which of the following reagents could be used to distinguish between propan-1-ol and propan-2-ol

Analyze the question: Propan-1-ol (CH₃CH₂CH₂OH) is a primary alcohol and propan-2-ol (CH₃CH(OH)CH₃) is a secondary alcohol. We need a reagent that gives a different observable result with each.

Concept:

• Acidified K₂Cr₂O₇ oxidizes both primary and secondary alcohols. Both turn the solution from orange to green, so it can't tell them apart. (The products differ, an aldehyde/acid vs a ketone, but the colour change is the same.)
• The iodoform test (I₂ / OH⁻) is positive only for compounds containing the CH₃CH(OH)– group (or CH₃C=O). Propan-2-ol has this group, but propan-1-ol does not.
• CH₃COOH (esterification) needs a catalyst and heat, and both alcohols form esters. It doesn't distinguish them clearly.
• NaCl does not react with either alcohol.

Solution:

1. Add I₂ and NaOH to propan-1-ol and warm: no yellow precipitate forms.
2. Add I₂ and NaOH to propan-2-ol and warm: the alcohol is first oxidized to propanone (CH₃COCH₃). The methyl ketone is then iodinated and cleaved, giving a pale yellow precipitate of iodoform (CHI₃) with a characteristic antiseptic smell.

CH₃CH(OH)CH₃ → CH₃COCH₃ → CHI₃ (yellow ppt) + CH₃COO⁻

Answer: B (I₂ / OH⁻).

Exam tip: Whenever a question asks you to distinguish between alcohols, look for the CH₃CH(OH)– group. If it is present, the iodoform test (yellow precipitate) is positive. Ethanol and propan-2-ol give it, while methanol and propan-1-ol do not.