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VITEEE Chemistry Test - 2 - JEE MCQ


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30 Questions MCQ Test VITEEE: Subject Wise and Full Length MOCK Tests - VITEEE Chemistry Test - 2

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VITEEE Chemistry Test - 2 - Question 1

To form propane-3-ethylpentan-3-ol, the reagents needed are

Detailed Solution for VITEEE Chemistry Test - 2 - Question 1

To form propane-3-ethylpentan-3-ol, the following reagents are necessary:

  • Reagent A: CH3CH2MgBr + CH3COCH2CH3

  • Reagent B: CH3MgBr + CH3CH2CH2COCH2CH3

  • Reagent C: CH3CH2MgBr + CH3CH2COCH2CH3

  • Reagent D: CH3CH2CH2MgBr + CH3COCH2CH3

This reaction involves the use of Grignard reagents, which are crucial in forming new carbon-carbon bonds. The correct choice for achieving the desired alcohol structure is Reagent C.

VITEEE Chemistry Test - 2 - Question 2

Aromatic aldehydes in the presence of cyanide ion as catalyst, are converted into acyloins. This reaction is called

Detailed Solution for VITEEE Chemistry Test - 2 - Question 2

Aromatic aldehydes can undergo a transformation in the presence of a cyanide ion catalyst, leading to the formation of acyloins. This chemical reaction is specifically known as benzoin condensation.

  • In this reaction, two molecules of aromatic aldehyde combine.
  • The cyanide ion acts as a catalyst, facilitating the reaction.
  • The product, acyloin, contains both an alcohol and a ketone functional group.

The significance of the benzoin condensation lies in its ability to create complex organic molecules, which are essential in various fields such as pharmaceuticals and materials science.

VITEEE Chemistry Test - 2 - Question 3

When vapours of isopropyl alcohol are passed over heated copper, the major product obtained is

Detailed Solution for VITEEE Chemistry Test - 2 - Question 3

When vapours of isopropyl alcohol are passed over heated copper, the major product obtained is:

  • The reaction primarily produces acetone.
  • Isopropyl alcohol undergoes a process called dehydrogenation when exposed to heat and copper.
  • This process removes hydrogen from the alcohol, leading to the formation of ketones, particularly acetone.
  • Acetone is a common solvent and is widely used in industries and laboratories.
VITEEE Chemistry Test - 2 - Question 4

Which alkene gives the same product by the Markownikoff's and anti-Markownikoff's method

Detailed Solution for VITEEE Chemistry Test - 2 - Question 4

Alkenes can react differently depending on the method used, such as Markownikoff's or anti-Markownikoff's.

The alkene that produces the same product through both methods is:

  • β -butylene

This is significant because:

  • Markownikoff's rule states that during addition reactions, the more substituted carbon will receive the new substituent.
  • Anti-Markownikoff's method leads to the less substituted carbon receiving the substituent.

However, for β -butylene, both methods yield the same product due to its symmetrical structure, which allows for equal distribution of the substituent regardless of the approach taken.

VITEEE Chemistry Test - 2 - Question 5

Which of the following has a derived unit?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 5

Derived units are units that are defined from other units, using mathematical relationships. In contrast, base units measure fundamental quantities. Here’s a brief overview of derived units:

  • Unit of mass (e.g., kilogram) is a base unit.
  • Unit of time (e.g., second) is also a base unit.
  • Unit of length (e.g., metre) is a base unit.
  • Unit of volume (e.g., litre) is a derived unit, typically calculated as length cubed (e.g., cubic metre).

In summary, among the options listed, the unit of volume is the only one that qualifies as a derived unit.

VITEEE Chemistry Test - 2 - Question 6

How many monochlorobutanes will be obtained on chlorination of n-butane?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 6

VITEEE Chemistry Test - 2 - Question 7

The ratio of the molar amounts of H₂S needed to precipitate the metal ions from 20 mL each of 1M Cd(NO₃)₂ and 0.5 M CuSO₄ is

Detailed Solution for VITEEE Chemistry Test - 2 - Question 7

As Cd2+ and Cu2+, both are dipositive ions but Cu2+ has one half of the concentration compared to that of the Cd2+. Thus, the molar amount of the H2S used to precipitate the metals ions Cd2+ and Cu2+ is 2 : 1.

