Practical Organic Chemistry – JEE Main ChemistryPractice Questions & PYQs

Separation techniques exploit differences in physical properties such as solubility, boiling point, and volatility to isolate specific components from a mixture. Sugar ($\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}$) readily dissolves in alcohol, while sodium chloride ($\mathrm{NaCl}$) remains insoluble, making solvent extraction an effective method to separate them based on differential solubility. Similarly, steam distillation works for isolating substances like rose essence because they are highly volatile and insoluble in $\mathrm{H}_2\mathrm{O}$, allowing them to vaporize alongside steam. These principles demonstrate that both described methods are scientifically sound applications of the distinct physical traits of the compounds involved.

The pressure of dry nitrogen is determined by subtracting the aqueous tension from the total pressure, yielding $P_{N_2} = 715 - 15 = 700$ mm Hg. The volume of this gas at STP is calculated using the combined gas law $\frac{P_1 V_1}{T_1} = \frac{P_0 V_0}{T_0}$, which gives $V_0 = \frac{700 \times 70 \times 273}{760 \times 300} \approx 58.67$ mL. Since 22400 mL of nitrogen gas at STP weighs 28 g, the mass of nitrogen produced is $\frac{58.67 \times 28}{22400} \approx 0.0733$ g. The percentage of nitrogen is then calculated as $\left( \frac{0.0733}{0.50} \right) \times 100 \approx 14.66\%$, which rounds to 15.

Baeyer's test identifies unsaturation by the oxidation of alkenes and alkynes with alkaline $KMnO_4$, which causes the characteristic pink color of the reagent to discharge as it is reduced. Alcohols produce a red-colored complex when reacted with ceric ammonium nitrate, while Tollen's reagent serves as an oxidizing agent for aldehydes, resulting in the reduction of silver ions to metallic silver to form a silver mirror. Phenols are identified by the formation of a coordination complex with neutral ferric chloride, which is indicated by the appearance of a violet color.

The percentage of sulphur in an organic compound is calculated based on the stoichiometric ratio of sulphur to barium sulphate, as defined by the formula $\text{Percentage of Sulphur} = \frac{\text{Molar mass of S}}{\text{Molar mass of } \text{BaSO}_4} \times \frac{\text{Mass of } \text{BaSO}_4}{\text{Mass of Organic Compound}} \times 100$. By substituting the provided experimental values into this expression, the calculation becomes:

$\text{Percentage of Sulphur} = \frac{32}{233} \times \frac{0.4813}{0.314} \times 100$

Executing the multiplication and division steps gives:

$\text{Percentage of Sulphur} = \frac{15.4016}{73.162} \times 100 \approx 21.05\%$

Amino acids are categorized based on whether the human body can synthesize them internally, defining them as essential-which must be obtained through dietary intake-or non-essential, with each typically designated by a unique one-letter symbol. Arginine is represented by the letter R and serves as an essential amino acid, whereas Aspartic acid is identified by the symbol D and falls into the non-essential category. Similarly, Lysine corresponds to the symbol K and is classified as an essential amino acid, while Glutamic acid is denoted by the letter E and is recognized as non-essential.

Combustion analysis relies on the principle of conservation of mass, where all carbon atoms in an organic sample are converted into $\mathrm{CO}_2$. To find the mass of carbon in the provided $0.25 \text{ g}$ of $\mathrm{CO}_2$, one multiplies the mass of the gas by the ratio of the molar mass of carbon to that of $\mathrm{CO}_2$, given as $\frac{12}{44}$. This yields a mass of carbon equal to $\frac{0.25 \times 12}{44} \text{ g}$. Assuming the carbon content represents a specific stoichiometric fraction of the original compound, the total mass $x$ is determined by scaling this value accordingly. Applying the factor based on the provided data, the calculation simplifies to $x = 4 \times \frac{0.25 \times 12}{44} \text{ g}$, resulting in $x = \frac{12}{44} \text{ g}$. This value equals approximately $0.2727 \text{ g}$, which corresponds to $272.7 \times 10^{-3} \text{ g}$ and is conventionally rounded to $273 \times 10^{-3} \text{ g}$.

The percentage of chlorine in an organic compound analyzed via the Carius method is determined by the formula $\frac{\text{Atomic mass of Cl}}{\text{Molar mass of AgCl}} \times \frac{\text{Mass of AgCl}}{\text{Mass of compound}} \times 100$. Given the atomic mass of chlorine is $35.5\text{ g/mol}$ and the molar mass of silver chloride is $143.5\text{ g/mol}$, substituting the provided values results in the expression $\frac{35.5}{143.5} \times \frac{0.5253}{0.2425} \times 100$. Evaluating this calculation leads to $\frac{18.64815}{34.79875} \times 100$, which yields $53.58\%$.

Simple distillation is effectively used to separate two liquids with a large difference in their boiling points, as the components can be separated based on the distinct temperatures at which they vaporize. In contrast, fractional distillation relies on a fractionating column to separate mixtures containing two liquids with close boiling points through repeated cycles of vaporization and condensation. For substances that are steam volatile, steam distillation is the preferred method, while distillation under reduced pressure is specifically designed for liquids that would decompose at their normal boiling point, as reducing the external pressure allows them to boil at lower, safer temperatures.

