Practical Organic Chemistry – JEE Main ChemistryPractice Questions & PYQs

The weight of a chemical substance is determined by multiplying its molar mass by the amount in moles. For the antiviral compound $C_9O_5FIN_2H_{10}$, summing the constituent atomic weights results in a molar mass of 372 g/mol. Multiplying this molar mass by the 0.1 mol provided yields the total mass:

$\text{Mass} = 0.1 \times 372 = 37.2$

As 37.2 is numerically equivalent to $372 \times 10^{-1}$, the total weight of 0.1 mol of this compound is $372 \times 10^{-1}$ g.

Methanol (A) serves as the precursor that is transformed into acetic acid (B), a well-known food preservative and acidic component in cooking, via chemical carbonylation processes. Acetic acid is subsequently reduced by diborane ($B_2H_6$) to produce ethanol (C), which functions as an oxygenated petrol additive that enhances combustion efficiency and reduces vehicle emissions. Finally, ethanol undergoes acid-catalyzed intermolecular dehydration at $140^{\circ} \mathrm{C}$ in the presence of concentrated sulfuric acid to yield diethyl ether (D), a compound historically utilized as an inhalable anesthetic.

In the Dumas method, we determine the nitrogen content by applying the combined gas law to convert the experimental gas volume to standard temperature and pressure (STP) conditions. Since the nitrogen is collected over water, the first necessary adjustment is to determine the pressure of dry gas by subtracting the aqueous tension from the total pressure, yielding $715 \text{ mm Hg} - 15 \text{ mm Hg} = 700 \text{ mm Hg}$. We then convert the observed volume of $50 \text{ mL}$ at $300 \text{ K}$ and $700 \text{ mm Hg}$ to STP conditions ($273 \text{ K}$ and $760 \text{ mm Hg}$) using the relation $V_{\text{STP}} = \frac{700 \times 50 \times 273}{760 \times 300}$, which equals $41.9 \text{ mL}$. Given that $22400 \text{ mL}$ of nitrogen gas at STP has a mass of $28 \text{ g}$, the mass of the collected nitrogen is $\frac{41.9}{22400} \times 28$, resulting in approximately $0.052 \text{ g}$. Finally, dividing the mass of the nitrogen by the initial mass of the organic compound of $0.292 \text{ g}$ and multiplying by $100$ results in approximately $17.94\%$, which rounds to $18\%$ as the nearest integer.

Dumas' method determines the percentage of nitrogen in an organic compound by utilizing the ideal gas law to calculate the quantity of nitrogen gas evolved from a known mass of sample. Because the gas is collected over water, the total measured pressure includes the aqueous tension, which must be subtracted to isolate the partial pressure of the dry nitrogen. Subtracting the aqueous tension of 15 mm Hg from the total pressure of 715 mm Hg yields a dry nitrogen pressure of 700 mm Hg.

To determine the amount of nitrogen, we use the equation $n = \frac{PV}{RT}$ to calculate the number of moles. By using the dry pressure of 700 mm Hg, the volume of 60 mL, and the temperature of 300 K, we find the molar amount of nitrogen gas. Multiplying these moles by the molar mass of nitrogen, which is 28 g per mole, provides the mass of the nitrogen gas collected. Dividing this mass by the 0.5 gram mass of the original compound and multiplying by 100 results in a nitrogen content of 12.57 percent.

Steam distillation is utilized to separate substances like aniline from an aqueous mixture because they are volatile in steam and immiscible with water. Glycerol requires distillation at reduced pressure to prevent its thermal decomposition, which would occur if it were heated to its boiling point under atmospheric pressure. Fractional distillation is employed for separating crude oil into various fractions due to the components' close boiling points, whereas simple distillation is sufficient for mixtures like chloroform and aniline that possess a significant difference in boiling points.

Dumas' method for nitrogen estimation relies on determining the volume of nitrogen gas produced by decomposing an organic compound, then calculating its mass using ideal gas behavior relative to standard temperature and pressure (STP). The partial pressure of the dry nitrogen gas ($P_{N_2}$) is obtained by subtracting the aqueous tension from the observed pressure, $P_{N_2} = 900 - 15 = 885 \text{ mm Hg}$. Converting the observed volume to STP conditions using the combined gas law gives $V_{STP} = \frac{885 \times 150 \times 273}{760 \times 300} = 158.95 \text{ mL}$. The percentage of nitrogen is calculated using the molar volume of nitrogen gas: $\text{Percentage of } N = \frac{158.95 \times 28}{22400 \times 1} \times 100 = 19.87\%$. Rounding to the nearest integer results in 20%.