VITEEE Chemistry Test - 2 - Question 8

What happens when conc. H₂SO₄ is treated with sugar?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 8

When concentrated H2SO4 reacts with sugar, a chemical process occurs that can be described as follows:

  • Concentration: Concentrated sulphuric acid is a strong dehydrating agent.
  • Dehydration Reaction: The acid removes water from sugar (sucrose), leading to a dehydration reaction.
  • Carbon Formation: This process produces carbon, resulting in a black, charred substance.
  • Heat Generation: The reaction is exothermic, releasing heat.

Overall, the primary process that occurs when concentrated sulphuric acid is treated with sugar is dehydration, which ultimately transforms sugar into a carbon-rich material.

VITEEE Chemistry Test - 2 - Question 9

Which of the following acts as a source of energy in living systems?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 9

All of the options mentioned can act as sources of energy in living systems, but they serve different roles and are broken down differently within the body. Here's an explanation for each:

1. Fat: Fats are a major source of energy for the body, providing 9 calories per gram. They are stored in adipose tissue and can be broken down through a process called beta-oxidation to produce energy in the form of ATP when the body needs it.

2. Glucose: Glucose is a simple sugar and the primary source of energy for cells. It is the main product of carbohydrate digestion and is transported through the bloodstream to cells, where it is broken down via glycolysis and cellular respiration to produce ATP, the main energy currency for cells.

3. Fructose: Fructose is another simple sugar found in fruits, honey, and some vegetables. It can be converted into glucose in the liver and then used as a source of energy through the same pathways as glucose.

VITEEE Chemistry Test - 2 - Question 10

Which of the following represents proteins?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 10

Proteins are polymers composed of amino acids linked by peptide bonds, which are a type of amide bond (-CONH-). Thus, proteins are classified as polyamides.

  • Option A (Polyamides): Correct, as proteins consist of long chains of amino acids connected by amide (peptide) bonds.

  • Option B (Polythioethers): Incorrect; thioethers involve sulfur linkages, not relevant to protein structure.

  • Option C (Glycerides): Incorrect; glycerides are lipid molecules (esters of glycerol and fatty acids).

  • Option D (Polysaccharides): Incorrect; polysaccharides are carbohydrate polymers (e.g., starch, cellulose).

Proteins are represented by polyamides due to their peptide bond structure.

VITEEE Chemistry Test - 2 - Question 11

What will happen if LiAlH₄ is added to an ester ?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 11

When lithium aluminium hydride (LiAlH₄) is added to an ester, it undergoes a reduction reaction.

  • The reaction produces two molecules of alcohol.
  • This occurs because LiAlH₄ acts as a strong reducing agent.
  • Esters are typically converted into primary alcohols.
  • Therefore, the end products of this reaction are alcohols, not acids or other compounds.

In summary, adding LiAlH₄ to an ester results in:

  • Two units of alcohol.
VITEEE Chemistry Test - 2 - Question 12

Which of the following cannot reduce Fehling solution?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 12

Fehling solution is a chemical reagent used to test for reducing sugars. Not all substances can reduce it.

The following substances are considered:

  • Formic acid: A strong reducing agent, can effectively reduce Fehling solution.
  • Acetic acid: This is a weak acid that does not have reducing properties, so it cannot reduce Fehling solution.
  • Formaldehyde: A simple aldehyde that readily reduces Fehling solution.
  • Acetaldehyde: Another aldehyde that can also reduce Fehling solution.

Thus, of the options listed, acetic acid is the only one that cannot reduce Fehling solution.

VITEEE Chemistry Test - 2 - Question 13

Which of the following compounds on boiling with KMnO₄ and the subsequent acidification will not give benzoic acid ?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 13

When compounds are boiled with KMnO4 and subsequently acidified, some will not yield benzoic acid.

The following points explain why certain compounds behave differently:

  • Benzyl alcohol: This compound can be oxidised to benzoic acid through the reaction with KMnO4.
  • Acetophenone: This compound is also oxidisable, leading to the formation of benzoic acid upon treatment with KMnO4.
  • Anisole: This compound does not yield benzoic acid when boiled with KMnO4. The methoxy group is stable under these conditions, preventing oxidation to benzoic acid.
  • Toluene: This compound can be oxidised to benzoic acid by KMnO4.

