See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material and first reaction: The starting material is 4-methoxytoluene (p-methoxytoluene, anisole with para methyl group). The reaction conditions are Li/EtOH/NH3(l), which are classic Birch reduction conditions. Step 2 - Birch reduction regioselectivity: For electron-donating groups (EDG) like OMe on an aromatic ring, Birch reduction reduces the positions NOT bearing the substituent. The double bonds remain conjugated with the EDG substituent. For p-methoxytoluene (OMe at C1, Me at C4), the Birch reduction leaves the double bonds at C1-C2 and C3-C4 (i.e., the double bonds remain in conjugation with OMe). This gives 1-methoxy-4-methylcyclohexa-2,5-diene. Wait - more precisely: EDG directs reduction at C2,C5 (positions not bearing OMe), leaving double bonds at C1-C6 and C3-C4, giving a 1,4-diene where OMe is on a double-bond carbon. The product P is 2-methoxy-5-methylcyclohexa-2,5-diene or equivalently 1-methoxy-4-methylcyclohexa-2,5-diene - a non-conjugated 1,4-cyclohexadiene with OMe on one sp2 carbon and Me on the other sp2 carbon at para position. This matches option (c) and (d) for P: the cyclohexadiene ring with MeO and Me on double-bond carbons. Step 3 - Ozonolysis with 1 equivalent O3, reductive workup (Me2S): With 1 equivalent of ozone at -78°C followed by reductive workup (Me2S), only ONE double bond is cleaved (the more electron-rich or more accessible double bond). In 1-methoxy-4-methylcyclohexa-2,5-diene, the double bond bearing the OMe group (enol ether double bond) is more reactive toward electrophilic ozone. Ozonolysis of an enol ether C=C gives an ester (methyl ester, MeO2C-) on the OMe-bearing carbon and an aldehyde (OHC-) on the other carbon of that double bond. Since the ring is intact (cyclic), cleavage of one double bond opens to give a linear dialdehyde/ketoaldehyde - but since it's cyclic, cleavage of one double bond gives a ring-opened product with two carbonyl groups at the ends, keeping the rest of the ring. Actually with a cyclic diene and 1 equiv O3: cleavage of the enol ether double bond (C1=C2 bearing OMe) gives an open-chain product with MeO2C- at one end (from the carbon bearing OMe) and CHO at the other end, with the remaining double bond (C4=C5 bearing Me, i.e., the trisubstituted alkene) intact. This gives a product with a cyclohex-ene fragment containing Me, plus MeO2C and OHC termini. The reductive workup (Me2S) keeps any peroxide as aldehyde (not further oxidized). The product Q has: a six-membered ring with one remaining double bond, Me substituent, MeO2C group and OHC group. This matches option (d): P = 1-methoxy-4-methylcyclohexa-2,5-diene and Q = a cyclohexene ring with Me, bearing MeO2C and OHC groups. Step 4 - Why not other options: (a) P is shown as aromatic (4-methoxytoluene) - this would mean no Birch reduction occurred, which is wrong. (b) Q has COMe (ketone) instead of ester - ozonolysis of enol ether gives methyl ester not methyl ketone, so wrong. (c) Q has CHO and COMe - same issue, COMe is wrong for enol ether ozonolysis product. (d) Q correctly shows MeO2C (methyl ester from enol ether carbon) and OHC (aldehyde from other carbon), with the remaining double bond and Me group intact in the ring. This is correct. Therefore, the correct answer is D.