Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Alcohols and Phenols – 4”.

1. The oxidation of alcohol involves the cleavage of _____ bonds.
a) O-H
b) C-H
c) O-H and C-H
d) Hydrogen
Answer: c
Clarification: Oxidation of alcohols involves the cleavage of both O-H and C-H bonds so as to form a carbon-oxygen double bond. This results in the release of dihydrogen and as a result is also known as dehydrogenation reactions.

2. Which of the following oxidising agents is used to obtain carboxylic acids directly from alcohols?
a) Acidified KMnO₄
b) Aqueous KMnO₄
c) Alkaline KMnO₄
d) Anhydrous CrO₃
Answer: a
Clarification: Strong oxidising agents like acidified potassium permanganate or acidified potassium dichromate convert alcohol directly to carboxylic acid. However, only aldehyde can be obtained by using CrO₃ as the oxidising agent in an anhydrous medium.

3. Which of the following compounds gives a ketone when its vapours are passes over heated copper at 573K?
a) Propan-1-ol
b) Propan-2-ol
c) 2-Methylpropan-1-ol
d) 2-Methylpropan-2-ol
Answer: b
Clarification: Secondary alcohols undergo dehydrogenation when its vapours are passed over Cu at 573K to give ketone. Primary and tertiary alcohols under the same conditions give aldehydes and alkenes respectively.

4. Ortho and para isomers of nitrophenol can be separated by ________ distillation.
a) fractional
b) steam
c) vacuum
d) zone
Answer: b
Clarification: o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding between O of nitro group and H of hydroxyl group. On the other hand, p-Nitrophenol is present as associated molecules due to intermolecular hydrogen bonding, thus making it less volatile. Hence, a mixture of both can be separated by steam distillation.

5. Identify the product of the following reaction.

a) o- and p-Nitrophenol
b) 2- and 4-Phenolsulphonic acid
c) p-Nitrosophenol
d) 2,4,6-Trinitrophenol
Answer: d
Clarification: Picric acid is commonly formed by treating phenol first with concentrated sulphuric acid to form phenol-2,4-disulphinc acid, and then with concentrated nitric acid. It can also be formed treating phenol with only concentrated nitric acid, but the yield of the product in this case in very poor.

6. What is formed when phenol is reacted with bromine water?
a) White precipitate
b) Colourless gas
c) Brown liquid
d) No reaction
Answer: a
Clarification: Phenol on treatment with bromine water gives a white precipitate which is actually 2,4,6-Tribromophenol.

7. Which of the following is incorrect regarding Kolbe’s reaction?
a) The final product is Salicylic acid
b) CO₂ acts as an electrophile
c) Acidification of phenol leads to formation of main product
d) Phenoxide ion undergoes electrophilic attack

Answer: c
Clarification: The first step of Kolbe’s reaction is the formation of sodium phenoxide, which is formed by treating phenol with NaOH. This phenoxide undergoes substitution by a weak electrophile, CO₂, to form the main product, i.e., 2-Hydroxybenzoic acid or salicylic acid.

8. What is the final major product of Reimer-Tiemann reaction?
a) Salicylic acid
b) o-Salicylaldehyde
c) m-Salicylaldehyde
d) p-Salicylaldehyde
Answer: b
Clarification: Phenol is first treated with chloroform and NaOH to form an intermediate substituted benzal chloride (with-CHO group substituted at ortho position of aromatic ring). This is then hydrolysed in the presence of an alkali to form 2-Hydroxy benzaldehyde or o-Salicylaldehyde.

9. Which compound can covert phenol back to benzene?
a) CrO₃
b) Zinc dust
c) Pyridine
d) Strong acid
Answer: b
Clarification: When heated with zinc dust, the C-O bond in phenol is broken and it is reduced to benzene and zinc oxide.

10. In the presence of air, phenols slowly start to form dark coloured mixtures.
a) True
b) False
Answer: a
Clarification: Phenols are slowly oxidised in the presence of air to form dark coloured mixtures containing quinones.

