Which of the following will not give yellow precipitate when treated with \(I_2\) in presence of \(NaOH\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → CH3–OH
The yellow precipitate is iodoform, CHI3. \( \ce{I2} \) and NaOH give sodium hypoiodite, which converts a CH3CO– group (methyl ketones and ethanal) into iodoform. Hypoiodite is also an oxidising agent, so an alcohol with the CH3CH(OH)– group is first oxidised to a methyl ketone and then gives the test. With propan-2-ol, for example:
\( \displaystyle \ce{CH3CH(OH)CH3 + NaOI -> CH3COCH3 + NaI + H2O} \)
\( \displaystyle \ce{CH3COCH3 + 3NaOI -> CHI3 v + CH3COONa + 2NaOH} \)
Option (1) CH3–CH(OH)–CH3 — Gives a yellow precipitate (iodoform). Propan-2-ol has the \( \ce{CH3CH(OH)} \)– group and is oxidised to propanone, as shown above.
Option (2) CH3–CH(OH)–CH2–CH3 — Gives a yellow precipitate (iodoform). Butan-2-ol is oxidised to butanone, \( \ce{CH3COC2H5} \), a methyl ketone.
Option (3) CH3–OH — Does not give a yellow precipitate. Methanol has no carbon next to the carbinol carbon; it is oxidised only to methanal, HCHO, which has no \( \ce{CH3CO} \)– group.
Option (4) CH3–CH(OH)–CH(CH3)–CH3 — Gives a yellow precipitate (iodoform). 3-Methylbutan-2-ol carries \( \ce{CH3CH(OH)} \)– and is oxidised to 3-methylbutan-2-one, \( \ce{CH3COCH(CH3)2} \).
NCERT (2026-27 edition):
Oxidation of Methyl Ketones by Haloform Reaction – “Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (methyl ketones) are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound. The methyl group is converted to haloform.”