Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “P-Block Elements – Hydrogen Chloride”.

1. Hydrogen chloride is manufactured by the ‘salt cake’ method.
a) True
b) False
Answer: a
Clarification: Hydrogen chloride is manufactured by the ‘salt cake’ method. In this method, concentrated sulphuric acid is added to rock salt in order to form sodium sulphate and hydrogen chloride.

2. Which of the following scientists first prepared hydrogen chloride by heating common salt with sulphuric acid?
a) Rudolf Glauber
b) Joseph Priestley
c) Humphry Davy
d) Antoine Lavoisier
Answer: a
Clarification: Rudolf Glauber prepared hydrogen chloride in 1648 by heating common salt with concentrated sulphuric acid. Humphry Davy, in 1810, established its true nature and showed that it is a compound of hydrogen and chlorine.

3. How many stages is the ‘salt cake’ method performed in?
a) 1
b) 3
c) 2
d) 4
Answer: c
Clarification: The ‘salt cake’ method is an endothermic reaction. It is performed in two stages. The first stage involves the reaction between rock salt and sulphuric acid at 420K and in the second stage, the product of the first stage is heated to a temperature of 823K so that it further reacts with sulphuric acid.

4. Hydrogen chloride can be obtained from the heavy organic chemical industry.
a) True
b) False
Answer: a
Clarification: Large amount of impure HCl are obtained in recent years as a by-product from the heavy organic chemical industry. For example, HCl is produced in the conversion of 1,2-dichloroethane to vinyl chloride.

5. Which of the following is formed in the first stage of the ‘salt cake’ method?
a) Sodium sulphate
b) Sodium bisulphate
c) Sodium Chloride
d) Hydrogen
Answer: b
Clarification: In the first stage of the ‘salt cake’ method, solid sodium chloride reacts with concentrated sulphuric acid at 420K and gets coated with insoluble sodium bisulphate. This prevents further reaction and hence, the name ‘salt cake’ was given to this method.

6. Which of the following can be used to dry HCl gas?
a) Sodium hydroxide
b) Potassium hydroxide
c) Ammonium hydroxide
d) Concentrated sulphuric acid
View Answer

Answer: d
Clarification: The hydroxides are basic in nature and hence, they will react with hydrogen chloride gas. Therefore, concentrated sulphuric acid is used as a drying agent for hydrogen chloride gas since it does not react with it.

7. What is the boiling point of hydrogen chloride?
a) 168 K
b) 189 K
c) 198 K
d) 150 K
Answer: b
Clarification: Hydrogen chloride is a colourless and pungent smelling gas. It can be easily liquefied to form a colourless liquid. The boiling point of hydrogen chloride is 189 K and its freezing point is 150 K.

8. Which of the following does not react with hydrochloric acid to give chlorine as a product?
a) KMnO₄
b) MnO₂
c) K₂Cr₂O₇
d) Mg
Answer: d
Clarification: Concentrated hydrochloric acid reacts with strong oxidising agents to yield chlorine gas. KMnO₄, MnO₂, K₂Cr₂O₇, PbO₂ are examples of oxidising agents which release chlorine on reaction with hydrochloric acid.

9. Which of the following reacts with hydrochloric acid to give hydrogen as a product?
a) Zinc
b) Sodium carbonate
c) Sodium sulphate
d) Sodium nitrate
Answer: a
Clarification: Active metals such as zinc, magnesium, iron and aluminium react with hydrochloric acid to produce their corresponding metal chlorides with the liberation of hydrogen gas. Hydrochloric acid decomposes salts of weaker acids like carbonates, sulphides and so on.

10. Which of the following is hydrochloric acid not used for?
a) Manufacture of metal chlorides
b) Preparing TNT
c) Preparation of aqua regia
d) Extracting glue from bones
Answer: b
Clarification: Hydrochloric acid reacts with metals to form their respective chlorides and so, is used in the manufacture of metal chlorides. Aqua regia is made using hydrochloric acid and nitric acid. Hydrochloric acid is also used for extracting glue from bones.

