JEE Main 2024 question paper pdf with solutions- Download April 9 Shift 2 Chemistry Question Paper pdf

Step 1: Understanding the reaction.

KCN in DMSO gives a strong nucleophilic SN2 reaction. The attack occurs from the backside, causing inversion of configuration at the carbon where Br was attached.


Step 2: Analyzing the stereochemistry.

Since SN2 causes complete inversion, the CN group must appear on the opposite side of the original Br orientation. Among the given structures, only option (B) shows correct stereochemical inversion.


Step 3: Conclusion.

Thus, product with inverted stereochemistry at the reacting carbon is formed → structure (B).
Quick Tip: SN2 reactions always proceed with backside attack and cause complete inversion of configuration.

Step 1: Determine the molecular orbital configurations.

For O₂: total electrons = 16 → \(\pi^*\) has 2 electrons.

For O₂₊: one electron removed → \(\pi^*\) = 1 electron.

For O_2–: one added → \(\pi^*\) = 3 electrons.


Step 2: Add all \(\pi^*\) electrons.

1 (O₂₊) + 2 (O₂) + 3 (O_2–)
= 4 electrons.


Step 3: Conclusion.

Total \(\pi^*\) electrons in all species combined = 4.
Quick Tip: Remember that O₂ has two unpaired electrons in \(\pi^*\) orbitals, forming the basis of its paramagnetism.

Step 1: Identify the element with Z = 99.

Z = 99 corresponds to Einsteinium (Es), which belongs to the actinide series.


Step 2: Apply the filling order for actinides.

For actinides, electrons fill the 5f subshell first, followed by 7s.
The general pattern around this region shows increasing filling of 5f orbitals.


Step 3: Einsteinium configuration.

Einsteinium has the configuration:
[Rn] 5f¹¹ 7s².


Step 4: Conclusion.

Thus, the correct configuration is [Rn] 5f¹¹ 7s².
Quick Tip: For actinides, the 5f subshell fills progressively; Es (Z = 99) naturally reaches 5f¹¹.

Step 1: Identify functional groups.

The given molecule contains a nitrile (–CN) and an ester group (–CO₂CH₃).
Grignard reagent MeMgBr reacts strongly with both.


Step 2: Reaction with MeMgBr (excess).

(1) Grignard adds twice to the nitrile, converting it into a ketone, then into a tertiary alcohol.

(2) Grignard also reacts with the ester:
Ester + 2 MeMgBr → tertiary alcohol after hydrolysis.


Step 3: Identify the final product.

Both functional groups ultimately form alcohols.
The major product contains a tertiary alcohol attached to the benzene ring and is shown in structure (C).


Step 4: Conclusion.

Thus, the final major product is structure (C), containing the tertiary alcohol.
Quick Tip: Excess Grignard + nitrile or ester always gives a tertiary alcohol after hydrolysis.

Step 1: Identify stereogenic elements.

The given molecule contains:
(1) A C=C double bond with two different substituents → 1 geometrical (E/Z) center.

(2) Two chiral carbons (each bonded to four different groups).


Step 2: Total stereoisomer formula.

For a molecule with one C=C (E/Z) + two chiral centres:

Total stereoisomers = \(2^{n} \times\) (E/Z possibilities)
\(n = 2\) chiral carbons → \(2^{2} = 4\)

E/Z possibilities = 2

Total = \(4 \times 2 = 8\) stereoisomers.


Step 3: Conclusion.

Thus, the compound can exist in 8 stereoisomeric forms.
Quick Tip: For stereochemistry: multiply \(2^n\) (chiral centres) by the number of geometrical (E/Z) isomers.

Step 1: Rule for Friedel–Crafts reaction.

Friedel–Crafts alkylation/acylation does NOT occur with:
• Strongly deactivating groups (–NO₂)

• Basic groups (–NH₂) which get protonated by AlCl₃ and form salts


Step 2: Analyzing each compound.

(a) Nitrobenzene → Strongly deactivated → Does NOT undergo FC reaction.

(b) Aniline → NH₂ forms salt with AlCl₃ → Does NOT undergo FC reaction.

(c) Benzene → Undergoes FC easily.

(d) Toluene → Activating group → Undergoes FC easily.

(e) Phenoxy derivatives → O–alkyl bond is stable, benzene ring activated → FC occurs.

(f) Isopropylbenzene → Ring is activated → FC occurs.


Step 3: Conclusion.

Only (a) and (b) fail to give Friedel–Crafts reaction → Total = 2 compounds.
Quick Tip: Friedel–Crafts fails with –NO₂ (strongly deactivating) and –NH₂ (forms AlCl₃ salt).

Step 1: Analyse each complex.

(P) [Ni(CO)₄]: CO is a strong field, neutral ligand. Ni is in 0 oxidation state → forms tetrahedral complex → hybridisation = sp³. So P → 2.

(Q) [Co(NH₃)₆]Cl₃: Co is in +3 oxidation state, NH₃ is strong ligand → inner orbital octahedral → hybridisation = d²sp³. So Q → 3.

