JEE Main 2024 question paper pdf with solutions- Download Jan 30 Shift 1 Chemistry Question Paper pdf

Step 1: Understanding the reaction.

The given reaction involves the hydrogenation of acyl chloride over a catalyst, palladium or barium sulfate. The key reaction is the reduction of an acyl chloride to an aldehyde. This is a characteristic feature of the Rosenmund reaction.

Step 2: Analyzing the options.

- (1) Etard reaction: This involves the oxidation of a methyl group on an aromatic ring to form an aldehyde. It is not related to acyl chloride reduction.

- (2) Stephen's reaction: Involves the reduction of an aldehyde to a primary alcohol using tin and hydrochloric acid.

- (3) Wolff Kishner reduction: This reduces a carbonyl group to a methylene group (CH2) by using hydrazine and a base.

- (4) Rosenmund reaction: This is the reduction of acyl chlorides to aldehydes using hydrogen and palladium/barium sulfate. This matches the given reaction.

Step 3: Conclusion.

Therefore, the reaction is correctly identified as the Rosenmund reaction. Quick Tip: For reduction of acyl chlorides to aldehydes, the reaction is known as the Rosenmund reaction.

Step 1: Understanding the Fehling test.

The Fehling test is used to detect reducing sugars. Reducing sugars contain free aldehyde or ketone groups that can reduce Fehling's solution. Sucrose is a non-reducing sugar because it does not have a free aldehyde or ketone group.

Step 2: Analyzing the options.

- (1) Lactose: It is a disaccharide with a free aldehyde group and can reduce Fehling’s solution.

- (2) Maltose: A disaccharide with a free aldehyde group and is a reducing sugar.

- (3) Sucrose: This disaccharide does not have a free aldehyde or ketone group, hence it is non-reducing.

- (4) Glucose: It is a monosaccharide with a free aldehyde group and is a reducing sugar.

Step 3: Conclusion.

Sucrose is the non-reducing sugar and will not give a positive Fehling test. Quick Tip: Sucrose is a non-reducing sugar because it does not have a free aldehyde or ketone group.

Step 1: Understanding the reaction.

The given reaction involves the following steps:
1. Grignard reagent (Bromobenzene reacts with magnesium in dry ether) forms phenyl magnesium bromide.
2. This phenyl magnesium bromide reacts with carbon dioxide, leading to the formation of a carboxylate intermediate.
3. The addition of acid converts this intermediate into benzoic acid.
4. After the final step with ammonia and a base, the product formed is an amine.

Step 2: Analyzing the options.

- (1) Aniline: The final product is an amine group, so aniline is formed.

- (2) Acetanilide: It would require an acylation step, but that is not involved in this reaction.

- (3) Benzamide: This would require an additional acylation reaction with ammonia, not occurring in this reaction.

- (4) Benzoic acid: It is an intermediate in this reaction, but the final product is aniline.

Step 3: Conclusion.

The final product of the reaction is aniline. Quick Tip: Grignard reagents, when reacted with CO2, form carboxylate intermediates, which then convert to benzoic acid, and further treatment with ammonia leads to aniline formation.

Step 1: Understanding stability.

The stability of cyclic compounds depends on aromaticity, which follows Hückel’s rule. For a compound to be aromatic, it must have a conjugated \(\pi\)-system with \(4n+2\) \(\pi\)-electrons, where \(n\) is a non-negative integer.

Step 2: Analyzing the options.

- (1) Cyclopropene: This compound is not aromatic. Its instability is due to its strained three-membered ring.

- (2) Benzene: Benzene is highly stable and aromatic, but we are looking for the most stable among the given options, which involves understanding how other ions and compounds interact.

- (3) Cycloheptatrienyl cation: This ion has 6 \(\pi\)-electrons and follows the \(4n+2\) rule for aromaticity. This makes it the most stable among the given compounds.

- (4) Cyclopropyl cation: Although cyclopropyl cation has some stability due to delocalization, it is still less stable than the cycloheptatrienyl cation.

Step 3: Conclusion.

