Question

In the reaction sequence the product B is:
$2C{{H}_{3}}CHO\xrightarrow{O{{H}^{-}}}A\xrightarrow{\Delta }B$
(A) $C{{H}_{3}}-CH=CH-CHO$
(B) $C{{H}_{3}}COC{{H}_{3}}$
(C) $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}$
(D) $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-OH$

Answer 1

The given reaction belongs to aldol condensation reactions. In the question, at the presence of hydroxide ions, two moles of acetaldehyde will condense to form a beta-hydroxy alcohol as compound A and on dehydration A is converted into compound B which is an alpha beta-unsaturated aldehyde.

Complete answer:
- Let’s start with the concept of aldol condensation. It is a type of condensation reaction in organic chemistry in which an enolate ion or an enol reacts with a carbonyl compound (aldehydes or ketones) to form a β-hydroxy ketone or β-hydroxyaldehyde, followed by dehydration to form a conjugated enone.
- The first part of the reaction mechanism involves the aldol reaction and the second part involves dehydration or an elimination reaction in which an alcohol molecule or a water molecule is being eliminated.
- The aldol addition product is usually dehydrated by two mechanisms. One is a strong base like potassium hydroxide or sodium hydride, potassium t-butoxide in an enolate mechanism, and the other is in an acid-catalyzed enol mechanism.
- In the first step of the reaction in the presence of hydroxide ions ($O{{H}^{-}}$), two moles of acetaldehyde ($C{{H}_{3}}CHO$) will condense to form a beta-hydroxy alcohol (A) and the corresponding reaction is given below
$2C{{H}_{3}}CHO\xrightarrow{O{{H}^{-}}}C{{H}_{3}}-CH\left( OH \right)C{{H}_{2}}-CHO$
This compound A or the beta-hydroxy alcohol undergoes dehydration (removal of water molecule) and an alpha beta-unsaturated aldehyde (B) is formed. The reaction which represents the dehydration is given below
$C{{H}_{3}}-CH\left( OH \right)C{{H}_{2}}-CHO\xrightarrow[-{{H}_{2}}O]{\Delta }C{{H}_{3}}-CH=CH-CHO$
The reaction can be summarized as follows
$2C{{H}_{3}}CHO\xrightarrow{O{{H}^{-}}}C{{H}_{3}}-CH\left( OH \right)C{{H}_{2}}-CHO\xrightarrow{\Delta }C{{H}_{3}}-CH=CH-CHO$

Therefore the answer is option (A). $C{{H}_{3}}-CH=CH-CHO$.

Note:
Keep in mind that, along with aldol condensation reactions there is a class of reactions called crossed aldol condensation reactions. They use two different aldehyde or ketone reactants. Because there are two or more two different carbonyl electrophiles, and possible enolate nucleophiles, they usually give a mixture of multiple condensation products.