See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material and first reagent (Mg): The starting material is (CH3O)2CHCH2CH2CH2Br, a dimethyl acetal with a terminal alkyl bromide. Treatment with Mg (in dry ether) converts the alkyl bromide into a Grignard reagent: (CH3O)2CHCH2CH2CH2MgBr. Step 2 - Reaction with formaldehyde (H2C=O): The Grignard reagent attacks formaldehyde (H2C=O). The nucleophilic carbon of the Grignard adds to the carbonyl carbon of formaldehyde, giving after workup a primary alcohol extended by one carbon: (CH3O)2CHCH2CH2CH2CH2OH (as the magnesium alkoxide, then protonated). Step 3 - H3O+/heat (acid hydrolysis of the acetal): The acetal group (CH3O)2CH- is hydrolyzed under aqueous acid conditions with heat. An acetal hydrolyzes to an aldehyde: (CH3O)2CH- becomes H-C(=O)-. This converts the molecule to H-C(=O)-CH2CH2CH2CH2OH, which is 5-hydroxypentanal (an aldehyde with a terminal hydroxyl group). Step 4 - Identify the product: The final product is H-C(=O)-CH2CH2CH2CH2OH, which matches option (c). Why other options fail: - Option (a) shows a methyl ester group (CH3O-C=O), which would require different chemistry. - Option (b) shows a methyl ketone (CH3-C=O), which is not produced here. - Option (d) shows a dialdehyde, which would require oxidation of the terminal alcohol, which does not occur under these conditions. Therefore, the correct answer is C.