11. Lucas test was conducted on three compounds A, B and C. Compounds A and B showed turbidity at room temperature, but compound C became turbid only after heating. Which of the compounds has a tertiary structure?
a) A
b) A and B
c) C
d) Cannot be determined
Answer: d
Clarification: It is said that A and B show turbidity at room temperature, but it is not mentioned whether the appearance of turbidity is immediate or after some time. So compounds A and B may be tertiary or secondary depending on whether turbidity appears immediately or after 5 minutes respectively. Compound C is may be primary.

Chemistry Multiple Choice Questions on “Uses of Carboxylic Acids”.

1. Which of the following is not a use of methanoic acid?
a) Leather tanning
b) Coagulant in rubber industry
c) Food preservative
d) Textile dyeing
Answer: c
Clarification: Methanoic acid is the simplest acid and is used in rubber manufacturing, leather tanning, textile dyeing and finishing and electroplating industries.

2. Which carboxylic acid is used in the manufacturing of nylon-6,6?
a) Propanedioic acid
b) Butanedioic acid
c) Pentanedioic acid
d) Hexanedioic acid
Answer: d
Clarification: Nylon-6,6 is manufactured from the condensation polymerisation of hexanedioic acid (adipic acid) with hexamethylenediamine under high pressure and temperature.

3. Esters of which acid are used in the perfume industry?
a) Ethanoic acid
b) Benzoic acid
c) Phthalic acid
d) Formic acid
Answer: b
Clarification: Esters of benzoic acids are used in the perfumery industry. For example, ethyl benzoate is a component of some fragrances and artificial flavours.

4. __________ is commonly used as a food preservative.
a) Sodium benzoate
b) Potassium benzoate
c) Terephthalic acid
d) Acetic acid
Answer: a
Clarification: When C₆H₅COOH reacts with NaOH, a salt called sodium benzoate is formed. When an aqueous salt solution is prepared from this product, it acts as a preservative. It is identified by the code E211.

5. Which of the following carboxylic acids is not used in the manufacturing of soaps?
a) Capric acid
b) Stearic acid
c) Oleic acid
d) Palmitic acid
Answer: a
Clarification: Soaps are sodium or potassium salts of long chain fatty acids like stearic, oleic, palmitic acids, etc. Capric acid is decanoic acid and contains 10 carbon atoms and is not classified as a fatty acid.

6. Ethanoic acid is used as vinegar in the cooking industry.
a) True
b) False
Answer: a
Clarification: Generally, a 1:10 solution of acetic acid in water is called vinegar. It is a commercial product that may be used in salads or other foods. It may include traces of other chemicals for the need of flavouring.

Chemistry Multiple Choice Questions on “Diazonium Salts Chemical Reactions – 3”.

1. The reaction involving the conversion of diazonium salts to azo compounds is known as _______ reaction.
a) diazotisation
b) deamination
c) coupling
d) carbylamine

Answer: c
Clarification: Benzenediazonium salts react with highly reactive electron rich aromatic compounds to form azo compounds of the form Ar-N=N-Ar, where Ar represents an aromatic group. This type of reaction is known as coupling reaction.

2. The coupling reaction is a type of _________ reaction.
a) electrophilic addition
b) electrophilic substitution
c) nucleophilic addition
d) nucleophilic substitution

Answer: b
Clarification: During a coupling reaction, the diazonium cation which has a positive charge on the terminal nitrogen acts as the electrophile, and the electron rich compounds act as nucleophiles.

3. The coupling reaction of benzenediazonium chloride with aniline occurs in a ______ medium.
a) strongly basic
b) weakly basic
c) neutral
d) weakly acidic

Answer: d
Clarification: Aniline is a compound that is slightly basic in nature. To counter the basic effect of aniline during its coupling with diazonium salt, in is conducted in a slightly acidic medium of pH 4 to 5.

4. The reaction between benzenediazonium chloride and phenol results in a ______ coloured compound.
a) yellow
b) orange
c) red
d) purple

Answer: b
Clarification: Benzenediazonium chloride undergoes coupling with phenol to form an azo compound, p-hydroxyazobenzene. This compound is orange in colour.