Chemistry Multiple Choice Questions on “Isomerism in Coordination Compounds – 1”.

1. Two or more compounds that have the same chemical formula, but different arrangement of atoms are called _______
a) isotopes
b) isotones
c) isomers
d) allotropes
Answer: c
Clarification: Isotopes are forms of one element due to different number of neutrons in each atom. Isotones are forms of different elements that have the same number of neutrons and allotropes are the different physical forms in which a given element can exist.

2. Which type of isomerism exhibits compounds with same chemical formula and bonds but different spatial arrangement?
a) Optical isomerism
b) Linkage isomerism
c) Structural isomerism
d) Solvate isomerism
Answer: a
Clarification: Stereoisomers have the same chemical formula and binds but have different spatial arrangement. Optical isomerism is a type of stereoisomerism.

3. Which of the following is not a subdivision of structural isomerism?
a) Geometrical isomerism
b) Linkage isomerism
c) Coordination isomerism
d) Ionisation isomerism
Answer: a
Clarification: Geometrical isomerism is a type of stereoisomerism. Linkage, coordination and ionisation isomers show structural isomerism with different bonds.

4. Geometrical isomerism can be observed in some homoleptic complexes.
a) True
b) False
Answer: b
Clarification: Geometrical isomerism takes place only in heteroleptic ligands as there is more than one type of bonding group that can be arranged in different possible manners to give rise to isomers.

5. A coordination complex [MX₂L₂], has a CN=4 and two unidentate ligands X and L. When the two L ligands are arranged opposite to each other in its geometry, it is called _______ isomer.
a) cis
b) trans
c) fac
d) mer
Answer: a
Clarification: The shape of the complex will be square planar as the CN is 4 and isomerism cannot be observed in tetrahedral shapes (also CN=4). When a particular ligand in this type of complex is arranged opposite to each other, it is a trans isomer, else it is called a cis isomer.

6. How many geometrical isomers are possible in [Al(C₂o₄)₃]^3-?
a) 0
b) 2
c) 3
d) 4
Answer: a
Clarification: The entity shown has a CN=6 as oxalate is a bidentate ligand. The structure of the entity will be same no matter which positions in the geometry each of the oxalate ligands occupy because their relative positions will remain the same in each case.

7. A tetrahedral compound of type [MP₂Q₂] has two geometrical isomers.
a) True
b) False
Answer: b
Clarification: Tetrahedral complexes do not form geometric isomers as the relative positions of all the ligands will always be same because of the geometric properties of a tetrahedron.

8. Identify the trans isomer of [Pt(NH₃)₂Cl₂] from the following.
a)
b)
c)
d)
Answer: b
Clarification: The compound [Pt(NH₃)₂Cl₂] forms two geometric isomers, one cis and one trans. When the two Cl ligands are arranged adjacent to each other, a cis isomer is formed and when the two Cl ligands are opposite to each other, a trans ligand is formed.

9. How many geometrical isomers are possible in a complex of type [MA₂(D)₂], where A is unidentate and D is didentate?
a) 0
b) 2
c) 3
d) 4
Answer: b
Clarification: In a complex of such type, the two A ligands can be arranged either adjacent to or opposite each other to form cis and trans isomers respectively, making it a total of two possible isomers.

10. The square planar complex [MABCD] is known to form three isomers, two cis and one trans. Shown below are the two cis isomers of the complex. Identify the third trans isomer.

a)
b)
c)
d)
Answer: a
Clarification: If you observe the two given cis isomers, it can be seen that they are formed by simply swapping the positions of B and C, but both of them are adjacent to each other in both isomers. So, to obtain the trans isomer, the ligands B and C should lie opposite to each other. The other incorrect options are the same cis isomers but are depicted in different orientations.

