Category: Class 12 Chemistry Objective Questions

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

1. Which of the following is the coordination entity in K₂[Zn(OH)₄]?
a) K⁺
b) Zn²⁺
c) OH^–
d) [Zn(OH)₄]^2-
Answer: d
Clarification: A coordination entity consists of a central metal atom bonded to a fixed number of atoms/molecules. Here, K⁺ is the counter ion, OH is the molecule bonded to the central atom which is Zn.

2. The central atom/ion of a coordination complex is also referred to as ________
a) Lewis acid
b) Lewis base
c) Bronsted-Lowry acid
d) Bronsted-Lowry base
Answer: a
Clarification: A Lewis acid is a species that can accept an electron pair. All cations are Lewis acids. Since the central atom of a coordination complex is metal and always accept electrons, it is a Lewis acid.

3. Identify the Lewis acid in K₃[Al(C₂O₄)₃].
a) K⁺
b) Al
c) Al³⁺
d) [Al(C₂O₄)₃]^3-
Answer: c
Clarification: Aluminium ion is the Lewis acid as it can accept 3 electrons from the donor atom to form the complex [Al(C₂O₄)₃]^3-.

4. Which of the following is the central atom/ion in [CoCl(NH₃)₅]²⁺?
a) Co
b) Co²⁺
c) Co³⁺
d) Cl^–
View Answer

Answer: c
Clarification: The central ion in the given complex ion is cobalt as it accepts electrons to form a bond with Cl atom and ammonia molecules. Since the primary valence of Co in this compound is +3, the ion in Co³⁺.

5. Which of the following cannot be a ligand?
a) Ni²⁺
b) Cl^–
c) H₂O
d) NH₃
Answer: a
Clarification: The ions/molecules bound to the central atom/ion is called a ligand. Ni²⁺ is a metal ion, and according to Werner the secondary valences can be satisfied only by neutral molecules or negative ions. Cl^–, H₂O and NH₃ are all possible ligands.

6. How many donor atoms can EDTA^4- ligand bind through?
a) 2
b) 4
c) 6
d) 8
Answer: c
Clarification: EDTA^4- is a hexadentate ligand which can bind through two nitrogen and four oxygen atoms to a central metal ion, giving it a total of 2 + 4 = 6 donor atoms.

7. A protein can exist in a coordination entity as a ligand.
a) True
b) False
Answer: a
Clarification: A protein is a macromolecule that consists of atoms that may donate electrons to a metal ion to form a coordinate bond, and hence act as a ligand.

8. Which is the donor atom in the coordinate bond shown below?

a) N
b) O1
c) O2
d) Both N and O
Answer: a
Clarification: This is nitrito-N, formed by ambidentate ligand NO^2- ion when it bonds through its N atom to the central metal ion. The other form is nitrito-O, which is formed when NO^2- coordinates through the O1 atom.

9. The SCN^– ion is a unidentate ligand.
a) True
b) False
Answer: a
Clarification: SCN^– is an ambidentate ligand that can either bond through S atom or N atom to a central atom/ion. But, at a time there is only a single donor atom, either S or N, making it a unidentate ligand. For a fact, all ambidentate ligands are unidentate.

10. Coordination number is a characteristic of which of the following?
a) Central atom
b) Ligand
c) Coordination entity
d) Coordination compound
Answer: a
Clarification: Coordination number is also known as the secondary valence of a central metal ion in a complex and is defined as the number of donor atoms it is directly bonded to. Hence, coordination number is a quantity associated with the metal ion.

11. What is the coordination number of chromium in K₃[Cr(C₂O₄)₃]?
a) 1
b) 2
c) 3
d) 6
Answer: d
Clarification: There are three oxalate groups attached to the central ion in the given compound, but since oxalate is a didentate ligand, there are a total of 6 donor atoms to which the metal is directly bonded. Thus, the CN of Cr in K₃[Cr(C₂O₄)₃] is 6.

12. Identify the coordination sphere in the compound K₄[Fe(CN)₆].
a) K⁺
b) Fe²⁺
c) [Fe(CN)₆]^4-
d) CN^–
Answer: c
Clarification: The coordination sphere refers to the square bracket within which the central atom/ion and the groups attached to it are enclosed. This is a non-ionisable group as compared to the entity outside the bracket, which is ionisable and is known as the counter ion.

