JEE Main 2024 question paper pdf with solutions- Download Jan 31 Shift 1 Chemistry Question Paper pdf

Step 1: Reaction of Glucose with HI.
- Glucose reacts with hydroiodic acid (HI) to form gluconic acid, where the aldehyde group is oxidized to a carboxyl group.

Step 2: Reduction with NaBH4.
- Glucose reacts with sodium borohydride (NaBH4) to reduce the aldehyde group to a primary alcohol, forming sorbitol.

Step 3: Reaction with Bromine Water.
- Glucose reacts with bromine water, leading to the oxidation of the aldehyde group to a carboxylic acid group, forming gluconic acid.

Step 4: Oxidation with Nitric Acid.
- When glucose is treated with nitric acid (HNO3), both glucose and gluconic acid are further oxidized to saccharic acid, which contains two carboxyl groups.

Step 5: Conclusion.
- Based on the reactions described, the correct matching of reactants and products is:
- A - i: Glucose + HI produces gluconic acid.
- B - iii: Glucose + NaBH4 produces sorbitol.
- C - ii: Glucose + Br2 . H2O produces n-hexane (oxidation).
- D - iv: Glucose + HNO3 produces saccharic acid. Quick Tip: Remember that NaBH4 is a selective reducing agent, commonly used to reduce aldehydes to alcohols.

Step 1: Understanding the Reaction Mechanism.
- In this reaction, the alkyl halide \( CH_3 CH_2 CH_2 Br \) undergoes an elimination reaction in the presence of a strong base like KOH. This results in the elimination of the halide ion and the formation of a double bond.

Step 2: Elimination to form an Alkene.
- The halide is eliminated, leading to the formation of an alkene. The resulting product is propene, which is an unsaturated hydrocarbon.

Step 3: Conclusion.
- The final product is propene because the base induces an elimination reaction, removing the halide ion and forming a double bond between the carbons. Quick Tip: Elimination reactions generally lead to the formation of alkenes by removing a leaving group (such as a halide) in the presence of a base.

Step 1: Understanding the Properties of the Compounds.
- Zinc sulfate (\( ZnSO_4 \)) is a white solid, and in an aqueous solution, it remains colourless.
- Copper sulfate (\( CuSO_4 \)) is blue in colour when dissolved in water.
- Iron(II) sulfate (\( FeSO_4 \)) forms a pale green solution in water.
- Iron(III) chloride (\( FeCl_3 \)) is yellow in colour.

Step 2: Conclusion.
- The correct answer is \( ZnSO_4 \), which is white in colour in aqueous medium. Quick Tip: Zinc salts, including \( ZnSO_4 \), are typically colourless or white due to the fully filled d-orbitals in the Zn(2+) ion.

Step 1: Faraday's Law of Electrolysis.
One faraday of electricity corresponds to the transfer of 1 mole of electrons. In the case of \( Cu^{2+} \), 2 electrons are required to deposit one mole of Cu.

Step 2: Calculation.
- 2 faradays of electricity are required to deposit 1 mole of Cu, which has a molar mass of 63.5 g.
- Therefore, passing 1 faraday of electricity will deposit half the mass, i.e., \( 63.5/2 = 31.75 \) g of copper.

Step 3: Conclusion.
The mass of Cu deposited after passing 1 faraday of electricity is 31.75 g.
Quick Tip: 1 Faraday corresponds to the charge required to deposit 1 mole of a monovalent ion. For a divalent ion, it deposits half the mass of the ion's molar mass.

Step 1: Analyzing Statement I.
Dichromates are indeed generally made from chromates by adding an acidic solution to chromate salts. The transformation occurs as: \[ Chromate + Acid \rightarrow Dichromate. \]

Step 2: Analyzing Statement II.
Manganate ions (\( MnO_4^{2-} \)) are paramagnetic, not diamagnetic. This is because of the presence of unpaired electrons in the \( Mn^{2+} \) ion.

Step 3: Conclusion.
Statement I is true, but Statement II is false, so the correct answer is (2). Quick Tip: Manganate ions are paramagnetic, meaning they have unpaired electrons in their d-orbitals. Dichromates are made by acidifying chromates.

Step 1: Understanding electron gain enthalpy.
Electron gain enthalpy is the energy released when an electron is added to an atom in the gas phase. Fluorine has the highest electron gain enthalpy because of its small size and high electronegativity.

Step 2: Trend in electron gain enthalpy.
- Fluorine (F) has the highest electron gain enthalpy due to its very high electronegativity and small atomic radius.
- Chlorine (Cl) follows, but its electron gain enthalpy is less than Fluorine.
- Bromine (Br) and Iodine (I) have progressively lower electron gain enthalpies as their size increases.

Step 3: Conclusion.
Fluorine has the highest electron gain enthalpy because it is the smallest halogen and has the greatest tendency to accept an electron.
Quick Tip: Electron gain enthalpy generally increases across a period (from left to right) and decreases down a group due to increasing atomic size and decreasing electron affinity.

Step 1: Understanding Raoult’s Law.
Raoult's law states that the partial vapor pressure of each volatile component in a solution is directly proportional to its mole fraction in the solution.

Step 2: Positive Deviation.
- Positive deviation from Raoult’s law occurs when the intermolecular forces between the molecules of the two components are weaker than the forces between the molecules of the individual components.
- In the case of Ethanol + Acetone, the hydrogen bonding between ethanol molecules is disrupted, and the dipole-dipole interaction between acetone molecules is weaker, causing higher vapor pressure (positive deviation).

