Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Surface Chemistry – Colloids Classification”.

1. Colloids are classified into lyophobic and lyophilic colloids.
a) True
b) False
Answer: a
Clarification: Based on the interaction between the dispersed and dispersion medium, colloids are classified into two types, lyophilic and lyophobic colloids. Lyophilic colloids are those in which the dispersed phase has high affinity for dispersion medium and lyophobic colloids are those in which the dispersed phase has no affinity or least affinity for the dispersion medium.

2. Which of the following is not an example of lyophilic colloids?
a) Starch solution
b) Gelatin
c) Gum
d) Silver solution
Answer: d
Clarification: Starch solution, gelatin and gum are colloids in which the dispersed phase has very high affinity for the dispersion medium and therefore they are categorized as lyophilic colloids. Silver solution on the other hand is an example of lyophobic colloids.

3. Which of the following is not an example of lyophobic colloids?
a) Gold solution
b) Sulphur solution
c) NaCl solution
d) Blood
Answer: c
Clarification: Gold solution, Sulphur solution and blood are colloids in which the dispersed phase has very low affinity for the dispersion medium and once the dispersed phase and dispersion medium are separated we cannot get the solution directly by remixing the two phases. Therefore they are categorized under lyophobic colloids.

4. Colloids are classified into multi-molecular, macro-molecular and associated colloids.
a) True
b) False
Answer: a
Clarification: Based on the nature of dispersed phase, colloids are classified into multi-molecular, macro-molecular and associated colloids. In multi-molecular colloids a large number of atoms aggregate to form a particle of colloidal size, in macro-molecular colloids molecules dissolve in suitable solvents and give rise to particles of colloidal size. Colloids which behave as a normal strong electrolyte at low concentration but exhibits colloidal properties at higher concentration are known as associated colloids.

5. Which of the following is a characteristic of a multi-molecular colloid?
a) Large number of molecules combine to form a particle of colloidal size
b) A large number of atoms aggregate to form a particle of colloidal size
c) Starch solution is an example of multi-molecular colloid
d) Multi-molecular colloids are normally of the lyophilic type
Answer: b
Clarification: In this type of colloidal solution, a large number of atoms or small molecules aggregate to form a particle of colloidal size, these are normally of lyophobic type. Starch solution is an example of lyophilic colloids and hence is not an example for multi-molecular colloids.

6. Which of the following is false regarding macro-molecular colloids?
a) Protein solution is an example for macro-molecular colloids
b) Man-made macro-molecules like polythene can form such colloids
c) Silver solution can form macro-molecular colloids
d) These are normally of lyophilic type
Answer: c
Clarification: Macro-molecules dissolve in a suitable solvents, gives rise to particles of colloidal size. These are normally of lyophilic type and silver solution is an example for lyophobic and multi-molecular colloids.

7. Which of the following is not an example of associated colloids?
a) Sodium stearate
b) Potassium stearate
c) Gum
d) Detergents
Answer: c
Clarification: Some substances at low concentration behave as true solutions. As the concentration of the solution increases, it turns to be a colloidal solution. These type of colloids are knows as associated solution. Soaps like sodium stearate potassium stearate and detergents are examples of associated colloids whereas gum is an example for lyophilic colloids.

8. Which of the following colloids cannot be formed by direct mixing?
a) Lyophilic colloids
b) Lyophobic colloids
c) Macro-molecular colloids
d) Associated colloids
Answer: b
Clarification: Lyophobic colloids such as metal solutions like gold and silver solution, sulphur solution and blood cannot be prepared directly by mixing. These are prepared by special methods and are irreversible in nature. Once the dispersed phase and medium are separated it is not possible to get the solution by remixing the two phases.

9. Which of the following colloids is most stable?
a) Starch solution
b) Blood
c) Sulphur solution
d) Silver solution
Answer: a
Clarification: Starch solution is most stable among the following because it is a lyophilic colloid and the solutions are highly stable due to the high interaction between the dispersed phase and the dispersion medium.

10. In which of the following, the dispersed phase and medium can be separated by evaporation?
a) Ferric hydroxide solution
b) Sulphur solution
c) Metal in water
d) Starch solution
Answer: d
Clarification: In starch solution, if the dispersed phase and the dispersion medium are separated by evaporation, the colloidal solution can be regained by remixing the dispersed phase and dispersion medium. Hence these are called reversible solutions are highly stable because of the high interaction between the dispersed phase and medium.

Chemistry Multiple Choice Questions on “P-Block Elements – Phosphorus – Allotropic Forms”.

1. Which allotrope of phosphorus is the most stable?
a) White phosphorus
b) Red phosphorus
c) Black phosphorus
d) Phosphine
Answer: c
Clarification: Black phosphorus is thermodynamically, the most stable allotrope of phosphorus and does not burn in air even up to 673 K. It has a sharp melting point of 860 K. Like graphite, it is fairly a good conductor of electricity.

2. Which allotrope of phosphorus is the most reactive?
a) White phosphorus
b) Metal phosphorus
c) Red phosphorus
d) Beta-black phosphorus
Answer: a
Clarification: The three allotropic forms of phosphorus differ widely in their chemical reactivity in which white phosphorus is the most reactive while black and red phosphorus are less reactive. White phosphorus is made up of discrete P₄ tetrahedra which are subjected to very high angular strain as the angles is 60 degrees. This high angular strain makes white phosphorus unstable and highly reactive.

