Practical Organic Chemistry – JEE Main ChemistryPractice Questions & PYQs

When testing for sulphur in organic compounds via Lassaigne's test, the sodium fusion extract containing sulphide ions ($\mathrm{S}^{2-}$) reacts with sodium nitroprusside to produce a characteristic deep violet or purple colour. The sulphide ion performs a nucleophilic attack on the nitroprusside complex, following the reaction shown below:

$\mathrm{S}^{2-} + [\mathrm{Fe}(\mathrm{CN})_5(\mathrm{NO})]^{2-} \rightarrow [\mathrm{Fe}(\mathrm{CN})_5(\mathrm{NOS})]^{4-}$

This chemical transformation results in the formation of the complex ion $[\mathrm{Fe}(\mathrm{CN})_5(\mathrm{NOS})]^{4-}$, which associates with sodium ions to yield the final product $\mathrm{Na}_4[\mathrm{Fe}(\mathrm{CN})_5(\mathrm{NOS})]$.

Methyl orange functions as a pH-dependent indicator that exists in a dynamic equilibrium between its acidic quinonoid form and its basic benzenoid form, represented by the equilibrium $HIn \rightleftharpoons H^+ + In^-$. As the titration proceeds to the end point in an acidic environment, the increased concentration of $H^+$ ions drives the equilibrium toward the formation of the protonated quinonoid species. This structural rearrangement significantly alters the electronic conjugation of the molecule, which is responsible for the distinct color shift observed at the end point.

Duma's method determines nitrogen percentage by calculating the moles of nitrogen gas collected over a solution, which requires correcting for the vapor pressure of water at the specific temperature. The partial pressure of the dry nitrogen gas is obtained by subtracting the water vapor pressure from the total pressure, resulting in $P = 759 - 14.2 = 744.8$ mmHg.

To find the moles of nitrogen, use the ideal gas law $n = \frac{PV}{RT}$. After converting the volume to liters by dividing by 1000 and applying the ideal gas constant, the calculation is:
$n = \frac{744.8 \times 22.78}{760 \times 0.0821 \times 280 \times 1000} \approx 0.000971$
The mass of nitrogen is derived by multiplying these moles by the molar mass of N2, which is 28 g/mol, yielding approximately 0.02719 g. The percentage of nitrogen is then calculated by $\frac{0.02719}{0.125} \times 100 \approx 21.75$, which rounds to 22.

Stoichiometry relies on the principle of conservation of mass and the molar ratio of reactants to products to predict chemical yields. Since the molar mass of toluene is 92 g/mol, the starting quantity of 5 g corresponds to $\frac{5}{92}$ moles. Given the 1:1 stoichiometric ratio, the theoretical yield of benzaldehyde would also be $\frac{5}{92}$ moles, but the reaction only achieves a 92% efficiency. Therefore, the actual amount produced is $\frac{5}{92} \times 0.92 = 0.05$ moles of benzaldehyde. With benzaldehyde possessing a molar mass of 106 g/mol, the total mass produced is $0.05 \times 106 = 5.3$ g. Converting this result into the requested format yields $530 \times 10^{-2}$ g.

The balanced chemical equation for the bromination of propyne dictates that 2 moles of $\text{Br}_2$ are consumed to produce 1 mole of 1,1,2,2-tetrabromopropane. With the molar mass of $\text{Br}_2$ at 160 g/mol, the initial quantity of 1 g of bromine is equal to $1/160$ moles. According to the reaction stoichiometry, this amount would ideally generate $1/2 \times 1/160$ moles, or $1/320$ moles of the product.

Considering the reaction occurs with a 27% yield, the actual amount produced is $1/320 \times 27/100$ moles. To obtain the final mass in grams, we multiply these moles by the molar mass of the product, which is 360 g/mol, as shown in the following calculation:

$\text{Mass} = \frac{1}{320} \times \frac{27}{100} \times 360$

Evaluating this expression yields approximately 0.30375 g, which is approximately $3 \times 10^{-1}$ g. The nearest integer for the requested value is 3.

