See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting alcohol is H2C=CH-CH2-CH2-C(OH)(CH2CH3) where the carbon bearing OH also bears an ethyl group and is connected to the butenyl chain (CH2CH2CH=CH2). This is a tertiary allylic alcohol (the OH is on a tertiary carbon, but the double bond is terminal at C1-C2 of the chain, i.e., homoallylic position relative to the OH carbon). Step 2 - Reaction with acetone (CH3COCH3) / TsOH: TsOH is an acid catalyst. Acetone reacts with the tertiary alcohol under acid catalysis. Since the alcohol is tertiary and acetone is a ketone, this forms a mixed ketal (actually a tertiary alkyl vinyl ether or more precisely a tertiary ether by acid-catalyzed reaction). More precisely, under TsOH, the tertiary alcohol reacts with acetone to form a 2,2-dimethyl-1,3-dioxolane type protection OR simply forms a tertiary alkyl isopropylidene ketal. The product (A) is the isopropylidene ether: the OH is protected as -O-C(CH3)2... but since it is a monoalcohol (not a diol), it forms a mixed tertiary ether: R-O-CMe2-OH, or via the enol ether mechanism: R-O-C(CH3)=CH2. Actually, with TsOH and acetone, a tertiary alcohol undergoes acid-catalyzed etherification with the enol form of acetone to give R-O-C(CH3)=CH2 (isopropenyl ether). So (A) = H2C=CH-CH2-CH2-C(Et)(CH2-O-C(CH3)=CH2... wait, the OH is on the quaternary carbon. Actually there is only one OH. Under acid with acetone, the reaction of a tertiary alcohol with acetone/TsOH gives the tert-alkyl isopropenyl ether via an SN1-type mechanism or via protonation of acetone enol. So (A) = H2C=CH-CH2-CH2-C(Et)(... )-O-CMe=CH2, i.e., the terminal alkene is still present plus a new isopropenyl ether. Step 3 - Ozonolysis of (A) with O3 / (CH3)2S (reductive workup): Ozonolysis cleaves C=C double bonds to give carbonyl compounds. Compound (A) contains TWO double bonds: (i) the terminal alkene H2C=CH- from the original starting material, and (ii) the isopropenyl ether double bond -O-C(CH3)=CH2. Ozonolysis of H2C=CH- (terminal, monosubstituted) gives HCHO (formaldehyde, which is the HCHO shown in the product) + an aldehyde at the other carbon. Ozonolysis of -O-C(CH3)=CH2 gives HCHO (from =CH2 end) + an ester/ketal fragment -O-C(CH3)=O which is actually -O-C(=O)-CH3 (methyl ester) -- wait, ozonolysis of a vinyl ether R-O-CR'=CH2 gives R-O-C(=O)-R' (an ester) + HCHO. So the isopropenyl ether -O-C(CH3)=CH2 upon ozonolysis gives -O-C(=O)-CH3... but that gives an ester, not matching option (a). Let me reconsider: With TsOH and excess acetone (2-methoxypropene equivalent), a tertiary alcohol can form a cyclic protecting group if there are two OH groups, but here there is only one. Alternatively, the reaction could form a hemiketal or the tertiary carbocation could rearrange. More likely: TsOH catalyzes the reaction of the terminal alkene (allyl system) with acetone via a Prins-type cyclization or the alcohol attacks acetone to give a mixed ketal on both the OH and the allylic system. Actually, re-reading: the molecule has a terminal alkene AND a tertiary OH separated by two CH2 groups. Under TsOH/acetone, a Prins-type cyclization could occur forming a cyclic ketal where the alkene and the alcohol react with acetone. This would give a 6-membered ring (1,3-dioxane derivative) or similar. The Prins cyclization of a homoallylic alcohol with a carbonyl compound under acid gives a tetrahydropyran. Here: the OH and terminal alkene are 1,4-related (OH on C5, alkene on C1-C2, with C3-C4 as CH2CH2), so a 6-membered ring could form. Under Prins conditions with acetone, (A) would be a 1,3-dioxane (cyclic acetal) incorporating the two oxygens from acetone and the tethered