Aldehydes and ketones, having atleast one methyl group linked to the carbonyl carbon atom (methyl ketones), are oxidized by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound is converted to haloform. This oxidation does not affect a carbon-carbon double bond, if present in a molecule.

Which of the following statements is true?

(A)\(CH_3CHO\), and \(CH_3CH_2CHO\) can be distinguished by Tollen’s test.

(B)  \(CH_3CH_2CHO\), and \(CH_3CCOCH_3\) can be distinguished by Fehling’s test.

(C)  \(CH_3CHO\), and \(C_6H_5CHO\) can be distinguished by Fehling’s test.

(D)\(HCOOH\), and \(CH_3CHO\) can be distinguished by by \(NaHCO_3\) text.

Choose the correct answer from the options given below:

(B), (C) only

The correct answer is (2) (B), (C) only.

(A) Ethanal (also known as acetaldehyde, \(CH_3CHO\)) and propanal (also known as propionaldehyde, \(CH_3CH_2CHO\)) cannot be easily distinguished by Tollen's test.
Tollen's test, also known as the silver mirror test, is used to detect the presence of aldehydes but does not differentiate between different aldehydes. In this test, a solution of Tollens' reagent, which is ammoniacal silver nitrate (\(Ag(NH_3)_2NO_3\)), is added to the aldehyde. If an aldehyde is present, it will reduce the silver ions in the Tollens' reagent to form a silver mirror on the inner surface of the test tube or container. The reaction involves the oxidation of the aldehyde to a carboxylic acid and the reduction of silver ions to silver metal.
The generic reaction for Tollen's test with an aldehyde can be represented as follows:
\[RCHO + 2Ag(NH_3)_2^+ + 3OH^- \rightarrow RCOOH + 2Ag + 4NH_3 + 2H_2O\]
Both ethanal and propanal are aldehydes, and they will both give a positive Tollen's test result by forming a silver mirror. Therefore, Tollen's test alone cannot be used to distinguish between ethanal and propanal. To differentiate between these two aldehydes, you would need to use other chemical tests or analytical methods that take advantage of their distinct properties or reactions with specific reagents.

(B)\(CH_3CH_2CHO\) reduces Fehling's solution to red ppt. of \(Cu_2O\) but \(CH_3CCOCH_3\) being ketone does not

\(\underset{Propanal}{CH_3CH_2CHO} + \underbrace{2Cu^{2+}+5OH^{-}}_{\text{Fehling's solution}}  \longrightarrow \underset{\text{Propanoate ion}}{CH_3CH_2COO^-} + \underset{\underset{(\text{Red ppt.})}{\text{Cuprous oxide}}}{Cu_2O↓} + 3H_2O\)

(C) Although \(CH_3CHO\) and \(C_6H_5CHO\) are both aldehydes yet aliphatic aldehyde reduces Fehling's solution but aromatic aldehydes do not. Thus, acetaldehyde gives a red ppt. of \(Cu_2O\) with Fehling's solution but benzaldehyde does not.

\(\underset{Acetaldehyde}{CH_3CHO} + \underbrace{2Cu^{2+}+5OH^{-}}_{\text{Fehling's solution}}  \longrightarrow \underset{\text{Acetate ion}}{CH_3COO^-} + \underset{\underset{(\text{Red ppt.})}{\text{Cuprous oxide}}}{Cu_2O↓} + 3H_2O\)

(D) The \(NaHCO_3\) (sodium bicarbonate) test is used to distinguish between aldehydes and ketones in organic chemistry. Aldehydes and ketones react differently with sodium bicarbonate, which helps in their differentiation. However, both formaldehyde and acetaldehyde are aldehydes, and they will show similar reactions with \(NaHCO_3\). Therefore, the \(NaHCO_3\) test alone cannot be used to distinguish between formaldehyde and acetaldehyde since both will exhibit similar behavior when treated with sodium bicarbonate.

(E) The sodium bicarbonate \((NaHCO_3)\) test for carboxylic acids cannot easily distinguish between formic acid (\(HCOOH\)) and acetic acid (\(CH_3COOH\)) because both acids react similarly with sodium bicarbonate. In this test, carboxylic acids generally react with sodium bicarbonate to produce effervescence (bubbling) due to the release of carbon dioxide gas. This reaction occurs because carboxylic acids can undergo a reaction with sodium bicarbonate to form a carboxylate salt, carbon dioxide, and water. The general reaction is represented as follows:
\[RCOOH + NaHCO3 \rightarrow RCOONa + CO_2 + H_2O\]
Both formic acid and acetic acid are carboxylic acids and will exhibit a similar reaction when treated with sodium bicarbonate. Therefore, relying solely on the \(NaHCO_3\) test is insufficient for distinguishing between formic acid and acetic acid.
To differentiate between these two acids, alternative chemical tests or analytical methods must be employed, taking advantage of their unique properties or reactions with specific reagents.