Category: Engineering Chemistry Objective Questions

Engineering Chemistry Multiple Choice Questions on “Cracking – 2”.

1. How can we increase the ageing of alumina-silica catalyst?
a) By making the contact of catalyst with heavy metals
b) By making the contact of catalyst with lighter metals
c) By making the contact of catalyst with heavy non-metals
d) By making the contact of catalyst with lighter non-metals
Answer: b
Clarification: The process of ageing can be accelerated, if the catalyst comes into the contact with compounds containing sulphur, vanadium, nickel etc. The ageing can also be increased if the conditions are more rigid.

2. How can we regenerate alumina-silica catalyst?
a) By increasing the particles of coke on the catalyst
b) By increasing the amount of catalyst
c) By decreasing the particles of coke on the catalyst
d) By making it react with suitable anion or cation
Answer: c
Clarification: The catalyst can be regenerated to store their activity and reactivity by burning of the coke deposited on the catalyst particles. The catalyst used in the catalytic reactions are very expensive, due to which their regeneration is necessary.

3. What happens when the crude petroleum fractions are made in contact of catalyst in cracking process?
a) Solid substances are obtained
b) Liquid substances are obtained
c) Gaseous substances are obtained
d) Solid, liquid and gaseous substances are obtained
Answer: d
Clarification: Gases, gasoline, coke and other liquid products are formed when a catalyst comes in contact with crude petroleum fractions in cracking process. This happens due to the various reactions such as dehydrogenation, isomerization etc.

4. Catalytic cracking process are more expensive to run than the thermal cracking process.
a) True
b) False
Answer: a
Clarification: The catalyst used in these processes are expensive. This type of cracking gives more stable and superior product. The yields obtained by this type of cracking are higher.

5. Which compounds can be obtained from higher percentile yields of butyne-butylene fractions?
a) Gasoline
b) Coke
c) Super-Gasoline
d) Diesel
Answer: c
Clarification: The octane ratings of gasoline fuel is 87 where as the octane ratings of super-gasoline is 93. These fuels are more suitable than gasoline fuels.

6. In which form does the catalyst are used in fluid-bed cracking?
a) In the form of oxides
b) In the form of vapours
c) Liquid form
d) In the form of fine powder
Answer: d
Clarification: In this process, the catalyst in the form of fine powder is circulated through the cracking reactor with the help of oil vapours. The catalyst accelerates and directs the cracking and also acts as a heat transfer medium.

7. In which process does the oil vapours are heated to their cracking temperatures?
a) Thermal cracking
b) Fixed-bed cracking
c) Fluid-bed cracking
d) Coking
Answer: b
Clarification: In fixed-bed cracking the oil vapours are heated to their cracking temperatures and are passed on to the fixed catalyst bed. When the catalyst gets carbonised, it is reactivated by burning off the carbon deposited.

8. At what temperature does the oil is vaporised and heated in fixed-bed cracking process?
a) 400-500^oC
b) 700-800 ^oC
c) 100-200 ^oC
d) More than 1000 ^oC
Answer: a
Clarification: The heavy oil charge is passed through a heater, where the oil is vaporised and heated to 400-500 ^oC. The hot vapours are then passed over fixed bed catalyst. The pressure is about 1-5 kg/cm².

9. How much percentile of low molecular weight hydrocarbons are formed in fixed-bed catalytic cracking?
a) 15-20 %
b) 50-60 %
c) 30-40 %
d) 0-5 %
Answer: c
Clarification: About 30-40 % of the charge is converted into low molecular weight hydrocarbons, conforming the composition of catalyst. About 4 % of carbon is formed during this process.

10. What happens to the heavy gas oil fractions in the cracking process?
a) It escapes out
b) Condensed at the bottom of the column
c) It gets dissolved in the solution
d) It blocks the air holes
Answer: b
Clarification: The cracked vapours of the oil enters the fractionating column where the gasoline vapours and other gaseous products are recovered from the top, while the heavy gas oil fractions are condensed at the bottom of the column.

11. What happens in the stabilizer of fixed-bed catalytic cracking?
a) Coke is obtained in it
b) Coal gas is formed in it
c) Butane is formed
d) Gasoline is recovered
Answer: d
Clarification: The condensed vapours of the catalytic system are send to the stabilizer, where the dissolved gases are removed and gasoline is recovered. The light gases produced in cracking are no longer allowed to escape into the air.