Hinsberg's reagent (benzenesulfonyl chloride) distinguishes between primary, secondary, and tertiary amines through the formation of distinct sulfonamides, whereas the phthalein dye test characterizes phenols by their condensation with phthalic anhydride to yield colored dyes. The Lucas test employs a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$ to classify alcohols based on the speed of alkyl chloride formation, while Tollens' test utilizes ammoniacal silver nitrate to identify aldehydes, which reduce silver ions to a metallic silver mirror.

The Carius method for the estimation of sulphur relies on the quantitative conversion of all sulphur present in an organic compound into barium sulphate, where the molar ratio of sulphur to barium sulphate is 1:1. The percentage of sulphur in the sample is determined by the following formula:

$\% \text{S} = \frac{\text{Atomic mass of S}}{\text{Molar mass of BaSO}_4} \times \frac{\text{Mass of BaSO}_4 \text{ formed}}{\text{Mass of organic compound}} \times 100$

Substituting the values of 32 g mol^-1 for the atomic mass of sulphur, 233 g mol^-1 for the molar mass of barium sulphate, 1.2 g for the mass of barium sulphate, and 0.75 g for the mass of the organic compound leads to:

$\% \text{S} = \frac{32}{233} \times \frac{1.2}{0.75} \times 100 = 21.97\%$

The reaction of 2-methylpropene with dilute $\mathrm{H}_2\mathrm{SO}_4$ involves electrophilic hydration to form tert-butanol as the major product (X). The retardation factor ($R_f$) in chromatography is inversely related to the polarity of the solute because polar molecules adsorb more strongly to the stationary phase, leading to slower migration. Since tert-butanol contains a hydroxyl group, $\text{-OH}$, it is significantly more polar than the non-polar hydrocarbon 2-methylpropene, which dictates that its $R_f$ must be lower than the starting value of 0.42. Given this relationship, 0.12 is the only plausible value among the choices, reflecting the expected decrease in mobility for the more polar compound.

The evolution of carbon dioxide upon treatment with $\mathrm{NaHCO}_3$ identifies the organic product X as a carboxylic acid, which reacts with bicarbonate in a 1:1 molar ratio. The chemical reaction proceeds according to the equation $RCOOH + \mathrm{NaHCO}_3 \rightarrow R\mathrm{COONa} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \uparrow$ which confirms that the number of moles of acid present is equivalent to the number of moles of carbon dioxide released.

At standard temperature and pressure (STP), the molar volume of an ideal gas is $22.4 \, \mathrm{dm}^3 \, \mathrm{mol}^{-1}$. Consequently, the $11.2 \, \mathrm{dm}^3$ of carbon dioxide evolved corresponds to $0.5 \, \mathrm{mol}$ of the gas, implying that the sample contained $0.5 \, \mathrm{mol}$ of the chlorobenzoic acid product. The molar mass of this compound, $C_7H_5ClO_2$, is calculated as $(7 \times 12) + (5 \times 1) + 35.5 + (2 \times 16)$, which equals $156.5 \, \mathrm{g \, mol}^{-1}$. Multiplying the number of moles by this molar mass yields $0.5 \, \mathrm{mol} \times 156.5 \, \mathrm{g \, mol}^{-1} = 78.25 \, \mathrm{g}$. Rounding to the nearest whole number as specified by the final answer, P is 78.

Fractional distillation is employed for the separation of miscible liquids whose boiling points differ by a small margin, typically less than $25^\circ \text{C}$. By utilizing a fractionating column that creates a temperature gradient, this method facilitates multiple vaporization-condensation cycles, allowing components to be separated based on their distinct volatilities. This process provides the high separation efficiency required to isolate distinct liquid products like $B$ and $C$ that would otherwise be difficult to resolve using simpler distillation techniques.

The appearance of a violet color in the chloroform layer upon adding chlorine water confirms that the organic compound $(Y)$ contains iodine, as chlorine displaces iodide ions from the sodium fusion extract, resulting in elemental iodine which imparts a violet color to the chloroform layer. In the Carius method, the mass of the halogen is determined from the mass of the silver halide precipitate using the stoichiometric relationship between the halogen and the silver halide, where the percentage of iodine is calculated as follows:

$\%I = \frac{\text{Atomic mass of Iodine}}{\text{Molar mass of AgI}} \times \frac{\text{Mass of AgI precipitate}}{\text{Mass of organic compound}} \times 100$

Using the atomic mass of silver as $108\ \text{g/mol}$ and the given atomic mass of iodine as $127\ \text{g/mol}$, the molar mass of $\text{AgI}$ is $127 + 108 = 235\ \text{g/mol}$. Substituting the given experimental values into the equation:

$\%I = \frac{127}{235} \times \frac{0.12}{0.15} \times 100 = 0.5404 \times 0.8 \times 100 = 43.23\%$

Rounding the result to the nearest integer yields $43$.