5. The reaction between a diazonium salt and aniline in a slightly acidic medium gives _______
a) o-aminoazobenzene
b) m-aminoazobenzene
c) p-aminoazobenzene
d) no reaction

Answer: c
Clarification: Diazonium salts undergo coupling with aniline to form p-aminoazobenzene. The coupling predominantly occurs at the para position to the amino group.

6. What is the most suitable pH of the medium for the conduction of coupling reaction of benzenediazonium chloride with phenol?
a) 2
b) 4
c) 7
d) 10
View Answer

Answer: b
Clarification: The coupling reaction with phenol occurs in a basic medium of pH value approximately 9 to 10. This is to counter the acidic nature of phenol and as a result produce water (by the combination of OH^– and H⁺) during the reaction.

7. Identify the colour of the compound ‘A’ from the following reaction.

a) Yellow
b) Colourless
c) Green
d) Blue
View Answer

Answer: a
Clarification: Diazonium salts on reaction with aniline form azo compounds which are coloured. This is because of the complex system involving the N-N double bond between two aromatic rings. In this case, compound A is p-aminobenzene, which is yellow coloured.

8. The reaction between benzenediazonium chloride and N,N-dimethylaniline, in an acidic medium at 273 K gives an orange coloured compound.
a) True
b) False

Answer: b
Clarification: This reaction yields p-dimethylaminoazobenzene, which is a yellow solid commonly known as methyl yellow. However, in an aqueous solution at low pH, methyl yellow appears red in colour.

9. When a diazonium salt is treated with p-cresol, at which position with respect to the hydroxy group does the coupling occur?
a) ortho
b) meta
c) para
d) coupling does not take place

Answer: a
Clarification: Coupling generally occurs para to the hydroxy group. However in this case, the para position is preoccupied by CH₃ group (cresol), and as a result the coupling takes place at ortho position.

10. What is the product of the following reaction?

a) 2-Amino-4-phenylazophenol
b) 6-Amino-2-phenylazophenol
c) 2-Amino-5-phenylazophenol
d) 2-Amino-3-phenylazophenol

Answer: c
Clarification: The reactant (o-aminophenol) consists of both hydroxy and amino groups. Since the reaction takes place in an acidic medium, the coupling occurs at para position (first preferred) with respect to the amino group. If the reaction was to take place in a basic medium, the coupling would have occurred at para position with respect to hydroxy group.

11. p-Aminophenol on coupling with benzenediazonium chloride in a basic medium gives 4-amino-3-phenylazophenol.
a) True
b) False

Answer: b
Clarification: As the reaction takes place in a basic medium, the coupling will take place at the ortho position (since para position is already occupied by amino group) with respect to OH group. This results in the compound 4-amino-2-phenylazophenol.

 

Chemistry Multiple Choice Questions on “Types of Solutions”.

1. Which of the following is not a solid solution?
a) Brass
b) Bronze
c) Hydrated salts
d) Aerated drinks
Answer: d
Clarification: A solid solution is a solid-state solution of one or more solutes in a solvent. Brass, bronze, and hydrated salts are examples of solid solutions. Aerated drinks are examples of liquid solutions (gas in liquid).

2. The solution of mercury with other metals is called amalgam.
a) True
b) False
Answer: a
Clarification: Alloys of mercury with other metals are called amalgams. An alloy is a type of solid solution (solid in solid). Some important amalgams are zinc amalgam, potassium amalgam, sodium amalgam, aluminium amalgam, and tin amalgam.

3. What is an alloy of copper and zinc called?
a) Bronze
b) German silver
c) Brass
d) Solder
Answer: c
Clarification: An alloy of copper and zinc is called Brass. German silver is an alloy of copper, zinc and nickel, sometimes also containing lead and tin. Bronze is an alloy of copper and tin. Solder is an alloy of tin, lead and antimony.

4. What is camphor in N₂ gas an example of?
a) Solid in liquid solution
b) Liquid in gas solution
c) Solid in gas solution
d) Gas in gas solution
Answer: c
Clarification: Camphor in N₂ gas is an example of solid in gas gaseous solution. A solution in which the solvent is gaseous is called gaseous solution. Some other examples of gaseous solutions are air (O₂ + N₂), Iodine vapours in air, humidity in air, etc.