11. Which of the following do not show geometrical isomerism? (Assume all ligands are unidentate)
a) Square planar [MXL₃]
b) Square planar [MX₂L₂]
c) Octahedral [MX₂L₄]
d) Octahedral [MX₃L₃]
Answer: a
Clarification: In square planar complexes, the type [MXL₃] does not have any isomers as there is no pair of ligands that can be arranged adjacent to or opposite each other to form cis or trans forms respectively. All possible combinations result in the same spatial arrangement.

12. In the coordination entity [Co(NH₃)₃(NO₂)₃], if all three N atoms of the amine ligands occupy adjacent positions at the corners of an octahedral face, the geometrical isomer formed is known as _______ isomer.
a) cis
b) trans
c) fac
d) mer
Answer: c
Clarification: Octahedral entities of the type [MA₃B₃] form two types of isomers based on whether the same ligand groups occupy adjacent positions on the octahedral face (fac) or positions around the meridian of the octahedron (mer).

13. Which of the following compounds has a meridional isomer?
a) [Fe(NO)₅Br]⁺
b) [Al(CO)₃(NO₂)₃]
c) [K(NH)₃)₄(NO)₂]⁺
d) [Fe(H₂O)₂(CO)₂(NO)₂]³⁺
Answer: b
Clarification: Fac-mer isomerism is only exhibited in octahedral coordination entities of the type [MA₃B₃] depending on the positions of the similar ligands.

14. Optical isomers are also known as __________
a) structural isomers
b) facial isomers
c) meridional isomers
d) enantiomers
Answer: d
Clarification: Optical isomers are different forms of the same complex that are mirror images of each other, and which cannot be superimposed. They are chiral complexes also known as enantiomers.

15. The optical isomer that rotates the plane of polarised in the clockwise direction is called ______
a) trans
b) dextro
c) mer
d) laevo
View Answer

Answer: b
Clarification: The optical isomers are of two types based on the direction they rotate the plane of polarised light in a polarimeter. If it rotates it to the right, it is called dextro and if it rotates it to the left, it is called laevo.

Chemistry Aptitude Test for IIT JEE Exam on “Haloalkanes & Haloarenes – Polyhalogen Compounds”.

1. Which of the following is not a use of methylene chloride?
a) Paint removal
b) Propellant in aerosols
c) Metal cleaning
d) Antiseptic
Answer: d
Clarification: Methylene chloride can harm the human central nervous system and application on the skin may result in intense burning and redness.

2. Which of the following is an effect of exposure to low levels of dichloromethane in the air?
a) Slightly impaired vision
b) Dizziness
c) Nausea
d) Numbness in fingers
Answer: a
Clarification: Dizziness, nausea and numbness are caused due to higher levels of dichloromethane in the air. Lower levels cause slight impairment of hearing and vision.

3. The oxidation of trichloromethane results in the formation of a poisonous gas called ________
a) carbon monoxide
b) carbonyl chloride
c) hydrogen sulphide
d) phosphine
Answer: b
Clarification: Trichloromethane is slowly oxidized in air in the presence of light to form an extremely poisonous gas, carbonyl chloride, also called as phosgene.

4. Chloroform was previously used as a general anaesthetic in surgery.
a) True
b) False
Answer: a
Clarification: It was stop being used as an anaesthetic due to its toxicity and harmful effects on the human body. It was replaced by safer options like ether.

5. What is the reason for iodoform to be used as an antiseptic?
a) Due to its unpleasant odour
b) Due to its melting point
c) Due to its solubility in alcohol
d) Due to the liberation of free iodine
Answer: d
Clarification: The antiseptic properties of iodoform are due to the liberation of free iodine and not due the properties and characteristics of iodoform itself.

6. Which of the following polyhalogen compounds in useful in extinguishing fires?
a) Dichloromethane
b) Trichloromethane
c) Triiodomethane
d) Tetrachloromethane
Answer: d
Clarification: Carbon tetrachloride is used as a cleaning fluid, degreasing agent, spot remover and a fire extinguisher.

7. Carbon tetrarchloride may indirectly cause skin cancer.
a) True
b) False
Answer: a
Clarification: When CCl₄ is released into the atmosphere, it causes the depletion of the ozone layer, which increases exposure to UV rays leading to skin cancer, eye diseases and disorders.