Chemistry Multiple Choice Questions on “Haloarenes Physical Properties”.

1. Which of the following haloalkanes exist as liquids at room temperature?
a) Methyl chloride
b) Methyl bromide
c) Ethyl chloride
d) Propyl bromide
Answer: d
Clarification: The lower members like methyl chloride, methyl bromide and ethyl chloride exist as gases at room temperature. The higher members exists as liquids or solids.

2. For a particular alkyl group R, what is the correct order of boiling points of the following compounds?
a) RBr > RCl > RF
b) RF > RCl > RBr
c) RCl > RBr > RF
d) RF > RBr > RCl
Answer: a
Clarification: As the size of halogen atom increases, the magnitude of van der Waal forces increases, that is why RF has the lowest boiling point as it is the smallest halogen atom with weakest van der Waal attraction.

3. What is the correct relation between boiling points of the following alkyl iodides, where R=methyl, R’=ethyl and R”=propyl?
a) RI > R’I > R”I
b) R”I > R’I > RI
c) RI = R’I = R”I
d) R”I = R’I > RI
Answer: b
Clarification: The forces of attraction become stronger as the molecules getter bigger in size and number of electrons increases. Hence, the methyl halides will have the lowest boiling points among all alkyl groups.

4. If chloromethane has a boiling point of 250K, what will be the probable boiling point of methane?
a) 110K
b) 250K
c) 275K
d) 315K
Answer: a
Clarification: Haloalkanes are polar in nature and have higher intermolecular forces of attraction in terms of dipole-dipole and van der Waal forces. Hence, the boiling points of haloalkanes are higher than their respective hydrocarbons.

5. Which of the following has the highest boiling point?
a) Chloromethane
b) Dichloromethane
c) Trichloromethane
d) Tetrachloromethane

Answer: d
Clarification: The boiling points increase with the number of halogen atoms as the molecular masses increase and so do the intermolecular forces of attraction.

6. The boiling points of isomeric alkyl halides _______ with increase in branching.
a) increase
b) decrease
c) remains same
d) cannot be determined
View Answer

Answer: b
Clarification: The branching of compounds makes it more compact and reduces the surface area, which in turn van der Waals attraction forces and thus, decrease the boiling point.

7. If the compound shown has a boiling point of 346K, predict the boiling point of n-Butyl bromide?

a) 300K
b) 325K
c) 350K
d) 375K
Answer: d
Clarification: The compound shown is a highly branched isomer of n-Butyl bromide, i.e., tert-Butyl bromide. The n-Butyl bromide structure has no branches and therefore has a significantly higher boiling point than tert-Butyl bromide.

8. What is the correct order of boiling of isomeric dichlorobenzenes?
a) ortho>meta>para
b) para>meta>ortho
c) para>ortho>meta
d) meta>ortho>para
Answer: c
Clarification: Para isomers have the highest melting points due to their symmetry and ease with which they can fit into a crystal lattice compared to meta and ortho isomers.

9. The dipole moment of chloromethane is higher than that of fluoromethane.
a) True
b) False
Answer: a
Clarification: Although the F atom is much more electronegative than Cl, the very small size of F compared to Cl results in the product of charge and distance, i.e., dipole moment, of CH₃F to be lower than that of CH₃Cl.

10. What is the value of diploe moment of p-Dichlorobenzene?
a) 0
b) 1
c) >1
d) Answer: a
Clarification: The para isomer of dichlorobenzene is symmetrical and the two equal dipoles on opposite sides cancel and result in zero dipole moment.

11. Calculate the dipole moment of m-Dichlorobenzene if the dipole moments of C-Cl bonds are 1.72D.
a) 0
b) 1.72D
c) 2.54D
d) 2.98D
Answer: b
Clarification: The angle between the two halogen atoms in meta isomer is 120° and let the dipole moments of C-Cl bonds be d=1.72D and let d’ be the dipole moment of the compound. So, by law of parallelogram of forces,
(Rightarrow ,d’ = sqrt{d^2 + d^2 + 2d^2.cos(120°)})
(Rightarrow ,d’ = sqrt{1.72^2 + 1.72^2 – (2 ,x, 0.5 ,x, 1.72^2)})
(Rightarrow ,d’ = sqrt{1.72^2})
(Rightarrow ,d’ = 1.72D)

12. Which of the following has the highest density?
a) CH₃Cl
b) CH₂Cl₂
c) CHCl₃
d) CCl₄
Answer: d
Clarification: The density of haloalkanes increases with the number of halogen atoms present in because of increase in molecular weight.

