SRMJEEE Chemistry Mock Test - 9 - JEE MCQ

To determine the identity of compound Y, we can follow these steps:

• Compound X is reduced by LiAlH₄, a strong reducing agent, which typically converts carbonyl compounds into alcohols.

• This suggests that Y is likely an alcohol derived from a carbonyl compound (X).

• When vapours of Y are passed over freshly reduced copper at 300°C, X is reformed. This indicates that Y can undergo dehydration or oxidation to regenerate X.

• Common alcohols that can be oxidised to yield ketones or aldehydes include:

• Primary alcohols—oxidised to aldehydes.

• Secondary alcohols—oxidised to ketones.

• Among the options presented, CH₃CH₂OH (ethanol) is a primary alcohol that can be oxidised to yield acetaldehyde (CH₃CHO).

• Thus, the compound Y that undergoes this reaction and can be oxidised back to X is most likely ethanol.

Schiff's reagent is a chemical compound used primarily in organic chemistry for detecting aldehydes. It is composed of rosaniline hydrochloride and is sensitive to the presence of these functional groups.

• It is notable for its vibrant colour change when it reacts with aldehydes.
• The reagent is often employed in histology and microscopy to identify specific cellular components.
• Rosaniline, the active ingredient, provides the characteristic colour associated with the reagent.

The element responsible for the oxidation of water to O₂ in biological processes is manganese (Mn).

Manganese plays a crucial role in photosynthesis by facilitating the splitting of water molecules. This process is essential for releasing oxygen into the atmosphere.

• Manganese is a key component of the oxygen-evolving complex found in photosystem II.
• It aids in the conversion of light energy into chemical energy.
• This reaction is vital for sustaining life on Earth, as it generates the oxygen we breathe.

The compound formed in the borax bead test for Cu²⁺ ions in an oxidising flame is:

• Copper(II) borate is typically formed during this test.
• This compound appears as a blue bead when heated.
• The presence of Cu²⁺ ions is crucial for this reaction.
• The borax bead test is useful for identifying metal ions in minerals.

The boiling points of the compounds can be compared based on their structure and intermolecular forces.

• Isobutane has a branched structure which generally leads to lower boiling points due to reduced surface area and weaker van der Waals forces.
• I-butyne, being an alkyne, experiences stronger dipole interactions, resulting in a higher boiling point.
• I-butene, an alkene, has a moderate boiling point influenced by its double bond.
• n-butane is a straight-chain alkane, typically exhibiting higher boiling points than branched alkanes due to increased surface area.

In summary, among these compounds, isobutane has the minimum boiling point due to its branched structure and weaker intermolecular forces.

The reacting ion in the nitration of benzene is

• Nitration involves the introduction of a nitro group into benzene.
• The process typically uses nitric acid and sulphuric acid.
• The key reactive ion is the nitronium ion (NO₂⁺).
• This ion acts as an electrophile, attacking the electron-rich benzene ring.
• The result is the formation of nitrobenzene.

The sun is classified under the G spectrum. This classification is based on its temperature and spectral characteristics.

• G-type stars are characterised by their moderate temperatures, typically ranging from 5,300 to 6,000 degrees Celsius.
• They emit a balanced amount of visible light and have a yellowish appearance.
• This classification helps astronomers understand the sun's properties and its place in the universe.

The most stable carbonium ion is the tertiary carbonium ion.

Tertiary carbonium ions are the most stable due to the following reasons:

• They have three alkyl groups attached to the positively charged carbon.
• These alkyl groups provide electron-donating effects, which help stabilise the positive charge.
• This stability is greater than that of primary or secondary carbonium ions.

In contrast:

• Methyl carbonium ions have no alkyl groups, making them very unstable.
• Primary carbonium ions have one alkyl group, offering some stability, but less than tertiary.
• Secondary carbonium ions have two alkyl groups, providing moderate stability.

Overall, the tertiary carbonium ion is the most stable due to the combined stabilising effects of the surrounding alkyl groups.

To determine which compounds generate a carbene upon exposure to ultraviolet light, let's examine each compound:

• CH₂N₂: This compound, known as diazomethane, readily decomposes under UV light to produce a carbene.
• CH₂ = C = O: Also known as ketene, this compound does not typically generate a carbene directly from UV exposure.
• CH₂I₂: This compound, dibromomethane, can also produce a carbene when exposed to UV light, through a homolytic cleavage of the carbon-iodine bond.

Based on this analysis, the compounds that generate a carbene when exposed to UV light are:

• CH₂N₂
• CH₂I₂

Thus, the correct answer includes compounds 1 and 3.

The molecule with the highest dipole moment among the given options is NF₃.

