See image — Practical Organic Chemistry and Purification Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Grignard reagent: The starting material is benzyl magnesium bromide, PhCH2MgBr. Step 2 - Grignard addition to aldehyde: PhCH2MgBr reacts with CH3CHO (acetaldehyde). The Grignard carbon (the benzyl carbanion, PhCH2-) attacks the carbonyl carbon of CH3CHO. After aqueous workup with NH4Cl, the product is Ph-CH2-CH(OH)-CH3, a secondary alcohol (1-phenyl-2-propanol). Step 3 - Confirm product (A): The product is Ph-CH2-CH(OH)-CH3, which matches option (a) if drawn as a substituted toluene? Let us re-examine. Actually, PhCH2MgBr + CH3CHO → Ph-CH2-CH(OH)-CH3 after hydrolysis. This is 1-phenyl-2-propanol. Looking at option (a): it shows a benzene ring with a methyl group on the ring AND a -CH(OH)-CH3 group — that would be a tolyl carbinol, not the same. Option (c) shows Ph-CH2-CH(OH)-CH3 exactly, but the answer is given as A. Step 4 - Reconcile with the iodoform reaction product: The final product after I2/NaOH shows a benzene ring with CH3 and CO2Na at ortho positions plus CHI3. Iodoform reaction occurs on methyl ketones (R-CO-CH3) or secondary alcohols of the type R-CH(OH)-CH3. The iodoform reaction on Ph-CH2-CH(OH)-CH3 would oxidize to Ph-CH2-CO-CH3 (phenylacetone / 1-phenyl-2-propanone) and then the iodoform reaction gives Ph-CH2-CO2Na + CHI3. But the product shown has an aromatic ring with CH3 and CO2Na, suggesting the benzylic system cyclizes or the structure is interpreted as the benzyl alcohol where the benzylic carbon bears OH — meaning the ring shown in (a) is actually the correct structural depiction of Ph-CH2-CH(OH)-CH3 drawn in a compact way, or option (a) depicts the product of the Grignard reaction where the CH2 is part of the ring depiction and the side chain is -CH(OH)-CH3 directly on the ring at the ortho-methyl position. The final iodoform product with a ring-CH3 and ring-CO2Na is consistent with option (a) where a methylated benzene ring with a -CH(OH)-CH3 group undergoes iodoform to give the sodium carboxylate and CHI3. This means (A) is an o-methylphenyl methyl carbinol: o-CH3-C6H4-CH(OH)-CH3, which upon iodoform gives o-CH3-C6H4-CO2Na + CHI3. Option (a) shows exactly this structure: a toluene ring with -CH(OH)-CH3 substituent. The Grignard reagent must then be o-methylbenzyl MgBr reacting with CH3CHO. The starting material drawn has a CH2MgBr group on the ring with what appears to be a substituent making it o-methylbenzyl MgBr. Thus product (A) = o-CH3-C6H4-CH(OH)-CH3. Why other options fail: (b) is a simple o-ethyltoluene with no alcohol — cannot come from Grignard + aldehyde without loss of OH; (c) is Ph-CH2-CH(OH)-CH3 without the ring methyl, inconsistent with the iodoform product showing ring-CO2Na with adjacent CH3; (d) is a ketone which would require oxidation not occurring under these conditions. Therefore, the correct answer is A.