Therefore, among the listed compounds, Anisole is the one that will not produce benzoic acid upon boiling with KMnO4 and acidification.

VITEEE Chemistry Test - 2 - Question 14

The law of mass action was given by

Detailed Solution for VITEEE Chemistry Test - 2 - Question 14

The law of mass action is a principle in chemistry that describes how the rate of a reaction is related to the concentration of the reactants. It was developed by Guldberg and Waage in the 19th century. Their work laid the foundation for understanding chemical equilibria.

  • Key Concept: The rate of a reaction increases with the concentration of the reactants.
  • Importance: This law helps predict how changes in concentration affect the speed of chemical reactions.
  • Applications: It is widely used in chemical engineering, environmental science, and biochemistry.

Guldberg and Waage's law is essential for understanding dynamic systems where reactants and products are constantly interchanging. Their findings highlight the significance of concentration in achieving equilibrium in chemical reactions.

VITEEE Chemistry Test - 2 - Question 15

The heats of dissociation (k.cals/mole) of methane and ethane are 360 and 620 respectively. From these, C - C bond energy (k.cals/mole) in ethane can be evaluated as

Detailed Solution for VITEEE Chemistry Test - 2 - Question 15

VITEEE Chemistry Test - 2 - Question 16

A Carnot engine, working between 300 K to 600 K has a work output of 800 J-cycle⁻1. What is the amount of heat energy supplied to the engine from the source per cycle?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 16

VITEEE Chemistry Test - 2 - Question 17

In which of the following neutralization reaction, the heat of neutralisation will be highest?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 17

In a neutralisation reaction, the heat of neutralisation varies based on the reactants involved.

Key factors affecting the heat of neutralisation include:

  • Type of Acid and Base: Strong acids and bases typically release more heat when they react compared to weak acids and bases.
  • Complete Ionisation: Strong acids and bases ionise completely in solution, leading to a more exothermic reaction.
  • Reaction Products: The nature of the products formed can influence the overall energy change during the reaction.

Among common neutralisation reactions:

  • HCl and NaOH: This is a strong acid-strong base reaction, resulting in a significant heat release.
  • NH4OH and H2SO4: This involves a weak base and a strong acid, which typically produces less heat than strong acid-strong base reactions.
  • CH3COOH and KOH: Here, acetic acid is a weak acid, leading to a lower heat release.
  • H2CO3 and KOH: Carbonic acid is also a weak acid, resulting in less heat compared to strong acid-strong base reactions.

Therefore, the reaction between HCl and NaOH produces the highest heat of neutralisation due to the complete ionisation of both reactants, resulting in a more energetic reaction.

VITEEE Chemistry Test - 2 - Question 18

During the deepening of the colour of a dye, the absorption shifts towards red. This shifting is called

Detailed Solution for VITEEE Chemistry Test - 2 - Question 18

During the deepening of the colour of a dye, the absorption shifts towards red. This shifting is called

  • Bathochromic shift: This refers to the red shift in absorption, indicating a longer wavelength.
  • Hypsochromic shift: This indicates a shift towards shorter wavelengths, or a blue shift.
  • Hyperchromic shift: This is an increase in absorption intensity without a shift in wavelength.
  • Auxochromic shift: This occurs when a substance changes the colour of a dye without affecting the position of the absorption.

The correct term for the shift towards red in dye absorption is the Bathochromic shift.

VITEEE Chemistry Test - 2 - Question 19

The IUPAC name of the compound having the formula (CH₃)₃C-CH =CH₂ is

Detailed Solution for VITEEE Chemistry Test - 2 - Question 19

VITEEE Chemistry Test - 2 - Question 20

[Co(NH₃)₄Cl₂]NO₂ and [Co(NH₃)₄ClNO₂]Cl exhibit which type of isomerism

Detailed Solution for VITEEE Chemistry Test - 2 - Question 20

Isomerism Types: The compounds [Co(NH3)4Cl2]NO2 and [Co(NH3)4ClNO2]Cl can exhibit different forms of isomerism.