Lassaigne's test relies on the fundamental principle of converting elements covalently bonded within an organic molecule into water-soluble ionic species by fusing the sample with molten sodium metal, which confirms the validity of the first statement. The second statement is incorrect because when both nitrogen and sulfur are present, the fusion process produces sodium thiocyanate, which reacts with ferric ions to form a blood-red complex rather than the Prussian blue precipitate, which is represented by the formula $\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3$ and is characteristic solely of nitrogen-containing compounds.

In the Carius method, we rely on the conservation of mass, specifically the principle that all sulphur in an organic compound is quantitatively converted into barium sulphate. This stoichiometric relationship implies that the molar amount of sulphur atoms corresponds exactly to the molar amount of the barium sulphate precipitate formed, because each molecule of barium sulphate contains one atom of sulphur. We first determine the molar mass of barium sulphate as $137 + 32 + 4 \times 16 = 233$ grams per mole. Given that we obtained 0.40 g of this precipitate, we calculate the moles present as $\frac{0.40}{233}$ moles. The mass of sulphur is subsequently found by multiplying these moles by the atomic mass of sulphur, 32 grams per mole, which equates to 0.055 g. Finally, calculating the percentage of sulphur in the 0.20 g sample involves dividing the mass of sulphur by the total sample mass and multiplying by 100, which results in 27.5 percent, or $275 \times 10^{-1}$ percent.

Simple distillation relies on a significant difference in boiling points; for instance, chloroform, with a boiling point of 334 K, is easily separated from aniline, which has a much higher boiling point of 547 K. Fractional distillation is the standard method for separating components with closer boiling points, such as the various hydrocarbons found in diesel and petrol. Distillation under reduced pressure is specifically employed for substances that decompose at their normal boiling temperatures, making it ideal for extracting glycerol from spent-lye. Finally, steam distillation is utilized for separating organic compounds that are immiscible with water and volatile in the presence of steam, which is the classic method for purifying aniline from a mixture with water.

Solvent extraction is governed by acid-base chemistry, which alters the solubility of components to drive them into specific phases. When 5 M $NaOH$ is introduced into the funnel, it converts the benzoic acid into a water-soluble salt that partitions into the aqueous layer. Since aniline acts as a base and chlorobenzene is neutral, neither reacts with the hydroxide ions, causing both to remain retained within the organic ethyl acetate phase.

The Carius method relies on the principle of stoichiometry, where the halogen in an organic compound is quantitatively precipitated as a silver halide, such that the moles of bromine in the sample equal the moles of silver bromide formed. We calculate the molar mass of silver bromide by summing the atomic masses of silver and bromine, represented by $\mathrm{M_{AgBr} = 108 + 80 = 188\ g\ mol^{-1}}$. Given the precipitate mass of 0.15 g, the number of moles of silver bromide is found using $\mathrm{n_{AgBr} = \frac{0.15}{188}}$. Since each mole of silver bromide contains one mole of bromine, the mass of bromine is calculated as $\mathrm{Mass_{Br} = \frac{0.15 \times 80}{188}\ g}$. Finally, determining the percentage by mass in the initial 0.25 g sample involves the expression $\mathrm{\%Br = \frac{0.15 \times 80 \times 100}{188 \times 0.25}}$, which simplifies to 25.5%, or $255 \times 10^{-1} \%$.

Glycerol undergoes thermal decomposition before reaching its normal boiling point, necessitating the use of vacuum distillation to lower the boiling temperature and enable safe purification. Steam distillation is effective specifically for compounds that are immiscible in water and possess high vapor pressure; since aniline is immiscible rather than miscible, the description provided for its purification method is fundamentally flawed. Ethanol and water form a constant-boiling azeotrope, which prevents simple distillation from achieving complete separation, thereby requiring the application of azeotropic distillation techniques. Finally, verifying purity through the mixed melting point method remains a standard practice, as an identical melting point upon combining an unknown sample with a pure reference confirms the sample's purity.