13. Haloalkanes are more soluble in water than in alcohols.
a) True
b) False
View Answer

Answer: b
Clarification: Haloalkanes are very slightly soluble in water due to the strong hydrogen bonds between water molecules. However, they dissolve much more easily in organic solvents as the strength of new forces is similar to that of existing intermolecular attractions.

14. Which of the following has the lowest boiling point?
a) Bromomethane
b) Bromoform
c) Chloromethane
d) Dibromomethane
Answer: c
Clarification: Since boiling point increases with increases in molecular mass, chloromethane has the lowest molecular mass and hence the lowest boiling point.

Chemistry Question Papers for Class 12 on “Nomenclature and Structure of Carbonyl Groups – 2”.

1. Acetone is the common name of which of the following ketones?
a) Dimethyl ketone
b) Ethyl methyl ketone
c) Diethyl ketone
d) Methyl n-propyl ketone
View Answer

Answer: a
Clarification: In the common system, the ketones are named by listing the alkyl groups present in it in alphabetical order. The simplest dimethyl ketone is also known as acetone.

2. Which of the following will be the main group when present in a compound together?
a) -CHO
b) -CO
c) -OH
d) -Cl
Answer: a
Clarification: If a compound contains both aldehyde and ketone groups, then the former is considered as the principal functional group, and the ketone is regarded as substituent. Thus, it is named as a derivative of alkanal.

3. _________ ketones are generally named by adding the name of the acyl group as prefix to the word phenone.
a) Ethyl methyl
b) Dialkyl
c) Alkyl phenyl
d) Diphenyl
Answer: c
Clarification: For example, the ketone with COCH₃ group attached to a benzene ring is named as acetophenone. Also, diphenyl ketone as benzophenone.

4. What is the IUPAC name of CH₃-CO-CH₂-CH₂-CH₃?
a) Butan-1-one
b) Butan-2-one
c) Pentan-1-one
d) Pentan-2-one
Answer: d
Clarification: The carbon of the CO ketonic group should be included in the parent chain, and the numbering starts from the end nearer to the carbonyl group. Hence in the given compound, there are five carbon atoms in the parent chain and the carbonyl carbon is second one.

5. What is the name of the shown compound?

a) Ethane-1,2-dione
b) Butane-1,2-dione
c) Butanone
d) Butane-2,3-dione
Answer: d
Clarification: This is a symmetrical dimethyl diketone. There are four carbons in the parent chain, including the two carbons of ketonic group and numbering can start from either end of the chain.

6. What is the correct IUPAC name of α-Methylcyclohexanone?
a) Methylcyclohexanone
b) Methylcyclohexan-2-one
c) 2-Methylcyclohexanone
d) Cyclomethylhexan-2-one
Answer: c
Clarification: Alpha represents the presence of a substituent on the carbon next to the carbon of the CO group. The ketone group is given the lower number in case of substituted ketones hence the number 2 is associated with the methyl group in this case.

7. What is the IUPAC name of diisopropyl ketone?
a) 1,3-Diisopropylpropan-2-one
b) 2,4-Dimethylpentan-3-one
c) 2-Methyl-1-(1-methylethyl)propan-1-one
d) 1,3-Dimethylpropan-2-one
Answer: b
Clarification: The formula of diisopropyl ketone is (CH₃)₂CHCOCH(CH₃)₂ which consists of 7 carbon atoms. The longest parent chain including the keto group is of 5 carbon atoms, leaving out one C atom from the methyl group at each end. This is a symmetric ketone and the methyl groups are present at the second and fourth carbon starting from any one end and the C-O double bond is present at the third carbon.

8. Identify the correct IUPAC name of the following compound.

a) 3-Methylcyclopent-2-en-1-one
b) 1-Methylcyclopent-1-en-3-one
c) Cyclo-1-methyl-pent-1-en-3-one
d) Cyclo-3-methyl-pent-2-en-1-one
Answer: a
Clarification: This is a cyclic ketone with a double bond and a substituted methyl group. In this compound the keto carbon will be given the lowest number. Hence, the double bond is at second carbon and methyl group at the third carbon. It is named as a substituted cyclopentanone.