The cycloheptatrienyl cation is the most stable due to its aromaticity. Quick Tip: Aromaticity provides additional stability to cyclic compounds with conjugated double bonds and a \(4n+2\) \(\pi\)-electron system.

Step 1: Understanding diamagnetism.

Diamagnetic substances are those that do not have any unpaired electrons. In other words, they have a completely paired electron configuration, which results in no net magnetic moment.

Step 2: Analyzing the options.

- (1) Ni\(^{2+}\), Cu\(^{2+}\): Both of these ions have unpaired electrons, making them paramagnetic.

- (2) Eu\(^{3+}\), Gd\(^{3+}\): These ions also have unpaired electrons, so they are paramagnetic.

- (3) Cu\(^+\), Zn\(^{2+}\): Cu\(^+\) has a d\(^{10}\) configuration, and Zn\(^{2+}\) has a d\(^{10}\) configuration, both of which are fully paired, making them diamagnetic.

- (4) Ce\(^{4+}\), Pr\(^{3+}\): Both of these ions have unpaired electrons, so they are paramagnetic.

Step 3: Conclusion.

Cu\(^+\) and Zn\(^{2+}\) are both diamagnetic, as they have fully paired electrons. Quick Tip: Diamagnetic ions have completely paired electrons and show no magnetic moment.

Step 1: Understanding degeneracy in orbitals.

Degenerate orbitals are orbitals with the same energy level. In the case of hydrogen atoms, orbitals in the same principal quantum number (n) are degenerate (i.e., have the same energy), but only in s, p, and d orbitals of the same energy level.

Step 2: Analyzing the statements.

- Statement-I: In the case of hydrogen, the 3p and 3d orbitals do not have the same energy. The 3d orbitals are higher in energy compared to the 3p orbitals. Therefore, statement-I is incorrect.

- Statement-II: The definition of degeneracy is correct. Orbitals with the same energy level are degenerate, but this does not apply to the 3p and 3d orbitals in hydrogen.

Step 3: Conclusion.

Therefore, statement-I is incorrect, but statement-II is correct. Quick Tip: For the hydrogen atom, orbitals of the same principal quantum number (n) are degenerate, but p and d orbitals of the same level have different energies.

Step 1: Understanding the reaction.

The first reaction involves the dehydrohalogenation of a halide to form a conjugated alkene. The second reaction generates the sodium salt of the conjugated alkene.

Step 2: Analyzing the options.

- (1) CH\(_3\)-C=CH: This represents the initial conjugated alkene. The second part does not match the expected product.

- (2) CH\(_3\)-C=Na: This is an intermediate compound that forms after sodium reacts with the alkene. The second part is not correct.

- (3) CH\(_3\)-C=CH: The first part shows the formation of a conjugated alkene, and the second part correctly shows the product with sodium addition. This is the correct option.

- (4) CH\(_3\)-C=CH: This part is correct, but the second part of the product does not match the expected outcome.

Step 3: Conclusion.

The correct sequence of reactions is option (3). Quick Tip: Dehydrohalogenation reactions lead to the formation of conjugated alkenes, and the addition of sodium forms sodium salts of conjugated alkenes.

Step 1: Understanding the assertion.

As we move down the group in the periodic table, the atomic and ionic radii increase. For example, from arsenic (As) to bismuth (Bi), the covalent radius does increase, but only marginally because the addition of electron shells makes the increase less noticeable.

Step 2: Understanding the reason.

The reason is generally correct. The covalent and ionic radii do increase as you move down a group in the periodic table, but the rate of increase can be slow due to the relativistic effects and electron shielding.

Step 3: Conclusion.

Both (A) and (R) are correct, but the reason does not fully explain the assertion as it is not a major contributor to the small observed increase from As to Bi. Quick Tip: Relativistic effects and electron shielding play important roles in determining the covalent and ionic radii in heavier elements.

Step 1: Understanding the electronic configurations.

We are matching the metal ions with their correct electronic configurations.

Step 2: Analyzing the options.

- (1) Mn\(^{2+}\): Mn\(^{2+}\) has the configuration 3d\(^5\) 4s\(^0\) after losing two electrons from the 4s orbital.

