See image — Practical Organic Chemistry and Purification Chemistry Question

💡 Solution & Explanation

Concept: To differentiate 1,1-dichloroethane (CH3CHCl2) and 1,2-dichloroethane (ClCH2CH2Cl), we need a reagent that reacts with one but not the other, producing a distinguishable product. Step 1: Consider what happens when each compound is treated with aqueous KOH (hydrolysis). - 1,1-dichloroethane (CH3CHCl2) undergoes hydrolysis with aq. KOH. The geminal diol intermediate (CH3CH(OH)2) formed is unstable and loses water to give acetaldehyde (CH3CHO), an aldehyde. - 1,2-dichloroethane (ClCH2CH2Cl) undergoes hydrolysis with aq. KOH to give ethylene glycol (HOCH2CH2OH), a diol with no carbonyl group. Step 2: After treatment with aq. KOH, add 2,4-DNP (2,4-dinitrophenylhydrazine reagent). - Acetaldehyde (from 1,1-dichloroethane) is an aldehyde/carbonyl compound and will give a positive 2,4-DNP test, forming a yellow/orange precipitate (2,4-dinitrophenylhydrazone). - Ethylene glycol (from 1,2-dichloroethane) has no carbonyl group and will NOT react with 2,4-DNP, giving no precipitate. Step 3: Why option (a) fails: If 2,4-DNP is added first (before aq. KOH), neither compound has yet been converted to a carbonyl compound, so no differentiation occurs at that stage. The order matters. Step 4: Why options (c) and (d) fail: - NaHSO3 reacts with aldehydes and methyl ketones but cannot directly differentiate these two dichloroethanes without prior hydrolysis. - Lucas reagent (ZnCl2/conc. HCl) differentiates primary, secondary, and tertiary alcohols, not vicinal or geminal dihalides. Therefore, the correct sequence is aq. KOH first (to hydrolyze and generate acetaldehyde from 1,1-dichloroethane) then 2,4-DNP (to detect the carbonyl compound), which is option (b). Therefore, the correct answer is B.