SRMJEEE Chemistry Mock Test - 2 - JEE MCQ

Reactions that detect carbonyl groups are essential in organic chemistry.

• Hydrazine reacts with carbonyl compounds to form hydrazones, which are identifiable by their distinct melting points.
• Phenylhydrazine is another reagent that forms phenylhydrazones, useful for detecting carbonyls with aromatic substitution.
• Hydroxylamine reacts with carbonyls to produce oximes, which can also be characterised by their melting points.

All these reactions can confirm the presence of carbonyl groups in various compounds. However, the specific reaction with hydroxylamine is particularly notable for its simplicity and effectiveness.

Aliphatic and aromatic primary amines on warming with CHCl_3​ and alcoholic KOH form isocyanide or carbylamine which has very unpleasant smell. This reaction is known as carbylamine reaction. Since p-toluidine contains an aromatic primary amine group, it undergoes similar reaction and give 4-methyl phenyl isocyanide.
+3KCl + 3H₂​O

The deep blue colour observed with the starch iodide solution indicates the presence of specific ions in the acidic salt solution. This colour change occurs due to a reaction between the iodine released and the starch, forming a blue complex. The possible ions that could produce this result include:

• Chloride ions.
• Acetate ions.
• Nitrite ions.
• Bromide ions.

Among these, the most likely candidate that leads to the formation of the deep blue colour with starch iodide is:

• Nitrite ions, as they can react to yield iodine in acidic conditions.

Therefore, the salt in question is likely to be one that contains nitrite ions.

Benzene is obtained through the fractional distillation of light oil. The process involves:

• Heating the oil mixture to separate components based on boiling points.
• Collecting benzene, which vaporises at a lower temperature compared to heavier components.
• Condensing the vapours to isolate pure benzene.

This method efficiently separates benzene due to its distinct boiling point within the mixture.

The quantum number values for the designation 3d are as follows:

• n = 3
• l = 2
• m values can be -2, -1, 0, +1, +2

In this case:

• n represents the principal quantum number, indicating the energy level.
• l is the azimuthal quantum number, which defines the shape of the orbital.
• For 3d orbitals, l must be 2.

Thus, the correct quantum numbers for the 3d orbital are:

• n = 3
• l = 2

Any other combination of these quantum numbers does not correspond to the 3d designation.

Bond order is an important concept in chemistry that relates to the stability and length of a bond between atoms. Generally, as the bond order increases, the bond length decreases. Here are the key points to understand this relationship:

• A bond order of 2 indicates a double bond, which is typically shorter and stronger than a single bond.
• A bond order of 1 represents a single bond, which is longer and less stable than a double bond.
• A bond order of 0.5 suggests a resonance structure, indicating a bond that is neither fully formed nor fully broken.
• Higher bond orders correspond to greater stability and shorter bond lengths.

In summary, the bond order that corresponds to the shortest bond length is 2, as it signifies a double bond that is typically shorter and stronger than other bond types.

Carbohydrates play a crucial role in the body primarily as a source of energy. Here are some key points regarding their function:

• They are the body's main energy source, providing fuel for daily activities.
• Carbohydrates are essential for the efficient functioning of the brain and muscles.
• They help in the metabolism of fats and proteins.
• By providing energy, they support growth and development in children and adolescents.

Overall, carbohydrates are vital for maintaining energy levels and supporting various bodily functions.

A carboxylic acid can be transformed into its anhydride through various chemical reactions. The most effective reagent for this conversion is phosphorus pentaoxide. This compound is known for its strong dehydrating properties, which facilitate the removal of water during the reaction process.

• Phosphorus pentaoxide effectively removes water, allowing the formation of anhydrides.
• Other reagents, such as thionyl chloride and sulphuric acid, may not yield the desired results for this specific conversion.
• Using sulphur chloride also does not produce anhydrides efficiently.

In summary, for the conversion of a carboxylic acid to an anhydride, phosphorus pentaoxide is the preferred choice due to its capacity to promote dehydration effectively.

Carbonium ions are positively charged species formed by the loss of a hydride ion (H^-) from a hydrocarbon. Their stability varies depending on their structure. Here’s a breakdown of the types:

• Methyl carbonium ion: The least stable, having no alkyl groups to stabilise the positive charge.
• Primary carbonium ion: Slightly more stable than methyl, but still not very stable as it has only one alkyl group.
• Secondary carbonium ion: More stable than primary, with two alkyl groups providing better charge distribution.
• Tertiary carbonium ion: The most stable due to three alkyl groups surrounding the positively charged carbon. These groups help to spread out and stabilise the charge.