(R) [Ni(CN)₄]^2-: CN^- is strong ligand → square planar complex → hybridisation = dsp². So R → 1.

(S) [CoF₆]^3-: F^- is weak ligand → high spin octahedral → hybridisation = sp³d². So S → 4.


Step 2: Match with the given options.

The correct matching P → 2, Q → 3, R → 1, S → 4 corresponds exactly to option (1).


Step 3: Conclusion.

Thus, the correct match is given in option (1).
Quick Tip: Strong-field ligands (like CN^- and NH₃) often produce inner orbital complexes (d²sp³ or dsp²), while weak-field ligands (like F^-) form outer orbital complexes (sp³d²).

Step 1: Analyse the resonance contributors.

(I) Contains a neutral resonance form with a stable conjugated system → most stable.

(II) Contains charge separation but the positive charge is adjacent to a double bond → moderately stable.

(III) Contains a less favourable charge distribution (negative on carbon and positive on oxygen), which is highly unstable → least stable.


Step 2: Apply resonance stability principles.

Neutral structures are more stable than charge-separated ones.

Among charge-separated forms, structures with proper charge placement (negative on more electronegative atoms) are more stable.

Here, (II) is better than (III) because the charge distribution in (III) is highly unfavourable.


Step 3: Conclusion.

Thus, the stability order follows: I > II > III.
Quick Tip: Neutral resonance structures are always more stable than charge-separated forms. Also, negative charges prefer electronegative atoms, while positive charges prefer electropositive atoms.

Step 1: Understand second ionisation energy.

Second ionisation energy refers to the energy required to remove an electron from a positively charged ion. The higher the second ionisation energy, the more difficult it is to remove the second electron, implying a stable electron configuration before the second electron is removed.


Step 2: Ionisation energy and electron configuration.

Sc (Z = 21): [Ar] 3d¹ 4s² → second ionisation removes 4s² electrons, which are weakly bound.

Ti (Z = 22): [Ar] 3d² 4s² → second ionisation energy is high due to stable 3d² configuration.

V (Z = 23): [Ar] 3d³ 4s² → second ionisation energy is moderate.

Cr (Z = 24): [Ar] 3d⁵ 4s¹ → very stable due to half-filled 3d⁵ configuration; second ionisation energy is higher.

Mn (Z = 25): [Ar] 3d⁵ 4s² → stable 3d⁵ configuration but second ionisation energy is moderate.


Step 3: Conclusion.

The element with the maximum second ionisation energy is Ti⁺, as it has the most stable electron configuration (3d²) before the second ionisation.
Quick Tip: Elements with stable d-orbital configurations (like Ti⁺ with 3d²) tend to have higher second ionisation energies because the electron is more tightly bound.

Step 1: Understand glucose properties.

- Glucose is an aldohexose, meaning it is a 6-carbon sugar with an aldehyde group at the C1 position, making it an Aldohexose (Option 1 is correct).

- Glucose has several isomeric forms in aqueous medium due to its ability to cyclize and form an equilibrium of anomers (Option 2 is correct).

- Glucose is soluble in water not only because of its aldehyde group but because of its numerous hydroxyl groups (not just the aldehyde) that form hydrogen bonds with water molecules. Therefore, the presence of the aldehyde group alone is not the primary reason for solubility (Option 3 is incorrect).

- Glucose is a reducing sugar because the aldehyde group can reduce other compounds (Option 4 is correct).


Step 2: Conclusion.

The incorrect statement is (3), as glucose is soluble in water primarily due to the hydroxyl groups and not just the aldehyde functional group.
Quick Tip: Glucose's solubility in water is primarily due to its multiple hydroxyl groups, which can form hydrogen bonds with water molecules.

Step 1: Understand the Tollen's test.

Tollen's test is used to identify aldehydes. It involves the reduction of silver ions (Ag⁺) to metallic silver, forming a silver mirror on the test tube.


Step 2: Identify compounds that give a positive Tollen's test.

- Acetone (I) is a ketone and does not give a positive Tollen's test.

- Formaldehyde (II), Formic acid (III), and Benzaldehyde (V) are all aldehydes, which will give a positive Tollen's test.

- Acetic acid (IV) is a carboxylic acid and does not give a positive result.


Step 3: Conclusion.

The correct answer is (1), as all listed compounds except acetone give a positive Tollen's test.
Quick Tip: Only aldehydes (not ketones or carboxylic acids) give a positive Tollen's test.

Step 1: Understand the reaction.

When an aromatic halide (in this case, bromobenzene) reacts with alcoholic potassium cyanide (KCN), the nucleophilic cyanide (CN^-) will replace the halide (Br) via a nucleophilic substitution reaction (SN2).


Step 2: Analyze the compounds.

The compound is a bromobenzene derivative with a methyl group (CH₃) and a halogen (Br). The reaction with KCN will replace the Br group with a CN^-, resulting in the formation of a phenylacetonitrile derivative. The product will be a nitrile compound (CN-CH₃).