9. What is the IUPAC name of the following compound?

a) 6-(2-oxocyclohexyl)-hexan-3-one
b) 2-(3-oxopentyl)-cyclohexan-1-one
c) 2-(4-oxopentyl)-cyclohexan-1-one
d) 5-(2-oxocyclohexyl)-hexan-3-one
Answer: b
Clarification: The cyclic ketone is considered as the main group with the 3-oxopentyl keto group substituted at alpha carbon.

10. What is the approximate planar bond angle between the carbonyl C-O bond and the bond between the carbonyl carbon and the atom attached to it?
a) 60°
b) 90°
c) 120°
d) 135°
Answer: c
Clarification: The bond angles associated with the carbonyl group is approximately 120°, which is similar to that of a trigonal coplanar structure.

11. The C-O bond in carbonyls is polarised, and hence the carbonyl carbon and carbonyl oxygen act as _________ and ________ respectively.
a) electrophile; nucleophile
b) nucleophile; electrophile
c) Lewis base; Lewis acid
d) electrophile; Lewis acid
View Answer

Answer: a
Clarification: The polarity of the C-O bond is due to the higher electronegativity of O than C. This means that oxygen is a nucleophilic centre and carbon is an electron seeking centre.

12. Ketones are more polar than ethers.
a) True
b) False
Answer: a
Clarification: Ketones are carbonyl compounds which show neutral as well as dipolar resonance structures, which is not possible in ethers.

Chemistry Question Papers for Class 12,

Chemistry Multiple Choice Questions on “Amines Nomenclature – 2”.

1. What is the common name of the simplest aromatic amine?
a) Aniline
b) Benzylamine
c) Benzenamine
d) Aminobenzene
Answer: a
Clarification: The simplest amine is C₆H₅NH₂ and is known as aniline. This name is also accepted as the IUPAC name. Its actual IUPAC name is benzenamine, or it may also be written as aminobenzene.

2. What is the substituent group in 2-methylbezebamine?
a) Methyl group
b) Amino group
c) Benzene ring
d) Methyl and amino groups
Answer: a
Clarification: 2-Methylbenzenamine is an aromatic amine which is a derivative of benzenamine with the presence of a methyl group as substituent at the ortho position.

3. Identify the substituent in p-anisidine?
a) CH₃
b) OCH₃
c) NH₂
d) COCH₃
Answer: b
Clarification: p-Anisidine is better known as 4-methoxybenzenamine, which has a methoxy group at para position to the amino group. The common name anisidine is derived from the common names anisole and aniline.

4. What is the correct IUPAC name of the compound which has a benzene ring substituted with amino group at one position and a bromine atom at a position para to it?
a) 4-Aminobromobenzene
b) 4-Bromobenylamine
c) 4-Bromoaniline
d) p-Bromoaniline
Answer: c
Clarification: This compound is an aromatic compound which is a bromo derivative of aniline and is named such that aniline is principle group. The position of bromine is indicated by number 4 which corresponds to the para position.

5. What is the IUPAC name of the shown compound?

a) N-Methyl-N-ethylbenzenamine
b) N-Methyl-N-phenylethanamine
c) N-Ethyl-N-methylbezenamine
d) N-Ethyl-N-phenylmethanamine
Answer: c
Clarification: There are three substituents viz., methyl group, ethyl group and phenyl group. Since the phenyl group is the largest, it is considered as the main group with CH₃ and C₂H₅ acting as substituents. Also, ethyl is named before methyl due to alphabetical reasons.

6. What is the correct name of the compound with two amino groups around a benzene ring at opposite (para) positions?
a) Benzenediamine
b) 4-Aminobenzenamine
c) p-Aminoaniline
d) Benzene-1,4-diamine
Answer: d
Clarification: Both the amino groups (at para position to each other) are given equal importance and is named by giving the prefix di to the main compound. The prefix is also numbered 1 and 4, representing the positions at which amino groups are present in the benzene ring.

