Q.A student is given three unlabelled alcohols, one of which is 2-methylpropan-2-ol. Explain how the Lucas test lets the student identify it immediately, without waiting.

The iodoform (yellow precipitate) test is positive for alcohols with a CH3-CH(OH)- group (or CH3-CO- ketones); propan-2-ol qualifies. Correct option (c).

The iodoform test (I2 + NaOH) gives a yellow precipitate of triiodomethane (CHI3) with compounds containing a CH3-CO- group or a CH3-CH(OH)- group (which is first oxidised to CH3-CO- by the reagent).

• Methanol (CH3OH): no CH3-CHOH- unit -> negative.
• Propan-1-ol (CH3CH2CH2OH): the carbinol carbon has no methyl attached -> negative.
• Propan-2-ol (CH3-CHOH-CH3): has CH3-CH(OH)- -> positive (yellow CHI3). …

The iodoform test is positive for compounds having a $CH_3CH(OH)-$ or $CH_3CO-$ group; among the options only isopropanol qualifies.

The iodoform test (with $I_2$/NaOH) is given by compounds containing either a methyl ketone group ($CH_3CO-$) or a secondary alcohol group with a methyl attached to the carbinol carbon ($CH_3CH(OH)-$), since NaOH/I₂ first oxidises such an alcohol to the corresponding methyl ketone, which is then haloform-cleaved to give the pale-yellow iodoform ($CHI_3$) precipitate, plus the sodium salt of the resulting carboxylic acid.

Checking the options:

• (a) $CH_3CH_2CH_2OH$ (n-propanol): a primary alcohol, no $CH_3CH(OH)-$ group — negative. …