Step 3: Conclusion.

The correct product is CN-CH₃ (option 3).
Quick Tip: When an aromatic halide reacts with KCN in alcoholic solution, the cyanide ion (CN^-) replaces the halogen atom, forming a nitrile group.

Step 1: Understanding the properties.

- Size increases down the group. So, Tl has the largest size and B has the smallest size (P → 1).

- Ionization enthalpy decreases down the group as the atomic size increases, so Tl has the lowest ionization enthalpy and B has the highest (Q → 2).

- Melting point decreases as we move down the group, with B having the highest melting point and Tl the lowest (R → 3).

- Ionic radius increases down the group, with Tl having the largest ionic radius and B having the smallest (S → 4).


Step 2: Conclusion.

Thus, the correct order is P → 1; Q → 2; R → 3; S → 4.
Quick Tip: For Group 13 elements, remember that the size increases down the group, and ionization enthalpy, melting point, and ionic radius all decrease down the group.

Step 1: Understand the concept of R_f.

R_f (B) and R_f (A) refer to the retardation factor of substances A and B in chromatography, calculated by the ratio of the distance traveled by the substance to the distance traveled by the solvent front. The given relation indicates a proportional relationship between the two.


Step 2: Analyze the data from the image.

The given data indicates the distances moved by A and B on the chromatogram. The distance moved by B is 9 cm, and the distance moved by A is 5 cm. The formula for the retardation factor is given by:

R_f (B) = n × R_f (A), so we substitute:

9 = n × 5. Solving for n, we get: n = 9/5 = 1.8.


Step 3: Conclusion.

Thus, the value of X in the equation n = (X) × 10^-1 is 18.
Quick Tip: In chromatography, the retardation factor (R_f) is a measure of how far a substance moves relative to the solvent front. A higher R_f indicates a faster-moving substance.

Step 1: Understand the equation for work done.

In this process, \( pv^3 = C \) is given. The work done \( W \) for a process is given by: \[ W = \int_{V_1}^{V_2} P \, dV \]
From the equation \( pv^3 = C \), we have: \[ P = \frac{C}{V^3} \]
Substitute into the equation for work done: \[ W = \int_{V_1}^{V_2} \frac{C}{V^3} \, dV \]

Step 2: Integrate and calculate.

Performing the integration gives us the work done. Given that the temperature increases from 300 K to 330 K, we can use the ideal gas law to relate pressure and volume with temperature.

Using the relationship, the work done is found to be 250 J.


Step 3: Conclusion.

Thus, the correct answer is (2) 250 J.
Quick Tip: For processes like \( pv^3 = C \), the work done can be calculated by integrating the pressure with respect to volume using the ideal gas law.

Step 1: Use the relationship between \( \Delta H \) and \( \Delta S \).

The equilibrium temperature \( T \) in a reaction is related to the change in enthalpy \( \Delta H \) and the change in entropy \( \Delta S \) by the equation: \[ \Delta G = \Delta H - T \Delta S \]
At equilibrium, \( \Delta G = 0 \), so: \[ 0 = \Delta H - T \Delta S \implies T = \frac{\Delta H}{\Delta S} \]

Step 2: Apply the given values.

From the question, \( \Delta H = x \) and \( \Delta S = y \). Thus, the equilibrium temperature is \( T = \frac{x}{y} \).


Step 3: Conclusion.

The correct answer is (2) \( \frac{x}{y} \).
Quick Tip: For equilibrium temperature, use the equation \( T = \frac{\Delta H}{\Delta S} \) to find the temperature at which the reaction is at equilibrium.

Step 1: Use the relation for strong electrolytes.

For strong electrolytes, the limiting molar conductivity (\( \lambda_m^0 \)) is related to the conductivity at any concentration (\( \lambda_m \)) through the equation: \[ \lambda_m - \lambda_m^0 = A \lambda_C \]
where \( \lambda_C \) is the molar conductivity at the given concentration and \( A \) is a constant that depends on the electrolyte and its concentration.


Step 2: Apply the given options.

The correct relation for strong electrolytes is given by option (1) as: \[ \lambda_m - \lambda_m^0 = A \lambda_C \]

Step 3: Conclusion.

Thus, the correct answer is (1) \( \lambda_m - \lambda_m^0 = A \lambda_C \).
Quick Tip: For strong electrolytes, the difference between molar conductivity at any concentration and the limiting molar conductivity is proportional to the concentration of the electrolyte.

Step 1: Understand the diagram.

The work done is the area enclosed in the PV diagram. The area for a triangular process can be calculated as: \[ W = \frac{1}{2} \times Base \times Height \]
Given the pressure and volume changes, the work done in the cyclic process is computed to be 250 J.

Step 2: Conclusion.

Thus, the work done in the process is 250 J. 250 J.
Quick Tip: For cyclic processes, the area enclosed by the process on the PV diagram gives the work done during the cycle.