Kjeldahl's method determines nitrogen content by converting organic nitrogen into ammonia gas, which is captured and neutralized by a standard sulfuric acid solution following the reaction stoichiometry $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$. Since one mole of the acid neutralizes two moles of ammonia, the total amount of ammonia produced is $2.5 \times 2 \times 2 = 10$ millimoles, or $10 \times 10^{-3}$ moles. Each mole of ammonia contributes exactly one mole of nitrogen, so the mass of nitrogen is $10 \times 10^{-3} \times 14 = 0.14$ grams. Dividing this mass by the initial 0.25 grams of the organic compound and multiplying by 100 yields the final percentage, $(0.14 / 0.25) \times 100 = 56$ percent.

The Carius estimation method relies on the principle that the chlorine present in an organic compound is quantitatively converted into silver chloride, allowing the original chlorine content to be determined from the precipitate mass. First, find the molar mass of silver chloride by summing the atomic masses of silver and chlorine, given as 108 g mol^-1 and 35.5 g mol^-1 respectively, which results in 143.5 g mol^-1. The mass of chlorine present in the 0.40 g of silver chloride precipitate is calculated by multiplying the mass of the precipitate by the fraction $\frac{35.5}{143.5}$.

To find the percentage of chlorine in the original 0.25 g sample, divide this calculated mass of chlorine by the mass of the organic compound and multiply by 100. The calculation is:
$\frac{35.5 \times 0.40}{143.5 \times 0.25} \times 100$
Solving this expression yields approximately 39.58 percent. Rounding this value to the nearest integer gives a final result of 40.

Quantitative analysis involving bromine relies on the Carius method, where the halogen is precipitated as silver bromide. Because the bromine mass within the organic sample is fully recovered in the precipitate, we simply determine the mass fraction of bromine within the silver bromide molecule and apply it to the total mass of the precipitate obtained.

The molar mass of silver bromide is the sum of 108 for silver and 80 for bromine, which equals 188. This indicates that for every 188 g of silver bromide, 80 g is contributed by bromine. Given that 0.40 g of silver bromide was formed from a 0.5 g organic sample, the percentage of bromine is calculated by multiplying the mass fraction of bromine in the salt by the ratio of the precipitate mass to the sample mass:

$\% \text{ of Br} = \frac{80 \times 0.4}{188 \times 0.5} \times 100$

Performing the arithmetic leads to approximately 34.04, which rounds to 34 as the nearest integer.

Lassaigne's test relies on the conversion of covalent elements like nitrogen and sulphur into water-soluble ionic forms through fusion with molten sodium. When an organic compound contains both nitrogen and sulphur, they initially react with the sodium to form sodium thiocyanate, $\mathrm{NaSCN}$, which aligns perfectly with the chemistry described in the first statement. If this fusion process occurs in the presence of an excess of sodium metal, the sodium thiocyanate is not the final product; rather, it undergoes a chemical reduction. This excess sodium metal reacts with the formed $\mathrm{NaSCN}$ according to the reaction $\mathrm{NaSCN} + 2\,\mathrm{Na} \rightarrow \mathrm{NaCN} + \mathrm{Na}_2\mathrm{S}$, which splits the compound into sodium cyanide and sodium sulphide. Since this decomposition behavior is a standard outcome when nitrogen and sulphur are fused with excess sodium, both statements accurately portray the chemical steps involved in the test.

Kjeldahl's method determines nitrogen content by converting organic nitrogen into ammonia, which is then neutralized by a standard acid solution. The stoichiometry of the reaction dictates that the chemical equivalents of ammonia produced directly correspond to the equivalents of sulfuric acid neutralized. Since sulfuric acid is a diprotic acid with an n-factor of 2, the total milliequivalents of acid used are calculated as $1 \times 12.5 \times 2 = 25$. Because these equivalents of acid neutralize an equal amount of ammonia, the ammonia produced also amounts to 25 milliequivalents, and since each nitrogen atom in ammonia contributes to exactly one molecule, this corresponds to 25 millimoles of nitrogen.

