Practical Organic Chemistry – JEE Main ChemistryPractice Questions & PYQs

The retardation factor, $R_f$, is defined as the ratio of the distance a solute moves from the baseline to the distance moved by the solvent front. For the spots in this measurement, the $R_f$ values are $R_f = \frac{4}{8}$ for A and $R_f = \frac{6}{8}$ for B. Calculating the ratio of the $R_f$ of B to the $R_f$ of A gives $\frac{(R_f)_B}{(R_f)_A} = \frac{6/8}{4/8} = 1.5$. Because the relationship is defined as $x \times 10^{-1}$, the equation $1.5 = x \times 10^{-1}$ shows that $x$ is 15.

The Kjeldahl's method is an analytical technique used for the estimation of nitrogen in organic compounds, which involves digesting the sample with concentrated $\mathrm{H}_2\mathrm{SO}_4$. Because this digestion reaction proceeds slowly on its own, the process requires an additive to increase the reaction rate. $\mathrm{CuSO}_4$ is introduced into the mixture specifically to serve as a catalytic agent, which effectively accelerates the breakdown of the organic nitrogen into ammonium sulfate.

Chromatography is a purification technique governed by the differential affinity of solutes for the stationary phase versus the mobile phase, a relationship expressed quantitatively as the $R_f$ value. This retardation factor is defined as the ratio of the distance traveled by a solute to the distance traveled by the solvent front, making it a fractional measure of migration. Because polar compounds typically interact more strongly with the stationary phase and retard their movement, they exhibit a lower rate of travel compared to non-polar compounds, which favor the mobile phase. Based on this principle, it is a consistent fact that non-polar compounds have a higher $R_f$ value than polar compounds, confirming that a polar compound will naturally have a smaller $R_f$ than a non-polar one.

Separation techniques rely on the distinct physical properties of components within a mixture, such as boiling point differences, thermal stability, and miscibility. Standard distillation effectively separates substances with significantly different boiling points, such as chloroform and aniline, while fractional distillation handles complex mixtures like crude oil where components have boiling points close to one another. Substances that are immiscible with water and possess high vapor pressure at temperatures below their boiling point, like aniline, are best purified using steam distillation. Finally, distillation under reduced pressure is the method of choice for heat-sensitive materials like glycerol, which would otherwise decompose before reaching their normal boiling point.

Determining the oxidation state of chromium in chromium pentoxide, $\mathrm{CrO}_5$, requires analyzing its unique "butterfly" molecular structure, which contains two peroxo linkages. In this configuration, the central chromium atom is bonded to four oxygen atoms in peroxo groups, each possessing an oxidation state of -1, and one terminal oxo group with an oxidation state of -2. By setting the sum of these oxidation numbers to zero, the equation $\mathrm{Cr} + 4(-1) + 1(-2) = 0$ reveals that chromium retains an oxidation state of +6.

The retardation factor, denoted as $R_f$, characterizes the movement of a substance relative to the solvent front in chromatography, defined by the following relationship:
$R_f = \frac{\text{Distance travelled by organic compound}}{\text{Distance travelled by solvent}}$
Substituting the given measurements of 3.5 cm for the organic compound and 5 cm for the solvent, the division 3.5 divided by 5 equals 0.7. Expressing this result in the required format yields $7 \times 10^{-1}$.

Potassium hydrogen phthalate serves as an ideal primary standard for the standardization of sodium hydroxide solutions because it is a stable, non-hygroscopic solid that allows for precise concentration measurements that do not change over time. During this specific acid-base titration between the weak acid potassium hydrogen phthalate and the strong base sodium hydroxide, the equivalence point occurs at a slightly alkaline pH, making phenolphthalein-which effectively indicates an endpoint within the pH range of $8.3$ to $10.1$-an entirely suitable choice for the procedure.

Methyl orange acts as a weak base indicator in solution, which directly contradicts the classification provided in the first statement. Regarding structural colouration, the quinonoid form exhibits a more intense colour due to its extended conjugation compared to the benzenoid form, rendering the second statement false as well. Consequently, because both premises misrepresent the fundamental chemical properties and structural behaviour of the indicator, neither statement holds true.

For a substance to function effectively as a primary standard in volumetric analysis, it must maintain a stable, unchanging concentration in solution, a requirement met by $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ but not by $\mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7$, whose concentration is prone to variation. Despite this practical advantage in standardization, $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ is physically less soluble in water than $\mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7$, making the assertion that it has higher solubility incorrect.