12. Which type of raw materials are formed from the light gases produced by cracking?
a) Synthetic rubber
b) Freezing agents
c) Acids
d) Inorganic compounds
Answer: a
Clarification: These light gases form valuable raw materials for the synthesis of new types of fuel. The other materials such as antifreeze, plastics and synthetic rubber are obtained from these gases.

13. Which of the following is the main disadvantage of hydrogenation cracking?
a) It cannot achieve complete cracking
b) It creates holes in the reactor
c) It is expensive and intricate
d) It requires more human effort
Answer: c
Clarification: In hydrogenation cracking, the heavy oil is cracked in the presence of hydrogen under high temperature and pressure. Complete cracking can be achieved by recycling.

14. In fluid-bed catalytic cracking, cracking takes place on ___________________
a) The side walls of the turbulent
b) The surface of the turbulent
c) Upper portion of the turbulent
d) Instruments of the turbulent
Answer: b
Clarification: Cracking takes place on the surface of the turbulent catalyst bed as it circulates with the oil vapours in the reactor at a temperature of 530 ^oC and at a pressure of about 3 to 5 kg/cm².

15. Fluid-bed catalytic cracking products contain high proportions of aromatic and iso-paraffin compounds than thermal cracking process.
a) True
b) False
Answer: a
Clarification: The products obtained in fluid-bed catalytic cracking are distributed in different chambers due to which it has higher proportions of aromatic and iso-paraffin compounds than thermal cracking. This process is used by the industries for the cracking process.

Engineering Chemistry Multiple Choice Questions on “Lime Soda Process”.

1. What is the molecular formula of lime?
a) CaCO₃
b) Al(OH)₃
c) Ca(OH)₂
d) Mg(OH)₂
Answer: c
Clarification: The molecular formula of lime is Ca(OH) ₂. It is a white caustic alkaline substance consisting of calcium oxide, which is obtained by heating limestone and which combines with water with the production of much heat.

2. What is the molecular formula of soda?
a) Na₂CO₃
b) Al₂(SO₄)₃
c) Mg(OH) ₂
d) Ca(OH) ₂
Answer: a
Clarification: The molecular formula of soda is Na₂CO₃. It is also called washing soda or soda ash.

3. Which of the following is not a precipitate in lime-soda process?
a) CaCO₃
b) Fe(OH) ₃
c) Al(OH) ₃
d) Al₂(SO₄)₃
Answer: d
Clarification: Al₂(SO₄)₃ is not a precipitate in lime-soda process. All the other options are the precipitate in lime-soda process.

4. Which of the following is not used as a coagulant in lime-soda process?
a) Alum
b) Calcium bicarbonate
c) Aluminium sulphate
d) Sodium aluminate
Answer: b
Clarification: Calcium bicarbonate is not used as a coagulant in lime-soda process. Alum, aluminium sulphate and sodium aluminate are used as coagulants in lime-soda process.

5. What is the function of coagulant?
a) It helps in the formation of fine precipitate
b) It helps in the formation of coarse precipitate
c) It helps in increasing the solubility
d) It helps in increasing the boiling point
Answer: b
Clarification: Coagulants helps in the formation of coarse precipitate. C?oagulation is a process of addition of coagulant to destabilize a stabilized charged particle.

6. For the softening of one mole of magnesium bicarbonate, the number of mole of lime required is?
a) 2
b) 3
c) 4
d) 5
Answer: a
Clarification: For the softening of one mole of magnesium bicarbonate, the number of mole of lime required is two. The ratio of bicarbonate to lime in lime-soda process is 1:2.

7. The residual hardness in lime-soda process is ___________
a) 0-2 ppm
b) 5-15 ppm
c) 15-50 ppm
d) 50-70 ppm
Answer: c
Clarification: The residual hardness in lime-soda process is 15-50 ppm. Lime soda softening cannot produce a water at completely free of hardness because of the low solubility of CaCO₃ and Mg(OH)₂.

8. In lime soda process, capital cost is ___________ whereas operational expenses are __________
a) High, low
b) High, high
c) Low, low
d) Low, high
Answer: d
Clarification: In lime soda process, capital cost is low whereas operational expenses are high. It is because the chemicals and reagents used are costly.