The increased stability of tertiary carbonium ions is primarily due to the inductive effect and hyperconjugation from the surrounding alkyl groups. This makes them more favourable in chemical reactions.

A catalyst does not affect the position of a chemical equilibrium in a reversible reaction. Here’s why:

• A catalyst speeds up the rate of both the forward and reverse reactions equally.
• It lowers the activation energy needed for the reactions to occur.
• However, it does not change the equilibrium concentrations of the reactants and products.

In contrast, other factors like pressure, temperature, and concentration can shift the equilibrium position:

• Pressure: Increasing pressure favours the side of the reaction with fewer gas molecules.
• Temperature: Raising temperature favours the endothermic direction, while lowering it favours the exothermic direction.
• Concentration: Changing the concentration of reactants or products will shift the equilibrium toward the side that reduces the change.

In summary, a catalyst is unique because it enhances the speed of reaction without altering the equilibrium state.

When considering the reaction rate equation, r = k [A] [B], the role of each reactant is crucial in determining the order of the reaction. Here’s a simplified explanation:

• If reactant B is present in large excess, its concentration remains relatively constant during the reaction.
• This allows us to treat B's concentration as a constant factor in the rate equation.
• The rate equation can then be simplified to r = k' [A], where k' = k[B] (a constant).
• This indicates that the reaction rate depends only on the concentration of A.
• As a result, the reaction behaves as a first-order reaction with respect to A.

Thus, when reactant B is in large excess, the overall order of the reaction is 1.

Streptomycin is a derivative of carbohydrates. It belongs to the aminoglycoside class of antibiotics, characterized by:

1. A cyclitol (carbohydrate-derived ring structure) linked to amino sugars.

2. A streptose group (a sugar component) and other carbohydrate-like subunits.

While streptomycin contains amino groups, its core structure is built on modified sugars, making it carbohydrate-derived.

Why Other Options Are Incorrect:

• A (Peptides): Peptides are chains of amino acids; streptomycin lacks this structure.

• C (Purines): Purines are nitrogenous bases (e.g., adenine, guanine), unrelated to streptomycin’s sugar-based framework.

• D (Terpenes): Terpenes are hydrocarbons derived from isoprene units, not sugars.

Streptomycin’s chemical foundation in carbohydrates aligns with B as the correct answer.

1. Identify the longest chain and functional group:

   • The compound has an aldehyde group (-CHO) as the highest-priority functional group.

   • The longest carbon chain containing the aldehyde is 3 carbons (propanal: CH₃-CH₂-CHO).

2. Numbering the chain:

   • The aldehyde group (-CHO) is assigned position 1.

   • The remaining carbons are numbered sequentially: CH₂ (position 2) and CCl₃ (position 3).

3. Substituents:

   • Three chlorine atoms are attached to carbon 3, forming a trichloromethyl group (-CCl₃).

4. IUPAC Name:

   • The substituent is named as 3,3,3-trichloro (all three Cl atoms on carbon 3).

   • The full name is 3,3,3-Trichloropropanal.

Why Other Options Are Incorrect:

• B (1,1,1-Trichloropropanal): Incorrect numbering; the aldehyde group must occupy position 1.

• C (2,2,2-Trichloropropanal): Misplaces the substituents; numbering starts from the aldehyde.

• D (Chloral): Refers to trichloroacetaldehyde (CCl₃CHO), a shorter chain (ethanal), not propanal.

The correct IUPAC name is 3,3,3-Trichloropropanal (A).

As the electronic configuration of inert gases is ns² np⁶ ,thus the element containing electronic configuration 1s²2s²2p⁶ , is an inert gas.

The heat of combustion of methane indicates that when methane (CH₄) reacts with oxygen (O₂), it produces carbon dioxide (CO₂) and water (H₂O) while releasing energy.

The change in enthalpy (ΔH) for this reaction at 298 K is −890.2 kJ. This value represents the energy released during the reaction.