7. Identify the correct name of the shown compound.

a) 4-Amino-1,1-Dimethylcyclohexane
b) 4,4-Dimethylcyclohexan-1-amine
c) 4,4-Dimethylaniline
d) N-(4,4-Dimethylcyclohexyl)amine
Answer: b
Clarification: This is an aliphatic primary amine which is cyclic in structure. The two methyl groups at C-4 position act as substituents of the cyclohexyl ring with the amino group at C-1 position.

8. p-tert-Butyl aniline is a tertiary amine.
a) True
b) False
Answer: b
Clarification: p-tert-Butyl aniline is a derivative of aniline (which is a primary amine) with a tert-Butyl substituent ((CH₃)₃C) at para position. Hence, it is a primary amine.

9. Identify the incorrect name of the compound formed when two hydrogen atoms of ammonia are replaced by phenyl groups.
a) Diphenylamine
b) N-Phenylaniline
c) N-Phenylbenzenamine
d) N,N-Diphenylaniline
Answer: d
Clarification: It is a secondary aromatic amine that will be formed. One of the phenyl groups will be considered as a substituent on nitrogen atom (N-phenyl) and the other will act as aniline with the N atom.

10. How many isomeric amines are possible of the formula C₄H₁₁N.
a) 5
b) 8
c) 10
d) 11
Answer: b
Clarification: Eight isomeric amines with the formula C₄H₁₁N are possible. Four of them are primary amines, three are secondary and one of them is a tertiary amine.

11. What is the IUPAC name of CH₃-NH-CH₂-CH(NO₂)-CH₃ ?
a) N-Methyl-2-nitropropanamine
b) 1-Aminomethyl-2-nitropropane
c) N-(2-Nitropropyl)methanamine
d) N-Methyl-2-methylnitropropanamine
Answer: a
Clarification: This compound contains all substituent functional groups, and is named as a derivative of the compound containing the functional group which occurs first in alphabetical order, i.e., amino group.

12. Identify the correct naming of H₂NCH₂CH₂OH.
a) 2-Hydroxyethanamine
b) 2-Aminoethanol
c) Ethane-2-hydroxy-1-amine
d) 1-Aminoethan-2-ol
Answer: b
Clarification: In case of multiple functional groups, the amino group is considered as substituent and the other is regarded as the principle functional group and is given the lower carbon position. In this case, the compound is named as a derivative of ethanol.

13. Which of the following names are incorrect according to IUPAC?
a) 1-Phenylethanamine
b) N-Methylaniline
c) N-Butylaminoethane
d) Propane-1,2-diamine
Answer: c
Clarification: The larger alkyl group should be considered as parent chain, which in this case is butyl group. Thus, the correct name will be N-ethylbutanamine.

14. If NH₂ and NO₂ exist in the same compound, it will be named as a derivative of the amino compound.
a) True
b) False
Answer: a
Clarification: Since amino group (NH₂) comes alphabetically ahead of nitro group (NO₂), the compound is named as a derivative of amine with nitro as a substituent.

Chemistry Quiz for IIT JEE Exam on “Biomolecules Carbohydrates – 3”.

1. Which of the following tests does glucose give?
a) Tollen’s test
b) 2,4-DNP test
c) Schiff’s test
d) Addition product with NaHSO₃
Answer: a
Clarification: Despite having a CHO group, glucose does not react with NaHSO₃ to form an addition product and give either Schiff’s test or 2,4-DNP test like other aldehydes. However, it forms a silver mirror on treatment with Tollen’s reagent. These observations indicate the existence of glucose in other forms and gave way to the cyclic structure of glucose.

2. The α-D-glucose and β-D-glucose isomers of glucose are known as _______
a) enantiomers
b) stereoisomers
c) anomers
d) glycomers
Answer: c
Clarification: In the cyclic structure of glucose, the pair of optical isomers that differ in the configuration only around the C₁ carbon (anomeric carbon) are called anomers.

3. The hydroxyl group at which carbon is involved in ring formation in glucose?
a) C₃
b) C₄
c) C₅
d) C₆
Answer: c
Clarification: Monosaccharides undergo intramolecular reaction to form cyclic hemiacetal structures. Generally in aldoses, the C₄ or C₅ carbon is involved in cyclisation. In glucose, the hemiacetal (six-membered ring) is formed between the CHO group and the OH group on the C₅ carbon.

