501. Which compound gives a positive Tollens’ test?
ⓐ. \(CH_3COCH_3\)
ⓑ. \(CH_3CH_2OH\)
ⓒ. \(CH_3CHO\)
ⓓ. \(CH_3COOH\)
Correct Answer: \(CH_3CHO\)
Explanation: \(CH_3CHO\) is ethanal, an aldehyde. Aldehydes give Tollens’ test because they are readily oxidised to carboxylate ions. \(CH_3COCH_3\) is a ketone and usually does not give Tollens’ test. \(CH_3CH_2OH\) is an alcohol and \(CH_3COOH\) is already a carboxylic acid.
502. Which pair can be distinguished using Tollens’ reagent?
ⓐ. Ethanol and propanol
ⓑ. Ethanal and propanone
ⓒ. Ethanoic acid and propanoic acid
ⓓ. Benzene and toluene
Correct Answer: Ethanal and propanone
Explanation: Ethanal is an aldehyde and gives a positive Tollens’ test. Propanone is a simple ketone and usually does not reduce Tollens’ reagent. Therefore the silver mirror appears with ethanal but not with propanone. This difference allows the pair to be distinguished using Tollens’ reagent.
503. Which reagent gives a brick-red precipitate with many aliphatic aldehydes?
ⓐ. Fehling’s solution
ⓑ. \(2,4\)-DNP
ⓒ. \(NaBH_4\)
ⓓ. Ammoniacal silver nitrate
Correct Answer: Fehling’s solution
Explanation: Fehling’s solution is an alkaline copper reagent. Many aliphatic aldehydes reduce \(Cu^{2+}\) to \(Cu_2O\), which appears as a brick-red precipitate. The aldehyde itself is oxidised to a carboxylate ion. Simple ketones generally do not give this test.
504. Which observation indicates a positive Fehling’s test?
ⓐ. Silver mirror deposit
ⓑ. Violet colour
ⓒ. White fumes
ⓓ. Brick-red precipitate
Correct Answer: Brick-red precipitate
Explanation: Fehling’s solution contains \(Cu^{2+}\) ions in alkaline medium. Aldehydes reduce \(Cu^{2+}\) to cuprous oxide, \(Cu_2O\). \(Cu_2O\) is observed as a brick-red precipitate. A silver mirror is associated with Tollens’ reagent, not Fehling’s solution.
505. Which compound is most likely to give Fehling’s test?
ⓐ. Benzaldehyde
ⓑ. Propanone
ⓒ. Ethanal
ⓓ. Acetophenone
Correct Answer: Ethanal
Explanation: Ethanal is an aliphatic aldehyde and generally reduces Fehling’s solution to give brick-red \(Cu_2O\). Propanone and acetophenone are ketones and usually do not give Fehling’s test. Benzaldehyde is an aromatic aldehyde and generally does not reduce Fehling’s solution under the usual test conditions. The test is therefore especially useful for many aliphatic aldehydes.
506. Which statement correctly compares Tollens’ and Fehling’s tests?
ⓐ. Both are used only for carboxylic acids.
ⓑ. Both involve oxidation of aldehydes under mild conditions.
ⓒ. Fehling’s test gives silver mirror, while Tollens’ gives brick-red \(Cu_2O\).
ⓓ. Both are reduction tests for converting aldehydes into alcohols.
Correct Answer: Both involve oxidation of aldehydes under mild conditions.
Explanation: Tollens’ and Fehling’s tests both depend on the easy oxidation of aldehydes. In Tollens’ test, \(Ag^+\) is reduced to metallic silver. In Fehling’s test, \(Cu^{2+}\) is reduced to \(Cu_2O\), a brick-red precipitate. The aldehyde is oxidised in both cases, usually to a carboxylate under alkaline conditions.
507. Which inorganic product gives the brick-red precipitate in Fehling’s test?
ⓐ. \(CuCl_2\)
ⓑ. \(Ag\)
ⓒ. \(CuO\)
ⓓ. \(Cu_2O\)
Correct Answer: \(Cu_2O\)
Explanation: Fehling’s solution contains \(Cu^{2+}\) ions in alkaline medium. Many aliphatic aldehydes reduce \(Cu^{2+}\) to \(Cu_2O\). Cuprous oxide, \(Cu_2O\), appears as a brick-red precipitate. Metallic silver is formed in Tollens’ test, not in Fehling’s test.
