VITEEE Chemistry Test - 10 - JEE MCQ

Ethylene can be produced by treating concentrated H₂SO₄ with a suitable alcohol at high temperatures (160-170°C).

• The reaction involves dehydration of an alcohol, which results in the formation of ethylene.
• Among the given options, the most suitable alcohol for this reaction is one that can effectively undergo dehydration.
• C₂H₅OH (ethanol) is a viable candidate as it readily dehydrates to form ethylene when treated with concentrated sulphuric acid.
• Other options like CH₃OH (methanol) and C₃H₇OH (propyl alcohol) are less likely to yield ethylene through this method.
• Therefore, the correct choice for the alcohol X, which produces ethylene when treated with concentrated H₂SO₄, is C₂H₅OH.

Aromatic aldehydes can undergo a specific reaction when treated with sodium or potassium hydroxide. This process involves the disproportionation of aldehydes, resulting in the formation of both an alcohol and an acid. This reaction is referred to as the Cannizzaro reaction.

• Occurs in the presence of strong bases like sodium or potassium hydroxide.
• Generates an alcohol and a carboxylic acid from the aldehyde.
• Typical for aromatic aldehydes that lack alpha-hydrogens.

The carbonyl compound reacts with hydrogen cyanide to form a cyanohydrin, which upon hydrolysis yields a racemic mixture of α-hydroxy acid.

• The process begins with a carbonyl compound.
• When it reacts with hydrogen cyanide, a cyanohydrin is produced.
• Hydrolysis of the cyanohydrin results in a racemic mixture of α-hydroxy acid.
• The key factor is that the carbonyl compound must have a suitable structure to enable this reaction.

In this case, the suitable carbonyl compounds include:

• Formaldehyde Produces a different product.
• Acetaldehyde: Reacts to form a racemic mixture.
• Acetone: Also leads to a racemic mixture.
• Therefore, the correct choice must be identified based on the nature of the products formed.

The alkene that yields the same product through both Markownikoff's and anti-Markownikoff's methods is:

• Propylene

This is because:

• In reactions, propylenes can lead to the same product regardless of the method used.
• Both methods involve the addition of H and X (where X can be a halogen) to the double bond.
• For propylene, the product does not vary significantly between the two approaches.

The Wurtz reaction involves the interaction of alkyl halides with sodium in dry ether. This process is used to create longer carbon chains through the coupling of two alkyl groups. Here are key points about the reaction:

• The reaction typically requires dry ether as a solvent.
• Alkyl halides are reactants that undergo coupling.
• Sodium serves as a reducing agent, facilitating the reaction.
• The process is useful in organic synthesis for forming larger molecules.

The reaction involves the addition of HBr to an alkene.

• The initial compound is an alkene, CH₂=CH-CH₃.
• When HBr reacts with the alkene, it undergoes a process called electrophilic addition.
• This type of reaction involves an electrophile (H⁺) attacking the double bond, resulting in the formation of a carbocation.
• Subsequently, the bromide ion (Br⁻) attacks the carbocation, leading to the final product, CH₃CHBr-CH₃.

This reaction is characteristic of alkenes and is a common method for adding halogens to organic compounds.

The law of multiple proportions states that when two elements combine to form more than one compound, the ratios of the masses of one element that combine with a fixed mass of the other element are simple whole numbers.

To identify pairs that illustrate this law, consider the following:

• NH₃ and HCl: These are not multiple compounds of the same elements.
• H₂S and SO₂: These compounds contain different elements and do not exemplify the law.
• CuO and Cu₂O: These are both copper oxides and demonstrate different ratios of copper to oxygen.
• CS₂ and FeSO₄: These compounds do not share the same elements in differing ratios.

Among the options, CuO and Cu₂O best illustrate the law of multiple proportions as they represent different compounds formed by varying amounts of the same elements.

Enzymes are primarily composed of:

• Proteins that have a specific structure.
• These proteins facilitate biochemical reactions in living organisms.
• They are not made from carbohydrates or combinations of carbohydrates and nitrogen.

In summary, enzymes are essential proteins that play a crucial role in various biological processes.

Cellulose is a type of carbohydrate known as a polymer, specifically made up of repeating units of D-glucose.