9. In lime-soda process, no exchange of ions occurs.
a) True
b) False
Answer: a
Clarification: In lime-soda process, no exchange of ions occurs. Recarbonation is used to stabilize the water. The excess lime and magnesium hydroxide are stabilized by adding carbon dioxide, which also reduces the pH from 10.8 to 9.5.

10. Which of the following is the second step for calculating the amount of lime and soda?
a) Identification of lime and soda for different salts
b) Calculation of CaCO₃ equivalents
c) Calculation of lime requirements
d) Calculation of soda requirements
Answer: b
Clarification: Calculation of CaCO₃ equivalents is the second step for calculating the amount of lime and soda. First step is to identify the lime and soda for different salts. and the last two steps are the calculation of lime and soda requirements.

Engineering Chemistry Multiple Choice Questions on “Characteristics of Electroplating Wastes”.

1. Under what condition will impurities form under cathode?
a) When temperature exceeds critical temperature
b) When pressure exceeds critical pressure
c) When the amount of electrolyte is increased enormously
d) When water content is increased
Answer: b
Clarification: Since during electroplating, the standard critical temperature and pressure are maintained. If the temperature is increased, then only the polarities of the electrode change, but if the pressure is increased, the impurities will shift from anode to cathode.

2. Which of the following is not the characteristic of impurities in electroplating?
a) Are formed near cathode
b) Are mostly noble metals
c) Are formed near anode
d) Increase the strength of solution
Answer: a
Clarification: During electroplating, the impurities are mostly noble metals (especially in the case of alkali and alkaline earth metals). That is why due to higher density at anode, the impurities are formed near anode.

3. What causes an increase in the amount of formation of impurities at a higher rate?
a) Increase in concentration
b) Decrease in concentration
c) Deterioration in the amount of molecules
d) Destruction in the electrode
Answer: c
Clarification: During electroplating process, the amount of molecules plays an important role in the determination of electrolysis of absorbent (water or any other fluid). Thus decrease in the amount of molecules will increase the formation of impurities.

4. Determine the coefficient of relative constant from the following data.
Alpha = 23
C = 43 g
T = 100 K
a) 10
b) 7
c) 9.89
d) 8
Answer: c
Clarification: We know that the relative constant can be calculated from the formula: C/T X alpha. Therefore from the given data, we can calculate the constant as 43/100×23 = 9.89.

5. Determine the coefficient of the relative constant from the following data.
Alpha = 1
C = 3 g
T = 10 K
a) 2
b) 0
c) 1
d) 0.3
Answer: d
Clarification: We know that the relative constant can be calculated from the formula: C/T X alpha. Therefore from the given data, we can calculate the constant as 3/100×1=0.3.

6. What will be the observation when the electrolyte concentration is increased to double times the previous concentration?
a) The Composition increases 4 times
b) The composition increases 2 times
c) The composition increases 8 times
d) The composition does not change
Answer: d
Clarification: Since during electroplating the concentration of the electrolyte is independent of external factors like temperature of the system, pressure and specific gravity, it does not have anything to do with composition.

7. Determine the curational capacity of zinc aluminium from the following.
a) Noblular transmission
b) Isotropic mass activation
c) Semi concentric circulation
d) Isobaric transmission
Answer: b
Clarification: During electroplating process, the contents like concentration, composition, electronic configuration of molecules are purely based on the ingenious capacity of electrolysis process, the curational capacity is isotropic in nature.

8. Calculate the electron concentration.
Alpha: 34
C = 3
T = 23 K
a) 11.3
b) 23
c) 8
d) 0
Answer: a
Clarification: We know that the electron concentration is given by the formula concentration/alpha value. Therefore in this problem, from the given data, we have the value of electron concentration as 34/3=11.3.

9. Calculate the value of alpha from the following.
Electron concentration: 4
C=2
a) 0
b) 8
c) 2
d) 1
Answer: b
Clarification: We know that the electron concentration is given by the formula concentration/alpha value. Therefore alpha = electron concentration x C. Therefore in this problem, from the given data, we have the value of alpha as 4×2=8.

10. The emulsified impurities are found near ____________
a) Cathode
b) Anode
c) Nowhere
d) At electrolyte site
Answer: a
Clarification: Since during the process of electroplating, the higher density molecules adhere to the anode and lower density molecules adhere to cathode. Since emulsified impurities are low density molecules, they stick towards cathode.

Engineering Chemistry Multiple Choice Questions on “Functional Materials”.