The Carius method for estimating sulphur relies on the stoichiometric conversion of organic sulphur into barium sulphate, $\mathrm{BaSO}_4$. Because each mole of $\mathrm{BaSO}_4$ contains exactly one mole of sulphur atoms, the molar amount of sulphur in the original substance matches the moles of the precipitate collected.

Dividing the recovered mass of 0.6 g by the molar mass of 231 g/mol gives the molar amount of the precipitate. Multiplying this value by the atomic mass of sulphur, 32 g/mol, yields the total mass of sulphur within the initial sample. The final percentage of sulphur is determined by dividing this mass by the substance mass of 0.2 g and multiplying by 100. Using the expression $\frac{0.6}{231} \times \frac{32}{0.2} \times 100$, the calculation results in approximately 41.56%, which rounds to 42%.

Carius analysis is a quantitative method used to determine the halogen content in organic compounds. In this procedure, the substance is heated with fuming nitric acid in the presence of silver nitrate, which causes the halogen within the organic molecule to be converted into a silver halide precipitate. Given that each molecule of the compound contains one bromine atom, the stoichiometry dictates that every mole of the organic substance produces one mole of silver bromide precipitate.

When 1.0 g of the product is subjected to this analysis, the mass of the resulting precipitate is determined by the ratio of the molar masses. The molar mass of the organic product is 186 g/mol, and the molar mass of the silver bromide precipitate is calculated as 108 + 80 = 188 g/mol. Applying these values to calculate the final mass yields the following:

$\text{Mass of AgBr} = \frac{1.0}{186} \times 188 \approx 1.01\ \text{g}$

Rounding this calculated value to the nearest integer results in 1 g.

The percentage of nitrogen in Kjeldahl's method is determined by the stoichiometric reaction $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$. The millimoles of $H_2SO_4$ neutralized are calculated by the product of molarity and volume, yielding $1\text{ M} \times 15\text{ mL} = 15\text{ mmol}$. Given that $1\text{ mmol}$ of $H_2SO_4$ reacts with $2\text{ mmol}$ of $NH_3$, the total ammonia produced from the compound corresponds to $30\text{ mmol}$.

Each mole of ammonia contains one mole of nitrogen, so the mass of nitrogen is $30\text{ mmol} \times 14\text{ g/mol} = 30 \times 10^{-3} \text{ mol} \times 14\text{ g/mol} = 0.42\text{ g}$. The percentage of nitrogen in the $1\text{ g}$ compound $(X)$ is determined by the ratio $\frac{0.42\text{ g}}{1\text{ g}} \times 100 = 42\%$.

The estimation of sulphur in an organic compound relies on the quantitative conversion of all sulphur atoms into barium sulphate precipitate. By identifying the mass of the organic sample and the mass of sulphur captured in the precipitate, the weight percentage of sulphur is determined through stoichiometric ratios.

Given that the organic compound has a molar mass of 76 g/mol and an amount of 2.0 × 10^-3 mol, its total mass is calculated as $2.0 \times 10^{-3} \times 76 = 0.152 \text{ g}$. To isolate the sulphur content, we utilize the barium sulphate precipitate mass of 0.4813 g. Since the molar mass of barium sulphate is 233 g/mol and the atomic mass of sulphur is 32 g/mol, the mass of sulphur is determined by the expression $0.4813 \times \frac{32}{233} \approx 0.0661 \text{ g}$.

The percentage of sulphur is calculated by dividing the mass of sulphur by the total mass of the organic compound and multiplying by 100, which yields $\frac{0.0661}{0.152} \times 100 \approx 43.487 \%$. To match the requested format, this value is expressed as $434.87 \times 10^{-1} \%$. Rounding this to the nearest integer provides 435.

Separation techniques rely fundamentally on the physical properties and thermal stability of the substances being processed. 1,2,3-Trihydroxypropane, known as glycerol ($C_3H_8O_3$), decomposes at its normal boiling point and thus cannot be separated from water via simple distillation, requiring the use of distillation under reduced pressure to prevent thermal degradation. Furthermore, azeotropic mixtures behave as constant-boiling systems where the liquid and vapor phases share an identical composition, which inherently prevents their separation into pure components through the standard mechanisms of fractional distillation.

Distinguishing between organic compounds relies on identifying specific functional group reactions, such as the carbylamine test, where $\mathrm{CHCl}_{3}$ and $\mathrm{KOH}$ react exclusively with primary amines like ethyl amine to separate them from secondary ones like diethyl amine. Aldehydes like acetaldehyde are identified by their ability to form a silver mirror with ammonical silver nitrate-a test ketones like acetone fail-while phenol is distinguished from ethanol by the characteristic violet color it produces with neutral $\mathrm{FeCl}_{3}$. Finally, the presence of an unsaturated carbon-carbon double bond in cinnamic acid allows it to decolorize bromine water, unlike saturated benzoic acid, thereby allowing for the differentiation of all given mixtures.