The $\mathrm{Br}_2$ water test is primarily used to detect unsaturation in organic compounds, as the characteristic reddish-orange color of the bromine solution disappears during the reaction. When testing for alcohols, ceric ammonium nitrate is employed to produce a distinct red color, whereas phenolic compounds are typically identified using ferric chloride, which results in a range of colors including blue, green, violet, or red. Finally, the 2, 4-DNP test serves as a specific indicator for aldehydes and ketones by producing a yellow, orange, or orange-red precipitate.

Essential oils found in flowers are unique organic compounds that are volatile when exposed to steam yet remain immiscible with water at room temperature. This specific combination of properties makes steam distillation the ideal separation technique, as it allows the oils to vaporize alongside steam and be recovered through condensation without needing to dissolve in the water. By passing steam through the raw plant material, these volatile components are carried away in the vapor phase, providing a clean and efficient method for extraction.

Steam distillation is the specific method utilized to purify substances that are steam volatile and insoluble in water. Since the mixture contains non-volatile impurities, the introduction of steam allows the target compound to vaporize at a temperature lower than its boiling point, where the sum of the partial pressures of the water and the substance reaches atmospheric pressure. This approach successfully separates the desired product from the non-volatile components without exposing the substance to excessively high temperatures.

Identifying organic functional groups relies on specific chemical reactions that produce observable changes, such as color shifts or the formation of precipitates. Bayer's test is the standard method for detecting unsaturation in organic molecules, while the ceric ammonium nitrate test is utilized to confirm the presence of an alcoholic-OH group. The phthalein dye test is specifically associated with the identification of phenol compounds, and Schiff's test serves as a sensitive indicator for aldehydes, turning a characteristic pink upon reaction.

To qualify as a primary standard, a substance must be stable and non-hygroscopic, as absorbing moisture from the atmosphere would compromise the accuracy of its mass and concentration. Although primary standards must react instantaneously and stoichiometrically, the inclusion of hygroscopic behavior in statement (C) renders that entire claim incorrect, and reagents such as $\mathrm{KMnO}_4$ and $\mathrm{NaOH}$ are classified as secondary standards because they lack the required long-term stability and precise composition. Therefore, the statements identifying primary standards as hygroscopic and categorizing $\mathrm{KMnO}_4$ and $\mathrm{NaOH}$ as suitable primary standards are fundamentally incorrect.

The chemical principle behind Kjeldahl's method is the stoichiometric relationship between the acid and the base involved; specifically, one mole of sulfuric acid neutralizes two moles of ammonia because the balanced equation is $\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NH}_3 \rightarrow (\mathrm{NH}_4)_2\mathrm{SO}_4$. First, determine the amount of acid used by multiplying its volume and molarity, which gives 20 millimoles of $\mathrm{H}_2\mathrm{SO}_4$. According to the reaction stoichiometry, this amount of acid requires twice as many millimoles of ammonia, resulting \in 40 millimoles of $\mathrm{NH}_3$. Since each mole of ammonia contains one mole of nitrogen, convert these 40 millimoles into 0.04 moles of nitrogen. Multiplying this by the atomic mass of nitrogen, 14 g/mol, gives a total mass of 0.56 g. Finally, dividing this mass by the 1 g sample size and multiplying by 100 reveals the percentage of nitrogen in the compound is 56%.

The iodoform test produces a yellow precipitate of triiodomethane ($CHI_3$) specifically in the presence of methyl ketones or precursors that can be converted into methyl ketones under basic conditions, such as gem-dihalides. Structure B, $CH_3CH_2-C(I)_2-CH_3$, undergoes nucleophilic substitution and hydrolysis in the presence of aqueous $KOH$ to yield 2-butanone ($CH_3CH_2-C(=O)-CH_3$), which contains the essential $CH_3-C(=O)-$ moiety required to form iodoform upon treatment with potassium hypoiodite.

Column chromatography and thin layer chromatography operate by allowing the components of a mixture to interact with a solid stationary phase, separating them based on their specific rates of adhesion or differential adsorption. In contrast, paper chromatography primarily relies on the partition principle, where substances distribute themselves between a liquid mobile phase and a stationary liquid phase held within the cellulose fibers of the paper. Thus, the techniques that rely fundamentally on the mechanism of differential adsorption are restricted to column and thin layer chromatography.