1. Select the incorrect statement from the following option.
a) Nitinol is a non-magnetic material
b) It has a shape memory strain of about 8.5%
c) It has lower ductility than other shape memory alloys like Au-Cd
d) It possess excellent corrosion resistant
Answer: c
Clarification: Nitinol has a higher ductility than other shape memory alloys like Au-Cd and can freely revert back to its original shape. All the other options are correct. It is a non-magnetic material and possess excellent corrosion resistant.

2. The materials which offer no resistance to the passage of electricity are ___________
a) Superconductors
b) Semiconductors
c) Optical fibre
d) Smart material
Answer: a
Clarification: Superconductors offer no resistance to the passage of electricity. Semi-conductor is a substance, usually a solid chemical element or compound, that can conduct electricity under some conditions but not others, making it a good medium for the control of electrical current.

3. The critical temperature for the cuprate La₂CuO₄ is ___________
a) 40 K
b) 90 K
c) 125 K
d) 35 K
Answer: d
Clarification: The critical temperature for the cuprate La₂CuO₄ is 35 K. Cuprate loosely refers to a material that can be viewed as containing copper anions.

4. Identify the class of the superconductor to which TiO and NbO belong.
a) Simple compounds
b) Nonstoichiometric crystals having defects
c) Charge-transfer compounds
d) Molecular crystals
Answer: b
Clarification: TiO and NbO are the examples of non-stoichiometric crystals having defects class of superconductors.

5. The dielectric material that shows spontaneous and reversible dielectric polarisation are ___________
a) Piezoelectrics
b) Pyroelectrics
c) Ferroelectrics
d) None of the mentioned
Answer: c
Clarification: Piezoelctrics are those dielectric materials in which pressure can produce an electrical response and electrical forces produce a mechanical response. Pyroelectrics are those dielectric in which polarisation changes with temperature. Ferroelectrics are those dielectric materials that show spontaneous and reversible dielectric polarisation.

6. What is the principle on which an optical fibre works?
a) Total internal reflection
b) Farraday’s law of induction
c) Doppler Effect
d) Hooke’s law
Answer: a
Clarification: Due to total internal reflection, light can be transmitted over many kilometers without any appreciable decrease in the intensity by optical fibres.

7. Optical fibre can be safely used for signal transmission near the nuclear installation because ___________
a) There is no leakage of signals
b) Plastic cladded silica fibres are resistant to nuclear radiations
c) Plastic cladded silica fibres are not resistant to nuclear radiations
d) There is no electromagnetic interference
Answer: b
Clarification: Optical fibre can be safely used for signal transmission near the nuclear installation because the plastic cladded silica fibres are resistant to nuclear radiation. Due to total internal reflection, light can be transmitted over many kilometers without any appreciable decrease in the intensity of optical fibres.

8. The correct sequence for the procedure adopted for making a pure elemental semiconductor consist of ___________
a) Preparation of ultrapure Si or Ge >> Doping >> Preparation of single crystals of Si or Ge
b) Doping >> Preparation of ultrapure Si or Ge >> Preparation of single crystals of Si or Ge
c) Preparation of ultrapure Si or Ge >> Preparation of single crystals of Si or Ge >> Doping
d) Preparation of single crystals of Si or Ge >> Preparation of ultrapure Si or Ge >> Doping
Answer: c
Clarification: The correct sequence for the procedure adopted for making a pure elemental semiconductor consist of: Preparation of ultrapure Si or Ge>> Preparation of single crystals of Si or Ge>> Doping

9. Which of the following is not an application of a transistor?
a) Tunnel diodes
b) Thermistors
c) Amplifiers
d) None of the mentioned
Answer: d
Clarification: All the mentioned options are the applications of a transistor. A transistor is a semiconductor device used to amplify or switch electronic signals and electrical power. It is used for tunnel diodes, thermistors and amplifiers.

10. The amplification factor for transistor when used as an amplifier will be ___________
a) 25
b) 50
c) 75
d) 100
Answer: b
Clarification: The amplification factor for transistor when used as an amplifier will be 50. Modern transistor audio amplifiers of up to a few hundred watts are common and relatively inexpensive.

Engineering Chemistry Questions and Answers for Entrance exams focuses on “Particle Size Measurement”.