8. Which of the following is not a freon?
a) CCl₃F
b) CCl₂F₂
c) CBr₂F₂
d) CClF₃
Answer: c
Clarification: Freons are chlorofluorocarbon compounds of methane and ethane. The compound CBr₂F₂ contains bromine, and hence is not a freon.

9. Freon 12 is manufactured from the Swarts reaction of which compound?
a) Dichloromethane
b) Trichloromethane
c) Tetrachloromethane
d) Dichlorodifluoromethane
Answer: c
Clarification: Freon 12 (CCl₂F₂) is obtained by heating CCl₄ with antimony fluoride in the presence of antimony pentachloride by Swarts reaction.

10. Which of the following is not a characteristic of CHClF₂?
a) Non-toxic
b) Non-flammable
c) Non-corrosive
d) Non-liquefiable
Answer: d
Clarification: CHClF₂ is a freon refrigerant also known as R-22. It is extremely stable, unreactive, non-toxic, non-corrosive and an easily liquefiable gas.

11. Which of the following is not a harmful effect of DDT?
a) It is non-biodegradable
b) Cannot be metabolized by animals
c) Highly toxic towards fish
d) Causes air pollution
Answer: d
Clarification: The chemical stability and fat solubility of DDT is a big drawback. It was banned due to its harmful effect on animals and fish.

12. Which of the following is not a use of carbon tetrachloride?
a) Manufacturing of refrigerants
b) Manufacturing of pharmaceuticals
c) Degreasing agent
d) Insecticide
Answer: d
Clarification: CCl₄ is produced in large quantities for use in the making of refrigerants and propellants for aerosol cans. It is also used as feedstock in pharmaceutical manufacturing and synthesis of chlorofluorocarbons.

Chemistry Aptitude Test for IIT JEE Exam,

Chemistry Written Test Questions and Answers for Class 12 on “Aldehydes and Ketones Chemical Reactions – 2”.

1. Catalytic hydrogenation of _______ gives propan-2-ol.
a) formaldehyde
b) acetaldehyde
c) propionaldehyde
d) acetone
Answer: d
Clarification: Reduction of acetone in the presence of catalysts like Ni, Pt or Pd converts it into propan-2-ol, which is a secondary alcohol. It can also be reduced by the reaction with LiAlH₄ or NaBH₄. Aldehydes on similar reaction give primary alcohols.

2. What is the product of Clemmensen reduction on acetophenone?
a) Benzaldehyde
b) Methyl benzene
c) Ethyl benzene
d) Benzophenone
Answer: c
Clarification: The carbonyl group of acetophenone is reduced to CH₂ group and results in the formation of a hydrocarbon when subjected to Clemmensen reduction, i.e., reaction with Zn-Hg and concentrated HCl.

3. A compound on treatment with hydrazine followed by heating up to 473K with KOH in ethylene glycol gives propane. Identify the compound.
a) Methanal
b) Ethanal
c) Propanal
d) Acetone
Answer: d
Clarification: Aldehydes and ketones are reduced to respective hydrocarbons when treated with hydrazine followed by heating with KOH/NaOH in high boiling solvents. This is known as Wolff-Kishner reduction. The CO group of acetone is reduced to CH₂, and propane is obtained.

4. Acetaldehyde is oxidised to ______ in the presence of nitric acid.
a) methanoic acid
b) ethanoic acid
c) acetone
d) ethanol
Answer: b
Clarification: Aldehydes are easily oxidised to carboxylic acids with same number of carbon atoms, in the presence of oxidising agents like HNO₃, KMnO₄, etc. This is because of the presence of H atom next to CO group, which can be easily converted to OH group with any bond cleavage.

