SRMJEEE Chemistry Mock Test - 10 - JEE MCQ

When ethyl hydrogen sulphate is heated with excess of alcohol at 410K, the product obtained is:

• The reaction involves heating ethyl hydrogen sulphate with excess alcohol.
• At 410K, this process promotes the formation of various products.
• The primary product formed is diethyl sulphate.
• Diethyl sulphate is an important alkylating agent in organic synthesis.

To determine the alkene formed from the ylide and carbonyl pair:

• Consider the structure of the ylide: CH₃CH₂CH₂CH═PPh₃.
• Recognise that it reacts with 2-butanone, a ketone.
• The reaction typically involves a Wittig reaction, leading to alkene formation.
• Identify the potential alkene products based on the ylide and carbonyl structure.
• The product will be formed by connecting the carbon chain from the ylide with the carbonyl's carbon.

The resulting alkene is likely to be 3-methyl-3-heptene, as it correctly accounts for the structural components from both reactants. Therefore, the answer is:

• A: 3-methyl-3-heptene

The following monosaccharides are available:

• Glucose - a hexose, not a pentose.
• Fructose - also a hexose, not a pentose.
• Arabinose - a pentose, containing five carbon atoms.
• Galactose - a hexose, not a pentose.

Conclusion: Among the listed monosaccharides, arabinose is the only pentose.

Hydrolysis of chloroform with aqueous potassium hydroxide results in the formation of:

• Hydrolysis Process: Chloroform (CHCl₃) reacts with potassium hydroxide (KOH) in water.
• Key Reaction: The reaction primarily leads to the formation of formic acid (HCOOH).
• Final Product: The main product of this hydrolysis is formic acid, which may dissociate to form potassium formate (HCOOK) in the presence of KOH.

Thus, the significant outcomes of this reaction include both formic acid and potassium formate. However, the dominant product is HCOOH.

AgCl is soluble in:

• Aqua regia: A mixture of nitric and hydrochloric acids capable of dissolving noble metals, but not AgCl.
• HCl: Hydrochloric acid does not dissolve AgCl.
• Aq.NH₃: Aqueous ammonia can dissolve AgCl by forming a complex ion.
• H₂SO₄: Sulphuric acid does not affect the solubility of AgCl.

In summary, the only option that effectively dissolves AgCl is aqueous ammonia.

The reaction of benzene with alkyl halide in the presence of anhydrous AlCl₃ is known as:

• Friedel Craft's reaction

This reaction is a significant method in organic chemistry for alkylating aromatic compounds. Here are some key points:

• It involves the formation of a carbocation from the alkyl halide, facilitated by AlCl₃.
• This carbocation then attacks the benzene ring, leading to the substitution of a hydrogen atom.
• The overall process increases the carbon chain length of the aromatic compound.

The number of unpaired electrons in Ni²⁺

To determine the number of unpaired electrons in Ni²⁺, follow these steps:

• Electron Configuration: Nickel (Ni) has an atomic number of 28. Its electron configuration is [Ar] 3d⁸ 4s².
• Ion Formation: When Ni loses two electrons to form Ni²⁺, it typically loses the two 4s electrons first.
• Resulting Configuration: The electron configuration for Ni²⁺ becomes [Ar] 3d⁸.
• Unpaired Electrons: In the 3d subshell, there are 8 electrons. When distributed, this results in:
• 4 paired electrons
• 2 unpaired electrons

Thus, Ni²⁺ has 2 unpaired electrons.

Mixture of sand and sulphur may best be separated by

• Using the dissolving method in carbon disulfide (CS₂) and then filtering the mixture is effective.

• This process allows the sulphur to dissolve while the sand remains solid.

• After filtering, the sand can be collected, leaving behind the dissolved sulphur.

The greater the s- character in an orbital, the shorter is the bond and greater is its strength. Thus, bond formed by sp hybridisation , as in acetylene is stronger as compare to ethane and ethene , because it has maximum bond energy.

