Explain what is meant by the Lucas reagent, and describe how it is used to distinguish a primary, a secondary and a tertiary alcohol, with the underlying mechanistic reason for the difference in reaction time.

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The Lucas reagent (conc. HCl + anhydrous $\text{ZnCl}_2$ ) converts an alcohol to the corresponding alkyl chloride, which is insoluble in the aqueous mixture and so appears as a visible turbidity/cloudiness. Mechanistically, $\text{ZnCl}_2$ coordinates to the alcohol's oxygen, turning $-\text{OH}$ into a good leaving group, and the alcohol ionises to a carbocation ( $S_N1$ ) before chloride captures it. Since this ionisation step is rate-determining, the reaction rate directly tracks carbocation stability: a tertiary alcohol forms the most stable tertiary carbocation almost instantly, giving immediate turbidity at room temperature; a secondary alcohol forms a less stable secondary carbocation more slowly, giving turbidity after roughly 5-10 minutes; a primary alcohol would have to form the least stable primary carbocation, which essentially does not ha …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$Q. Which will give yellow precipitate with iodine and alkali?$
(a) Methanol
(b) Propanol
(c) Propan-2-ol
(d) 2-Methyl-propan-2-ol
›Reveal solutionSolution
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). …
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CBSE 2016Set ANNUAL1 markMCQ
$Q. Which of the following compounds will respond to iodoform test?$
(a) CH3CH2CH2OH
(b) CH3CH(OH)CH3
(c) CH3OCH2CH3
(d) CH3OH
›Reveal solutionSolution
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. …
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Q.Explain what is meant by the Lucas reagent, and describe how it is used to distinguish a primary, a secondary and a tertiary alcohol, with the underlying mechanistic reason for the difference in reaction time.

Concept understanding — Distinguishing Primary, Secondary and Tertiary Alcohols

Q.Explain what is meant by the Lucas reagent, and describe how it is used to distinguish a primary, a secondary and a tertiary alcohol, with the underlying mechanistic reason for the difference in reaction time.

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