12 Chemistry Unit-18 Carbonyl Compounds

Tollens's reagent is ammoniacal silver nitrate. This reagent is used to test for aldehydes. When an aldehyde is present, it will react with Tollens's reagent to form a silver mirror on the inside of the test tube. This reaction is known as the Tollens's test and is used as a qualitative test for the presence of aldehydes.

Schiffs reagent gives a pink color with acetaldehyde.

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Methyl ketones are usually characterized by the iodoform test. This test is used to identify compounds that contain a methyl group attached to a carbonyl group. In the iodoform test, the compound is treated with iodine and a base, resulting in the formation of a yellow precipitate of iodoform. Methyl ketones, such as acetone, undergo this reaction and produce iodoform, which confirms the presence of a methyl ketone. Therefore, option 2 is the correct answer.

Acetaldehyde is an aldehyde compound. When heated with Fehling solution, it undergoes oxidation and reduces the Cu2+ ions present in the Fehling solution to Cu+ ions. The Cu+ ions then react with the excess OH- ions from the Fehling solution to form a reddish-brown precipitate of Cu2O (copper(I) oxide). Therefore, the correct answer is option 1, Cu2O.

Formaldehyde polymerizes to give paraformaldehyde.

The correct answer for this question is option 4, benzaldehyde. Hydrogenation of benzoyl chloride in the presence of Pd on BaSO4 leads to the reduction of the carbonyl group in benzoyl chloride to an aldehyde group, resulting in the formation of benzaldehyde.

Tertiary butyl alcohol is obtained by the action of methyl magnesium iodide on acetone (CH3COCH3). This reaction is known as a Grignard reaction, where the methyl group from the methyl magnesium iodide adds to the carbonyl carbon of acetone, resulting in the formation of tertiary butyl alcohol.

The correct answer is option c) aldehydes and ketones undergo nucleophilic substitution. This statement is incorrect because aldehydes and ketones undergo nucleophilic addition reactions rather than nucleophilic substitution reactions. In nucleophilic addition, the nucleophile attacks the carbonyl carbon and forms a new bond, whereas in nucleophilic substitution, the nucleophile replaces a leaving group. Therefore, option c) is the wrong statement.

Aldol is a compound that contains both an aldehyde group and an alcohol group. The correct answer, option 3, 3-hydroxy butanal, fits this description as it has both a hydroxy group (alcohol) and an aldehyde group. Option 1, 2-hydroxy butanol, is incorrect because it does not contain an aldehyde group. Option 2, 3-hydroxy butanol, is also incorrect because it does not contain an aldehyde group. Option 4, 2-hydroxy butanal, is incorrect because it does not contain an aldehyde group. Therefore, the correct answer is option 3, 3-hydroxy butanal.

The correct answer is option 2, CH3CHO. A cynohydrin is a compound that contains a hydroxyl group (-OH) and a cyanide group (-CN) bonded to the same carbon atom. When this compound is hydrolyzed, it breaks down into lactic acid. Among the given options, only CH3CHO (acetaldehyde) has the potential to form a cynohydrin with a hydroxyl group and a cyanide group bonded to the same carbon atom. Therefore, CH3CHO is the compound X.

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Acetaldehyde does not undergo the Cannizzaro reaction because it does not have an alpha hydrogen atom. The Cannizzaro reaction is a disproportionation reaction in which an aldehyde with an alpha hydrogen atom is oxidized to a carboxylic acid and reduced to an alcohol. Acetaldehyde does not have an alpha hydrogen atom, so it cannot undergo this reaction.

The chain isomer of 2-methyl-poropanal is butanal. This is because the compound 2-methyl-poropanal has a methyl group attached to the second carbon in the chain. In the chain isomer butanal, there is no methyl group attached to the second carbon, instead, there is a hydrogen atom. Therefore, butanal is the correct chain isomer of 2-methyl-poropanal.

The formation of a cyanohydrin from a ketone involves the addition of a nucleophile (cyanide ion) to the carbonyl carbon of the ketone. This process is known as nucleophilic addition.

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The correct IUPAC name for the given compound is 4-methylpent-3-en-2-one. This name indicates that the compound is a ketone (ending in -one), with a methyl group (CH3) attached to the fourth carbon atom of the pentane chain. The double bond is located between the third and fourth carbon atoms (hence the 3-en part), and the second carbon atom is attached to an additional methyl group.

Benzophenone does not give the iodoform test because it does not have a methyl group attached to a carbonyl carbon. The iodoform test is a chemical reaction that is used to detect the presence of a methyl ketone or a compound that can be oxidized to a methyl ketone. In this test, the compound is treated with iodine and sodium hydroxide, and if a methyl ketone is present, yellow precipitate of iodoform is formed. Benzophenone does not contain a methyl group attached to a carbonyl carbon, so it does not undergo this reaction and does not give the iodoform test.

The compound used in the preparation of the tranquilizer, sulphonal is acetone.

Isopropyl alcohol (also known as 2-propanol) reacts with air over a silver catalyst at 520 K to produce acetone. Therefore, option 3, acetone, is the correct answer.

In the reduction of acetaldehyde using LiAlH4, the hydride ion acts as a nucleophile. This is because nucleophiles are electron-rich species that donate a pair of electrons to form a new bond with an electron-deficient atom, typically a carbon atom in this case. The hydride ion, H-, is a strong nucleophile that attacks the carbonyl carbon in acetaldehyde, forming a new carbon-hydrogen bond and converting the carbonyl group into a primary alcohol.

Bakelite is a type of plastic that is created through the reaction between phenol and methanal. This reaction forms a polymer known as phenol formaldehyde resin, which is then molded and hardened to create Bakelite. The other options, such as formaldehyde and NaOH, aniline and NaOH, and phenol and chloroform, do not result in the formation of Bakelite.

Benzaldehyde does not reduce Fehling solution because it is an aromatic aldehyde and does not have an active hydrogen atom on the carbonyl group. Fehling solution is a test for the presence of reducing sugars, which are compounds that can donate electrons to the Fehling solution and cause it to be reduced. Aldehydes with active hydrogen atoms on the carbonyl group, such as formaldehyde, acetaldehyde, and propionaldehyde, can donate electrons and reduce Fehling solution. However, benzaldehyde does not have an active hydrogen atom on the carbonyl group, so it cannot donate electrons and does not reduce Fehling solution.

2-butanol is the correct answer because it can be oxidized to give ethyl methyl ketone. Oxidation of 2-butanol involves the loss of two hydrogen atoms from the carbon adjacent to the hydroxyl group, resulting in the formation of a carbonyl group. This carbonyl group is characteristic of ketones, and in this case, it specifically forms ethyl methyl ketone.

The compound in option 4, CH3-CH2-CH2-OH, is a primary alcohol. On strong oxidation, primary alcohols are converted into carboxylic acids. Therefore, on strong oxidation, CH3-CH2-CH2-OH will give propionic acid.