Practical Organic Chemistry – JEE Main ChemistryPractice Questions & PYQs

The molarity of a sodium hydroxide solution is determined by the law of equivalence, which states that at the point of complete neutralization, the equivalents of the base must equal the equivalents of the acid. First, the average volume of the base is calculated by taking the mean of the five burette readings: (4.5 + 4.5 + 4.4 + 4.4 + 4.4) / 5 = 4.44 mL. The relationship between the components is defined by $N_{1}V_{1} = N_{2}V_{2}$ where N represents normality and V represents volume. Oxalic acid is a diprotic acid with an n-factor of 2, so its normality is calculated as 1.25 M \times 2, equaling 2.5 N. Substituting the known values into the equation $S_{1} = \frac{V_{2} \times N_{2}}{V_{1}}$ where $S_{1}$ is the molarity of the base, the calculation 10.0 times 2.5 divided by 4.44 yields approximately 5.63. Since the n-factor of sodium hydroxide is 1, its molarity is the same as its normality, and rounding 5.63 to the nearest integer gives 6.

The concept of percentage yield evaluates reaction efficiency by comparing the actual mass of product recovered against the theoretical maximum predicted by chemical stoichiometry. In the nitration of benzene, the molar ratio between benzene and nitrobenzene is one-to-one, meaning every mole of starting material should ideally convert into an equivalent amount of product.

To determine the theoretical yield, first calculate the number of moles of benzene by dividing the mass of 3.9 g by the molar mass of 78 g/mol, which gives $\frac{3.9}{78} = 0.05$ mol. Since the reaction follows a one-to-one stoichiometry, this indicates that 0.05 mol of nitrobenzene is the theoretical production limit. The expected mass is obtained by multiplying these moles by the molar mass of nitrobenzene, 123 g/mol, calculated as:
$0.05 \times 123 = 6.15$
g. Finally, the percentage yield is found by dividing the actual mass of 4.92 g by the theoretical mass of 6.15 g and multiplying by 100, which is $\frac{4.92}{6.15} \times 100 = 80$ %.

According to the Law of Conservation of Mass, the total number of atoms of each element must remain identical on both the reactant and product sides of a balanced chemical equation. In the reactant $\mathrm{C_2H_7N}$, there are exactly 7 hydrogen atoms present. On the product side, hydrogen appears exclusively in the water component represented by $\frac{y}{2}\mathrm{H_2O}$, which accounts for $\frac{y}{2} \times 2$ hydrogen atoms. Equating the hydrogen count from both sides, we find that $y$ is equal to 7.

The Kjeldahl analysis method is fundamentally based on the stoichiometric reaction where ammonia, derived from the nitrogen content of an organic substance, is neutralized by a standard acidic solution such as sulfuric acid. The chemical equation $2\mathrm{NH}_3 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow (\mathrm{NH}_4)_2\mathrm{SO}_4$ demonstrates that two moles of ammonia are required to neutralize one mole of sulfuric acid.

The actual mass of nitrogen in the 0.8 g sample is calculated by taking 42% of that mass, which yields 0.336 g of nitrogen. Dividing this mass by the atomic weight of nitrogen, 14 g/mol, provides 0.024 moles of nitrogen, and subsequently 0.024 moles of ammonia. According to the reaction stoichiometry, the number of moles of sulfuric acid needed is half this amount, which is 0.012 moles. Given that the sulfuric acid solution has a concentration of 1 M, the required volume is determined by the equation $V = \frac{n}{M}$, leading to $V = \frac{0.012}{1} = 0.012$, which corresponds to 0.012 L or 12 mL of sulfuric acid.

This chemical transformation involves a precipitation reaction between ferric ions and ferrocyanide anions, governed by the requirement to balance their respective charges to form an insoluble coordination complex. By combining four $\mathrm{Fe}^{3+}$ cations with three $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ anions, the reaction yields ferric ferrocyanide, a compound recognized by its characteristic deep Prussian blue color. This specific stoichiometry is illustrated by the reaction equation:

$4\mathrm{FeCl}_3 + 3\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6] \rightarrow \mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3 + 12\mathrm{KCl}$

The identification of sulphide ions in qualitative chemistry often involves a characteristic color change produced through the reaction of sodium nitroprusside. When sulphide ions are added to the nitroprusside complex, $[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NO}]^{2-}$, a chemical transformation occurs represented by the equation $\mathrm{S}^{2-}+[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NO}]^{2-} \rightarrow [\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NOS}]^{4-}$. This specific reaction generates the complex $[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NOS}]^{4-}$, which is well-known for its deep violet color.

Determining the nitrogen content of an organic compound relies specifically on established techniques like Dumas's method or Kjeldahl's method, rather than the Carius method. The Carius method utilizes a sealed tube to digest organic matter for the estimation of halogens, sulfur, and phosphorus, meaning it is not used for nitrogen analysis. Within this protocol, any phosphorus present is oxidized and subsequently precipitated as $\text{Mg}_2\text{P}_2\text{O}_7$ by adding a magnesia mixture, which confirms the validity of the other procedures described.