4. Which form of glucose is obtained by crystallization from hot and saturated aqueous solution at 371K?
a) D-form
b) L-form
c) α-form
d) β-form
Answer: d
Clarification: When a saturated aqueous solution is heated to about 100°C and then crystallized, the β-form of glucose is obtained. On the other hand, a concentrated solution at room temperature on crystallization gives α-form.

5. When the α-form of glucose is dissolved in water and allowed to rest, what will be the percentage of β-form in this solution after some time?
a) 0
b) 36
c) 64
d) 100
Answer: c
Clarification: When either α or β forms of glucose are dissolved in water and are allowed to stand, an equilibrium mixture of both forms is formed (36% α-form and 64% β-form). This is due to a spontaneous change in specific rotation of an optically active compound with time to an equilibrium value. This phenomenon is called as mutarotation.

6. When the OH group at anomeric carbon lies on the same side as that of the OH group on C₃ carbon, it is known as ______
a) α-glucose
b) β-glucose
c) D-glucose
d) (+)-glucose
Answer: b
Clarification: The OH at C₃ carbon always lies on the side opposite to that of OH at C₂ and C₄ carbon in all structures of glucose. When the OH of anomeric carbon lies on the same side as that of the C₂ and C₄ carbon Oh group, it is known as α-form. Else, it is β-form.

7. Identify the compound from the Haworth projection shown.

a) α-D-(+)-glucopyranose
b) β-D-(+)-glucopyranose
c) α-D-(+)-glucofuranose
d) β-D-(+)-glucofuranose
Answer: a
Clarification: The α and β forms of glucose can be represented as a six-membered ring called pyranose, resembling the organic compound pyran. Since the OH group of anomeric carbon is at the opposite side to the OH at C₃ carbon, it is the α-form of glucopyranose.

8. Glucose can exist in both a straight chain and ring form.
a) True
b) False
Answer: a
Clarification: This is due to the observations where glucose in most cases behaves as an aldehyde and in some cases does not. This led to speculation of the existence of glucose in other forms where the CHO group was not present. The ring form of glucose was subsequently discovered.

9. Which of the following is incorrect with respect to fructose?
a) It is a ketohexose
b) It is present in honey
c) It has ‘L’ configuration
d) It is known as laevulose
Answer: c
Clarification: Fructose is an important ketohexose that naturally occurs in fruits and honey. It has a ketonic group at C₂ carbon with six carbon in a straight chain. It belongs to D-series (with respect to glyceraldehyde) and is laevorotatory.

10. What is the correct name of the following compound?

a) D-(+)-fructose
b) D-(-)-fructose
c) L-(+)-fructose
d) L-(-)-fructose
Answer: b
Clarification: Since the OH group on the carbon adjacent to the last CH₂OH group is on the right-hand side, it is of D configuration. Also, fructose is naturally a laevorotatory compound, hence the (-) sign.

11. The furanose structure of fructose is obtained by the interaction of groups at which carbon atoms?
a) C₁ and C₅
b) C₁ and C₆
c) C₂ and C₅
d) C₂ and C₆
Answer: c
Clarification: The fructofuranose structure is obtained by the internal ketal formation by combining the keto group at C₂ carbon and the OH group at C₅ carbon.

12. Fructose exists as both pyranose and furanose structures.
a) True
b) False
Answer: a
Clarification: In the free state, D-fructose exists as a six-membered ring (fructopyranose). However, in the combined state as a constituent in disaccharides, it exists in furanose form (five-membered hemiketal).

13. Identify the compound from the Haworth projection shown.

a) α-D-(+)-glucopyranose
b) β-D-(+)-glucopyranose
c) α-D-(-)-fructopyranose
d) β-D-(-)-fructopyranose
Answer: d
Clarification: This is the pyranose structure of fructose, where the C₂ keto group and C₆ OH group are involved in cyclisation. The α and β forms in this structure differ based on the arrangement of CH₂OH and OH groups around the C₂ carbon. Since in this compound, the OH at C₂ is in the same side as the OH at C₃, it is the β-form.

14. A solution having equal amount of D-glucose and D-fructose is called _____ sugar.
a) invert
b) fruit
c) brown
d) cane
Answer: a
Clarification: When sucrose is heated with water, it gives a mixture of glucose and fructose, called invert sugar. It is named as such because of the difference in optical activity of the initial sugar and the product mixture.