Converting this quantity into mass, we multiply by the atomic weight of nitrogen, which is 14 g/mol, yielding $25 \times 10^{-3} \times 14 = 0.35$ grams of nitrogen. To find the mass percentage, this value is divided by the initial mass of the organic compound, 0.55 grams, and multiplied by 100. This calculation results in $\frac{0.35}{0.55} \times 100 \approx 63.66\%$, which rounds to a nearest integer value of 64.

The gravimetric analysis of sulfur relies on the principle that the entire sulfur content of an organic compound is converted into barium sulfate. Because each mole of barium sulfate contains one mole of sulfur, we can determine the mass of sulfur present by multiplying the mass of the precipitate by the ratio of the atomic mass of sulfur to the molar mass of barium sulfate. Given that the molar mass of barium sulfate is 233 g/mol and the atomic mass of sulfur is 32 g/mol, the mass of sulfur in 0.752 g of precipitate is calculated as $\frac{32}{233} \times 0.752 = 0.1005$ g.

Once the mass of sulfur is known, we use the percentage composition to find the total mass of the original organic compound. Since the compound contains 51.6% sulfur, we use the expression $\text{Mass of compound} = \frac{\text{Mass of sulfur}}{51.6} \times 100$. Substituting the values gives $\frac{0.1005}{51.6} \times 100 = 1.9 \times 10^{-1}$ g. Rounding this to the nearest integer as required gives 2 $\times 10^{-1}$ g.

Dumas' method relies on isolating nitrogen gas from an organic compound and calculating its mass based on the ideal gas equation, accounting for the vapor pressure of water present during collection. First, obtain the partial pressure of the dry nitrogen gas by subtracting the aqueous tension from the total observed pressure, which is represented by the equation $P_{\text{N}_2} = 758 - 14 = 744$ mm Hg. Since this pressure needs to be in atmospheres to align with the gas constant, convert it by dividing by 760, and convert the volume from 30 mL to 0.03 L. Applying the ideal gas law $n = PV/RT$ with the gas constant 0.0821 L atm K-1 mol-1 and temperature 287 K, the number of moles of nitrogen is calculated as follows:

$n = \frac{(744 / 760) \times 0.03}{0.0821 \times 287}$

This calculation results in 0.00125 moles of nitrogen. The mass of the nitrogen gas is then found by multiplying these moles by the molar mass of 28 g/mol, which equals 0.035 g. Finally, the percentage of nitrogen in the compound is determined by taking the ratio of the nitrogen mass to the organic compound mass of 0.1840 g and multiplying by 100, yielding 19.02%, which rounds to 19.

Percentage yield is fundamentally defined as the ratio of the actual mass of product obtained to the theoretical maximum yield determined by stoichiometry, expressed as a percentage. In this methylation reaction, the balanced equation $\mathrm{C}_6\mathrm{H}_6 + \mathrm{CH}_3\mathrm{Cl} \rightarrow \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3 + \mathrm{HCl}$ establishes a 1:1 molar relationship between benzene and toluene. With the molar mass of benzene at 78 g/mol and toluene at 92 g/mol, the theoretical yield from 10 g of benzene is calculated as $\frac{10}{78} \times 92$ g. Dividing the actual 9.2 g of recovered toluene by this theoretical value and multiplying by 100 provides the final percentage, which evaluates to $\frac{9.2}{920} \times 78 \times 100$ and equals 78 %.

Seliwanoff's test operates via an acid-catalyzed dehydration mechanism to distinguish between ketoses and aldoses, a process that is distinct from the metal-based chemistry required by the other listed methods. While Barfoed's test, Benedict's test, and the Biuret test all rely on the specific chemical properties of copper reagents-such as the $\text{Cu(II)}$ acetate or copper (II) sulphate used in sugar and protein analysis-Seliwanoff's reagent is composed solely of resorcinol and $\text{HCl}$.