The redox titration between $\mathrm{KMnO}_{4}$ and $\mathrm{FeSO}_{4}$ in an acidic medium relies on the principle that the number of equivalents of the oxidizing agent must equal the number of equivalents of the reducing agent. In this reaction, $\mathrm{KMnO}_{4}$ acts as the oxidizing agent with an n-factor of 5, as the manganese atom is reduced from the +7 oxidation state to the +2 state. Conversely, $\mathrm{FeSO}_{4}$ acts as the reducing agent, with the iron atom undergoing oxidation from +2 to +3, which corresponds to an n-factor of 1.

The equivalence point can be expressed by the equation:
$M_{1}V_{1}n_{1} = M_{2}V_{2}n_{2}$
Substituting the given values, where $M_{1}$ is the unknown molarity of $\mathrm{KMnO}_{4}$, both volumes are 10 mL, and the molarity of $\mathrm{FeSO}_{4}$ is 0.1 M, the equation becomes:
$M_{1} \times 10 \times 5 = 0.1 \times 10 \times 1$
Solving this yields a molarity of 0.02 M for the $\mathrm{KMnO}_{4}$ solution. To find the strength in g/L, this molarity is multiplied by the molar mass of $\mathrm{KMnO}_{4}$, calculated as 39 + 55 + 4(16) = 158 g/mol. Multiplying 0.02 mol/L by 158 g/mol results in 3.16 g/L, which is equivalent to $316 \times 10^{-2}$ g/L.

To ensure an accurate test for halogens, the sodium extract must be acidified with nitric acid before the addition of silver nitrate, which prevents false results caused by common fusion impurities. If species such as sodium cyanide ($\mathrm{NaCN}$) or sodium sulfide ($\mathrm{Na}_2\mathrm{S}$) are present, they will react with silver nitrate and obscure the detection of halides, so nitric acid is used to decompose them via the reactions $\mathrm{NaCN} + \mathrm{HNO}_{3} \rightarrow \mathrm{NaNO}_{3} + \mathrm{HCN} \uparrow$ and $\mathrm{Na}_2\mathrm{S} + 2\mathrm{HNO}_{3} \rightarrow 2\mathrm{NaNO}_{3} + \mathrm{H}_2\mathrm{S} \uparrow$. By driving off these interfering substances as gases, the acidification step ensures that any precipitate formed is exclusively indicative of halide ions.

The Dumas method determines the amount of nitrogen in an organic compound by oxidizing it with copper(II) oxide, where the stoichiometry depends on the oxidation of carbon to carbon dioxide and hydrogen to water. The general combustion reaction for an organic compound $\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{N}_{z}$ is:
$\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{N}_{z} + \left(2x + \frac{y}{2}\right) \mathrm{CuO} \rightarrow x \mathrm{CO}_{2} + \frac{y}{2} \mathrm{H}_{2} \mathrm{O} + \frac{z}{2} \mathrm{N}_{2} + \left(2x + \frac{y}{2}\right) \mathrm{Cu}$

For N,N-dimethylaminopentane, which has the molecular formula $\mathrm{C}_{7}\mathrm{H}_{17}\mathrm{N}$, we have $x = 7$ and $y = 17$. Substituting these into the stoichiometric coefficient $2x + \frac{y}{2}$ results in 22.5, indicating that each mole of the compound requires 22.5 moles of $\mathrm{CuO}$. The molar mass of the compound is 115 g/mol, meaning a 57.5 g sample contains 0.5 moles of the substance. Multiplying this amount by the 22.5 moles of $\mathrm{CuO}$ required per mole of the compound yields 11.25 moles, which corresponds to $1125 \times 10^{-2}$ moles.

This specific chemical reagent, composed of sulphanilic acid and 1-naphthylamine in an acetic acid medium, acts as a selective indicator for nitrite ions. The process involves the diazotization of the sulphanilic acid by the $NO_{2}^{-}$ species under acidic conditions, which subsequently reacts with 1-naphthylamine to produce a characteristic red-colored azo dye. Because this chromogenic reaction is highly specific to the presence of the $NO_{2}^{-}$ anion, it serves as a reliable qualitative test for its identification.

The first aid principle for handling acid spills involves neutralizing the substance with a mild, safe base. Since toilet cleaning liquids contain approximately 10.5% w/v $\mathrm{HCl}$, applying aqueous sodium bicarbonate ($\mathrm{NaHCO}_3$) effectively counters the acidity. Using sodium hydroxide ($\mathrm{NaOH}$) is dangerous and must be avoided, as it is highly corrosive and would cause additional chemical burns to the skin.

In the Carius method for estimating bromine, we rely on the principle of conservation of mass, where all the bromine in the original organic sample is quantitatively converted into silver bromide. Because each mole of silver bromide contains one mole of bromine atoms, we first find the total molar mass of silver bromide by adding the atomic masses of silver and bromine, which gives $108 + 80 = 188$. Given that 1.88 g of silver bromide was obtained, dividing this mass by the molar mass using the equation $\frac{1.88}{188} = 0.01$ reveals that exactly 0.01 mol of silver bromide was produced.