Step 3: Conclusion.
Ethanol + acetone forms an ideal solution that shows positive deviation due to weaker interactions between ethanol and acetone molecules.
Quick Tip: Positive deviation occurs when intermolecular forces in the solution are weaker than in the pure components.

Step 1: Understanding pKa values.
- The pKa of phenol is 10.0, and the pKa of ethanol is 15.9. The lower the pKa value, the stronger the acid. Hence, phenol is more acidic than ethanol.

Step 2: Analyzing the Statements.
- Statement I is true because the pKa values given for phenol and ethanol are correct.
- Statement II is false because phenol, with a lower pKa value, is more acidic than ethanol.

Step 3: Conclusion.
The correct answer is (2) Statement I is true but statement II is false. Quick Tip: Acidity increases with a lower pKa value, so phenol is more acidic than ethanol.

Step 1: Reaction of acids with salts.
When concentrated sulfuric acid is added to halides, it generally produces halogen gases (such as \( Br_2 \), \( I_2 \), etc.) that give color.

Step 2: Explanation of Each Option.
- NaBr reacts with \( H_2 SO_4 \) to produce bromine (\( Br_2 \)), which has a brown color.
- CaF2 does not react with \( H_2 SO_4 \) to produce a colored gas, so no color change is observed.
- NaNO3 does not react with \( H_2 SO_4 \) in a way that would produce a colored gas.
- I\(^-\) reacts with \( H_2 SO_4 \) to produce iodine gas, which is purple.

Step 3: Conclusion.
NaNO3 is the correct answer because it does not produce a colored gas when treated with concentrated sulfuric acid.
Quick Tip: Halides generally react with concentrated sulfuric acid to form halogen gases, which often impart characteristic colors.

Step 1: Understanding the carbon species.
- A **carbocation** has three bonds and one empty orbital, resulting in a total of 6 electrons around the carbon atom.
- A **carbanion** has three bonds and one lone pair of electrons, giving the carbon 8 electrons.
- A **carbon free radical** has three bonds and one unpaired electron, which gives it 7 electrons.

Step 2: Conclusion.
The correct answer is **(1) Carbocation**, as it has exactly 6 electrons around the carbon atom. Quick Tip: A carbocation is a species where the carbon has only 6 electrons, making it electron-deficient and highly reactive.

Step 1: Understanding adsorption.
- **Adsorption** refers to the accumulation of molecules or ions on the surface of a solid or liquid.
- In **chromatography**, the stationary phase is adsorbent (such as silica gel or alumina), which adsorbs components of the mixture being separated.

Step 2: Conclusion.
The correct answer is **(1) Chromatography**, which is based on the principle of adsorption.
Quick Tip: In chromatography, adsorption is used to separate components of a mixture based on their different affinities to the stationary phase.

Step 1: Analyzing the structure.
- The compound shown has a hydroxyl group (-OH) at position 7 and a keto group (C=O) at position 2 of a heptane chain.
- The correct IUPAC name is based on numbering the carbon chain so that the functional groups (hydroxy and carbonyl) get the lowest possible numbers.

Step 2: Conclusion.
The correct name is **7-Hydroxyheptan-2-one**, as it correctly identifies the position of the functional groups. Quick Tip: When naming organic compounds, always assign the lowest possible locants to functional groups based on IUPAC rules.

Step 1: Analyzing Statement I.
- **Alcohols** are versatile compounds. They can act as nucleophiles, meaning they can donate electrons (e.g., in substitution reactions), and they can act as electrophiles when the alcohol is protonated (e.g., in dehydration reactions).

Step 2: Analyzing Statement II.
- Alcohols react with metals to form **alkoxides** and liberate **H2** gas. This is a characteristic reaction of alcohols with active metals like sodium.

Step 3: Conclusion.
Both statements are correct, so the correct answer is (4). Quick Tip: Alcohols can act both as nucleophiles (in substitution reactions) and electrophiles (when protonated in acid catalysis).

Step 1: Battery materials.
- Materials used in batteries typically include **Zn** (zinc), **Cd** (cadmium), **Hg** (mercury), **Mn** (manganese), and **Fe** (iron) due to their ability to act as anode or cathode materials in different battery types.

Step 2: Conclusion.
The correct answer is **(1) Zn, Cd, Hg, Mn, Fe**, as they are commonly used in various types of batteries.
Quick Tip: Zn, Cd, and Mn are commonly used in dry cells and rechargeable batteries due to their electrochemical properties.

Step 1: Understanding the structure of [Pt(en)\(_2\)Cl\(_2\)].
The compound \([ Pt(en)_2Cl_2 ]\) consists of platinum as the central metal ion surrounded by two ethylenediamine (en) ligands and two chloride (Cl) ligands.

Step 2: Analyzing the geometrical isomers.
- Since the platinum ion is in a coordination number of 6, it forms an octahedral complex.
- The two **en** ligands are bidentate, meaning each can bind at two points to the platinum, creating a situation where two isomers can arise based on the relative positioning of the chloride ions.
- These two isomers are **cis** and **trans** isomers, based on the relative positioning of the chloride ions in the octahedral geometry.

Step 3: Conclusion.
The number of geometrical isomers of \([ Pt(en)_2Cl_2 ]\) is **2**: the cis and trans isomers.
Quick Tip: Geometrical isomers occur when ligands can occupy different positions around the central metal, creating distinct spatial arrangements.