5. What are the major products of the oxidation of pentan-2-one with concentrated HNO3 at very high temperatures?
a) Butanoic acid and methanoic acid
b) Propanoic acid and ethanoic acid
c) Pentanoic acid
d) Butanoic acid and ethanoic acid
Answer: b
Clarification: Ketones undergo oxidation accompanied by C-C bond cleavage to give a mixture of carboxylic acids having lesser number of carbon atoms than the parent ketone. In case of petan-2-one (which is unsymmetrical), the cleavage occurs such that the CO group stays with the smaller alkyl group (Popoff’s rule). Hence, the bond between C-2 and C-3 will be broken to give a mixture of ethanoic acid and propanoic acid.

6. Four compounds A, B, C and D were heated individually with Tollen’s reagent. It was found that all compounds other than D formed a silver mirror on the inside of their test tubes. Identify the compound D.
a) Acetaldehyde
b) Butyraldehyde
c) Acetone
d) Benzaldehyde
Answer: c
Clarification: When aldehydes are heated with ammoniacal silver nitrate solution (Tollen’s reagent), it reduces the silver ions to metallic silver and forms a silver mirror. Ketones do not give this test as they are different to oxidise.

7. Which of the following is Tollen’s reagent?
a) Ammoniacal silver nitrate solution
b) Aqueous copper sulphate
c) Alkaline sodium potassium tartarate
d) Mixture of sodium carbonate, sodium citrate and Cu²⁺ complex
Answer: a
Clarification: Tollen’s reagent is prepared by adding ammonium hydroxide solution to silver nitrate solution till the grey precipitate of Ag₂O first formed just gets dissolved.

8. What is the proportion of Fehling solution A to Fehling solution B in the solution of Fehling’s reagent for conducting Fehling’s test?
a) 1:1
b) 1:2
c) 2:1
d) 1:3
Answer: a
Clarification: Two solutions, aqueous copper sulphate and alkaline sodium potassium tartarate are combined in equal proportions to result in Fehling’s solution. This is an important reagent in the conduction of Fehling’s test.

9. When an aldehyde is heated with Fehling’s solution, a reddish-brown precipitate is formed due to which compound?
a) CuS
b) AgBr
c) Cu₂O
d) CdS
Answer: c
Clarification: Fehling’s reagent has Cu²⁺ ions which oxidise aldehyde by adding an O atom and turning it into corresponding carboxylate ion. In this process, a reddish-brown compound is precipitated due to the formation of cuprous oxide.

10. Which of the following compounds can be distinguished from iodoform test?
a) Benzaldehyde and benzophenone
b) Benzaldehyde and formaldehyde
c) Acetophenone and acetaldehyde
d) Acetophenone and benzophenone
Answer: d
Clarification: Iodoform test involves the reaction of a compound with sodium hypoiodite to detect the presence of CH₃CO or CH₃CH(OH) group. A positive test forms CHI₃ (iodoform) in the form of a yellow precipitate. Acetaldehyde and acetophenone give yellow ppt in this test, whereas formaldehyde, benzophenone and benzaldehyde do not undergo this test due to the absence of CH₃CO group in their structures.

11. Which of the following does not give yellow ppt on iodoform test?
a) Propan-2-ol
b) Butan-2-ol
c) Pentan-2-one
d) Pentan-3-one
Answer: d
Clarification: Pentan-2-one consists of a CH₃CO groups and gives positive for iodoform test. Propan-2-ol and butan-2-ol pass iodoform test because of the presence of CH₃CH(OH) group which gets oxidised to CH₃O group during reaction. On the other hand, pentan-3-one consists of only ethyl groups on either side of the carbonyl carbon and is not oxidised by sodium hypoiodite to give iodoform.

12. Which of the following compound does not undergo aldol condensation?
a) Acetaldehyde
b) Propanal
c) Propanone
d) Benzaldehyde
Answer: d
Clarification: Aldehydes and ketones that have at least one α-hydrogen atom undergo aldol condensation, where the α-hydrogen of one molecule becomes attached to the carbonyl carbon of the second molecule to form aldols and ketols respectively.