This natural polymer serves several important functions:

• Structural support: Cellulose provides strength and rigidity to plant cell walls.
• Dietary fibre: It is a key component of dietary fibre in human nutrition, aiding in digestion.
• Biodegradability: Cellulose is biodegradable, making it environmentally friendly.

Overall, cellulose is vital for both ecological balance and human health.

The false statement among the options provided is:

• Hydrogen bonds play a crucial role in stabilising various protein structures:
• They stabilise the alpha helix structure.
• They also stabilise the beta helix structure.
• They are essential for maintaining the native structures of proteins.
• However, the alpha helix is not a part of the primary structure of proteins; it is a secondary structure.

When propanamide reacts with Br₂ and NaOH, the following occurs:

• Propanamide undergoes a reaction known as the Hofmann degradation.
• In this process, the amide is transformed into an amine through halogenation and hydrolysis.
• The reaction conditions (Br₂ and NaOH) facilitate the conversion of the carbonyl group into a nitrogen compound.
• As a result, the final product of this reaction is propylamine.

A carboxylic acid can be transformed into its anhydride using specific reagents. The most effective among these include:

• Thionyl chloride - commonly used for converting carboxylic acids to acyl chlorides, which can then be further processed into anhydrides.
• Sulphuric acid - acts as a dehydrating agent, facilitating the formation of anhydrides from carboxylic acids.
• Phosphorus pentoxide - known for its strong dehydrating properties, it can also promote the conversion of carboxylic acids to anhydrides.

In summary, the conversion process can be efficiently accomplished using thionyl chloride, sulphuric acid, or phosphorus pentoxide, with each reagent having its unique advantages in the chemical reaction.

When formic acid reacts with H₂SO₄, it undergoes a chemical transformation. The reaction leads to the formation of:

• Carbon monoxide (CO)
• This process involves dehydration, resulting in the release of water.
• Formic acid loses a water molecule to yield carbon monoxide as one of the products.

At the boiling point, a liquid and its vapour are in equilibrium, meaning:

• The molecules in both phases have the same kinetic energy. This is because they are at the same temperature.
• Potential energy and internal energy can differ between the two phases due to differences in intermolecular forces.
• Thus, while kinetic energy is equal, other energy types vary.

To determine the equilibrium constant (K), consider the following:

• The rate of the forward reaction is twice that of the reverse reaction.
• This implies that for every 2 units of products formed, 1 unit of reactants is produced in reverse.
• At equilibrium, the ratio of the rates of the forward and reverse reactions can be expressed as:
• Rate Forward = 2 × Rate Reverse
• Thus, at equilibrium:
• The equilibrium constant K can be derived from the relationship of these rates.
• Since K = [Products] / [Reactants], it reflects the ratio of the concentrations of products to reactants at equilibrium.
• Given the rates, we can conclude that K equals:
• K = 2

Therefore, the equilibrium constant K is 2.0.

To find the new intake temperature of the Carnot engine when its efficiency is increased:

• The efficiency of a Carnot engine is calculated using the formula: Efficiency = 1 - (T2/T1), where T1 is the intake temperature and T2 is the exhaust temperature.
• Given the initial efficiency of 20% (or 0.20) and an intake temperature (T1) of 450 K, we can find T2:
  • 0.20 = 1 - (T2/450)
  • Rearranging gives: T2 = 450 × (1 - 0.20) = 360 K.
• Now, with the increased efficiency of 30% (or 0.30), we need to find the new intake temperature (T1):
  • 0.30 = 1 - (360/T1)
  • Rearranging gives: T1 = 360 / (1 - 0.30) = 360 / 0.70.
  • T1 = 514.29 K.

The new intake temperature is approximately 514 K.

The enthalpy of a solution of KNO₃ is + 35.64 kJ.

This indicates:

• Absorption of heat during the dissolution process.
• This means the solution feels cooler to the touch as it takes in heat from the surroundings.
• Such a process is termed endothermic.

 Urea ((NH₂)₂CO) is considered a neutral fertilizer because it does not significantly alter the pH of the soil. Unlike ammonium nitrate, ammonium sulphate, and calcium ammonium nitrate, which can influence soil acidity, urea is relatively neutral in its effect on soil pH. Therefore, the correct answer is urea.

2,3-Dichloro butane exhibits optical isomerism.