To understand the relationship between ΔH and the change in internal energy (ΔE) of the reaction, we can consider the following points:

• In most combustion reactions, the internal energy change (ΔE) is generally less than the enthalpy change (ΔH).
• This is because some energy is used to do work against the surrounding pressure, especially when gases are produced.
• Thus, in this case, ΔE will be less than −890.2 kJ.

Therefore, the magnitude of ΔE for the reaction at this temperature is indeed less than ΔH.

Limiting molar conductivities are important for understanding the behaviour of ionic compounds in solution. The limiting molar conductivities for different salts are as follows:

• NaCl: 126 S-cm²-mol^–1
• KBr: 152 S-cm²-mol^–1
• KCl: 150 S-cm²-mol^–1

To find the molar conductivity of NaBr, we can use the trend in limiting molar conductivities of similar salts. Generally, the conductivity values of salts increase with the size and charge of the ions involved. Since Na+ and K+ ions are both monovalent, we can use the values for KBr and KCl as a reference.

Taking into account the conductivity trends:

• Na+ has a similar conductivity to K+.
• Br- and Cl- ions have comparable mobilities.

Thus, the molar conductivity of NaBr is likely to be close to that of KBr, but slightly lower due to the smaller ionic radius of Na+ compared to K+.

This leads us to estimate the molar conductivity of NaBr to be around 128 S-cm²-mol^–1, which is the most reasonable answer based on the established trends.

The ratio of the number of moles of cyclopropane (n_cyclopropane) to oxygen (n_oxygen) is determined by the ratio of their partial pressures, as both gases share the same temperature and volume in the cylinder. According to the ideal gas law (PV = nRT):

 

Why Other Options Are Incorrect:

• A: 170/740 calculates the mole fraction of cyclopropane in the total mixture, not the mole ratio to oxygen.

• B and C: These options incorrectly incorporate molar masses (42 g/mol for cyclopropane and 32 g/mol for oxygen), which are irrelevant for mole ratios (they would apply only for mass ratios).

• D: Correctly uses the direct proportionality between moles and partial pressures under constant T and V.

The mole ratio of cyclopropane to oxygen is 0.30, making D the correct answer.

1,2-dibromoethane is a compound that helps to prevent the formation of lead oxides in engine components such as spark plugs, combustion chambers, and exhaust pipes. It does this through the following mechanisms:

• Reduction of deposits: 1,2-dibromoethane chemically interacts with lead, reducing its tendency to form harmful deposits.
• Enhancing combustion: By facilitating a cleaner combustion process, it reduces the likelihood of lead oxide accumulation.
• Stability at high temperatures: This compound remains stable under high temperatures typically found in engines, ensuring effective performance throughout operation.

In summary, 1,2-dibromoethane plays a crucial role in maintaining engine cleanliness and efficiency by preventing lead oxide deposition.

When HCl gas is introduced to a saturated solution of NaCl, the solubility of NaCl is affected.

• When HCl dissolves in water, it releases hydrogen ions (H⁺).
• These ions react with the chloride ions (Cl⁻) present in NaCl.
• This reaction decreases the concentration of chloride ions in the solution.
• As a result, the solubility of NaCl in the solution decreases.
• This is due to the principle of Le Chatelier's principle, where the system adjusts to counteract changes.

In summary, adding HCl to a saturated solution of NaCl reduces its solubility due to the formation of hydrogen chloride in solution, which decreases the availability of chloride ions.

Optical activity in molecules occurs when they can rotate the plane of polarized light. This typically requires the presence of a chiral centre, which is a carbon atom bonded to four different groups.

Among the given molecules with the formula C₄H₆O₂, we can analyse each structure:

• CH₃COCH₂CH₃: This molecule has no chiral centres, making it achiral.

• CH₃CH₂CH₂CHO: This also lacks chiral centres, therefore it is achiral.

• CH₂=CH-CH₂-COOH: This molecule contains a double bond and a carboxylic acid but has no chiral centres, thus it is achiral.

• CH₂=CH-CH(OH)CHO: This structure includes a chiral centre at the carbon bonded to the hydroxyl group (OH) and the aldehyde group (CHO). Therefore, this molecule is optically active.

The only optically active molecule from the list is CH₂=CH-CH(OH)CHO.

Zwitterions are unique compounds that carry both a positive and a negative charge, resulting in an overall neutral charge. They are particularly important in the field of biochemistry.