508. Which compound is generally not expected to give Fehling’s test under usual conditions?
ⓐ. Ethanal
ⓑ. Benzaldehyde
ⓒ. Propanal
ⓓ. Cinnamaldehyde
Correct Answer: Benzaldehyde
Explanation: Fehling’s test is readily given by many aliphatic aldehydes such as methanal, ethanal, and propanal. Benzaldehyde is an aromatic aldehyde and generally does not reduce Fehling’s solution under the usual test conditions. This difference helps distinguish many aliphatic aldehydes from aromatic aldehydes. Tollens’ reagent, however, is usually reduced by aldehydes including benzaldehyde.
509. In Fehling’s test, what happens to the aldehyde?
ⓐ. It is reduced to a primary alcohol.
ⓑ. It is converted into a ketone.
ⓒ. It is oxidised to carboxylate ion.
ⓓ. It forms an oxime derivative.
Correct Answer: It is oxidised to carboxylate ion.
Explanation: Fehling’s solution is alkaline, so the aldehyde is oxidised to a carboxylate ion rather than directly written only as the free acid. At the same time, \(Cu^{2+}\) is reduced to \(Cu_2O\). The visible evidence is the brick-red precipitate of \(Cu_2O\). This redox change is the basis of the test.
510. Which pair gives positive Tollens’ test but only the first member usually gives positive Fehling’s test?
ⓐ. Propanone and ethanal
ⓑ. Acetone and acetophenone
ⓒ. Ethanoic acid and benzaldehyde
ⓓ. Ethanal and benzaldehyde
Correct Answer: Ethanal and benzaldehyde
Explanation: Ethanal is an aliphatic aldehyde and gives both Tollens’ and Fehling’s tests. Benzaldehyde is an aromatic aldehyde and usually gives Tollens’ test but not Fehling’s test under ordinary conditions. Tollens’ reagent is a more general mild oxidising test for aldehydes. Fehling’s solution is more characteristic for many aliphatic aldehydes.
511. Which test is closely related to Fehling’s test in detecting reducing behaviour of aldehydes?
ⓐ. Iodoform test
ⓑ. Sodium bicarbonate test
ⓒ. Ester test
ⓓ. Benedict’s test
Correct Answer: Benedict’s test
Explanation: Benedict’s test is related to Fehling’s test because both involve reduction of alkaline \(Cu^{2+}\) species. Reducing aldehydes can convert \(Cu^{2+}\) into \(Cu_2O\), giving a reddish precipitate. The exact complexing agents differ, but the redox idea is similar. It is not the same as iodoform or bicarbonate testing.
512. Which statement about Benedict-type recognition is most suitable?
ⓐ. It is based on formation of silver mirror only.
ⓑ. It detects all ketones by forming \(CO_2\).
ⓒ. It identifies carboxylic acids by effervescence.
ⓓ. It involves reduction of \(Cu^{2+}\) by aldehydes.
Correct Answer: It involves reduction of \(Cu^{2+}\) by aldehydes.
Explanation: Benedict-type tests rely on the reducing nature of suitable aldehydes. The aldehyde is oxidised while \(Cu^{2+}\) is reduced to \(Cu_2O\). This is conceptually similar to Fehling’s test. Carboxylic acid effervescence with bicarbonate and silver mirror formation belong to different tests.
513. Which reagent can oxidise aldehydes to carboxylic acids?
ⓐ. \(KMnO_4\)
ⓑ. \(NaBH_4\)
ⓒ. \(NH_2OH\)
ⓓ. \(NaHSO_3\)
Correct Answer: \(KMnO_4\)
Explanation: \(KMnO_4\) is an oxidising agent and can oxidise aldehydes to carboxylic acids. During this conversion, the \(CHO\) group changes into \(COOH\). \(NaBH_4\) reduces aldehydes to alcohols, while \(NH_2OH\) forms oximes. \(NaHSO_3\) gives bisulfite addition compounds.
514. Which compound is most resistant to mild oxidation among the following?
ⓐ. Methanal
ⓑ. Ethanal
ⓒ. Propanone
ⓓ. Benzaldehyde
Correct Answer: Propanone
Explanation: Aldehydes are readily oxidised to carboxylic acids by mild oxidants. Propanone is a ketone, and simple ketones usually resist mild oxidation. Oxidation of ketones generally requires stronger conditions and may involve carbon-carbon bond cleavage. This difference is a major chemical distinction between aldehydes and ketones.
515. Which statement correctly compares aldehydes and ketones toward mild oxidising agents?
ⓐ. Ketones oxidise more readily than aldehydes.