Isomerism is the phenomenon where compounds have the same molecular formula but differ in structure or arrangement. Here are the main types relevant to the given compounds:

  • Geometrical Isomerism: This occurs when ligands around a central metal ion are arranged differently in space, leading to distinct forms.
  • Optical Isomerism: This type arises when compounds can exist in two forms that are mirror images of each other, often affecting their interaction with light.
  • Linkage Isomerism: This occurs when a ligand can attach to the metal ion in different ways, leading to different structural forms.
  • Ionisation Isomerism: This form arises when different ions can be released from a complex in solution, leading to different ionic forms.

In the case of the mentioned cobalt complexes, they are primarily known to exhibit ionisation isomerism due to the presence of different ions that can be released.

VITEEE Chemistry Test - 2 - Question 21

Which of the following complexes is an outer orbital complex?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 21

To determine which complex is an outer orbital complex, we need to identify if the hybridization involves sp³d² (outer orbital) rather than d²sp³ (inner orbital). Outer orbital complexes are typically high-spin and use outer d-orbitals.

  1. [Ni(NH₃)₆]²⁺ (Option A):

    • Ni²⁺ has a d⁸ configuration.

    • NH₃ is a moderately strong field ligand.

    • In an octahedral field, d⁸ with NH₃ remains high-spin (t₂g⁶e₉², two unpaired electrons).

    • High-spin complexes use sp³d² hybridization, making this an outer orbital complex.

  2. [Mn(CN)₆]³⁻ (Option B), [Co(NH₃)₆]³⁺ (Option C), [Fe(CN)₆]³⁻ (Option D):

    • These complexes involve strong field ligands (CN⁻, NH₃), leading to low-spin configurations.

    • Low-spin complexes use d²sp³ hybridization (inner orbital complexes).

Only [Ni(NH₃)₆]²⁺ qualifies as an outer orbital complex due to its high-spin configuration and sp³d² hybridization.

VITEEE Chemistry Test - 2 - Question 22

Acetaldoxime reacts with P₂O₅ to give

Detailed Solution for VITEEE Chemistry Test - 2 - Question 22

Acetaldoxime undergoes a reaction with P₂O₅, resulting in the formation of specific organic compounds. Below are the possible products:

  • Methyl cyanide: A simple nitrile compound.
  • Ethyl cyanate: An ester of cyanic acid.
  • Ethyl cyanide: Also known as ethyl nitrile.
  • Mixture of all these: A combination of the above products.

The primary product of this reaction is typically methyl cyanide. The presence of other compounds may depend on specific reaction conditions.

VITEEE Chemistry Test - 2 - Question 23

If a salt bridge is removed between the two half cells, the voltage

Detailed Solution for VITEEE Chemistry Test - 2 - Question 23

If a salt bridge is removed between the two half-cells of an electrochemical cell, the voltage drops to zero.
The salt bridge is essential for maintaining ion flow between the half-cells, which balances charge and allows the redox reactions to continue. Without it:

  1. Charge Build-Up: The anode accumulates positive charge, and the cathode accumulates negative charge, halting electron flow.

  2. Open Circuit: The circuit becomes incomplete, preventing current flow.

  3. Voltage Measurement: While the theoretical potential difference still exists, the practical voltage (measurable under operating conditions) drops to zero because the cell cannot sustain the electrochemical reactions.

VITEEE Chemistry Test - 2 - Question 24

In which of the following solutions are ions present?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 24

In the context of solutions containing ions:

  • Sucrose in water does not produce ions. It dissolves as whole molecules, remaining neutral.

  • Sulphur in CS2 (carbon disulphide) also does not create ions, as both substances remain molecular without ionisation.

  • Cesium nitrate in water is the only option that produces ions. When dissolved, it separates into cesium ions (Cs+) and nitrate ions (NO3-).

  • Ethyl alcohol in water does not form ions either; it dissolves without breaking into charged particles.

The presence of ions is crucial for electrical conductivity and various chemical reactions in solutions. Therefore, among the options, only cesium nitrate in water contributes to the formation of ions.

