Aldehydes and Ketones Questions and Answers

Answer: D

Fehling's reagent is a mixture of aqueous \(CuSO_{4}\) and alkaline sodium potassium tartrate (printed 'tatarate').

Fehling's solution A is aqueous copper(II) sulphate; Fehling's solution B is alkaline sodium potassium tartrate (Rochelle salt). The tartrate complexes the \(Cu^{2+}\) and keeps it in solution in the alkaline mixture. Aldehydes reduce it to a red-brown precipitate of \(Cu_{2}O\).

Ammoniacal silver nitrate is Tollens' reagent, and 2,4-DNP is a test for carbonyl groups in general, not part of Fehling's reagent. KOH with tartrate alone lacks the copper.

Answer: E

The reaction is a cross aldol condensation.

Benzaldehyde has no α-hydrogen, while acetophenone has three. In dilute NaOH, the enolate of acetophenone adds to benzaldehyde and the product loses water to give benzalacetophenone (chalcone). An aldol condensation between two different carbonyl compounds is called a cross aldol condensation (the Claisen-Schmidt reaction).

'Aldol condensation' is the general family, but the specific name for two different partners is cross aldol. Cannizzaro is a disproportionation of aldehydes without α-hydrogen, and Wolff-Kishner and Clemmensen are reductions of C=O to CH₂.

Answer: C

Acetone is converted into 2-methylpropan-2-ol by \(\mathrm{CH_{3}MgI / H_{2}O}\).

A Grignard reagent adds an alkyl group to the carbonyl carbon. Methylmagnesium iodide adds a methyl group to acetone, and hydrolysis of the magnesium alkoxide gives a tertiary alcohol with three methyl groups on the carbinol carbon:

\[ \begin{aligned} &\mathrm{(CH_{3})_{2}CO + CH_{3}MgI} \\ &\mathrm{\rightarrow (CH_{3})_{3}COMgI} \\ &\mathrm{\xrightarrow{H_{2}O} (CH_{3})_{3}COH} \end{aligned} \]

\(\mathrm{LiAlH_{4}}\), \(\mathrm{NaBH_{4}}\) and \(\mathrm{Pd / H_{2}}\) only add hydrogen, reducing acetone to the secondary alcohol propan-2-ol; no new carbon is added. \(\mathrm{B_{2}H_{6} / H_{2}O_{2}, NaOH}\) is hydroboration-oxidation, a reaction of alkenes, not ketones.

Answer: B

The Etard reaction uses \(\mathrm{CrO_{2}Cl_{2} / CS_{2}}\), then \(\mathrm{H_{3}O^{+}}\).

In the Etard reaction toluene, or another methylarene, is oxidised by chromyl chloride in carbon disulphide (or \(\mathrm{CCl_{4}}\)). A brown chromium complex forms, and its hydrolysis gives benzaldehyde. The oxidation stops at the aldehyde instead of going on to benzoic acid.

The other reagents are other named routes to aldehydes. Side-chain chlorination with \(\mathrm{Cl_{2}}\) and light followed by hydrolysis goes through benzal chloride. CO and HCl with anhydrous \(\mathrm{AlCl_{3}}\) and CuCl is the Gattermann-Koch reaction. \(\mathrm{SnCl_{2}}\) and HCl reduce nitriles in the Stephen reaction, and DIBAL-H reduces nitriles and esters to aldehydes.

Answer: E

The conversion is the Rosenmund reaction.

In the Rosenmund reduction an acid chloride is hydrogenated over palladium on barium sulphate, poisoned with sulphur or quinoline so that the reduction stops at the aldehyde: C₆H₅COCl + H₂ → C₆H₅CHO + HCl.

The Etard reaction oxidises toluene to benzaldehyde with chromyl chloride; the Stephen reaction reduces a nitrile; the Gattermann-Koch reaction formylates benzene with CO and HCl; and the Gattermann reaction converts a diazonium salt using copper powder. None starts from benzoyl chloride.

Answer: B

The reaction is the Clemmensen reaction (Clemmensen reduction).

Zinc amalgam and concentrated hydrochloric acid reduce the C=O group of an aldehyde or ketone to CH₂, so propanone gives propane.

The Wolff-Kishner reduction does the same conversion but with hydrazine and KOH in ethylene glycol. Hofmann's reaction converts amides to amines, Kolbe's reaction converts phenol to salicylic acid, and the Cannizzaro reaction disproportionates aldehydes without α-hydrogen.

Answer: C

They are distinguished by Iodine and alkali.

Acetophenone, C₆H₅COCH₃, is a methyl ketone, so it gives the iodoform test: with iodine and sodium hydroxide it forms a yellow precipitate of CHI₃. Benzophenone, C₆H₅COC₆H₅, has no CH₃CO group and gives no precipitate.

Fehling's and Tollens' reagents test for aldehydes, and both compounds are ketones, so neither reacts. Lucas reagent distinguishes alcohols. Aqueous CrO₃ oxidises alcohols and aldehydes, not these ketones.

Answer: A

The product is C₆H₅CH = CHCOC₆H₅ (benzalacetophenone, chalcone).

This is a Claisen-Schmidt (cross aldol) condensation. Benzaldehyde has no \(\alpha\)-hydrogen, so it acts only as the electrophile. NaOH removes an \(\alpha\)-hydrogen from the CH₃ of acetophenone, and the carbanion adds to the aldehyde carbonyl. The \(\beta\)-hydroxy ketone formed loses water, because the resulting double bond is conjugated with both the ring and the C=O:

C₆H₅CHO + CH₃COC₆H₅ gives C₆H₅CH=CHCOC₆H₅ + H₂O.

Stilbene (C₆H₅CH=CHC₆H₅) has lost the carbonyl, and benzoin (C₆H₅CH(OH)COC₆H₅) comes from two benzaldehydes with cyanide. Benzil and deoxybenzoin are not formed here.

Answer: A

The product is 2-Methylpropan-2-ol.

The methyl group of the Grignard reagent adds to the carbonyl carbon of propanone, and hydrolysis gives a tertiary alcohol:

\[ \begin{aligned} &(CH_{3})_{2}C=O + CH_{3}MgBr \\ &\xrightarrow{H_{2}O} (CH_{3})_{3}C-OH \end{aligned} \]

A ketone always gives a tertiary alcohol with a Grignard reagent, which rules out the primary alcohols (butan-1-ol, 2-methylpropan-1-ol) and the secondary butan-2-ol. 2-Methylpropane is an alkane, not an alcohol.

Answer: D

It is the Gatterman-Koch reaction.

Benzene treated with carbon monoxide and hydrogen chloride in the presence of anhydrous \(AlCl_{3}\) (often with CuCl) gives benzaldehyde. This formylation is the Gattermann-Koch reaction.

The Etard reaction oxidises toluene with chromyl chloride; the Stephen reaction reduces nitriles with \(SnCl_{2}/HCl\); the Hell-Volhard-Zelinsky reaction halogenates the alpha carbon of acids; the aldol reaction joins two carbonyl compounds.