Pick the reactions which can be used to distinguish between n-propyl alcohol and isopropyl alcohol.
(A) Haloform reaction
(B) Dehydration
(C) Dehydrogenation with Cu/573 K
(D) Oxidation with \( CrO_3 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
n-Propyl alcohol, \( \ce{CH3CH2CH2OH} \), is a primary alcohol and isopropyl alcohol, \( \ce{CH3CH(OH)CH3} \), a secondary alcohol. A reaction distinguishes them only if the two give different, observable results.
(A) Haloform reaction — Correct. Isopropyl alcohol has the \( \ce{CH3CH(OH)-} \) group; iodine and NaOH oxidise it to acetone and then give a yellow precipitate of iodoform. n-Propyl alcohol does not.
\( \displaystyle \ce{CH3CH(OH)CH3 + 4I2 + 6NaOH -> CHI3 v + CH3COONa + 5NaI + 5H2O} \)
(B) Dehydration — Incorrect. Both alcohols lose water to give the same alkene, propene, so the products cannot tell them apart.
\( \displaystyle \ce{CH3CH2CH2OH} \xrightarrow{\text{conc. }\ce{H2SO4},\ \Delta} \ce{CH3CH=CH2 + H2O} \)
\( \displaystyle \ce{CH3CH(OH)CH3} \xrightarrow{\text{conc. }\ce{H2SO4},\ \Delta} \ce{CH3CH=CH2 + H2O} \)
(C) Dehydrogenation with Cu/573 K — Correct. The primary alcohol gives an aldehyde (propanal), the secondary alcohol a ketone (propanone); only propanal gives Tollens’ and Fehling’s tests.
\( \displaystyle \ce{CH3CH2CH2OH} \xrightarrow{\ce{Cu},\ 573\ \text{K}} \ce{CH3CH2CHO + H2} \)
\( \displaystyle \ce{CH3CH(OH)CH3} \xrightarrow{\ce{Cu},\ 573\ \text{K}} \ce{CH3COCH3 + H2} \)
(D) Oxidation with \(CrO_3\) — Correct. The primary alcohol is oxidised to propanal (and further to propanoic acid), the secondary alcohol only to propanone; the aldehyde or acid is easily told apart from the ketone.
\( \displaystyle \ce{CH3CH2CH2OH} \xrightarrow{\ce{CrO3}} \ce{CH3CH2CHO} \qquad \ce{CH3CH(OH)CH3} \xrightarrow{\ce{CrO3}} \ce{CH3COCH3} \)