JEE Main Chemistry Mock Test- 1 Free Online Test 2026

Alcohols are soluble in water. This is due to the hydroxyl group in the alcohol which is able to form hydrogen bonds with water molecules. Alcohols with a smaller hydrocarbon chain are very soluble. As the length of the hydrocarbon chain increases, the solubility in water decreases.

Alcohols with a hydroxyl group (-OH) form hydrogen bonds with water (H₂O), enhancing solubility.

In Cannizzaro's reaction, benzaldehyde undergoes a disproportionation reaction in the presence of a strong base like sodium hydroxide (NaOH). In this reaction:

• One molecule of benzaldehyde gets reduced to benzyl alcohol (C₆H₅CH₂OH).
• Another molecule of benzaldehyde gets oxidized to sodium benzoate (C₆H₅COONa).

This is a redox reaction where the aldehyde group (-CHO) is both reduced (to alcohol) and oxidized (to carboxylate).

Why not the other options?
A: Fittig's reaction: This reaction involves the formation of aryl alkanes by the reaction of halobenzene with alkyl halides in the presence of sodium metal. It doesn't produce benzyl alcohol.
C: Kolbe's reaction: This involves the decarboxylation of sodium phenoxide to form phenol, not benzyl alcohol.
D: Wurtz's reaction: This reaction involves the coupling of alkyl halides with sodium metal to form alkanes, not benzyl alcohol.
Thus, Cannizzaro's reaction is the correct process by which benzyl alcohol is obtained from benzaldehyde.

The peptide bond is a covalent linkage (CONH) formed between the carboxyl group (COOH) of one amino acid and the amin group (NH₂) of another, releasing water (H₂O).

A chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed. As a result, the amount of product formed immediately reacts backward to give reactants and so there is no change in concentration of reactant or product with the passage of time.

Chemical equilibrium is dynamic because the forward and backward reactions occur simultaneously at equal rates, maintaining constant concentrations.

Methyl orange is an azo dye formed by coupling the diazonium salt of sulfanilic acid with N,N-dimethylaniline. The diazonium salt is prepared by diazotizing sulfanilic acid with sodium nitrite and hydrochloric acid. Coupling with N,N-dimethylaniline forms the azo linkage, producing methyl orange, which is used as a pH indicator.

E_cell = E_right - E_left
= E_Cu²⁺/Cu - E_Zn²⁺/Zn
= (+0.34v) - (-0.76v)
=0.34+0.76
=1.1V

If rate is given by: rate =k[A]^x.  [B]^y , then the order = x + y.

This order can be a positive, negative no., zero or even fractional. It can be determined only experimentally and does not depend on stoichiometric coefficients of reactants.

The chemical formula for methane is CH₄ and its molecular mass is calculated as the sum of Mass of the carbon atom and Mass of hydrogen atoms. The molar mass of carbon is 12g/mol and the mass of hydrogen atom is 1g.

Molar mass of CH₄ = 12 + 4 × 1 = 16 g/mol.

Heat for 10 g = 560 kJ, so 1 g = 560 / 10 = 56 kJ/g.

For 16 g (1 mol) = 56 × 16 = 896 kJ/mol.

ΔH_comb = -896 kJ/mol (exothermic).

Electronegativity decreases down a group due to the increase in atomic radius and screening effect. Among the options (Carbon, Silicon, Lead, Tin), Carbon (C) has the highest electronegativity as it is at the top of Group 14.

The Ziegler-Natta catalyst is used for the polymerization of alpha-olefins (e.g., ethylene and propylene) to produce polymers like polyethylene and polypropylene. It consists of a transition metal compound (e.g., TiCl₄) and an organometallic compound (e.g., Al(C₂H₅)₃).

According to Faraday’s first law, one Faraday (96,485 C) deposits one equivalent weight of a substance. For a metal ion with valency n, one Faraday deposits 1/n moles.