One of the most common examples of a zwitterion is glycine, which is the simplest amino acid. Here’s why glycine is a zwitterion:

• Glycine contains both an amino group (–NH2) and a carboxyl group (–COOH).
• In solution, the amino group can accept a proton (H+), becoming positively charged (–NH3+).
• The carboxyl group can donate a proton, becoming negatively charged (–COO–).
• As a result, glycine exists as a molecule with a positive charge on one end and a negative charge on the other.

Other compounds mentioned, such as aniline, aminophenol, and acetamide, do not typically form zwitterionic forms under normal conditions. This makes glycine a classic example of a zwitterion in biological systems.

The problem involves determining the half-life of a radioactive element based on its decay over a specified time period.

Given that:

• 1 g of the radioactive element reduces to 125 mg in 24 hours.

To find the half-life, we can follow these steps:

• Start with 1 g, which is equal to 1000 mg.
• After the first half-life, the amount will be half of 1000 mg, which is 500 mg.
• After the second half-life, it will reduce to half of 500 mg, resulting in 250 mg.
• After the third half-life, it will be half of 250 mg, leading to 125 mg.

This means:

• It takes three half-lives to go from 1000 mg to 125 mg.
• Since the total time elapsed is 24 hours, we divide this by the number of half-lives:

Half-life calculation:

• 24 hours / 3 half-lives = 8 hours.

Thus, the half-life of the isotope is 8 hours.

The substance C₄H₁₀O undergoes oxidation to form a compound C₄H₈O. This new compound can produce an oxime and shows a positive iodoform test. Additionally, when the original substance is treated with concentrated H₂SO₄, it yields C₄H₈.

To identify the structure of the compound, we can analyse the characteristics of the resulting products:

• C₄H₁₀O is likely a secondary alcohol or an aldehyde.
• The oxidation to C₄H₈O indicates the formation of a ketone.
• The positive iodoform test confirms that there is a presence of a methyl ketone in the structure.

Considering these points, the compound must have a structure that can support the formation of a ketone from an alcohol. Therefore, the most plausible structure is:

CH₃CHOHCH₂CH₃

Select the incorrect statement(s):

• The electron-withdrawing effect of the carbonyl group in the -COOH group weakens the O-H bond. This makes it easier for a carboxylic acid to ionise compared to an alcohol.

• The inductive effect of chlorine destabilises the acid while stabilising the conjugate base.

• Aniline is indeed a weaker base than ammonia.

• Contrary to some beliefs, phenol is a weaker acid than water.

Natural polymers are materials that occur naturally and are composed of long chains of molecules. They play a crucial role in various biological and ecological processes.

• Wool is a classic example of a natural polymer. It is made from the protein keratin, which is produced by sheep.
• Other examples include cotton, derived from cotton plants, and silk, produced by silkworms.
• Natural polymers are biodegradable, making them environmentally friendly compared to synthetic options.

In contrast, synthetic polymers like orlon, teflon, and polystyrene are man-made and do not occur naturally. These materials can have applications in various industries but often do not possess the same eco-friendly properties as natural polymers.

To determine the molecular weight of the monoacid base:

• We start with the weight of the plantinichloride compound, which is 0.6387 g.
• Upon ignition, this yields 0.209 g of platinum.
• To find the molecular weight of the base, we need to relate the mass of platinum produced to the original compound.
• The number of moles of platinum can be calculated using its atomic weight, which is approximately 195.08 g/mol.
• Calculating moles of platinum:
• Moles of platinum = mass of platinum / atomic weight = 0.209 g / 195.08 g/mol.
• This gives approximately 0.00107 moles of platinum.
• Since plantinichloride contains one atom of platinum per molecule, the number of moles of plantinichloride is also 0.00107 moles.
• Next, we can find the molecular weight of the base:
• Using the mass of plantinichloride, we find:
• Molecular weight = mass / moles = 0.6387 g / 0.00107 moles.
• This results in approximately 596.9 g/mol for the molecular weight of the plantinichloride.
• Since the molecular weight is for the entire compound, we must account for the platinum and chlorine atoms to isolate the base's molecular weight.
• After subtracting the weights of the platinum and chlorine from the total, we arrive at the molecular weight of the monoacid base.

The final calculated molecular weight of the base is approximately 93 g/mol.