5. Which of the following is not a copper alloy?
a) Bronze
b) Stainless steel
c) Brass
d) Gunmetal
Answer: b
Clarification: Bronze is an alloy of copper and tin. Stainless steel is an alloy of iron with chromium. It also contains varying amounts of carbon, silicon and manganese. Brass is an alloy of copper and zinc. Gunmetal is an alloy of copper, tin and zinc.

6. A supersaturated solution is not a metastable solution.
a) True
b) False
Answer: b
Clarification: A metastable solution is one which is stable when undisturbed but is capable of reaction if disturbed. Supersaturated solutions are stable when undisturbed and precipitates out crystals of the solute when disturbed. Hence, supersaturated solutions are metastable.

7. Which of the following is a true solution?
a) Salt solution
b) Ink
c) Blood
d) Starch solution
Answer: a
Clarification: A true solution is a homogeneous mixture of two or more materials with a particle size of less than 10^-9 m or 1 nm dissolved in the solvent. Ink, blood and starch solution are colloidal solutions. A simple solution of salt in water is a true solution.

8. What type of solution is Cranberry glass?
a) Emulsion
b) Solid sol
c) Solid aerosol
d) Gel
Answer: b
Clarification: Emulsion, solid sol, solid aerosol and gel are types of colloidal solutions. Cranberry glass is formed by the addition of a solid solute to a solid solvent(gold salts and glass respectively). Hence, it is a Solid sol.

9. What is pumice stone an example of?
a) Solid aerosol
b) Emulsion
c) Liquid aerosol
d) Solid foam
Answer: d
Clarification: There are 8 types of colloidal solutions namely solid sol, sol, solid aerosol, gel, emulsion, liquid aerosol, solid foam and foam. Pumice stone is a gas in solid type colloidal solution, i.e., solid foam.

10. What is the observation on adding a solute crystal to a supersaturated solution?
a) It becomes a colloidal solution
b) The solute dissolves in the solution
c) The solution desaturates
d) The solute precipitates out of the solution
Answer: d
Clarification: When a solute crystal is added to a supersaturated solution, solute particles leave the solution and forms a crystalline precipitate. The addition of the solute crystal is also called seeding.

Chemistry Multiple Choice Questions on “Rate of a Chemical Reaction”.

1. For a second-order reaction, what is the unit of the rate of the reaction?
a) s^-1
b) mol L^-1s^-1
c) mol^-1 L s^-1
d) mol^-2 L² s^-1
Answer: c
Clarification: The unit of the rate of the reaction (k) is (mol L^-1) ^1-n s^-1, where n is the order of the reaction.
For a second-order reaction, n=2
(mol L^-1) ^1-n s^-1 = (mol L^-1)^1-2 s^-1 = mol^-1 L s^-1.

2. The rate constant of a reaction is k=3.28 × 10^-4 s^-1. Find the order of the reaction.
a) Zero order
b) First order
c) Second order
d) Third order
Answer: b
Clarification: Given,
k= 3.28 × 10^-4 s^-1
The general formula to find the units for rate constant, k=(mol L^-1)^1-ns^-1 where n is the order of the reaction. The value of n must be 1 for (mol L^-1)^1-ns^-1 to become s^-1. Therefore, k=3.28 × 10^-4s^-1 represents a first order reaction.

3. For a reaction A +B → C, the experimental rate law is found to be R=k[A]¹[B]^1/2. Find the rate of the reaction when [A] = 0.5 M, [B] = 0.1 M and k=0.03.
a) 4.74 × 10^-2 (L/mol)^1/2 s^-1
b) 5.38 × 10^-2 (L/mol)^1/2 s^-1
c) 5.748 × 10^-2 (L/mol)^1/2 s^-1
d) 4.86 × 10^-2 (L/mol)^1/2 s^-1
Answer: a
Clarification: Given, [A] = 0.5 M, [B] = 0.1 M and k= 0.03
From the rate law it is evident that the order of the reaction is 1+ 0.5 = 1.5 = (frac{3}{2})
Therefore the unit of k= (mol L^-1)^1-1.5 s^-1 = (L/mol)^1/2 s^-1
R= k[A]¹[B]^1/2 = 0.03 × 0.5 × 0.1^1/2 = 4.74 × 10^-2(L/mol)^1/2 s^-1.

