See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 3-iodo-3-methylbutanal, where the carbon bearing iodine also bears a 14C-labeled methyl group (14CH3). The structure is 14CH3-CH(I)-CH2-CHO. Step 2 - Base/heat treatment to give (A): Treatment of an alpha-iodo aldehyde (the iodine is beta to the aldehyde, on C3) with base under heat causes elimination of HI to give an alpha,beta-unsaturated aldehyde (enal). Specifically, base abstracts a proton beta to I, and elimination gives: 14CH3-CH=CH-CHO, which is (E)-but-2-enal (crotonaldehyde) with a 14C label on the methyl group (C4). This is compound (A). Step 3 - Aldol condensation of (A) with acetaldehyde (CH3CHO) under dilute base and heat: (A) is 14CH3-CH=CH-CHO (but-2-enal, 14C on methyl). Acetaldehyde (CH3CHO) acts as the nucleophilic enol component. Under dilute base, acetaldehyde forms an enolate at its alpha carbon. This enolate attacks the aldehyde carbonyl of (A) (crossed aldol). After aldol addition and dehydration (promoted by heat), we get a new alpha,beta-unsaturated aldehyde extending the conjugated chain. Alternatively, (A) as the Michael acceptor reacts with acetaldehyde enolate: CH3CHO enolate (-CH2CHO) attacks the beta carbon of (A) (1,4 addition in crossed aldol), but under dilute base conditions the standard aldol condensation applies: the enolate of CH3CHO attacks the carbonyl of (A), giving an aldol product that dehydrates to yield a dienal. The product of condensation of CH3CHO with 14CH3-CH=CH-CHO: The enolate of acetaldehyde (CH2=CHO-) attacks the carbonyl carbon of crotonaldehyde (14CH3-CH=CH-CHO). Aldol addition gives: 14CH3-CH=CH-CH(OH)-CH2-CHO, which upon dehydration gives: 14CH3-CH=CH-CH=CH-CHO (hexa-2,4-dienal with 14C on C6). The 14C label remains on the terminal methyl (originally from the starting material), which becomes C6 of the product chain, attached at the far end from the aldehyde. This matches option (a): CH3(14)-CH=CH-CH=CH-CHO where 14 is on the methyl carbon at the end of the diene chain. Step 4 - Why other options fail: (b) Places 14C on the methyl that came from acetaldehyde side, which is incorrect - the 14C came from the original starting material's methyl group and remains at the far end. (c) Introduces bromine which is not present in any reagent. (d) Shows a cyclic/branched structure inconsistent with simple aldol condensation of linear components. Therefore, the correct answer is A.