1. Which of the following is a method of particle size measurement?
a) Sieve analysis
b) Microscopic examination
c) Sedimentation analysis
d) All of the mentioned
Answer: d
Clarification: Sieve analysis, microscopic examination and sedimentation analysis are the three methods of particle size measurement.

2. The minimum size of particle which can be separated through sieve analysis is ____________
a) 75 microns
b) 100 microns
c) 44 microns
d) 20 microns
Answer: c
Clarification: The minimum size of particle which can be separated through sieve analysis is 44 microns. The results of sieve analysis are used to describe the properties of the aggregate and to see if it is appropriate for various civil engineering purposes such as selecting the appropriate aggregate for concrete mixes and asphalt mixes as well as sizing of water production well screens.

3. The instrument used for the examination of particle size ≥ 0.5 micron is ____________
a) Microscope
b) Electron microscope
c) Telescope
d) All of the mentioned
Answer: a
Clarification: The instrument used for the examination of particle size ≥ 0.5 micron is a microscope. Microscope is an optical instrument used for viewing very small objects.

4. Sedimentation techniques are particularly useful for the particle size ranging from ____________
a) 50-10 µm
b) 40-2 µm
c) 100-50 µm
d) 80-40µm
Answer: b
Clarification: Sedimentation techniques are particularly useful for the particle size ranging from 40-2 µm.

5. Sedimentation analysis is generally applicable to liquid dispersions containing _______________ % solids.
a) 5-10
b) 1-3
c) 2-5
d) 4-8
Answer: c
Clarification: Sedimentation analysis is generally applicable to liquid dispersions containing 2-5% solids. It is particularly useful for the particle size ranging from 40-2 µm.

6. Select the incorrect statement from the following option.
a) In the Coulter counter, a narrow orifice is immersed in an electrolyte
b) The resistance across the orifice is continuously noted on an oscilloscope
c) The pulses occurring on the oscilloscope are counted electronically
d) The magnitude of the pulse is inversely proportional to the volume of the electrolyte displaced
Answer: d
Clarification: The magnitude of the pulse is directly proportional to the volume of the electrolyte displaced and therefore to the volume of the particle. All the other options are correct.

7. The current generated in photoelectric cell represents the measure of the ____________
a) Turbidity
b) Adhesivity
c) Cohesivity
d) Viscosity
Answer: a
Clarification: The current generated in photoelectric cell represents the measure of the turbidity of the suspension. Turbidity is the cloudiness or haziness of a fluid caused by large numbers of individual particles that are generally invisible to the naked eye. It is the capacity of water to disperse the light.

8. Adsorption method for measurement of total surface area involves the evaluation of quantity of ____________
a) Helium gas
b) Oxygen gas
c) Nitrogen gas
d) Argon gas
Answer: c
Clarification: Adsorption method for measurement of total surface area involves the evaluation of quantity of nitrogen gas.

9. In photo-extinction method, a source of light is directed through a suspension of ____________
a) Kerosene in cement
b) Cement in kerosene
c) Cement in water
d) Water in cement
Answer: b
Clarification: In photo-extinction method, a source of light is directed through a suspension of cement in kerosene on to a photoelectric cell.

10. Fibrous fillers have a high capacity for absorbing energy because of their ____________
a) Shape
b) Size
c) Composition
d) Surface area
Answer: a
Clarification: Fibrous fillers have a high capacity for absorbing energy because of their shape. As a result, they impart excellent impart strength to the material.

areas of Engineering Chemistry for Entrance exams,

Engineering Chemistry Multiple Choice Questions on “Nucleophilic Substitution Reaction”.

1. The atom which defines the structure of a family of organic compounds and their properties is called ___________
a) Resonating structure
b) Homologous structure
c) Functional group
d) Nucleophile
Answer: c
Clarification: The atom which defines the structure of a family of organic compounds and their properties is called a functional group. Functional groups are specific groups of atoms or bonds within molecules that are responsible for the characteristic chemical reactions of those molecules.

2. The functional group in alkyl halide is ___________
a) Hydroxyl group
b) Halogen atom
c) Inert gas
d) All of the mentioned
Answer: b
Clarification: The functional group in alkyl halide is a halogen atom. Halide refers to halogen. Halogens are a group in the periodic table consisting of five chemically related elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At).

3. The halide ion is an extremely ___________
a) Weak base
b) Weak acid
c) Strong base
d) Strong acid
Answer: a
Clarification: The halide ion is an extremely weak base because hydrogen halide are acidic as they release a proton.

