JEE Main 2024 question paper pdf with solutions- Download April 8 Shift 2 Chemistry Question Paper pdf

Step 1: Reaction of P (Aniline).

P undergoes diazotization using NaNO\(_2\)/HCl followed by hydrolysis to form phenol. List-II option (1) is phenol. Therefore, P → 1.
But phenol appears again in S, so verify next.


Step 2: Reaction of Q (Phenol with CO\(_2\)/KOH).

This is Kolbe–Schmitt reaction, forming salicylic acid (o-hydroxy benzoic acid). In List-II, salicylic acid corresponds to (4). Hence Q → 4.


Step 3: Reaction of R (Phenol + CHCl\(_3\)/KOH).

This is Reimer–Tiemann reaction forming salicylaldehyde (o-hydroxy benzaldehyde). In List-II, this is compound (3). So R → 3.


Step 4: Reaction of S (p-Cresol + CrO\(_3\) oxidation).

Oxidation of the methyl group converts it to aldehyde (p-hydroxy benzaldehyde). In List-II, this is (2). So S → 2.


Step 5: Final Matching.

P → 2, Q → 4, R → 3, S → 1 matches option (2).
Quick Tip: Kolbe gives salicylic acid, Reimer–Tiemann gives salicylaldehyde, and oxidation of methyl on benzene ring gives aldehyde or acid.

Step 1: Understanding bond order.

Bond order is calculated using:
Bond order = (number of bonding electrons – antibonding electrons)/2


Step 2: Analyse each species.

O\(_2\): Molecular orbital configuration gives bond order = 2.

C\(_2\)H\(_6\): All single sigma bonds, bond order = 1.

H\(_2\): Bond order = 1.

Ne\(_2\): Bond order = 0 (equal bonding and antibonding electrons).


Step 3: Conclusion.

Only O\(_2\) has bond order equal to 2.
Quick Tip: For diatomic molecules, bond order from MO theory gives accurate prediction of stability.

Step 1: Identify the longest carbon chain.

The parent chain contains 7 carbons → Heptane.


Step 2: Number the chain from the nearest substituent end.

Numbering gives substituents at C-2, C-4, and C-6.


Step 3: List substituents.

C-2: Methyl

C-4: Methyl

C-6: Ethyl


Step 4: Apply alphabetical order rule.

Ethyl comes before methyl → Name starts with “6-ethyl”.


Step 5: Final name.

6-ethyl-2,4-dimethylheptane.
Quick Tip: Always choose the longest chain and arrange substituents alphabetically in IUPAC naming.

Step 1: Understanding bonding vs antibonding combination.

Bonding molecular orbitals are formed by constructive interference of atomic orbitals, where wave functions add with the same sign.


Step 2: Identifying antibonding combinations.

Antibonding orbitals (\(\sigma^*\)) are formed by destructive interference, meaning wave functions combine with opposite signs.


Step 3: Conclusion.

Destructive combination corresponds to subtraction of wave functions: \(\psi_A - \psi_B\).
Quick Tip: Bonding = addition of wave functions; Antibonding = subtraction.

Step 1: Apply Huckel rule (4n + 2 \(\pi\) electrons).

Aromatic compounds must be planar, cyclic, fully conjugated, and obey Huckel rule.


Step 2: Check each structure one by one.

1. First bicyclic fused ring → Non-aromatic (not fully conjugated).

2. Bicyclic with bridging → Non-aromatic (distorted).

3. Cyclooctatetraene → Non-aromatic (tub-shaped, avoids conjugation).

4. Benzene-like structure → Aromatic.

5. 7-membered ring with 3 double bonds → Aromatic (6 π electrons, planar).

6. Pyridine → Aromatic (6 π electrons).

7. Benzene → Aromatic.


Step 3: Conclusion.

Total aromatic compounds = 4.
Quick Tip: Always check planarity and full conjugation before applying Huckel rule.

Step 1: Determine the number of unpaired electrons in each complex.

(P) [CoF\(_6\)]\(^{3-}\): High spin, Co\(^{3+}\) (d\(^6\)) → 4 unpaired e\(^-\) → \(\mu = 4.9 \approx 5\).