13. How many products are formed from the aldol condensation reaction between ethanal and propanal?
a) 1
b) 2
c) 3
d) 4
Answer: d
Clarification: Ethanal and propanal both contain α-hydrogen atoms and undergo aldol condensation to form a mixture of four products. The self-aldol products of ethanal and propanal are but-2-enal and 2-methylpent-2-enal respectively. The two products formed from the cross-aldol condensation between one molecule of ethanal and one molecule of propanal are 2-methylbut-2-enal and pent-2-enal.

14. Identify the product A of the following reaction.

a) Phenol
b) Benzyl alcohol
c) Acetophenone
d) Benzophenone
View Answer

Answer: b
Clarification: Benzaldehyde does not contain any α-hydrogen and undergoes self- oxidation and reduction on treatment with conc. NaOH. One molecule of it is oxidised to carboxylic acid salt, and the other molecule is reduced to an alcohol (A). This is known as Cannizzaro’s reaction.

15. Formaldehyde cannot undergo aldol condensation as well as Cannizzaro’s reaction.
a) Trues
b) False
Answer: b
Clarification: Formaldehyde (HCHO) does not contain an α-hydrogen and hence cannot undergo self-condensation to form aldol product. However, it undergoes Cannizzaro’s reaction to given methanol and potassium formate.

Chemistry Written Test Questions and Answers for Class 12,

Chemistry Multiple Choice Questions on “Amines Chemical Reactions – 1”.

1. Amines are generally ______ in nature.
a) electrophilic
b) acidic
c) basic
d) neutral
Answer: c
Clarification: Amines behave as nucleophiles due to the presence of unshared pair of electrons on N atom. This also makes them proton acceptor, and react with acids to form salts.

2. Which of the following is associated with decrease in pK_b value of amines?
a) Increase in acidic strength
b) Increase in basic strength
c) Better proton donation
d) Better electron acceptor
Answer: b
Clarification: The basic strength of amines is expressed in terms of dissociation constant K_b. Greater the K_b value, stronger the base. It is more commonly expressed as pK_b=-logK_b. Hence smaller the value of pK_b, more is the basic strength of the amine.

3. The equilibrium constant of the reaction of amines with _____ is taken as a measure of its basic character.
a) an acid
b) a base
c) water
d) a Lewis base
Answer: c
Clarification: For the reaction,
RNH₂ + H₂O ↔ RNH₃⁺ + OH^–, the dissociation constant is,
K_eq = ([RNH₃⁺][OH^–])/([RNH₂][H₂O])
(Rightarrow) K_eq[H₂O] = ([RNH₃⁺][OH^–])/[RNH₂]
(Rightarrow) K_b = ([RNH³⁺][OH^–])/[RNH₂]

4. What is the product formed when ethanamine reacts with HBr?
a) NH₄Br
b) CH₃NH₃Br
c) CH₃CH₂NHBr
d) CH₃CH₂NH₃Br
Answer: d
Clarification: Since amines are basic in nature, they react with acids to form salts of ammonium. Ethanamine (CH₃CH₂NH₂) reacts with HBr in a reversible reaction to form ethylammonium bromide through addition.

5. The reaction between methylamine and hydrogen iodide results in the formation of a _______
a) colourless liquid
b) dark coloured gas
c) white solid
d) yellow liquid
View Answer

Answer: c
Clarification: Methylamine and hydrogen iodide are allowed to mix at very cold temperatures for about 2 hours. The resulting product is allowed to evaporate and methanaminium iodide (CH₃NH₃I) is obtained in the form of a white powder.

6. Which of the following is not produced on the reaction of methylammonium chloride with sodium hydroxide?
a) HCl
b) CH₃NH₂
c) H₂O
d) NaCl
Answer: a
Clarification: Amine salts on treatment with a base like NaOH gives back the parent amine along with a water molecule and a salt. For example, CH₃NH₃Cl with NaOH gives methanamine , water and sodium chloride.