Here are the key points regarding the dipole moments of the molecules:

• CO₂: This molecule is linear and has a dipole moment of zero because the two bond dipoles cancel each other out.
• BCl₃: This is also a symmetrical molecule with a zero dipole moment, as the individual bond dipoles cancel out.
• CCl₄: Similar to CO₂ and BCl₃, CCl₄ is tetrahedral and has a dipole moment of zero due to symmetrical arrangement.
• NF₃: This molecule has a net dipole moment because it is trigonal pyramidal, leading to non-cancellation of bond dipoles.

In summary, NF₃ exhibits the highest dipole moment due to its molecular shape and polar N-F bonds, while the other molecules have zero dipole moments owing to their symmetrical structures.

To determine which bond has the highest average bond energy, consider the following:

• S = O: This bond involves a double bond between sulfur and oxygen, which has a moderate bond energy.
• C = O: This carbon-oxygen double bond is known to have a high bond energy, typically around 745 kJ/mol.
• C = N: The bond energy for carbon-nitrogen double bonds is slightly lower than that of carbon-oxygen bonds.
• N = N: The nitrogen-nitrogen double bond is strong, but generally not as strong as the carbon-oxygen bond.

The bond with the highest average bond energy is the C = O bond.

Pieces of wood burn faster than a log of the same mass due to the following reasons:

• Increased surface area: Smaller pieces have a greater surface area compared to a log. This allows for more oxygen to reach the wood, enhancing the combustion process.
• Higher accessibility: The smaller size of pieces means they ignite and burn more quickly, as heat can penetrate them more easily.
• Density differences: Generally, logs have a higher density, which can slow down the burning rate compared to looser, smaller pieces.

The reaction N₂ + O₂ → 2NO is endothermic.

The forward reaction is influenced by various factors:

• Temperature: Lowering the temperature favours the endothermic reaction, shifting the equilibrium to the right.
• Pressure: Increasing pressure typically does not significantly affect this reaction since it involves the same number of gas molecules on both sides.
• Catalysts: Adding a catalyst does not shift the equilibrium; it merely speeds up the rate at which equilibrium is reached.

An exothermic reaction occurs when the reacting substances

• contain more energy than the products.
• result in the release of energy, often in the form of heat.
• lead to an increase in temperature of the surrounding environment.

In summary, during an exothermic reaction, the energy difference between the reactants and products results in a net release of energy. This is a key characteristic that defines such reactions.

Substances that lower body temperature are known as:

• Antipyretics: These medications help reduce fever and are commonly used in various medical situations.
• Analgesics: Primarily used for pain relief, they do not directly affect body temperature.
• Antibiotics: These treat bacterial infections but do not have a direct role in temperature regulation.
• Hypnotics: Used to induce sleep, they do not influence body temperature.

To determine which compound has carbon using only sp³ hybrid orbitals for bonding, consider the following:

• sp³ hybridization occurs when carbon forms four single bonds.
• Each carbon atom in the compound must be bonded to four other atoms with no double or triple bonds.
• Review the bonding in each option:
• HCOOH: Contains a carbon with a double bond to oxygen (C=O), indicating sp² hybridization.
• CH₃CHO: Features a carbon with a double bond to oxygen, so it also uses sp² hybrid orbitals.
• (CH₃)₃COH: The carbon bonded to three methyl groups and a hydroxyl group uses four single bonds, thus is sp³ hybridised.
• (NH₃)₂CO: Contains a carbon double bonded to oxygen, indicating sp² hybridisation.

Based on this analysis, the compound where carbon exclusively uses sp³ hybrid orbitals for bonding is (CH₃)₃COH.

C-Cl bond length is 0.176 nm; ionic radii of Na⁺ and F ^- are 0.095 and 0.136 nm respectively; C-C bond length is 0.154 nm.

To find the temperature at which the average speed of CH₄ molecules equals that of O₂ at 300 K, follow these steps:

• The average speed of gas molecules is related to temperature and molar mass.
• Use the formula for average speed: v = sqrt(3RT/M), where:
• v = average speed
• R = universal gas constant
• T = temperature in Kelvin
• M = molar mass of the gas
• Calculate the average speed of O₂ at 300 K using its molar mass (32 g/mol).
• Set the average speed of CH₄ (molar mass 16 g/mol) equal to that of O₂.
• Rearranging the formula allows you to solve for the unknown temperature of CH₄:
• v_O₂ at 300 K = sqrt(3R * 300 / 32)
• Set this equal to v_CH₄ = sqrt(3RT / 16)
• Solving the equation leads to the temperature for CH₄.

This calculation shows the relationship between temperature and molecular speed for different gases based on their molar masses.

The true statement about benzene is:

• Cyclic delocalisation of π electrons occurs in benzene, contributing to its stability and unique properties.