A carboxylic acid can be converted into its anhydride using the following reagents:

• Thionyl chloride - This reagent helps in the formation of acid chlorides, which can further react to form anhydrides.
• Sulphur chloride - Similar to thionyl chloride, it can facilitate the conversion process.
• Sulphuric acid - While it is a strong acid, it is not typically used for this conversion.
• Phosphorus pentaoxide - A dehydrating agent effective for converting carboxylic acids to anhydrides.

Among these options, thionyl chloride and phosphorus pentaoxide are the most effective for this conversion, while sulphuric acid is less suitable.

Intermolecular forces in solid hydrogen are primarily

• Van der Waals forces or London dispersion forces, which are weak attractions between molecules.
• Solid hydrogen does not exhibit covalent bonds, as its molecules are held together by weaker interactions.
• Hydrogen bonds are not significant in solid hydrogen, as they typically occur between hydrogen and more electronegative atoms.

Therefore, the main forces at play are the van der Waals forces.

To determine the enthalpy change (∆H) of the reaction per mole of copper produced:

• Mass of copper obtained: 3.175 g
• Molar mass of copper (Cu): 63.55 g/mol
• Calculate moles of copper produced:
• Moles of Cu = Mass / Molar mass = 3.175 g / 63.55 g/mol ≈ 0.0499 mol
• Heat evolved during the reaction: 20 J
• Calculate ∆H per mole of copper:
• ∆H = Heat evolved / Moles of Cu = 20 J / 0.0499 mol ≈ -400 J/mol

The enthalpy change of the reaction per mole of copper is approximately -400 J.

Le Chatelier's principle explains how a system at equilibrium responds to changes in conditions.

• It predicts the effects of changes in temperature, pressure, and volume.
• An increase in temperature shifts the equilibrium position to favour the endothermic reaction.
• Changes in pressure affect gaseous reactions, shifting the equilibrium towards the side with fewer moles of gas.
• Volume changes impact pressure; decreasing volume increases pressure, which can shift equilibrium.

The rate of reaction is influenced by various factors, including:

• Nature of the reactants: Different substances react at varying rates.
• Concentration of the reactants: Higher concentrations typically increase reaction rates.
• Temperature of the reaction: Increased temperature generally speeds up reactions.

However, the molecularity of a reaction refers to the number of reactant particles involved and does not directly influence the rate. Thus, it does not impact how quickly a reaction occurs.

2, 4, 6-Trinitrophenol is a chemical compound known for its properties as an acid dye. It is classified as:

• Acidic: This compound can donate protons (H⁺ ions), making it suitable for dyeing processes that require an acidic medium.
• Water-soluble: It dissolves in water, allowing it to easily penetrate materials during dyeing.
• High reactivity: Its structure allows it to react effectively with various substrates, enhancing colour adhesion.

Overall, 2, 4, 6-Trinitrophenol serves as an important acid dye in various applications.

To determine the order of stability among I. 1-butene, II. cis-2-butene, and III. trans-2-butene, consider the following points:

• 1-butene: This compound has less steric hindrance due to its terminal double bond, making it relatively less stable.
• cis-2-butene: The cis configuration leads to increased steric strain between the two methyl groups, resulting in lower stability compared to trans-2-butene.
• trans-2-butene: This is the most stable of the three due to the methyl groups being further apart, reducing steric hindrance.

This leads to the following order of stability: III > II > I.

Mulliken-Jaffe Electronegativity:
Electronegativity (EN) can be regarded as the average of the ionisation energy (IE) and the electron affinity (EA) of an atom.
EN = IP + EA/2

Pressure in a gas results from:

• Collisions of gas molecules with each other.
• The random movement of gas molecules throughout the space.
• Collisions of gas molecules against the walls of their container.

However, the intermolecular forces of attraction have minimal impact on gas pressure, as gases are typically far apart, reducing their influence.