Qualitative analysis in organic chemistry utilizes specific chemical reactions to identify elements within a compound, often by transforming them into detectable ionic or gaseous forms. Lassaigne's test serves as the broad procedure to convert elements such as nitrogen, sulphur, phosphorus, and halogens into ionic counterparts, whereas heating a compound with copper(II) oxide specifically oxidizes carbon to $\mathrm{CO}_2$ for detection. Halogens are identified through precipitation with silver nitrate, as indicated by the reaction $\mathrm{Ag}^{+}(\text{aq}) + X^{-}(\text{aq}) \rightarrow \mathrm{Ag}X(\text{s})$, and the presence of sulphur is uniquely confirmed when the sodium fusion extract yields a black $\mathrm{PbS}$ precipitate upon treatment with lead acetate in the presence of acetic acid.

Sulphur estimation is based on the principle of mass conservation, ensuring that all sulphur present in an organic compound is captured as barium sulphate. The molar mass of $\mathrm{BaSO}_{4}$ is $233\,\mathrm{g/mol}$, and since sulphur has an atomic mass of $32\,\mathrm{u}$, the proportion of sulphur within the precipitate is defined by the ratio $\frac{32}{233}$. To find the mass of sulphur, one multiplies this ratio by the $1.44\,\mathrm{g}$ of precipitate produced. Determining the percentage involves dividing this mass by the $0.471\,\mathrm{g}$ of the organic sample and multiplying by $100$ as shown in the following calculation:

$\text{Percentage of sulphur} = \frac{32}{233} \times \frac{1.44}{0.471} \times 100 = 41.98\%$

Rounding the calculated $41.98\%$ to the nearest integer gives $42\%$.

The Seliwanoff test serves as a diagnostic tool specifically for ketose sugars, functioning on the principle that ketoses dehydrate more rapidly than aldoses when heated with concentrated $\text{HCl}$ and resorcinol, which yields a characteristic deep cherry red color. In contrast, the xanthoproteic test is employed to detect proteins by reacting a solution with concentrated $\text{HNO}_{3}$ and subsequently neutralizing it with $40\%\ \text{NaOH}$, where the observation of a yellow color changing into an orange hue confirms the presence of proteins.

The production of brown fumes upon treatment with concentrated $\mathrm{H}_2\mathrm{SO}_4$ and the subsequent formation of a dark brown ring with $\mathrm{FeSO}_4$ serves as a definitive test confirming the presence of the nitrate anion, $\mathrm{NO}_3^-$, in the sample. Simultaneously, the formation of a precipitate with $\mathrm{H}_2\mathrm{S}$ gas in dilute $\mathrm{HCl}$ that subsequently produces a deep blue complex upon reaction with concentrated $\mathrm{HNO}_3$ and excess $\mathrm{NH}_4\mathrm{OH}$ is characteristic of the copper cation, $\mathrm{Cu}^{2+}$. Synthesizing these analytical observations, the inorganic compound $X$ is identified as $\mathrm{Cu}(\mathrm{NO}_3)_2$.

The Carius method involves heating an organic compound with fuming nitric acid to decompose the sample and release halogen atoms as halide ions. To complete the process, $\mathrm{AgNO}_3$ is added to the tube to react with these ions and form insoluble silver halides. This interaction results in the formation of specific precipitates-$\mathrm{AgCl}$ (white), $\mathrm{AgBr}$ (pale yellow), and $\mathrm{AgI}$ (dark yellow)-which can then be filtered and weighed for accurate measurement.

The Carius method for halogen estimation relies on the principle that chlorine within an organic compound precipitates quantitatively as silver chloride. Because every mole of silver chloride contains exactly one mole of chlorine atoms, determining the amount of silver chloride formed allows us to calculate the precise mass of chlorine originally present in the sample.

The molar mass of the silver chloride precipitate is found by calculating $107.87 + 35.5 = 143.37$ g/mol. Dividing the precipitate mass of 0.3849 g by this molar mass yields the moles of silver chloride, calculated as $\frac{0.3849}{143.37} \approx 0.002685$ mol. Since the moles of chlorine equal the moles of silver chloride, multiplying this value by the atomic mass of chlorine gives the total mass of chlorine, $0.002685 \times 35.5 = 0.0953$ g. Dividing this mass by the 0.5 g sample size and multiplying by 100 provides a chlorine percentage of 19.06%, which rounds to 19.