15. Fructose is commonly known as ______
a) dextrose
b) laevulose
c) pyranose
d) furanose
Answer: b
Clarification: Due to its laevorotatory behaviour, fructose is known as laevulose. On the other hand, glucose is known as dextrose for the opposite reason. Fructose can exist as furanose or pyranose depending on its state (free or combined).

Chemistry Multiple Choice Questions on “Therapeutic Action of Different Classes of Drugs – 1”.

1. Which of the following compounds help in controlling the acid production in the stomach?
a) Histamine
b) Cimetidine
c) Ranitidine
d) Omeprazole
Answer: a
Clarification: Histamine is the compound which is responsible for the secretion of pepsin and HCl in the stomach. Drugs like cimetidine (Tegamet), ranitidine (Zantac), omeprazole and lansoprazole prevent the interaction of histamine with the stomach wall receptors, resulting in release of lesser acid.

2. Which of the following is the least suitable antacid?
a) Magnesium hydroxide
b) Aluminium hydroxide
c) Sodium bicarbonate
d) Dihydroxy aluminium amino acetate
Answer: c
Clarification: All can be used as antacids. But when NaHCO₃ (and other hydrogen carbonates) are taken in excess, it increases the alkalinity of the stomach and stimulates the release to more acid to counter it. Metal hydroxides are better as they are insoluble and do not increase the pH above 7.

3. Which of the following drugs help in subsiding the effects of allergic reaction?
a) Benadryl
b) Dimetapp
c) Seldane
d) Nardil
Answer: d
Clarification: Benadryl, Dimetapp and Seldane are examples or antihistamines which are drugs that interfere with the main actions of histamine, which is a vasodilator and is released during allergic reactions. Nardil is an antidepressant drug.

4. Which of the following are neurologically inactive drugs?
a) Analgesics
b) Barbiturates
c) Antipyretics
d) Antihistamines
Answer: d
Clarification: Analgesics, antipyretics and barbiturates (tranquilizers) are drugs that interfere with the message transfer mechanism from the nerve to receptor and hence block neurological communication.

5. Which of the following tranquilizers cannot be part of sleeping pills?
a) Chlordiazepoxide
b) Amytal
c) Seconal
d) Nembutal
Answer: a
Clarification: Amytal, seconal and nembutal are examples of barbiturates or sleep-inducing drugs. These are hypnotic compounds barbituric acid and are important tranquilizers and components of sleeping pills. Chlordiazepoxide is a mild tranquilizer used only for relieving tension.

6. Which of the following compounds does not inhibit the enzymes which catalyse the degradation of noradrenaline?
a) Citalopram
b) Terfenadine
c) Iproniazid
d) Phenelzine
Answer: b
Clarification: Low levels of noradrenaline in the body reduces the signal-sending activity of neurons and lead to depression. Citalopram, iproniazid and phenelzine are antidepressants that inhibit the catalysis of the degradation of noradrenaline and helps treat depression.

7. Which of the following is not an analgesic?
a) Aspirin
b) Paracetamol
c) Morphine
d) Salvarsan

Answer: d
Clarification: Analgesics are drugs that are used to relieve pain without causing disturbances within the nervous system. Aspirin and paracetamol are non-narcotic analgesics, whereas morphine is a narcotic analgesic.

8. Which of the following analgesics have antipyretic properties?
a) Aspirin
b) Heroin
c) Codeine
d) Marijuana
Answer: a
Clarification: Aspirin is 2-acetoxybenzoic acid, is a non-addictive drug that blocks the production of prostaglandins and relive pain and inflammation. It also has the effect of lowering the body temperature during fever (antipyretic) and preventing platelet coagulation.

9. Which of the following is not an opiate?
a) Morphine
b) Valium
c) Thebaine
d) Codeine
Answer: b
Clarification: Opiates are a class of narcotic analgesics naturally obtained as alkaloids from the opium poppy plant. Morphine, codeine and thebaine are the opiates that have the maximum effect of the human nervous system.

10. Which of the following is not an antimicrobial?
a) Antiseptics
b) Antibacterial drugs
c) Anaesthetics
d) Disinfectants
Answer: c
Clarification: Antimicrobials are substances that are used to prevent and treat infections due to various microorganisms like fungi, virus, bacteria or parasites. They can be used for curing diseases and avoid them. Anaesthetics are drugs that produce insensitivity to pain.