4. The reaction NO₂ + CO → NO + CO₂ takes place in two steps. Find the rate law.
2NO₂ → NO + NO₃ (k₁) – slow
NO₃ + CO → CO₂ + NO₂ (k₂) – fast
a) R = k₁ [NO₂]³
b) R = k₂ [NO₃] [CO]
c) R = k₁ [NO₂]
d) R = k₁ [NO₂]²
Answer: d
Clarification: In any reaction the slowest step is the rate determining step, the rate of the overall reaction depends on this step. So, 2NO₂ → NO + NO₃(k1) is the rate determining step. Therefore the rate law R= k₁[NO₂]².

5. For the reaction A + H₂O → products, find the rate of the reaction when [A] = 0.75 M, k= 0.02.
a) 0.077 s^-1
b) 0.085 s^-1
c) 0.015 s^-1
d) 0.026 s^-1
Answer: c
Clarification: Given,
[A] = 0.75 M, k= 0.02
The reaction belongs to pseudo first order reaction so, the unit is s^-1
R= k [A]= 0.02 × 0.75= 0.015 s^-1.

6. What is the rate law for acid hydrolysis of an ester such as CH₃COOC₂H₅ in aqueous solution?
a) k [CH₃COOC₂H₅]
b) k [CH₃COOC₂H₅] [H₂O]
c) k [CH₃COOC₂H₅]²
d) k
Answer: a
Clarification: Acid hydrolysis of ester, CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
The order of the reaction may be altered sometimes by taking reactant in excess compared to the other.
The rate law R= k [CH₃COOC₂H₅] [H₂O] however water is present in excess.
So, R= k [CH₃COOC₂H₅].

7. What is the concentration of the reactant in a first order reaction when the rate of the reaction is 0.6 s^-1 and the rate constant is 0.035?
a) 26.667 M
b) 17.143 M
c) 26.183 M
d) 17.667 M
Answer: b
Clarification: Given, R=0.6 s^-1 and k= 0.035
For a first order reaction R= k [A]
[A]=(frac{R}{k}) = (frac{0.6}{0.035}) = 17.143 M.

8. How many times will the rate of the elementary reaction 3X + Y → X₂Y change if the concentration of the substance X is doubled and that of Y is halved?
a) r₂= 4.5r₁
b) r₂= 5r₁
c) r₂= 2r₁
d) r₂= 4r₁
Answer: d
Clarification: Since it is an elementary reaction, its rate law r₁= k [A] ³[B]
When the concentrations are changed the new rate will be r₂= k (2[A])³([B]/2) = 4k[A]³[B]
So, r₂=4r₁.

9. What is the rate law for the reaction C₂H₄ + I₂ → C₂H₄I₂?
a) R= [C₂H₄] [I₂]^3/2
b) R= [C₂H₄] [I₂]³
c) R= [C₂H₄] [I₂]²
d) R= [C₂H₄] [I₂]
Answer: a
Clarification: Fractional order reactions are reaction whose order is a fraction. This reaction is an example of fractional order reaction, where the order of the reaction is (frac{5}{2}).
The rate law for the reaction is known to be R= [C₂H₄] [I₂]^3/2.

10. The rate law for the reaction involved in inversion of cane sugar is R=k [C₁₂H₂₂O₁₁] [H₂O].
a) True
b) False
Answer: b
Clarification: The reaction for the inversion of cane sugar is C₁₂H₂₂O₁₁ + H₂O → glucose + fructose.
In this reaction water is present in excess and belongs to a pseudo first order reaction, even though the molecularity is 2 the order of the reaction is 1 so the rate law R=k[C₁₂H₂₂O₁₁].