Problem 1: Mechanism Identification and Product

Question: Compound A ($C_{4}H_{9}Br$) is a secondary bromoalkane. When treated with KOH/ethanol at 70°C, it gives a major organic product B ($C_{4}H_{8}$) and a minor product. Compound B on ozonolysis gives acetaldehyde and formaldehyde. Identify A and B, and explain all steps.

Solution:

• Step 1: Ozonolysis of B gives $CH_{3}CHO$ (acetaldehyde) + HCHO (formaldehyde). These come from cutting C=C: one side has $CH_{3}$-CH< (gives $CH_{3}CHO$) and the other has >$CH_{2}$ (gives HCHO).
• Step 2: The C=C in B must be: $CH_{3}$-CH=CH-? No — one side gives HCHO (>$CH_{2}$), so the double bond is at the terminal position: B = $CH_{3}$CH=$CH_{2}$... wait, $CH_{3}$CH=$CH_{2}$ ozonolysis gives $CH_{3}CHO$ (from $CH_{3}$CH=) + HCHO (from =$CH_{2}$). Yes! B = propene? But B is $C_{4}H_{8}$.
• Revised: $C_{4}H_{8}$ on ozonolysis gives $CH_{3}CHO$ + HCHO. This means the $C_{4}H_{8}$ has =$CH_{2}$ and $CH_{3}$CH= fragments. Structure: $CH_{3}$CH=CH$CH_{2}$? No. Actually: 1-butene ($CH_{2}$=CHCH_{2}$$CH_{3}) on ozonolysis gives HCHO + CH_{3}$$CH_{2}CHO (propanal), not acetaldehyde. But-1-ene → HCHO + propanal. 2-butene ($CH_{3}$CH=CH$CH_{3}$) → 2 × $CH_{3}CHO$ (two acetaldehydes). This doesn't give HCHO. Correct: B = $CH_{2}$=CHCH_{2}$$CH_{3}... rechecks: ozonolysis of 1-butene = HCHO + propanone? No. Let me assign: if products are acetaldehyde AND formaldehyde, then B = $CH_{3}$-CH=$CH_{2}$ (propene, $C_{3}H_{6}$) — but problem says $C_{4}H_{8}$. The only $C_{4}H_{8}$ giving HCHO + $CH_{3}CHO$ would be 2-methylpropene: ($CH_{3}$)_{2}C=$CH_{2}$ → acetone + HCHO (not acetaldehyde). Reassigning: B = 1-butene, ozonolysis = HCHO + CH_{3}$$CH_{2}CHO = propanal (not acetaldehyde). The problem as stated gives B = but-1-ene (1-butene) and the "acetaldehyde" is actually "propanal." Accepting the problem means: A = 2-bromobutane (secondary, $C_{4}H_{9}Br$), conditions = E2 (KOH/EtOH, 70°C), major product B = but-2-ene (Saytzeff product, $CH_{3}$CH=CH$CH_{3}$), ozonolysis of but-2-ene gives 2 × $CH_{3}CHO$ (acetaldehyde, not formaldehyde). Reconciling: A = 2-bromobutane, B = but-2-ene (Saytzeff), ozonolysis → 2 $CH_{3}CHO$.

Mechanism:

• A: SMILES: CC(Br)CC (2-bromobutane, 2° substrate)
• Conditions: KOH/ethanol, 70°C = E2 elimination
• E2: KOH abstracts β-H from C3 (Saytzeff: H from C3 gives more substituted but-2-ene vs H from C1 giving but-1-ene)
• Rate law: Rate = k[2-bromobutane][KOH] (bimolecular, concerted)
• Major product B: but-2-ene SMILES: CC=CC (internal, disubstituted) > but-1-ene (terminal, monosubstituted) by Saytzeff
• Ozonolysis of but-2-ene: $CH_{3}$CH=CH$CH_{3}$ + $O_{3}$/Zn → 2 $CH_{3}CHO$ (acetaldehyde)
• Answer: A = 2-bromobutane; B = but-2-ene (Saytzeff product)