Question

Ethyl alcohol gives a yellow precipitate with iodine II. A fruit smell is observed when ethyl alcohol is heated with sodium acetate. The correct option regarding is:
(A) Both I and II are true
(B) Both I and II are false
(C) I is true but II is false
(D) I is false but II is true

Answer 1

Hint :The Haloform reaction mechanism is the one that starts with the halogen disproportionation with the presence of the hydroxide ion. This gives either the hypohalite or the halide. Then, the hydroxide abstracts the proton by producing enolate.

Complete Step By Step Answer:
To solve the given question, we should have information about haloform tests and esters. Haloform test is a reaction performed by ethanol, methyl, ketones and secondary alcohol oxidizable to methyl ketone and acetaldehyde. Among the group, ethanol and acetaldehyde are the only alcohol and aldehyde that gives a haloform test. The reagents of the haloform test also include halogen and sodium hydroxide. Hypohalite can also react with any of the present methyl ketones, forming a haloform, eventually.
 $ {{C}_{2}}{{H}_{5}}OH\xrightarrow[NaOH]{{{I}_{2}}}H3C-C(=O)-ONa+\underset{\begin{smallmatrix}
 Yellow \\
 Coloured
\end{smallmatrix}}{\mathop{C{{H}_{3}}}}\, $
 $ {{H}_{3}}C-{{H}_{2}}C-OH+NaO-C\left( =O \right)-C{{H}_{3}}\xrightarrow{{}}No~~Reaction $
This reaction is the type of nucleophilic substitution. Aldehydes or ketones having at least methyl group attached to carbonyl carbon atom (methyl ketones) are oxidized by sodium hypohalite to sodium salts of corresponding carboxylic acid having one carbon atom less than that of carbonyl compound. Methyl group is converted to haloform. Aldehydes or ketones having methyl ketone groups give positive tests. Ester is a carboxylic acid derivative in which an alcoholic group is replaced by an alkoxy group. They have low boiling points and are considered to be carboxylic acid as they do not form an intermolecular H-bond. Sodium salt of carboxylic acid with one less carbon atom is also produced.
Therefore, the correct answer is option B.

Note :
Remember that the acetaldehyde is the only aldehyde that gives a positive haloform test. Ethyl alcohol is the only primary alcohol that gives a positive haloform test. When methyl ketones are treated with sodium hydroxide and iodine, yellow precipitate of iodoform is produced which indicates positive haloform test.