7. Identify X and Y respectively in the following reaction.

a) OH^–, HCl
b) HCl, OH^–
c) H₂O, OH^–
d) HCl, H₂O
Answer: b
Clarification: Aniline undergoes addition in the presence of HCl to give anilinium chloride (salt) which on action with OH^– ions (from a base like NaOH) gives back aniline.

8. What is the correct order of pK_b values of the following amines?
a) Methanamine > Ethanamine > Benzenamine
b) Benzenamine > Ethanamine > Methanamine
c) Ethanamine > Methanamine > Benzenamine
d) Benzenamine > Methanamine > Ethanamine
Answer: d
Clarification: The higher pK_b value means lower basic strength. This means that benzenamine has the lower basic strength among methanamine and ethanamine. This is because of the electron withdrawing nature of the aryl group. Also in primary amines, the increase in size of alkyl group increases +I effect leading to high electron density on N atom and as a result higher basicity.

9. The treatment of N,N-dimethylaniline with acetic acid gives no reaction.
a) True
b) False
Answer: b
Clarification: N,N-Dimethylaniline reacts with acetic acid to form a salt, N,N-dimethylanilinium acetate. This is due to the basic nature of amines.

10. If the pK_b value of ammonia is 4.75, predict the pK_b value of methanamine?
a) 3.38
b) 4.70
c) 8.92
d) 9.38
Answer: a
Clarification: Aliphatic amines are stronger bases than ammonia and hence have a much lower pK_b than ammonia. This is because of the electron donating nature of alkyl groups which increase the negative charge on nitrogen atom, making it less prone to electron acceptance.

11. If the K_b values of ammonia, methylamine and ethylamine are x, y and z respectively, identify the correct relation between x, y and z from the following.
a) x > y
b) y c) x > z
d) x > y > z
Answer: b
Clarification: Larger the value of K_b, stronger is the base. Since ethylamine is a stronger base than methylamine which is stronger than ammonia, due to the effect of alkyl groups, the K_b vale of methylamine will be less than that of ethylamine (y

12. Consider three gaseous alkylamines A, B and C or 1°, 2° and 3° respectively. What will be the correct order of their basicity?
a) A > B > C
b) C > B > A
c) B > A > C
d) B > C > A
Answer: b
Clarification: In the gaseous state, the solvation effect is missing and hence the expected order of basicity will be 3° > 2° > 1°. This is because as the number of alkyl groups increase, the +I effect strengthens and more negative charge is accumulated on the N atom. This makes the unpaired share mire available for sharing with the proton of the acid.

13. The basic strength of alkylamines does not depend on which of the following?
a) Number or alkyl groups
b) Size of alkyl groups
c) Physical state of the amine
d) Presence of an aromatic ring
Answer: d
Clarification: Since only basicity of alkylamines is in question the presence of aromatic ring is not considered. As the size and number of alkyl groups increases, the stability of ammonium ion (formed from the amine) increases due to dispersal of more positive charge by the +I effect of alkyl groups. The physical state, gaseous or aqueous, also determines the basicity as the hydration effect comes into play.

14. What is the correct order of basicity of aliphatic amines purely on the basis of solvation effect of the ammonium cation?
a) 1° > 2° > 3°
b) 3° > 2° > 1°
c) 2° > 1° > 3°
d) 2° > 3° > 1°
Answer: a
Clarification: In aqueous phase, the greater the size of the ion, lesser will be the hydration and less stabilised is the ion. Since the stability of ions is directly proportional to the basic strength of amines, primary amines will be the most basic and tertiary amines will be the least basic. This order is completely opposite to that based on the inductive effect of alkyl groups.

15. Stearic hinderance of alkyl groups has an effect on the basic character of amines.
a) True
b) False
Answer: a
Clarification: When the alkyl group is small, there is no stearic hinderance to hydrogen bonding. But when the alkyl group size or number increases, there will be hinderance to formation of hydrogen bonds, and this affects the order of basic strength of amines.

Chemistry Multiple Choice Questions on “Biomolecules – Enzymes”.