VITEEE Chemistry Test - 2 - Question 25

0.066 gram of metal was deposited when a current of 2 amperes is passed through a metal ion solution for 100 seconds. What is the electrochemical equivalent (in gram coulomb -1) of the metal?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 25


= 3.3 x 10-4 gm coulomb

VITEEE Chemistry Test - 2 - Question 26

A metal X on heating in nitrogen gas gives Y. Y on treatment with H₂O gives colourless gas which when passed through CuSO₄ solution gives blue colour. Y is

Detailed Solution for VITEEE Chemistry Test - 2 - Question 26

When metal X is heated in nitrogen gas, it produces Y. Y, when treated with water, generates a colourless gas. This gas, when passed through a copper(II) sulphate solution, turns blue.

The substance Y can be understood as follows:

  • Metal X reacts with nitrogen to form Y.
  • Upon treatment with water, Y releases a colourless gas.
  • This gas, upon interaction with CuSO4 (copper(II) sulphate), results in a blue colour.

This blue colour indicates the formation of copper(II) ions from the reaction with the colourless gas, which is likely ammonia (NH3). Thus, Y is likely magnesium nitride (Mg3N2), which decomposes in water to produce ammonia.

VITEEE Chemistry Test - 2 - Question 27

The ether that undergoes electrophilic substitution reaction is

Detailed Solution for VITEEE Chemistry Test - 2 - Question 27

The ether that undergoes electrophilic substitution is C₆H₅OCH₃ (anisole). The methoxy group (-OCH₃) attached to the benzene ring activates the aromatic ring toward electrophilic substitution by donating electron density through resonance. This activation directs electrophiles to the ortho and para positions of the benzene ring.

Key Points:

  • Electrophilic substitution requires an activated aromatic system.

  • Aliphatic ethers (options A, C, D) lack an aromatic ring and cannot undergo such reactions.

  • Anisole (B) contains a benzene ring activated by the methoxy group, enabling electrophilic substitution.

The correct answer is B: C₆H₅OCH₃.

VITEEE Chemistry Test - 2 - Question 28

The poisonous gas present in the exhaust fumes of cars is

Detailed Solution for VITEEE Chemistry Test - 2 - Question 28

Carbon monoxide (CO) is a toxic gas produced by the incomplete combustion of carbon-containing fuels.

Key points about carbon monoxide in car exhaust:

  • It is colourless and odourless, making it difficult to detect.
  • CO can cause serious health issues, including headaches, dizziness, and even death.
  • Vehicles that are poorly maintained or have faulty engines are more likely to emit high levels of CO.
  • Proper ventilation and regular maintenance can help reduce CO emissions.

Understanding the dangers of carbon monoxide is essential for safety and health.

VITEEE Chemistry Test - 2 - Question 29

When AgNO₃ is heated strongly the products formed are

Detailed Solution for VITEEE Chemistry Test - 2 - Question 29

When silver nitrate (AgNO3) is heated strongly, it decomposes to produce specific gases.

  • The primary gas formed is nitrogen dioxide (NO2).
  • In addition, oxygen (O2) is released during this process.

This decomposition reaction can be summarised as follows:

  • AgNO3 → NO2 + O2

It is important to note that nitric oxide (NO) is not a product of this specific reaction when silver nitrate decomposes under heat. The presence of nitrogen monoxide (NO) is associated with different reactions and conditions.

Thus, the key products of heating silver nitrate are nitrogen dioxide and oxygen.

VITEEE Chemistry Test - 2 - Question 30

Which of the following is not an acid salt?

Detailed Solution for VITEEE Chemistry Test - 2 - Question 30

Acid salts are formed when a strong acid reacts with a weak base, resulting in a salt that still has acidic properties. Here’s a brief overview of the provided compounds:

  • NaH₂PO₂ (Sodium dihydrogen phosphite) - This is an acid salt because it contains two hydrogen ions that can donate protons.
  • NaH₃PO₂ (Sodium hydrogen phosphite) - This is also an acid salt as it has three hydrogen ions available for reaction.
  • NaH₂PO₄ (Sodium dihydrogen phosphate) - This compound is another example of an acid salt, retaining one hydrogen ion.

None of these options qualify as non-acid salts, indicating that they all possess acidic characteristics. Therefore, the answer to the question is:

  • D: None of these - All options listed are acid salts.
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