JEE Main 2024 question paper pdf with solutions- Download Jan 31 Shift 2 Chemistry Question Paper pdf

Step 1: Understanding the properties of [NiCl₄]²⁻.

Nickel in [NiCl₄]²⁻ is in the +2 oxidation state. The electronic configuration of Ni²⁺ is [Ar] 3d⁸, which means there are unpaired electrons, making it paramagnetic. Thus, option (1) and (3) are incorrect.


Step 2: Understanding the properties of [Ni(CO)₄].

Nickel in [Ni(CO)₄] is in the zero oxidation state. Carbon monoxide (CO) is a strong field ligand and causes pairing of electrons in the 3d orbitals. This results in a diamagnetic complex. Therefore, option (2) is correct.


Step 3: Conclusion.

The correct answer is (2) because [Ni(CO)₄] is diamagnetic.
Quick Tip: For metal complexes, the type of ligand plays a crucial role in determining the magnetic properties of the complex. Strong field ligands like CO tend to cause pairing of electrons, making the complex diamagnetic.

Step 1: Understanding the reaction.

The reaction is a Markovnikov addition, where the halogen (Cl) adds to the more substituted carbon atom. The D is a deuterium (hydrogen isotope) that will remain unchanged in the reaction.

Step 2: Markovnikov's rule.

According to Markovnikov's rule, the electrophile (Cl) will add to the carbon that already has more substituents (in this case, a more substituted carbon). The product formed will have the Cl atom added to the more substituted carbon, while the D will remain on the other carbon.

Step 3: Conclusion.

The correct product formed is given in option (1).
Quick Tip: Markovnikov's rule applies to reactions of unsymmetrical alkenes with HX, where the more substituted carbon gets the halide.

Step 1: Understanding the reaction mechanism.

The given reaction is a Friedel-Crafts acylation reaction, where the acyl group from the acyl chloride reacts with the aromatic ring in the presence of an anhydrous catalyst (AlCl₃). The reaction forms a new bond between the acyl group and the benzene ring.


Step 2: Reaction details.

In this reaction, the carbonyl group from the acyl chloride will attack the benzene ring at the position that leads to the most stable intermediate. The reaction produces a substituted aromatic compound where the acyl group is attached to the ring.


Step 3: Conclusion.

The major product is the one formed by the Friedel-Crafts acylation, which results in a benzene ring with a substituted carbonyl group attached. This corresponds to option (1).
Quick Tip: In Friedel-Crafts acylation reactions, an acyl chloride reacts with an aromatic compound in the presence of an AlCl₃ catalyst to form an aryl ketone.

Step 1: Understanding the reaction.

The given reaction is the Gattermann Koch reaction, which involves the synthesis of aromatic aldehydes from aromatic compounds using carbon monoxide (CO) and hydrochloric acid (HCl) in the presence of a catalyst, such as anhydrous AlCl₃.


Step 2: Conclusion.

The reaction corresponds to the Gattermann Koch reaction, where the aromatic ring reacts with CO and HCl, leading to the formation of the aromatic aldehyde. Therefore, the correct answer is option (3).
Quick Tip: In the Gattermann Koch reaction, an aromatic compound is converted into an aldehyde using CO, HCl, and a Lewis acid like AlCl₃.

Step 1: Understanding the reaction.

The given reaction involves a bromobenzene compound. The first step of the reaction is nitration using concentrated HNO₃, which results in the formation of nitrobenzene substituted at the para position to the bromine.


Step 2: Reaction conditions.

The second step is carried out in the presence of NaOH at 573 K, which leads to the dehydrobromination of the compound, replacing the bromine with a hydroxyl group, resulting in the formation of a phenolic compound, i.e., the compound B.


Step 3: Conclusion.

The correct products are A = \(Br-C_6H_4-NO_2\) (bromonitrobenzene) and B = \(OH-C_6H_4-NO_2\) (hydroxy-nitrobenzene), corresponding to option (2).
Quick Tip: In the nitration reaction, bromine is a deactivating group at the meta position, which leads to the nitro group being placed at the para position. The hydroxylation reaction is performed in the presence of NaOH.

Step 1: Understanding the compound.

The given compound is 2,2-dibromo-1-phenylpentane, which means the structure has a phenyl group (Ph) attached to the first carbon of a pentane chain. Additionally, two bromine atoms are attached to the second carbon of the chain.


Step 2: Analyzing the options.

- Option (1) shows a structure where the bromine atoms are attached to different carbons, which does not match the given IUPAC name.

- Option (2) is the correct structure, with the two bromine atoms attached to the second carbon of the pentane chain and the phenyl group attached to the first carbon.

- Option (3) and (4) do not match the IUPAC name, as they either have incorrect positions for the bromines or missing substituents.


Step 3: Conclusion.

The correct IUPAC structure is option (2).
Quick Tip: When determining the IUPAC name or structure, carefully analyze the position of substituents and the parent chain to ensure correct naming.

Step 1: Understanding the reaction.

The reaction is an azo coupling reaction, which occurs between a diazonium salt (in this case, N₂Cl⁻) and an electron-rich aromatic compound. In this case, the methyl group (Me) is attached to the nitrogen of aniline, and the coupling occurs with a sulfonated benzene ring (C₆H₄SO₃Na). The result is the formation of an azo compound, which is the product formed in this reaction.


Step 2: Conclusion.

The correct product formed is methyl orange, and the correct structure matches option (1).
Quick Tip: Azo coupling reactions typically form products where the diazonium salt reacts with electron-rich aromatic compounds. The resulting product often has a characteristic bright color, as seen in methyl orange.

The work done during an isothermal reversible expansion is given by the formula: \[ w = -nRT \ln \left( \frac{V_2}{V_1} \right) \]

Taking the magnitude of the work, we have: \[ |w| = 2.303 \, nRT \log \left( \frac{V_2}{V_1} \right) \]

Step 1: Substitute the given values.

Here, \( n = 1 \) mol, \( R = 8.314 \, J/mol K \), \( T = 300 \, K \), \( V_2 = 100 \, L \), and \( V_1 = 10 \, L \).
\[ |w| = 1 \times 2.303 \times 8.314 \times 300 \log \left( \frac{100}{10} \right) \]

Step 2: Calculate the work.
\[ |w| = 2.303 \times 8.314 \times 300 \log 10 \] \[ |w| = 5744 \, J = 5.744 \, kJ \approx 6 \, kJ \]

Thus, the magnitude of the work done is approximately 6 kJ.
Quick Tip: For isothermal reversible processes, the work done is related to the ratio of the final and initial volumes by \( w = -nRT \ln \left( \frac{V_2}{V_1} \right) \). Use logarithms in base 10 to simplify the calculation.