1. Which of the following best describes a particular enzyme?
a) Chemical catalyst
b) Fibrous protein
c) Highly selective
d) Can be used for various reactions
Answer: c
Clarification: Enzymes are biological catalysts produced by living cells. They differ from other catalysts in being highly selective and specific. Almost all enzymes are globular proteins. They are very specific for a particular reaction and a particular substrate.

2. Enzymes are generally named after the ________
a) compound on which they work
b) compound which they form as product
c) medium in which they act
d) place from where they are derived
Answer: a
Clarification: Enzymes are generally name after the compound or class of compounds on which they work. For example, the catalyst that hydrolyses the reaction of starch (amylum) to glucose is called as amylase.

3. The enzyme which catalyses the conversion of proteins to amino acids is ______
a) invertase
b) urease
c) nuclease
d) protease
Answer: d
Clarification: Proteases are enzymes that helps in protein catabolism by hydrolysis of peptide bonds. Pepsin and trypsin are the most common proteases, found in the human stomach and pancreas respectively. Both of them are major digestive enzymes.

4. Which of the following is a substrate specific enzyme?
a) Maltase
b) Carboxylase
c) Hexokinase
d) Carbonic anhydrase
View Answer

Answer: a
Clarification: Substrate specific enzymes are those which can act only on one particular compound to give a product(s). For example, maltase acts only on maltose to break the glycosidic linkage between the two glucose units.

5. Enzymes are regarded as ______
a) biocatalysts
b) messengers
c) inhibitors
d) antibodies
Answer: a
Clarification: Enzymes are biological catalysts produced by living cells which catalyse the biochemical reactions in living organisms. Chemically, enzymes are naturally occurring simple or conjugate proteins.

6. Enzymes are basically ______
a) polysaccharides
b) sugars
c) polypeptides
d) pyrimidine bases
Answer: c
Clarification: Almost all enzymes are globular proteins, which are nothing but very long chains of amino acids residues (>100) and higher molecular mass, or polypeptides.

7. Cellulose is not digestible by humans due to the absence of which of the following enzymes?
a) Amylase
b) Urease
c) Cellulase
d) Invertase
Answer: c
Clarification: The enzyme cellulase hydrolyses cellulose into glucose and thereby digests it. However, the human stomach does not produce any enzyme capable of digesting cellulose, and hence it cannot be digested.

8. Enzymes reduce the magnitude of activation energy for a reaction.
a) True
b) False
Answer: a
Clarification: For example, the activation energy for acid hydrolysis of sucrose is 6.22kJ/mol, while the activation energy is only 2.15kJ/mol when it is hydrolysed in the presence of sucrase.

9. The hydrolysis of lactose can be catalysed only by the enzyme lactase. Also, lactase is only able to work on lactose and no other compound.
a) True
b) False
Answer: a
Clarification: Enzymes are highly specific in nature. Almost all biochemical reactions ae controlled by its own specific enzyme.

10. The prosthetic groups which get attached to the enzyme at the time of reaction are called _____
a) cofactors
b) coenzymes
c) messengers
d) inhibitors
Answer: b
Clarification: Most active enzymes are associated with some non-protein components required for their activity. These are called prosthetic groups and they may be cofactors or coenzymes.

11. Albinism is caused by the deficiency of which enzyme?
a) Phenylalanine hydroxylase
b) Streptokinase
c) Prolidase
d) Tyrosinase
Answer: d
Clarification: Deficiency of tyrosinase results in insufficient production of melanin. This causes a condition of white skin and hair called albinism. It can be prevented by including the required supply of enzyme through diet.

12. Identify the correct statement about enzymes.
a) Enzymes increase the activation energy of a reaction
b) Enzymes need to be used in excess compared to the reagent to catalyse the reaction
c) Enzymes work only at their optimum temperature and pH
d) The activity of enzymes cannot be affected by other compounds
Answer: c
Clarification: Enzymes decrease the activation energy of a reaction. Extremely small quantities of enzymes can increase the rate of reaction by thousands. Enzyme action can be inhibited by compounds known as enzyme inhibitors.