alcohol... Actually the Prins reaction of a homoallylic alcohol with ketone/acid gives a 1,3-dioxane. The product (A) would be a cyclic 1,3-dioxane with the acetone-derived dimethyl substitution at C2, and the ethyl group and CH2 pendant from the ring carbons. Step 4 - Re-examining with ozonolysis giving HCHO: The ozonolysis of (A) gives (B) + HCHO. HCHO comes from a terminal =CH2. Looking at option (a): it shows an aldehyde H-C(=O)-CH2-CH2-C(Et)(CH2-O-CMe2)(O-CMe2). This structure has the quaternary carbon bearing Et, a -CH2-O-CMe2 group, and an -O-CMe2 group (a hemiketal ether from acetone). This is consistent with: the original terminal alkene H2C=CH- was cleaved by ozonolysis to give HCHO + the chain aldehyde OHC-CH2-CH2-C(Et)(...). The acetone-derived protecting group on the tertiary OH is -O-C(CH3)2- which upon ozonolysis of the isopropenyl ether gives the ketal fragment. Actually if (A) is the isopropenyl ether R-O-C(CH3)=CH2, ozonolysis gives R-O-C(=O)CH3 (ester) which does NOT match (a). But option (a) shows -O-C(CH3)2 intact, suggesting the ketal oxygen is not from ozonolysis of the protecting group double bond but rather the protecting group double bond is the one that gives HCHO. Ozonolysis of R-O-C(CH3)=CH2: the =CH2 end gives HCHO, and the R-O-C(CH3)= end gives R-O-C(=O)-CH3, an ester. This gives option (c) pattern (formate-like) if the other alkene is also cleaved, but option (a) shows -O-CMe2 (dimethyl substituted carbon still attached to O without a carbonyl). Alternative: If (A) is the THP (Prins product), a cyclic 1,3-dioxane with no remaining double bond, then there is no double bond to cleave. So A must retain a double bond. Most likely interpretation: TsOH/acetone converts the tertiary OH to -O-CMe2OH (hemiketal, unstable) or more practically the reaction of the tertiary alcohol with 2-methoxypropene (formed from acetone/TsOH) gives R-O-C(CH3)=CH2 (the isopropenyl ether). Then ozonolysis of BOTH double bonds: terminal alkene H2C=CH- → HCHO + OHC- (the stated HCHO), and isopropenyl ether -O-C(CH3)=CH2 → -O-C(=O)CH3 + HCHO. But the question says only HCHO is the coproduct (B) + HCHO, implying one equivalent of HCHO total. This means only one double bond is cleaved, or both give HCHO but only one is shown separately. Actually re-reading: (B) + HCHO, so HCHO is one fragment, and (B) is the other. With two double bonds cleaved: terminal alkene H2C=CH- → HCHO + -CHO (incorporated in B), and -O-C(CH3)=CH2 → HCHO + -O-C(=O)CH3 (ester in B). This would give B as: OHC-CH2-CH2-C(Et)(-O-COCH3) with a -CH2- fragment... this is getting complex. Let me try a cleaner approach: Looking at option (a) directly. Option (a) is: OHC-CH2-CH2-C(Et)(CH2-O-CMe2... specifically CH2-O-C(CH3)2 as a pendant group, and -O-C(CH3)2 as the main chain ether. The structure has CMe2 connected to TWO oxygens (one from the CH2-O and one from the C-O on the quaternary carbon), making it a cyclic acetal fragment (neopentyl glycol ketal type). Wait—looking at structure (a) again: the quaternary carbon has: Et, CH2-O-CMe2, O-CMe2, and CH2CH2CHO. The CMe2 is bonded to both an O from CH2-O and an O from the direct C-O linkage. So CMe2 is part of a cyclic structure: -C(quaternary)-O-CMe2-O-CH2-C(quaternary)- forming a 5-membered ring (1,3-dioxolane). So (A) is a 1,3-dioxolane (cyclic acetal) formed between the tertiary OH and the CH2OH... but there's no CH2OH in the starting material. Wait—perhaps the starting material I'm reading is wrong. Let me re-read: H2C=CH-CH2-CH2-C(OH)(CH2-CH3) with also a CH2OH group? The structure shows: C bearing OH, with CH2-CH3 as one substituent AND the main chain connects to CH2-CH2-CH=CH2. There appear to