Question
Question: \( MeCHO\xrightarrow[\Delta ]{NaOH}A\xrightarrow[\Delta ]{KOH}B. \) Here B is: (A) \( Me{{\left...
MeCHONaOHΔAKOHΔB.
Here B is:
(A) Me(CH=CH)3CHO.
(B) MeCH−CHCHO
(C) Me(CH=CH)2−CHO
(D) Me(CH=CH)4−CHO
Solution
Hint : We know that the aldol condensation and its procedure and the condition on which molecule it is applicable. The presence of an aldehyde with -hydrogen, in presence of a base shows aldol condensation to form an unsaturated compound then we have to write the complete reaction of the acetaldehyde undergoing aldol condensation.
Complete Step By Step Answer:
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 takes place by the hydroxide ion. The ion formed as a product is known as enolate ion which will further react with the unreacted aldehyde and yields alkoxide ion.
Further, the protonation of alkoxide ions will take place by water which will yield the final product of aldol condensation of the acetaldehyde.
CH3CHONaOHΔCH3−CH=CH−C(=O)−HKOHΔCH3−(CH=CH)3−CHO.
Therefore, the correct answer is option A.
Note :
Remember that the alpha hydrogen is the hydrogen atom present on the adjacent carbon to the carbonyl group, which is present in the acetaldehyde but not in formaldehyde. Also, a cross aldol condensation reaction takes place due to presence of two different aldehydes in the reaction and 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.