Aldehydes and Ketones Questions and Answers

Answer: A

The product is (A), bicyclo[3.3.1]nonan-9-one.

At −78 °C a strong base removes the more accessible proton, on C6, the unsubstituted α-carbon, giving the kinetic enolate. That enolate carbon attacks the carbon bearing the tosylate at the end of the three-carbon chain on C2 in an intramolecular S_N2. The new ring contains C6, C1, C2 and the three chain carbons: a six-membered ring, which closes easily. C2 and C6 become bridgeheads joined by the carbonyl carbon, the original C3-C5 and the new chain: bicyclo[3.3.1]nonan-9-one.

Alkylation from the more substituted enolate at C2 would have to close a four-membered ring, giving the spiro[3.5] ketone (B), which is much slower. (C) would be O-alkylation, and (D) does not correspond to any bond this substrate can form.

Answer: C

The product is the fused furan ketone of option C, 6,7-dihydrobenzofuran-4(5H)-one.

NaOH makes the enolate of the 1,3-dione, which is alkylated at the central carbon (C-2) by bromoacetaldehyde, giving a 2-(2-oxoethyl) dione. The enol oxygen of one carbonyl then adds to the aldehyde, and loss of water closes the aromatic furan ring. The carbon that was C-2 is now the furan carbon next to the CH, and the remaining ketone stays on the ring carbon adjacent to it.

Options A, B and D place the C=O away from the ring carbon that bears the furan CH, which does not match this mechanism.

Answer: A

The answer is option A: I is 2-ethyl-2-methylcyclohexanone and II is 2-ethyl-6-methylcyclohexanone.

KH is a small, reversible base at room temperature, so the more substituted, thermodynamic enolate forms and is alkylated at the more substituted carbon, giving the quaternary 2-ethyl-2-methyl product (I). LDA at −78°C is bulky and irreversible, so it removes the proton from the less hindered side and gives the kinetic enolate; ethylation at C-6 gives 2-ethyl-6-methylcyclohexanone (II).

Option B reverses the two products, C proposes an enol ether for I (O-alkylation is not the major path here) and D claims both conditions give the same compound, which ignores the kinetic versus thermodynamic control the question is about.

Answer: A

The reagent sequence is thexylborane; CO; H₂O₂/NaOH.

An unsymmetrical ketone from two different alkenes is made by Brown's thexylborane method. Thexylborane (ThxBH₂) hydroborates the first alkene, then the second, giving a mixed trialkylborane ThxBRR'. Carbon monoxide inserts and the two primary alkyl groups R and R' migrate from boron to carbon; alkaline hydrogen peroxide then oxidises the boron intermediate to the ketone RCOR'. The thexyl group does not migrate, so the ketone carries exactly the two groups from the alkenes.

The printed product shows (CH₂)₅ on both sides, but anti-Markovnikov hydroboration of 1-hexene gives a hexyl group and of 1-butene a butyl group, so the actual product is hexyl butyl ketone; the method is the same. Without CO (option B) the borane is only oxidised to alcohols. 9-BBN has a single B-H and can add only one alkene, and triphenylborane has no B-H and cannot hydroborate at all.

Answer: C

The product is the ring-contracted unsaturated ester in option C.

This is a Favorskii rearrangement of an α,α'-dibromo ketone. Ethoxide removes the acidic proton on the α' carbon, and the enolate closes onto the other α carbon, expelling bromide to form a cyclopropanone. Ethoxide adds to the cyclopropanone C=O and the ring opens, so the ring shrinks by one carbon and carries a COOEt group. With two bromines available, the second one is lost as HBr, which creates a C=C next to the ester, giving the α,β-unsaturated ester drawn in C.

A and B are the saturated esters, which would arise from a monobromo ketone (one bromine only). D is simply the ketone, with both bromines lost and no rearrangement, which is not what ethoxide does to a dibromo ketone.