- (2) V\(^{+}\): The electron configuration for V\(^{+}\) should be 3d\(^3\) 4s\(^2\), not matching the given configuration.

- (3) Cr\(^{3+}\): Cr\(^{3+}\) has 3d\(^5\) 4s\(^0\) as the correct configuration after losing three electrons from the 4s and 3d orbitals. This matches the given electronic configuration.

- (4) Fe\(^{2+}\): Fe\(^{2+}\) has the configuration 3d\(^6\) 4s\(^2\), not fitting the description.

Step 3: Conclusion.

The correct match is (3). Quick Tip: When matching electronic configurations, always consider the loss of electrons from the 4s orbital before the 3d orbital.

Step 1: Identifying the reaction.

When sulphide ions react with dilute sulfuric acid, hydrogen sulfide (H\(_2\)S) gas is evolved. This gas reacts with lead acetate paper and forms lead sulfide (PbS), which turns the paper black.

Step 2: Analyzing the options.

- (1) Sulphite: Sulphites react with acids to form sulfur dioxide (SO\(_2\)), which does not turn lead acetate paper black.

- (2) Sulphide: This is the correct answer as sulphides release H\(_2\)S gas, which reacts with lead acetate to form PbS, turning the paper black.

- (3) Sulphate: Sulphates do not release H\(_2\)S gas, so this is incorrect.

- (4) Thiosulphate: Thiosulphates release sulfur dioxide (SO\(_2\)) upon reaction with acids, not H\(_2\)S, and thus would not turn the paper black.

Step 3: Conclusion.

The correct answer is (2), as sulphides release hydrogen sulfide gas, which forms lead sulfide. Quick Tip: When sulphides react with dilute acids, they release hydrogen sulfide gas (H\(_2\)S), which turns lead acetate paper black due to the formation of PbS.

Step 1: Understanding work done in a cyclic process.

In a cyclic process, work done is equal to the area enclosed by the curve on a pressure-volume (P-V) diagram. This is represented by the area inside the figure.

Step 2: Calculating the area.

The area inside the P-V diagram is given by the formula for the area of a triangle: \[ Area = \frac{1}{2} \times Base \times Height \]
Substitute the values from the figure to find the work done: \[ Area = \frac{1}{2} \times 20 \times 20 = 200 \, J \]

Step 3: Conclusion.

The work done in the cyclic process is 200 J. Quick Tip: To find work done in a cyclic process on a P-V diagram, calculate the area enclosed by the curve.

Step 1: Understanding hybridization.

When 2s and 2p orbitals of a single atom are mixed, they form hybrid orbitals. The number of hybrid orbitals formed is equal to the number of atomic orbitals mixed.

Step 2: Analyzing the options.

- When 2s and 2p orbitals mix, they form 4 hybrid orbitals, which are sp\(^3\). This is the maximum number of hybrid orbitals that can be formed.

Step 3: Conclusion.

The maximum number of hybrid orbitals formed is 4. Quick Tip: When mixing s and p orbitals, the number of hybrid orbitals formed equals the number of orbitals mixed.

Step 1: Calculating the volume of the silver plate.

Volume of the plate \( V = Area \times Thickness = 0.05 \, cm^2 \times 0.05 \, cm = 0.0025 \, cm^3 \).

Step 2: Determining the number of moles of silver.

The density of silver is 7.9 g/cm\(^3\), so the mass of the plate is: \[ Mass = Density \times Volume = 7.9 \times 0.0025 = 0.01975 \, g. \]
Molar mass of silver is 107.9 g/mol, so the number of moles of silver is: \[ Moles = \frac{0.01975}{107.9} = 1.83 \times 10^{-4} \, mol. \]

Step 3: Calculating the number of atoms.

Using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol), the number of atoms is: \[ Atoms = 1.83 \times 10^{-4} \times 6.022 \times 10^{23} = 1.1 \times 10^{19}. \]

Step 4: Conclusion.

The number of atoms is approximately \( 11 \times 10^{19} \). Quick Tip: Use the formula for volume and density to calculate mass, then use the molar mass and Avogadro's number to find the number of atoms.