To deposit one gram atomic weight (1 mole), the valency must be 1.

• For Na⁺ (valency 1), 1 Faraday is required to deposit 1 mole (23 g).
• For NaCl, Na⁺ (valency 1) requires 1 Faraday to deposit 1 mole (23 g).
• For Fe²⁺ (valency 2), 1/2 Faraday deposits 1 mole (31.75 g).
• For CuSO₄, Cu²⁺ (valency 2) deposits 1/2 mole (31.75 g).
• For BaCl₂, Ba²⁺ (valency 2) deposits 1/2 mole (68.5 g).
• For AlCl₃, Al³⁺ (valency 3) deposits 1/3 mole (9 g).

Thus, NaCl is correct.

CH₃COONa is a salt of a weak acid (acetic acid, CH₃COOH) and a strong base (NaOH). It undergoes anionic hydrolysis:

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

This produces OH⁻, resulting in a basic solution with pH ≈ 9.

Other options: NH₄NO₃ (acidic), CH₃COONH₄ (neutral), CaCO₃ (insoluble).

Answer: Correct (B: CH₃COONa).

Given pH = 4, calculate [H⁺] using pH = -log₁₀[H⁺].

[H⁺] = 10⁻ᵖᴴ = 10⁻⁴ = 0.0001 M.

Since HCl is a strong acid, [HCl] = [H⁺] = 0.0001 M.

CH₃COOH (acetic acid) is a weak electrolyte because it partially ionizes in water, unlike strong electrolytes (NaCl, HCl, K₂SO₄), which fully dissociate.

Acidity of phenols depends on the stability of the phenoxide ion. Electron-withdrawing groups (e.g., −NO₂, −Cl) stabilize the phenoxide ion, increasing acidity. Electron-donating groups (e.g., −CH₃) destabilize the phenoxide ion, decreasing acidity.

• o-Cresol (CH₃−C₆H₄−OH) has a −CH₃ group, which is electron-donating, making it the least acidic.
• p-Nitrophenol (NO₂−C₆H₄−OH) and p-Chlorophenol (Cl−C₆H₄−OH) have electron-withdrawing groups, increasing acidity.
• Phenol (C₆H₅OH) is intermediate in acidity.

Answer: Correct (A: o-Cresol).

A Lewis acid accepts an electron pair, typically having vacant orbitals or a positive charge. A Lewis base donates an electron pair, typically having lone pairs or pi electrons.

FeCl₃, BF₃, and SiF₄ are Lewis acids due to vacant orbitals in Fe, B, and Si, respectively.

C₂H₂ (acetylene) has a triple bond with pi electrons, which can be donated, making it a Lewis base, not a Lewis acid.

Answer: Correct (A: C₂H₂).

Polymers are macromolecules with high molecular weights, expressed as number-average or weight- average molecular mass due to varying chain lengths. They typically:
(A) Carry no charge (true for neutral polymers).
(B) Have high viscosity due to long chains (true).
(C) Scatter light due to large size (true, causing turbidity).
(D) Have high molecular weight (false, as polymers have high molecular weights).
Answer: Correct (D: Polymers have low molecular weight).

In a CsCl structure, the body diagonal of the cubic unit cell is √3 × a, where a is the edge length.
Given a = 404 pm, body diagonal = √3 × 404 ≈ 699.728 pm.
The distance between Cs⁺ and Cl⁻ is half the body diagonal:
Distance = 699.728 / 2 ≈ 349.864 pm ≈ 350 pm.
Answer: Correct (A: 350 pm)

Schottky defects occur in ionic compounds where cations and anions have similar sizes, minimizing lattice energy disruption.
Among NaCl, ZnS, AgI, and ZnO, NaCl has the closest size match between Na⁺ (102 pm) and Cl⁻ (181 pm), leading to the highest probability of Schottky defects.
Answer: Correct (D: NaCl).