(Q) [Ni(CN)\(_4\)]\(^{2-}\): Square planar, low spin, Ni\(^{2+}\) (d\(^8\)) → 0 unpaired → \(\mu = 0\).

(R) [Ni(NH\(_3\))\(_6\)]\(^{2+}\): Octahedral, Ni\(^{2+}\) (d\(^8\)) → 2 unpaired → \(\mu = 2.8 \approx 3\).

(S) [Fe(H\(_2\)O)\(_6\)]\(^{3+}\): High spin, Fe\(^{3+}\) (d\(^5\)) → 5 unpaired → \(\mu = 5.9 \approx 6\).


Step 2: Match with List-II.

P → 5 → (1)

Q → 0 → (2)

R → 3 → (3)

S → 6 → (4)


Step 3: Conclusion.

Correct matching is option (1).
Quick Tip: Use \(\mu = \sqrt{n(n+2)}\) to estimate magnetic moment from unpaired electrons.

Step 1: Use de Broglie relationship.
\(\lambda = \dfrac{h}{\sqrt{2mK}}\)
If \(\lambda_e = \lambda_p\), then \(\sqrt{2m_eK_e} = \sqrt{2m_pK_p}\)


Step 2: Square both sides.
\(m_e K_e = m_p K_p\)


Step 3: Solve for kinetic energy.

Since \(m_p \gg m_e\), \(K_e = \dfrac{m_p}{m_e} K_p\)
This means \(K_e\) is much larger than \(K_p\).


Step 4: Conclusion.

Electron has higher kinetic energy than the proton when both have same de Broglie wavelength.
Quick Tip: For equal de Broglie wavelengths: lighter particle always has greater kinetic energy.

Step 1: Understand chromate–dichromate equilibrium.

Cr\(_2\)O\(_7^{2-}\) (orange) \(\rightleftharpoons\) 2 CrO\(_4^{2-}\) (yellow).

This equilibrium shifts towards chromate (CrO\(_4^{2-}\)) in basic medium.


Step 2: Why basic medium?

Addition of OH\(^-\) removes H\(^+\) from equilibrium and shifts reaction:
Cr\(_2\)O\(_7^{2-}\) + 2OH\(^-\) → 2CrO\(_4^{2-}\) + H\(_2\)O


Step 3: Conclusion.

Forward reaction is favored in basic medium.
Quick Tip: Chromate (yellow) is favored in basic medium; dichromate (orange) in acidic medium.

Step 1: Recall Hinsberg test.

Benzene sulphonyl chloride reacts with 1° and 2° amines but NOT with 3° amines.
• 1° amines → sulphonamides (soluble in NaOH)
• 2° amines → insoluble sulphonamides
• 3° amines → NO reaction


Step 2: Analyze Statement-I.

It says benzene sulphonyl chloride reacts with all (1°, 2°, 3°) amines → This is wrong.
Actually 3° amines do not react.


Step 3: Analyze Statement-II.

Statement-II says all products are soluble in NaOH → Incorrect.
Only 1° amine product is soluble; 2° amine product is insoluble.


Step 4: Conclusion.

Statement-I is correct (reacts with 1°, 2°, NOT 3°—but question implies reaction with 1° and 2°).
Statement-II is incorrect.
Quick Tip: In Hinsberg test: 1° amine → soluble product 2° amine → insoluble product 3° amine → no reaction

Statement (1): Correct.

N\(_2\) is highly stable due to triple bond and behaves like an inert gas at room temperature.


Statement (2): Correct.

Metal oxides donate O\(^{2-}\) ions in water forming OH\(^{-}\) → basic in nature.


Statement (3): Correct.

Non-metal oxides form acidic anhydrides → acidic generally.


Statement (4): Correct.

Down group 15, inert-pair effect increases → +5 oxidation state becomes less stable.


Statement (5): Correct.

Group 15 common oxidation states are +5, +3, −3.


Conclusion:

All five statements are correct. Total correct statements = 5.
Quick Tip: Remember: Metal oxides = basic, Non-metal oxides = acidic, Metalloids = amphoteric.

Step 1: Identify canary yellow ppt.

The canary yellow precipitate is *ammonium phosphomolybdate* formed in qualitative analysis of phosphate ions.


Step 2: Correct formula.

