{{\text{H}}_{2}}\text{CHO + }{{\text{H}}_{2}}\text{O}
-Now, the ion formed as a product is known as enolate ion which will further react with the unreacted aldehyde and yields alkoxide ion.
CH3CHO + CH2CHO  CH3 - CO - CH2CHO\text{C}{{\text{H}}_{3}}\text{CHO + }{}^{-}\text{C}{{\text{H}}_{2}}\text{CHO }\to \text{ C}{{\text{H}}_{3}}\text{ - C}{{\text{O}}^{-}}\text{ - C}{{\text{H}}_{2}}\text{CHO}
-Now, the protonation of alkoxide ion will take place by water which will yield the final product of aldol condensation of the acetaldehyde that is:
CH3 - CO - CH2CHO + H2 CH3CH(OH) - CH2CHO\text{C}{{\text{H}}_{3}}\text{ - C}{{\text{O}}^{-}}\text{ - C}{{\text{H}}_{2}}\text{CHO}\ \text{+ }{{\text{H}}_{2}}\text{O }\rightleftharpoons \text{ C}{{\text{H}}_{3}}\text{CH(OH) - C}{{\text{H}}_{2}}\text{CHO}
-In protonation, the hydrogen ion from the water will separate and attach to the oxygen ion of alkoxide ion.

Therefore, option (D) is the correct answer.

Note: When two different types of molecules react together such as aldehyde and ketone then the organic reaction is known as crossed aldol condensation. In the problem, the alpha position refers to the first carbon that is attached to the functional group.