Sulphanilic acid ($NH_2C_6H_4SO_3H$) contains a sulfonic acid group rather than a carboxylic acid group, which prevents it from undergoing the esterification test typically associated with carboxylic acids. When subjected to Lassigne's test, the compound's nitrogen and sulphur atoms react during sodium fusion to form sodium thiocyanate, which then combines with ferric ions to produce a blood-red coloration, confirming that the second statement is valid.

Lassaigne's test identifies elements in an organic compound by fusing the sample with molten sodium to convert covalently bonded nitrogen and halogens into water-soluble sodium salts. Specifically, the fusion process converts nitrogen into sodium cyanide, $\mathrm{NaCN}$, which requires a carbon source, and halogens into sodium halides, such as $\mathrm{NaCl}$. Because $\mathrm{CH}_3\mathrm{NH}_2\cdot\mathrm{HCl}$ possesses both the essential carbon-nitrogen structure and chlorine atoms, it successfully undergoes this transformation to yield positive results for both nitrogen and the halogen.

Qualitative chemical analysis relies on converting elements within an organic compound into specific ions, which are then identified by adding reagents that produce characteristic color changes or precipitates. Nitrogen detection involves forming the deep blue complex $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$, whereas sulphur is identified using sodium nitroprusside, $\mathrm{Na}_2[\mathrm{Fe}(\mathrm{CN})_5\mathrm{NO}]$, which turns purple in the presence of sulphide ions. Phosphorus reacts with ammonium molybdate, $(\mathrm{NH}_4)_2\mathrm{MoO}_4$, to yield a canary yellow precipitate, while halogens are confirmed by treating their ionic forms with silver nitrate, $\mathrm{AgNO}_3$, to produce silver halide precipitates.

The Carius method relies on the principle of stoichiometry to quantify the mass of a specific element by converting it into a measurable precipitate, in this case, silver bromide. Since each mole of silver bromide contains precisely one mole of bromine atoms, the amount of precipitate directly reflects the total bromine content in the original organic sample.

The moles of silver bromide produced are calculated using the relation $\frac{0.376}{188} = 0.002$. Because the molar ratio between silver bromide and bromine is 1:1, this confirms that 0.002 moles of bromine atoms originated from the compound. Multiplying this molar value by the atomic mass of bromine results \in a total bromine mass calculated as $0.002 \times 80 = 0.16$. The percentage composition is determined by dividing this mass by the initial sample mass of 0.400 g and multiplying by 100, which results in 40%.

The Dumas method determines nitrogen content by heating an organic compound with $\mathrm{CuO}$, which acts as an oxidizing agent to produce $\mathrm{CO}_2$, $\mathrm{H}_2\mathrm{O}$, and $\mathrm{N}_2$ through the reaction $\mathrm{C}_{x}\mathrm{H}_{y}\mathrm{N}_{z} + \left(2x + \frac{y}{2}\right)\mathrm{CuO} \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O} + \frac{z}{2}\mathrm{N}_2 + \left(2x + \frac{y}{2}\right) \mathrm{Cu}$. Because this combustion can sometimes result in the formation of unwanted nitrogen oxides, the evolved gaseous mixture is passed over heated copper gauze to reduce these oxides entirely into free nitrogen gas. This critical step ensures that only nitrogen gas is collected for measurement, providing an accurate quantitative analysis.

The Carius method is a standard analytical technique primarily employed for the quantitative estimation of sulfur and halogens in organic compounds. In this specific experimental sequence, heating a sulfur-containing organic compound with sodium peroxide within a Carius tube effectively oxidizes the sulfur atoms into sulfate ions. When barium chloride is subsequently added to this acidic solution, it facilitates the precipitation of barium sulfate, expressed as $\text{SO}_4^{2-} + \text{Ba}^{2+} \rightarrow \text{BaSO}_4$, which allows for the precise measurement of the sulfur content. Among the given choices, methionine serves as the suitable candidate because it is an amino acid that contains sulfur, thereby fulfilling the requirements for this particular quantitative analysis.