4. When the nucleophile :OR attacks the RX, the resultant product will be ___________
a) R – OH
b) ROR
c) R:CN
d) RNHR
Answer: b
Clarification: When the nucleophile :OR attacks the RX, the resultant product will be ROR. X will be separated.

5. Select the correct statement from the following option.
a) S_N2 reaction follows second order kinetics
b) No intermediate is involved in S_N2 mechanism
c) S_N2 reactions are one-step reaction
d) All of the mentioned
Answer: d
Clarification: S_N2 reaction follows second order kinetics and no intermediate is involved in it. So, S_N2 reactions are one-step reaction.

6. The reactivity order of alkyl halides in S_N2 is ___________
a) CH₃ X > 1⁰ > 2⁰ > 3⁰
b) CH₃ X > 2⁰ > 1⁰ > 3⁰
c) CH₃ X > 3⁰ > 1⁰ > 2⁰
d) CH₃ X > 3⁰ > 2⁰ > 1⁰
Answer: a
Clarification: The correct order of reactivity in alkyl halide is- CH₃ X > 1⁰ > 2⁰ > 3⁰. CH₃ X is most reactive whereas 3⁰ is least reactive.

7. S_N1 reaction involves heterolysis to form the carbocation as an intermediate.
a) True
b) False
Answer: a
Clarification: S_N1 reaction involves heterolysis to form the carbocation as an intermediate.

8. Which step in S_N1 reaction is a slow rate determining step?
a) Attack of nucleophile
b) Formation of a racemic mixture
c) Formation of a transition state
d) All of the mentioned
Answer: c
Clarification: The first step of formation of the transition state is a slow rate determining step in S_N1 reaction.

9. A low concentration of nucleophile favours the ___________
a) S_N2 reaction
b) S_N1 reaction
c) Both S_N1 and S_N1 reaction
d) None of the mentioned
Answer: b
Clarification: A low concentration of nucleophile favours the S_N1 reaction but high concentration of nucleophile favours the S_N2 reaction.

10. Which of the following reactions are favoured by polar aprotic solvent?
a) S_N1 reactions
b) S_N2 reactions
c) Both S_N1 and S_N1 reactions
d) None of the mentioned
Answer: b
Clarification: S_N1 reactions are favoured by polar protic solvents whereas S_N2 reactions are favoured by polar aprotic solvent.

11. Arrange the following in the decreasing order of leaving group in nucleophilic substitution reaction.
a) H^– > Cl^– > HO^– > Br^– > CH₃COO^–
b) Cl^– > Br^– > HO^– > H^– > CH₃COO^–
c) Cl^– > Br^– > CH₃COO ^– > HO^– > H^–
d) HO^– > CH₃COO ^– > H^– > Br^– > Cl^–
Answer: c
Clarification: The correct order is- Cl^– > Br^– > CH₃COO ^– > HO^– > H^–.

12. Reaction of alcohol with SOCl₂ is ___________
a) S_N1
b) S_N2
c) S_NAr
d) S_Ni
Answer: d
Clarification: Reaction of alcohol with SOCl₂ proceeds with retention of configuration via substitution nucleophilic internal (S_Ni) mechanism.

13. The nucleophilic substitutions do not occur in haloarenes because ___________
a) The carbon-halogen bond is much shorter
b) The carbon-halogen bond is stronger compared to that in haloalkanes
c) The lone pair of electrons on the halogen participates in delocalisation with the π-electrons of benzene ring
d) All of the mentioned
Answer: d
Clarification: The nucleophilic substitutions do not occur in haloarenes because the carbon-halogen bond is much shorter and stronger compared to that in haloalkanes and the lone pair of electrons on the halogen participates in delocalisation with the π-electrons of benzene ring.

14. Which of the following drastic condition is required for the substitution in haloarenes?
a) High temperature
b) High pressure
c) Strong concentrated reagent
d) All of the mentioned
Answer: d
Clarification: Drastics conditions like high temperature, high pressure and strong concentrated reagents are used to carry substitution in haloarenes.

15. The rate of nucleophilic substitution reactions are higher in the presence of ___________
a) Electron withdrawing groups
b) Electron releasing groups
c) Both electron withdrawing and releasing groups
d) None of the mentioned
Answer: a
Clarification: The rate of nucleophilic substitution reactions are higher in the presence of electron withdrawing groups.