Purification & Characterisation of Organic Compounds Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on purification and characterisation of organic compounds with step-by-step solutions covering Carius halogen and sulphur estimation, combustion analysis, distillation techniques and chromatography Rf values.
A curated set of JEE Main 2026 previous-year questions on purification and characterisation of organic compounds, each solved step by step so you can check both the final answer and the reasoning.
Solutions are AI-generated and pending review.
Solution
Step 1 — mass and % of bromine from AgBr. Molar mass of $\text{AgBr} = 108 + 80 = 188$.
$$m_{\text{Br}} = 3.36 \times \frac{80}{188} = 1.430\ \text{g}$$$$\%\,\text{Br} = \frac{1.430}{2.0}\times 100 = 71.5\%$$Step 2 — % of hydrogen by difference.
$$\%\,\text{H} = 100 - \%\,\text{C} - \%\,\text{Br} = 100 - 26.7 - 71.5 = 1.8\%$$Step 3 — mole ratio.
$$\text{C}:\text{H}:\text{Br} = \frac{26.7}{12} : \frac{1.8}{1} : \frac{71.5}{80} = 2.225 : 1.80 : 0.894$$Divide by the smallest (0.894):
$$= 2.49 : 2.01 : 1.00 \;\xrightarrow{\times 2}\; 5 : 4 : 2$$Step 4 — empirical formula. $\text{C}_5\text{H}_4\text{Br}_2$ (M = 224). Check: $\%\text{C}=\tfrac{60}{224}\times100=26.8\%$, $\%\text{Br}=\tfrac{160}{224}\times100=71.4\%$ — matches the data.
Total carbon atoms in the empirical formula $= 5$.
Answer: 5
Solution
Reaction: acid-catalysed hydration (Markovnikov) of 2-methylpropene:
$$\text{(CH}_3\text{)}_2\text{C}=\text{CH}_2 \xrightarrow{\text{dil. H}_2\text{SO}_4} \text{(CH}_3\text{)}_3\text{C-OH}$$The product (X) is 2-methylpropan-2-ol (tert-butanol), which carries a polar $-\text{OH}$ group and is therefore far more polar than the parent alkene.
Chromatography reasoning: In normal-phase (silica) TLC, a more polar compound adsorbs more strongly to the polar stationary phase and travels a shorter distance, so it has a lower $R_f$. Since (X) is more polar than 2-methylpropene, its $R_f$ must be smaller than 0.42.
Among the options, only $0.12 < 0.42$.
Answer: D
Solution
Statement I: 1,2,3-trihydroxypropane is glycerol, a high-boiling liquid (b.p. $\approx 290\,^{\circ}\text{C}$) that decomposes at/near its boiling point. Liquids that decompose on heating to their boiling point are purified by distillation under reduced pressure, not by simple distillation. Hence Statement I is false.
Statement II: An azeotrope boils at a constant temperature and distils over without change in composition, so its components cannot be separated by ordinary fractional distillation. Statement II is true.
So Statement I is false but Statement II is true.
Answer: D
Solution
Step 1 — mass of compound taken.
$$m_X = 2.0\times 10^{-3}\ \text{mol} \times 76\ \text{g mol}^{-1} = 0.152\ \text{g}$$Step 2 — sulphur in the BaSO$_4$ formed. Each mole of $\text{BaSO}_4$ contains 1 mole of S.
$$m_{\text{S}} = 0.4813 \times \frac{32}{233} = 0.06610\ \text{g}$$Step 3 — percentage of sulphur.
$$\%\,\text{S} = \frac{0.06610}{0.152}\times 100 = 43.49\%$$Step 4 — express as $\times 10^{-1}\%$.
$$43.49\% = 434.9 \times 10^{-1}\% \approx 435 \times 10^{-1}\%$$Answer: 435
Solution
Step 1 — mass of carbon from $CO_2$.
$$m_{\text{C}} = 0.25 \times \frac{12}{44} = 0.06818\ \text{g}$$Step 2 — relate to % carbon. Carbon is 25% of the sample mass $X$:
$$0.25 \, X = 0.06818 \implies X = \frac{0.06818}{0.25} = 0.2727\ \text{g}$$Cross-check with hydrogen. $m_{\text{H}} = 0.12 \times \tfrac{2}{18} = 0.01333\ \text{g}$; $X = \tfrac{0.01333}{0.0489} = 0.2727\ \text{g}$ — consistent.