The stoichiometry of a redox titration in an acidic medium is defined by a balanced chemical equation that ensures all reactant and product atoms are accounted for. This reaction is represented as:

$2 \mathrm{KMnO}_4 + 8 \mathrm{H}_2\mathrm{SO}_4 + 10 \mathrm{FeSO}_4 \cdot (\mathrm{NH}_4)_2\mathrm{SO}_4 \cdot 6 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{K}_2\mathrm{SO}_4 + 2 \mathrm{MnSO}_4 + 5 \mathrm{Fe}_2(\mathrm{SO}_4)_3 + 10 (\mathrm{NH}_4)_2\mathrm{SO}_4 + 68 \mathrm{H}_2\mathrm{O}$

By examining the coefficients in this balanced equation, it is evident that the reaction involving 2 molecules of potassium permanganate leads to the production of 68 molecules of water.

The condensation of phthalic anhydride with two equivalents of phenol, which is commonly referred to as carbolic acid, in the presence of concentrated $\text{H}_2\text{SO}_4$ yields phenolphthalein. This resulting compound serves as a well-known acid-base indicator that transitions between colorless and pink across different pH ranges, identifying 'X' as carbolic acid and 'Y' as phenolphthalein.

The core principle underlying this problem is the law of stoichiometry, which allows us to relate the amount of substance at the end of a reaction sequence back to the starting reactant. We begin by analyzing the titration step, where liberated iodine reacts with sodium thiosulfate according to the equation $\mathrm{I}_2 + 2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \rightarrow 2\mathrm{NaI} + \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6$ Given that 60.0 mL of 0.1 M $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$ is used, the total moles of thiosulfate equal 0.060 L multiplied by 0.1 mol/L, resulting \in 0.006 mol. Because two moles of thiosulfate consume one mole of iodine, the amount of iodine produced is exactly half the amount of the thiosulfate, which equals 0.003 mol of $\mathrm{I}_2$.

Tracing this chain backward, the chlorine gas responsible for liberating this iodine comes from the reaction $\mathrm{Cl}_2 + 2\mathrm{KI} \rightarrow \mathrm{I}_2 + 2\mathrm{KCl}$ This indicates that the 0.003 mol of $\mathrm{I}_2$ must have originated from 0.003 mol of $\mathrm{Cl}_2$. Consequently, the primary reaction $\mathrm{MnO}_2 + 4\mathrm{HCl} \rightarrow \mathrm{MnCl}_2 + \mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O}$ establishes a one-to-one molar relationship between the manganese dioxide and the chlorine gas, meaning 0.003 mol of $\mathrm{MnO}_2$ was present in the original sample.

With a molar mass for $\mathrm{MnO}_2$ of 87 g/mol, the actual mass of the compound in the mixture is 0.003 mol multiplied by 87 g/mol, totaling 0.261 g. Calculating the percentage of $\mathrm{MnO}_2$ in the initial 2.0 g sample involves dividing 0.261 g by 2.0 g and multiplying by 100. This calculation results in 13.05%, which rounds to 13 when expressed as the nearest integer.

The Dumas method for nitrogen estimation relies on the principle that the nitrogen present in an organic compound can be converted into elemental nitrogen gas, and the volume of this gas at standard temperature and pressure allows us to determine the total nitrogen content. Since one mole of any ideal gas occupies 22.4 L at STP, the molar volume helps us convert the 22.400 mL of nitrogen gas evolved into moles using the calculation $\frac{22.400}{22400} = 10^{-3}$ moles.

Because each nitrogen gas molecule contains two nitrogen atoms, the total amount of nitrogen atoms is $2 \times 10^{-3}$ moles. Given that the molar mass of nitrogen is 14 g/mol, the total mass of nitrogen atoms is $14 \times 2 \times 10^{-3} = 28 \times 10^{-3}$ g. The percentage of nitrogen in the 0.2 g sample is calculated by dividing this mass by the mass of the organic compound and multiplying by 100, which is represented by $\frac{28 \times 10^{-3}}{0.2} \times 100 = 14$, resulting in 14%.

Neutral ferric chloride acts as a versatile qualitative reagent used to identify both acetate ions and phenolic hydroxyl groups through distinct colorimetric responses. When acetate ions are mixed with the reagent, they yield a characteristic deep red precipitate. In contrast, phenolic compounds react with neutral ferric chloride to produce a violet-colored complex, such as $[\mathrm{Fe}(\mathrm{C}_6\mathrm{H}_5\mathrm{O})_6]^{3-}$, effectively distinguishing them from alcoholic hydroxyl groups that do not undergo this transformation.