This type of isomerism arises from the presence of chiral centres in the molecule.

• Chiral centres are carbon atoms bonded to four different groups.
• In 2,3-dichloro butane, both carbon atoms at positions 2 and 3 are chiral.
• This results in two non-superimposable mirror images, known as enantiomers.

Thus, the isomerism displayed by 2,3-dichloro butane is specifically optical isomerism.

Prussian blue is formed by combining an aqueous solution of Fe³⁺ salt with ferrocyanide.

• This reaction produces a deep blue pigment known as Prussian blue.
• Ferrocyanide is crucial in this process, as it contains the cyanide group.
• The resulting compound has various applications, including in art and medicine.

The reaction of RCN with sodium in alcohol typically produces:

• Primary amines such as RCH₂NH₂.
• This process involves the reduction of the nitrile group (RCN) to an amine.
• The sodium acts as a reducing agent, facilitating the conversion.

In summary, the key product from this reaction is RCH₂NH₂, a primary amine formed through the reduction of the nitrile compound.

The diamagnetic species is:

• The species that exhibit diamagnetism have all their electrons paired.
• To determine the diamagnetic nature of the given complexes, we need to assess the oxidation states and the field strength of the ligands:
• [Ni(CN)₄]²⁻: Nickel in this complex is in the +2 oxidation state. CN⁻ is a strong field ligand, leading to pairing of electrons. Thus, this complex is diamagnetic.
• [Ni(Cl)₄]²⁻: Nickel is again +2. However, Cl⁻ is a weak field ligand, resulting in unpaired electrons, making it paramagnetic.
• [CoCl₄]²⁻: Cobalt is in the +2 state, and with Cl⁻ as a weak field ligand, it also has unpaired electrons, indicating paramagnetism.
• [CoF₆]²⁻: Cobalt in +2 state with F⁻ as a weak field ligand leads to unpaired electrons, hence it is paramagnetic.

In conclusion, the only diamagnetic species among the options is [Ni(CN)₄]²⁻.

The colour of p-aminoazobenzene is:

• Orange is the characteristic colour of p-aminoazobenzene.
• This compound is known for its vibrant hue, widely used in dyes.
• It is important in various applications, including textiles and pigments.

The law of electrolysis was established by Michael Faraday.

Faraday's work laid the foundation for understanding how electricity interacts with chemical substances. Here are some key points about his contributions:

• Faraday's First Law: The mass of a substance altered at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
• Faraday's Second Law: The mass of a substance altered is also proportional to its equivalent weight, which is specific to each substance.
• His laws help in calculating the amount of material produced or consumed in electrochemical reactions.

To find the degree of dissociation of the weak acid:

• The equivalent conductivity of the weak acid at 0.1 M is stated to be 100 times less than that at infinite dilution.
• At infinite dilution, the conductivity represents complete dissociation of the acid.
• The degree of dissociation (α) can be calculated using the following relationship:
• Degree of dissociation is defined as the fraction of the acid that dissociates in solution.
• If the conductivity at 0.1 M is 1% of the infinite dilution value, the degree of dissociation can be derived as:
• α = (Conductivity at 0.1 M) / (Conductivity at infinite dilution)
• Given that 0.1 M weak acid conductivity is 1% (1/100) of the infinite dilution value, α = 0.01.

The degree of dissociation is thus 0.01.

When ether is exposed to air for some time, it can produce an explosive substance known as:

• The main explosive substance that forms is called peroxide.
• Peroxides can be highly unstable and potentially dangerous.
• Proper storage and handling of ether are essential to prevent the formation of these dangerous compounds.

The method used for purifying titanium is the Van-Arkel process.

• This method involves thermal decomposition of titanium tetraiodide.
• Titanium is deposited on a tungsten filament, resulting in high purity.
• It is particularly effective due to its ability to remove impurities.
• Van-Arkel is a well-known process specifically tailored for titanium purification.

The method for obtaining highly pure silicon used as a semiconductor material is:

• Zone refining is the preferred technique.
• This process involves melting a small region of the silicon, which helps remove impurities.
• As the molten zone moves along the silicon, it leaves behind the impurities, resulting in higher purity.
• Zone refining is effective in achieving the high purity levels required for semiconductor applications.