To calculate the standard cell potential (Eº) for the given electrochemical cell:

• Identify the standard reduction potentials:
• Sn²⁺ | Sn: Eº = -0.14 V
• Zn²⁺ | Zn: Eº = -0.74 V
• Determine the half-reactions:
• For Zn: Zn → Zn²⁺ + 2e^- (oxidation)
• For Sn: Sn²⁺ + 2e^- → Sn (reduction)
• Calculate the cell potential:
• Use the formula: Eº(cell) = Eº(reduction) - Eº(oxidation)
• Here, Eº(oxidation) for Zn = -0.74 V and Eº(reduction) for Sn = -0.14 V
• Thus, Eº(cell) = (-0.14 V) - (-0.74 V)
• Eº(cell) = -0.14 V + 0.74 V = 0.60 V

The standard cell potential (Eº) is 0.60 V.

To suppress the dissociation of acetic acid, the appropriate compound to add is sodium acetate.

• Sodium acetate acts as a source of acetate ions.
• Adding it shifts the equilibrium, reducing the dissociation of acetic acid.
• This process helps maintain a higher concentration of undissociated acetic acid in solution.
• Other compounds listed do not perform this function effectively.

Chain isomerism occurs when compounds with the same molecular formula have different arrangements of carbon atoms. Here’s a breakdown regarding the options provided:

• C₃H₈: This compound can exist in two forms (propane and propene) but does not exhibit chain isomerism due to its limited carbon atoms.
• C₄H₁₀: This formula allows for multiple structural arrangements, hence it can show chain isomerism.
• C₅H₁₂O: This formula includes an oxygen atom and allows for different structures, indicating potential isomerism.
• C₅H₁₀O: Similar to the previous compound, it can also form various structures due to the presence of the oxygen atom.

In conclusion, among the given options, C₃H₈ will not show chain isomerism.

In a (p, n) reaction, nucleus A transforms into nucleus B. Here’s what you need to know:

• Isotopes: A and B are isotopes if they have the same number of protons but different numbers of neutrons.
• Isobars: A and B are isobars if they have the same mass number but different atomic numbers.
• Isotones: A and B are isotones if they have the same number of neutrons but different numbers of protons.
• Atomic Number: If A has a higher atomic number than B, it indicates that A is heavier in terms of protons.

Each of these relationships defines how A and B relate to one another in terms of nuclear properties.

When benzenediazonium chloride reacts with water, it undergoes hydrolysis.

• This reaction leads to the formation of phenol.
• Phenol is represented by the formula C₆H₅OH.
• Other options such as carboxylic acid, benzene, or aniline are not produced in this reaction.

Thus, the primary compound formed is C₆H₅OH.

Which of the following is a bio-degradable polymer?

The only bio-degradable polymer in the given options is:

• Cellulose - a natural polymer derived from plants that can break down in the environment.

The other options are:

• PVC - a widely used plastic that is not biodegradable.
• Nylon-6 - a synthetic polymer that is also non-biodegradable.
• Polythene - a common plastic that does not decompose easily.

The method used for separating two immiscible liquids is the separating funnel. This technique is effective because it takes advantage of the differences in density between the liquids.

• Separating funnel: A device specifically designed for separating two layers of immiscible liquids.
• The lighter liquid will float on top of the denser liquid.
• You can carefully drain the bottom layer through the tap of the funnel.
• This method is simple and allows for effective separation without mixing the liquids further.

To determine the possible formula of the compound with elements A, B, and C, consider their oxidation states:

• Element A: Oxidation number is +2
• Element B: Oxidation number is +5
• Element C: Oxidation number is -2

Next, we need to balance the total positive and negative charges in the compound:

• The total positive charge must equal the total negative charge.
• Assuming the formula is in the form of A_xB_yC_z:

Let’s calculate the charges:

• Positive charge from A: +2x
• Positive charge from B: +5y
• Negative charge from C: -2z

Setting up the equation:

2x + 5y = 2z

Now, we can evaluate the options:

• Option A: A₃(BC₄)₂ gives 6 + 10 = 10 (valid)
• Option B: A₃(B₄C)₂ gives 6 + 20 = 4 (invalid)
• Option C: A (B) C₂ gives 2 + 5 = 4 (invalid)
• Option D: A₂(BC₃)₂ gives 4 + 10 = 6 (invalid)

Thus, the most viable formula based on the oxidation states is:

• A₃(BC₄)₂