This result implies that 0.01 mol of bromine atoms were originally present in the sample. By multiplying this molar amount by the atomic mass of bromine, we find the actual mass of bromine to be $0.01 \times 80 = 0.80$ g. To find the percentage by mass in the original compound, we take the ratio of this bromine mass to the initial mass of the organic compound, which is 1.6 g, and multiply by 100. Performing the final calculation using the formula $\frac{0.80}{1.6} \times 100$ confirms that the compound contains 50% bromine.

The estimation of halogen content using the Carius method is based on the principle that the percentage of a halogen in an organic compound is determined by the ratio of the mass of the halogen to the total mass of the compound. Using the provided experimental values of 0.08 g of bromine in 0.172 g of the organic sample, the experimental percentage of bromine is calculated as $\frac{0.08}{0.172} \times 100 \approx 46.51\%$. To determine the correct molecular structure, this experimental result is compared with the theoretical bromine content, derived by calculating the ratio of the atomic mass of bromine to the molar mass of each candidate structure, expressed as $\frac{80}{\text{Molar Mass}} \times 100$. Since the structure in option A is the only one that yields the theoretical value of $\frac{80}{172} \times 100 = 46.51\%$, it is the valid structure for the compound.

Biuret testing detects peptide linkages present in proteins like albumin, whereas the Molisch test identifies the presence of carbohydrates. Because Barfoed's reagent produces a positive result only for monosaccharides and tests negative for disaccharides, substance B must be a monosaccharide such as glucose, while substance A is identified as a disaccharide like lactose. Since substance C yields a positive result in the Biuret test but lacks carbohydrate properties, it is classified as a protein, which is consistent with the characteristics of albumin.

Qualitative analysis relies on identifying functional groups through specific chemical interactions, where the solubility and reactivity profile dictates the molecular identity. The compound's insolubility in dilute $\mathrm{HCl}$ combined with its solubility in $\mathrm{NaOH}$ solution clearly indicates that it is an acidic organic compound rather than a basic one. Furthermore, the rapid decolourisation of $\mathrm{Br}_2$/water confirms the presence of a strongly activating functional group, such as the hydroxyl group found in phenols, which undergoes electrophilic substitution to consume the bromine. These combined analytical results are characteristic of phenols, which exhibit both acidic properties and high reactivity toward bromine substitution.

The absence of a reaction with neutral ferric chloride indicates the compound lacks a phenolic hydroxyl group, while the negative result with Fehling's solution confirms the absence of aliphatic aldehydes. Reactivity with a Grignard reagent signifies the presence of an electrophilic carbonyl center, and a positive iodoform test specifically identifies the structural unit $\text{CH}_3\text{CO}-$ or $\text{CH}_3\text{CH(OH)}-$. These experimental observations collectively characterize the compound as a methyl ketone or a secondary alcohol possessing the requisite methyl group, consistent with the chemical structure provided in option D.

Lysine is an amino acid that reacts with ninhydrin to produce a characteristic purple or violet colored complex, providing a standard test for amino acids. Furfural reacts with 1-naphthol to form a colored product, which is a principle utilized in the detection of carbohydrates where furfural acts as an intermediate. Benzyl alcohol forms a red or yellow complex upon reaction with ceric ammonium nitrate, serving as a diagnostic test for the presence of an alcohol group. Styrene contains a carbon-carbon double bond that undergoes oxidation by potassium permanganate ($KMnO_4$), known as Baeyer's reagent, resulting in the decolorization of the purple solution through the formation of a diol.

Organic solubility analysis is based on the acid-base behavior of functional groups, where phenolic compounds such as m-Cresol react with strong bases like $NaOH$ to form water-soluble phenoxide ions, but lack sufficient acidity to dissolve in weaker bases like $NaHCO_3$. Since carboxylic acids like oleic acid dissolve in $NaHCO_3$ and basic amines like o-Toluidine dissolve in $HCl$, the distinctive behavior of m-Cresol-solubility in $NaOH$ but insolubility in $NaHCO_3$-serves as the definitive diagnostic characteristic for its identification.

This reaction sequence illustrates the standard qualitative analysis test used to confirm the presence of phosphorus in an organic compound. The initial fusion with $\mathrm{Na}_2\mathrm{O}_2$ converts the phosphorus into $\mathrm{Na}_3\mathrm{PO}_4$, which is subsequently converted to $\mathrm{H}_3\mathrm{PO}_4$ upon acidification. The characteristic yellow precipitate formed after the addition of $(\mathrm{NH}_4)_2\mathrm{MoO}_4$ is $(\mathrm{NH}_4)_3\mathrm{PO}_4 \cdot 12\mathrm{MoO}_2$, which serves as a definitive indicator that the original compound contained phosphorus.