Other statements are inaccurate:

• Benzene does not easily undergo addition reactions due to its stability from aromaticity.
• There is only one type of C ― C bond in benzene, characterised by a uniform bond length.
• Monosubstitution of the benzene ring leads to multiple isomers, but not three distinct substances.

Which of the following is not isomeric with diethyl ether?

To determine which compound is not an isomer of diethyl ether, we should first understand what isomerism entails. Isomers are compounds that have the same molecular formula but different structural arrangements.

• Diethyl ether has the molecular formula C₄H₁₀O.
• Examine each option to see if they share this formula:
• Methyl n-propyl ether (C₄H₁₀O) is isomeric with diethyl ether.
• Butan-1-ol (C₄H₁₀O) is also isomeric with diethyl ether.
• 2-methyl propan-2-ol (C₅H₁₂O) does not share the same formula and is therefore not isomeric.
• Butan-2-one (C₄H₈O) is also not isomeric, as it has a different number of hydrogen atoms.

In summary, while methyl n-propyl ether and butan-1-ol are isomers of diethyl ether, 2-methyl propan-2-ol is definitely not an isomer due to its different molecular formula.

pH indicators are substances used to determine the acidity or alkalinity of a solution. They change colour based on the pH level. Here are the main types of pH indicators:

• Salts of strong acids and strong bases typically do not act as pH indicators.
• Salts of weak acids and weak bases can provide some indication of pH but are less common.
• Common pH indicators include weak acids or weak bases that exhibit noticeable colour changes at specific pH levels.
• Strong acids and strong bases themselves do not serve as indicators.

In summary, effective pH indicators are generally either weak acids or weak bases, which change colour in response to varying pH levels.

To calculate the maximum activity of the radioactive substance that can be injected:

• The human body requires 0.01 μC of activity for 24 hours.
• The half-life of the substance is 6 hours.
• In 24 hours, there are 4 half-lives (24/6 = 4).
• Using the half-life formula, the remaining activity after n half-lives is given by:
• A = A0 * (1/2)^n
• Let A0 be the initial activity (maximum activity injected).
• After 4 half-lives, the required activity is:
• 0.01 μC = A0 * (1/2)^4
• This simplifies to:
• 0.01 μC = A0 / 16
• Thus, the initial activity is:
• A0 = 0.01 μC * 16 = 0.16 μC
• The maximum amount of the radioactive substance that can be injected is 0.16 μC.

Dow's reaction involves

• Electrophilic addition is a key mechanism where an electrophile reacts with a nucleophile to form a saturated compound.
• This reaction typically forms a new bond by the addition of atoms or groups to a double or triple bond.
• The process is significant in organic synthesis and is often utilised to create complex molecules.

Select incorrect statement:

• Homolytic fission leads to the creation of free radicals.

• Heterolytic fission generates a Lewis acid and a Lewis base.

• The completion of the octet is a key factor driving Lewis acid-base reactions.

• Sterically bulky groups decrease nucleophilicity.

To calculate the weight average molecular weight of the polymer:

• Identify the molecular weights and their corresponding percentages:
  • 20,000 g/mol - 30% of molecules
  • 30,000 g/mol - 40% of molecules
  • 60,000 g/mol - 30% of molecules
• Calculate the contribution to the weight average molecular weight:
  • For 20,000 g/mol: 20,000 * 0.30 = 6,000
  • For 30,000 g/mol: 30,000 * 0.40 = 12,000
  • For 60,000 g/mol: 60,000 * 0.30 = 18,000
• Add the contributions together:
  • Total contribution = 6,000 + 12,000 + 18,000 = 36,000
• Finally, divide by the total percentage (1.00):
  • Weight average molecular weight = 36,000 / 1.00 = 36,000

The weight average molecular weight of the polymer is 36,000 g/mol.

In paper chromatography, the following characteristics apply:

• Moving phase: This is typically a liquid that transports the substances.
• Stationary phase: This is usually a solid that remains fixed in place.

This setup allows for the separation of different components based on their movement through the stationary phase, enabling effective analysis of mixtures.

The oxidation number of Mn in KMnO₄ is +7.

To determine the oxidation state of manganese (Mn) in potassium permanganate (KMnO₄), follow these steps:

• Identify the oxidation states of the other elements:
  • Potassium (K) has an oxidation state of +1.
  • Oxygen (O) typically has an oxidation state of -2.
• Since there are four oxygen atoms, their total contribution is: -2 x 4 = -8.
• Set up the equation to find the oxidation state of Mn: +1 (K) + x (Mn) + (-8) (O) = 0.
• Solving for x (the oxidation state of Mn) gives: x = +7.

Thus, the oxidation number of manganese in KMnO₄ is confirmed to be +7.