KETONES General methods of preparation, properties, reactions, and uses.: NEET MCQs with Answers

Explanation: The Grignard carbanion adds to the carbonyl carbon of acetone, which already bears two methyl groups. After hydrolysis the carbinol carbon carries three methyls and one –OH, i.e. 2-methylpropan-2-ol (tert-butyl alcohol).

Explanation: Methyllithium is a very strong base and removes an acidic α-hydrogen of cyclopentanone. This gives the resonance-stabilised enolate, the cyclopentanonyl anion, with methane as the by-product.

Explanation: Primary alcohols and aldehydes are oxidised by converting the terminal carbon into –COOH, so the carbon skeleton is retained. A ketone has no hydrogen on its carbonyl carbon and is oxidised only under vigorous conditions that break a C–C bond next to the carbonyl group. Butan-2-one therefore gives acids with fewer than four carbon atoms.

Explanation: The compound is a methyl ketone, so with sodium hypoiodite it undergoes the haloform reaction: the CH₃ group is removed as iodoform and the carbonyl carbon becomes a carboxylate. The C=C bond is left intact and acidification gives but-2-enoic acid. Hot permanganate and ozonolysis both cleave the C=C bond, and Clemmensen reduction converts C=O to CH₂.

Explanation: The haloform reaction needs a base. Hydroxide ion removes the hydrogens of the CH₃–CO– group one by one so that iodine replaces all three, and then attacks the carbonyl carbon to split off CI₃⁻, which becomes iodoform. Acid, a Lewis acid or a reducing metal cannot bring about this sequence.

Explanation: Mercury(II)-catalysed hydration of an alkyne follows Markovnikov's rule, so the –OH adds to the carbon carrying the aryl group. The enol formed, Ar–C(OH)=CH₂, tautomerises at once to the keto form. The product is the methyl ketone CH₃–C₆H₄–CO–CH₃, 1-(4-methylphenyl)ethan-1-one.

Explanation: Oxidation to a carbonyl group needs a hydrogen on the carbon carrying the –OH. Butan-2-ol is a secondary alcohol with one such hydrogen and is oxidised to butanone. 2-Methylpropan-2-ol is tertiary and has no such hydrogen, while butan-1-ol and phenylmethanol are primary and give an aldehyde and then an acid.

Explanation: A secondary alcohol has one hydrogen on the carbon bearing the –OH group, and oxidation removes it to form a C=O flanked by two alkyl groups, which is a ketone. Strong oxidation of a primary alcohol gives a carboxylic acid. Hydrolysis of an ester gives an acid and an alcohol, and an acyl chloride with an alcohol gives an ester.

Explanation: The iodoform test is given by compounds containing CH₃–CO– or CH₃–CH(OH)– groups, since the latter are first oxidised to methyl ketones. Butan-2-ol, CH₃CH(OH)CH₂CH₃, has this grouping and gives yellow iodoform. Pentan-3-one has ethyl groups on both sides of the carbonyl, propan-1-ol has its –OH on a terminal CH₂, and benzophenone has no methyl group.

Explanation: One molecule of methanol first adds to the carbonyl group to give a hemiketal. With methanol in excess and dry acid present, the –OH of the hemiketal is protonated and lost as water, and a second methanol molecule takes its place. The final product is the ketal CH₃CH₂C(OCH₃)₂CH₃.

Explanation: Sodium borohydride is a mild hydride donor that reduces aldehydes and ketones to alcohols but does not normally reduce esters. Only the ketone carbonyl is therefore converted, to a secondary alcohol, while the ester group remains. Reduction of the ester would need a stronger reagent such as LiAlH₄, and conversion of C=O to CH₂ needs Clemmensen or Wolff–Kishner conditions.

Explanation: Ketones react with ethylene glycol in the presence of dry HCl to give cyclic products called ethylene glycol ketals. Dry HCl protonates the carbonyl oxygen, which raises the electrophilicity of the carbonyl carbon and helps the attack by the glycol. Acetals and ketals are hydrolysed by aqueous mineral acids to give back the aldehyde or ketone.

Explanation: A methyl ketone is oxidised by sodium hypohalite to the sodium salt of a carboxylic acid with one carbon less, and the methyl group becomes haloform. This oxidation does not affect a carbon-carbon double bond present in the molecule. The product is therefore CH₃CH=CH–COONa together with CHCl₃.

Explanation: The Grignard reagent adds to the C≡N group to give an imine salt, which on hydrolysis gives a ketone. The nitrile carbon becomes the C=O carbon, with the nitrile's alkyl group on one side and the Grignard alkyl group on the other. Pentan-3-one, CH₃CH₂COCH₂CH₃, has an ethyl group on each side, so it needs CH₃CH₂CN and CH₃CH₂MgBr. Ethanenitrile with propylmagnesium bromide and butanenitrile with methylmagnesium bromide both give pentan-2-one, while propanenitrile with methylmagnesium bromide gives butan-2-one.

Explanation: A dialkylcadmium replaces the chlorine of an acyl chloride by an alkyl group and the reaction stops at the ketone: 2 CH₃CH₂COCl + (CH₃)₂Cd → 2 CH₃CH₂COCH₃ + CdCl₂. Excess Grignard reagent attacks the ketone again and gives a tertiary alcohol. Hydrogen over Pd–BaSO₄ gives propanal, and LiAlH₄ gives propan-1-ol.