Step 3 — express as $\times 10^{-3}$ g.
$$X = 0.2727\ \text{g} = 272.7\times 10^{-3}\ \text{g} \approx 273\times 10^{-3}\ \text{g}$$Answer: 273
Solution
Track the transformations:
- (i) Sn/HCl then $\text{OH}^-$: reduces $-\text{NO}_2$ to $-\text{NH}_2$ → p-toluidine (4-methylaniline).
- (ii) $(\text{CH}_3\text{CO})_2\text{O}$: acetylates the amine to the acetamide $-\text{NHCOCH}_3$ (protecting/moderating the strong activator so bromination is monosubstituted). Now the ring carries $-\text{NHCOCH}_3$ and, para to it, $-\text{CH}_3$.
- (iii) $\text{Br}_2/\text{AlBr}_3$: the amide nitrogen is the dominant $o,p$-director. Its para position is blocked by $-\text{CH}_3$, so Br enters ortho to $-\text{NHCOCH}_3$.
- (iv) $\text{H}_3\text{O}^+$: hydrolyses the amide back to the free amine.
Product P = 2-bromo-4-methylaniline, $\text{C}_7\text{H}_8\text{BrN}$, with one Br per molecule.
$$M_P = 7(12) + 8(1) + 80 + 14 = 186\ \text{g mol}^{-1}$$Carius estimation (1 Br → 1 AgBr, $M_{\text{AgBr}} = 188$):
$$m_{\text{AgBr}} = \frac{1.0}{186}\times 188 = 1.01\ \text{g} \approx 1\ \text{g}$$Answer: 1
Solution
Statement I: Sucrose is appreciably soluble in ethanol, whereas NaCl is essentially insoluble in ethanol. Stirring the mixture with alcohol dissolves only the sugar; filtration leaves NaCl behind. This is a valid differential-solubility (extraction) separation. True.
Statement II: Steam distillation applies to substances that are steam-volatile and immiscible with water. The volatile, water-insoluble oils of rose petals distil over with steam and are then separated. True.
Both statements are true.
Answer: A
Solution
Step 1 — sulphur contained in the BaSO$_4$. One mole of $\text{BaSO}_4$ carries one mole of S.
$$m_{\text{S}} = 0.6 \times \frac{32}{231} = 0.0831\ \text{g}$$Step 2 — percentage of sulphur in the sample.
$$\%\,\text{S} = \frac{0.0831}{0.2}\times 100 = 41.56\% \approx 42\%$$Answer: 42
Solution
Match each technique to the situation it is designed for:
- A. Simple distillation — works when the components have a large difference in boiling points → II.
- B. Fractional distillation — needed when two liquids have close boiling points → IV.
- C. Steam distillation — used for steam-volatile, water-immiscible compounds → I.
- D. Distillation under reduced pressure — used for a liquid that decomposes at its boiling point (lowering pressure lowers the boiling point) → III.
So A-II, B-IV, C-I, D-III.
Answer: B
Solution
Statement I: As the mixed vapours rise through the fractionating column, the higher-boiling (less volatile) component condenses first and returns down the column, while the lower-boiling vapours continue upward. True.
Statement II: With each successive vaporisation–condensation step, the vapour moving up the column becomes richer in the low-boiling (more volatile) component, not the high-boiling one. False.
So Statement I is true but Statement II is false.
Answer: C