In pyridine (C₅H₅N), the nitrogen atom is sp²-hybridized. It forms three σ bonds (two with carbon atoms and one with hydrogen) and has a lone pair in an sp² orbital, contributing to the aromatic system.
(A) Not hybridized (false)
(B) sp (false)
(C) sp² (true)
(D) sp³ (false)
Answer: Correct (C: sp² hybridized).

Reaction is an aldol condensation of acetone (CH₃COCH₃) with itself in the presence of NaOH and heat. The reaction can lead to the formation of three possible products.

Breakdown of the products:

1. Aldol Addition Product:
   The reaction between two molecules of acetone can lead to the formation of a β-hydroxy ketone. This is the direct aldol product before dehydration.
   The structure would be:
   CH₃-CO-CH₂-CH₂-CO-CH₃ (4-hydroxy-4-methylpentan-2-one).

2. Aldol Condensation Product (cis and trans isomers):
   On further heating, the aldol addition product undergoes dehydration (loss of water) to form an α,β-unsaturated ketone (mesityl oxide). This product can exist as two isomers: cis and trans.
   The structures would be:
   CH₃-CO-CH=CH-CO-CH₃ (mesityl oxide), existing as cis and trans isomers.

Conclusion:
There are 3 possible products:
1 aldol addition product.
2 aldol condensation products (cis and trans forms).
Answer: 3

Elements reacting with NaOH to produce H₂ are amphoteric or form soluble complexes.

• Zn + 2NaOH → Na₂ZnO₂ + H₂
• Al + 2NaOH + 6H₂O → 2NaAl(OH)₄ + 3H₂
• Sn + 2NaOH + 2H₂O → Na₂SnO₃ + 2H₂

Pb, P, and S do not produce H₂ with NaOH. Thus, 3 elements (Zn, Al, Sn).

Answer: Correct (3).

Molecular formula from calculation comes to be C₃H₆O i.e., it stands for the following compounds.

Elemental analysis: C (62.07%), H (10.34%), O (27.59%). Moles: C = 62.07 / 12 ≈ 5.17, H = 10.34 / 1 ≈ 10.34, O = 27.59 / 16 ≈ 1.72.
Ratio: 5.17 : 10.34 : 1.72 ÷ 1.72 ≈ 3 : 6 : 1 → C₃H₆O.
Vapour density = 29, molecular weight = 29 × 2 = 58, matching C₃H₆O.
No reaction with Na, so not an alcohol. Forms dibromo product, indicating a double bond. Possible isomers (e.g., prop-1-ene-3-one, cyclopropanone) = 9.
Bromine reaction: 58 g of C₃H₆O reacts with 160 g of C₆Br₂.
2.9 g reacts with 160 / 58 × 2.9 ≈ 8 g of C₆Br₂.
Y = 9, X = 8, Y - X = 9 - 8 = 1.

Reactions producing nitrogen oxides:

1. Ag + conc. HNO₃ → NO₂
2. Cu + conc. HNO₃ → NO₂
3. Cu + conc. HNO₃ → NO₂
4. C + conc. HNO₃ → NO₂
5. Zn + conc. HNO₃ → NO₂
6. Zn + dil. HNO₃ → N₂O
7. P₄ + conc. HNO₃ → NO₂
8. S₈ + conc. HNO₃ → NO₂
9. Cu + dil. HNO₃ → NO

ii. Sn + dil. HNO₃ → NH₄NO₃ (no nitrogen oxide).
Total = 8.

White phosphorus (P₄) has a tetrahedral structure:

• Triangles (x): 4 triangular faces, so x = 4.
• Planes of symmetry (y): 6 planes, each passing through an edge and bisecting opposite edges, so y = 6.
• P-P bonds (z): Each P atom forms 3 bonds, total bonds = 4 × 3 / 2 = 6, so z = 6.

Calculate:
y + z / x = (6 + 6) / 4 = 12 / 4 = 3.