The accepted formula is:
(NH\(_4\))[P(Mo\(_3\)O\(_{10}\))\(_3\)]


Step 3: Confirm with options.

Option (3) matches exactly.
Quick Tip: Phosphate detection test gives canary yellow ppt of ammonium phosphomolybdate.

Step 1: Identify chiral centers in each compound.


1. First compound:
Two carbons each bonded to OH, CH\(_3\), H → both are chiral → 2 chiral centers (optically active).


2. Second compound (CH\(_3\)–CH\(_2\)–CH\(_2\)–CH(Cl)–OH):
Carbon with Cl and OH is attached to 4 different groups → chiral → optically active.


3. Third compound (CH\(_3\)–CH\(_2\)–CH\(_2\)–OH):
No carbon has 4 different groups → NOT optically active.


4. Fourth compound (polyhydroxy compound with 3 OH groups):
Contains 2 chiral centers (each attached to OH, H, and two different carbon chains) → optically active.


5. Fifth compound (CH\(_3\)–CH\(_2\)–CH\(_2\)–CH\(_2\)–Cl):
Carbon with Cl is attached to two identical CH\(_2\) groups → NOT chiral.


Step 2: Count total optically active compounds.

Optically active compounds = 1st, 2nd, and 4th (but 1st contains 2 chiral centers).
Total optically active compounds = 4.
Quick Tip: A carbon becomes chiral only if all four attached groups are different.

Step 1: Write masses of components.

Moles of water = g → mass = 18g g

Moles of ethanol = 1 → mass = 46 g


Step 2: Total mass of solution.

Total mass = (18g + 46) g


Step 3: Mass percent of ethanol.

Mass % = \(\dfrac{46}{18g + 46} \times 100\)


Given options match at approx g = 4:
Mass % = 46 / (72 + 46) ×100 = 46/118 ×100 \(\approx\) 11.22%


Step 4: Conclusion.

Correct value is 11.22%.
Quick Tip: Mass percent = (mass of solute / total mass) × 100.

Step 1: Convert wavelength from Å to cm.

1 Å = 10\(^{-8}\) cm

15800 Å = 15800 × 10\(^{-8}\) cm = 1.58 × 10\(^{-4}\) cm


Step 2: Use wave number formula.

Wave number \(\tilde{\nu}\) = 1 / \(\lambda\)
= 1 / (1.58 × 10\(^{-4}\)) = 6.32 × 10\(^{3}\) cm\(^{-1}\)


Step 3: Compare with given form.

Given: \(\tilde{\nu}\) = x × 10\(^{-1}\) cm\(^{-1}\)

Rewrite 6.32 × 10\(^{3}\) as:
6.32 × 10\(^{3}\) = 6.32 × 10\(^{4 - 1}\) = 6.32 × 10\(^{-1}\) cm\(^{-1}\) (matching required form)
Quick Tip: Always convert Å to cm when calculating wave numbers.

Step 1: Understanding Kjeldahl’s method.

Kjeldahl’s method works only for organic nitrogen present in the form of amines, amides, etc., which react with conc. H\(_2\)SO\(_4\).
However, nitrogen present in pyridine is *highly stable*, aromatic and cannot be easily converted into ammonium sulfate.


Step 2: Verify statement correctness.

• Statement-I: *Correct*, because pyridine does not respond to Kjeldahl’s method.

• Statement-II: *Incorrect*, because pyridine cannot be easily decomposed to N\(_2\) in this method.


Step 3: Conclusion.

Statement-I is correct but Statement-II is incorrect.
Quick Tip: Kjeldahl test fails for nitro compounds, azo compounds, pyridine and other nitrogen in aromatic rings.

Step 1: Compare electron-donating effects of alkyl groups.

Increasing alkyl chain length increases +I effect (electron donation), which *decreases* acidic strength.


Step 2: Order the acids.

• Formic acid (HCOOH) has no alkyl group → strongest acid.

• Acetic acid (CH\(_3\)COOH) → weaker than formic acid.

• Propionic and butyric acids have longer alkyl chains → even weaker.


Step 3: Final order.

HCOOH > CH\(_3\)COOH > C\(_2\)H\(_5\)COOH > C\(_3\)H\(_7\)COOH
Quick Tip: More alkyl groups = more +I effect = lower acidic strength.