be two substituents on the C: CH2-CH3 and from the drawing possibly also a CH2OH making it a diol? No, looking again: the structure is H2C=CH-CH2-CH2-C(-OH)(-CH2CH3) where the vertical bar above C means there's something above: looking at the image description again: the C-OH has a vertical bar above it indicating another substituent. Actually the structure written is: H2C=CH-CH2-CH2-C(OH)(?) where C also has CH2-CH3 below. The | above C-OH might mean the C is connected to the chain above making it: the chain is ...CH2-C(OH)-... with CH2CH3 as a branch below and the alkene chain to the left. So the compound is: pent-4-en-1-ol type with branching, specifically: 2-ethylpent-4-en-1-ol... no. Let me just accept: the alcohol is a tertiary alcohol with one OH, one ethyl group, and a butenyl chain (CH2CH2CH=CH2). Molecular formula contribution: the C bearing OH has: -OH, -CH2CH3, -(CH2)2CH=CH2, and one more group (since it needs 4 bonds). Looking at the image: there's a vertical bond above C-OH suggesting the chain continues above, but the leftmost part H2C=CH-CH2-CH2- is the chain. So it's quaternary C: OH, Et, CH2CH2CH=CH2, and... the | above might just be the bond to OH shown differently. If it's truly tertiary (3 C substituents + OH), the 4th substituent must be H (secondary) or another carbon group. Given the image shows | above C-OH, perhaps there's a methyl group above giving: C(OH)(CH3)(Et)(CH2CH2CH=CH2)—a quaternary alcohol (no H on the OH-bearing carbon). That makes it truly tertiary. Actually the most consistent reading: The starting material is 2-ethyl-2-methylpent-4-en-1-ol... no. Let me just accept the starting material as drawn with a tertiary C bearing OH, CH2CH3, CH2CH2CH=CH2, and possibly a CH2OH based on how option (a) makes sense. If the starting material has BOTH an OH and a CH2OH group (i.e., it's a diol: HO-CH2-C(OH)(Et)(CH2CH2CH=CH2)), then with acetone/TsOH it forms a cyclic 1,3-dioxolane (5-membered cyclic acetal) protecting the diol. This gives (A) = 2,2-dimethyl-1,3-dioxolane with the butenyl and ethyl substituents on C4 of the dioxolane, plus the terminal alkene intact. Ozonolysis of the terminal H2C=CH- then gives HCHO + OHC- fragment, where the CHO-containing fragment is (B) = OHC-CH2-CH2-C(Et)(CH2-O-CMe2-O) as a cyclic ketal. This matches option (a) perfectly: the dioxolane ring is drawn open in (a) as -CH2-O-CMe2-O- with CMe2 bridging two oxygens. The aldehyde comes from ozonolysis of the terminal alkene (H2C=CH- → HCHO + -CHO), and the cyclic acetal (dioxolane) is preserved. HCHO is the coproduct from the =CH2 end. Therefore the starting material must be a 1,2-diol (or at least have a CH2OH group alongside the tertiary OH): H2C=CH-CH2-CH2-C(OH)(Et)-CH2OH (a tertiary diol type: 2-ethyl-4-penten-1,2-diol... actually 2-ethylpent-4-ene-1,2-diol or similar). This diol reacts with acetone/TsOH to give the cyclic acetal (1,3-dioxolane), compound (A). Ozonolysis of the terminal alkene in (A) gives HCHO + the aldehyde (B). Option (a) shows B correctly as the aldehyde with the intact 1,3-dioxolane (drawn in open form showing both C-O bonds). Verification against other options: (b) shows a cyclic peroxide (1,2-dioxolane) which would require different reaction conditions and doesn't arise from standard ozonolysis/ketal chemistry. (c) shows a formate ester H-O-C(=O)- which would come from formic acid, not expected here. (d) has only one CH2 between the aldehyde and the dioxolane carbon, but the starting material has two CH2 groups (CH2CH2) between the terminal alkene and the quaternary carbon, so after ozonolysis there should be two CH2 groups, matching (a) not (d). Therefore, the correct answer is A.