ⓑ. Aldehydes and ketones both give silver mirror equally.
ⓒ. Aldehydes oxidise easily; ketones resist mild oxidation.
ⓓ. Ketones form carboxylate ions faster than aldehydes in Tollens’ test.
Correct Answer: Aldehydes oxidise easily; ketones resist mild oxidation.
Explanation: Aldehydes contain a carbonyl carbon attached to hydrogen and are easily oxidised to acids or carboxylates. Ketones have the carbonyl carbon attached to two carbon groups. This structure makes simple ketones much less susceptible to mild oxidation. Therefore Tollens’ and Fehling’s tests are generally positive for aldehydes but negative for ordinary ketones.
516. Which equation represents oxidation of propanal by a mild oxidant?
ⓐ. \[CH_3CH_2CHO + [O] \rightarrow CH_3CH_2COOH\]
ⓑ. \[CH_3CH_2CHO + 2[H] \rightarrow CH_3CH_2CH_2OH\]
ⓒ. \[CH_3CH_2CHO + HCN \rightarrow CH_3CH_2CH(OH)CN\]
ⓓ. \[CH_3CH_2CHO + NH_2OH \rightarrow CH_3CH_2CH=NOH + H_2O\]
Correct Answer: \[CH_3CH_2CHO + [O] \rightarrow CH_3CH_2COOH\]
Explanation: \(\textbf{Starting compound:}\)
Propanal is \(CH_3CH_2CHO\).
\(\textbf{Reaction type:}\)
Aldehydes are readily oxidised.
\(\textbf{Functional-group change:}\)
The \(CHO\) group changes into \(COOH\).
\(\textbf{Product:}\)
The product is \(CH_3CH_2COOH\), propanoic acid.
\(\textbf{Final equation:}\)
\[CH_3CH_2CHO + [O] \rightarrow CH_3CH_2COOH\]
517. Which test is most suitable for identifying a methyl ketone group?
ⓐ. Tollens’ test
ⓑ. Fehling’s test
ⓒ. Iodoform test
ⓓ. Sodium bicarbonate test
Correct Answer: Iodoform test
Explanation: The iodoform test is characteristic of compounds containing the \(CH_3CO-\) group or compounds that can be oxidised to such a group under the test conditions. Methyl ketones give a yellow precipitate of iodoform, \(CHI_3\), with iodine and alkali. Tollens’ and Fehling’s tests mainly detect aldehydes. Sodium bicarbonate test is used for carboxylic acids.
518. Which reagent set is used in the iodoform test?
ⓐ. \(AgNO_3/NH_3\)
ⓑ. \(I_2/NaOH\)
ⓒ. \(Cu^{2+}\) alkaline solution
ⓓ. \(NaHCO_3\)
Correct Answer: \(I_2/NaOH\)
Explanation: The iodoform test is carried out using iodine in alkali, commonly written as \(I_2/NaOH\). A positive test gives a yellow precipitate of iodoform, \(CHI_3\). The reaction is especially important for methyl ketones and a few related compounds. Ammoniacal silver nitrate belongs to Tollens’ test.
519. Which observation indicates a positive iodoform test?
ⓐ. Yellow precipitate of \(CHI_3\)
ⓑ. Silver mirror on the test tube
ⓒ. Brick-red precipitate of \(Cu_2O\)
ⓓ. Brisk effervescence of \(CO_2\)
Correct Answer: Yellow precipitate of \(CHI_3\)
Explanation: A positive iodoform test gives a yellow precipitate of iodoform, \(CHI_3\). This observation is associated with methyl ketones and compounds that can form the \(CH_3CO-\) group under the reaction conditions. Silver mirror formation belongs to Tollens’ test. Brick-red \(Cu_2O\) belongs to Fehling’s test.
520. Which compound gives a positive iodoform test because it is a methyl ketone?
ⓐ. \(CH_3CH_2CHO\)
ⓑ. \(CH_3COCH_3\)
ⓒ. \(C_6H_5CHO\)
ⓓ. \(CH_3CH_2COOH\)
Correct Answer: \(CH_3COCH_3\)
Explanation: \(CH_3COCH_3\), propanone, contains the methyl ketone group \(CH_3CO-\). Methyl ketones give the iodoform test with \(I_2/NaOH\). The yellow precipitate formed is \(CHI_3\). Propanal, benzaldehyde, and propanoic acid do not contain the methyl ketone group.