3. Red phosphorus is kept under water to protect it from air.
a) True
b) False
Answer: b
Clarification: White phosphorus is a highly reactive clement and must be stored underwater for safekeeping to prevent it from catching fire spontaneously in the air. In water, white phosphorus reacts with oxygen within hours or days. In water with low oxygen, white phosphorus may degrade to a highly toxic compound called phosphine, which eventually evaporates to the air and is changed to less harmful chemicals.

4. From which type of phosphorus is alpha -black phosphorus formed?
a) Phosphide
b) White phosphorus
c) Black phosphorus
d) Red phosphorus
Answer: d
Clarification: Alpha-black phosphorus is the most stable allotrope of black phosphorus. Alpha-black phosphorus is produced from red phosphorus. When red phosphorus is heated in a sealed tube at 803 K, it forms alpha-black phosphorus.

5. Beta-black phosphorus is prepared by heating white phosphorus.
a) True
b) False
Answer: a
Clarification: Yes, Beta-black phosphorus is prepared by heating white phosphorus at 473 K under high pressure(4000-12000 atm) in an inert atmosphere. It has layered structure in which each phosphorus atom is covalently bonded to three neighbouring phosphorus atoms.

6. Like white phosphorus, which phosphorus also exists as P₄?
a) Black phosphorus
b) Red phosphorus
c) Phosphine
d) Beta-black phosphorus
Answer: b
Clarification: Like white phosphorus, red phosphorus also exists as P₄ tetrahedra but these are joined together through covalent bonds to give a polymeric structure. Because of polymeric structure, its melting point (883 K) is much higher than that of white phosphorus(317 K).

7. Which allotrope of phosphorus does not catch fire easily?
a) White phosphorus
b) Alpha- black phosphorus
c) Beta- black phosphorus
d) Red phosphorus
Answer: d
Clarification: Red phosphorus is a relatively stable allotrope of phosphorus at room temperature. Its ignition temperature(543 K) is much higher than that of white phosphorus(303 K). As a result, it does catch fire easily.

8. White phosphorus can be reconverted to red phosphorus.
a) True
b) False
Answer: a
Clarification: Yes, red phosphorus sublimes on heating giving vapours which are the same as by white phosphorus. When these vapours are condensed, white phosphorus is obtained. This gives us simple method of reconverting red phosphorus into white phosphorus.

9. Which allotrope of phosphorus is also called yellow phosphorus?
a) Black phosphorus
b) Red phosphorus
c) Beta- black phosphorus
d) White phosphorus
Answer: d
Clarification: White phosphorus on exposure to light, it turns yellow. It glows greenish in the dark (when exposed to oxygen) and is highly flammable and pyrophoric (self-igniting) upon contact with air. Therefore it is called yellow phosphorus.

10. What is the allotrope of phosphorus in which P-atom completes its octet?
a) Alpha-black phosphorus
b) Beta-black phosphorus
c) White phosphorus
d) Red phosphorus
Answer: c
Clarification: White phosphorus exists as P₄ units. The four sp³ hybridized phosphorus atoms lie at the corners of a regular tetrahedron with an angle of 60 degree. Each phosphorus atom is linked to three other P-atoms by covalent bonds so that each p-atom completes it octet.

Chemistry Multiple Choice Questions on “P-Block Elements – Interhalogen Compounds”.

1. Which is the least basic trihalide of nitrogen?
a) NI₃
b) NBr₃
c) NCl₃
d) NF₃
Answer: d
Clarification: NF₃ is the least basic trihalide of nitrogen because it has different size and the bond ismorepolar towards nitrogen as it is more electronegative than the remaining halides, which makes electron availabilitymoreon nitrogen(in NF₃) than the remaining halides. Thus, it can donate electrons easily and therefore NF₃ is least basic.

2. Which of the following is true about interhalogen compounds?
a) They have unpaired electrons
b) They are highly stable
c) They are diamagnetic
d) They are paramagnetic
Answer: c
Clarification: Interhalogen compounds are diamagnetic in nature. This is because they have bond pairs and lone pairs. The A-X bond in interhalogen compounds is much weaker than the X-X bond in halogens, except for the F-F bond.

3. What do asthma patients use for respiration?
a) O₂ and H₂
b) O₂ and He
c) O₂ and Ar
d) O₂ and Ne
Answer: b
Clarification: Helium-Oxygen (80:20 or 70:30) mixture provides a dramatic benefit for asthma patients with severe exacerbations. Helium is about 10% as dense as room air and, consequently, travels more easily down narrowed passages.

4. Interhalogen compounds are more reactive than halogen compounds.
a) True
b) False
Answer: a
Clarification: Interhalogen compounds are formed by ionic bonds due to differences in electronegativity. They are more reactive because they can dissociate easily as they have weaker bonds as compared to pure halogen compounds possessing covalent bonds that are formed by the sharing of electrons.

5. Which is the correct statement regarding halogens?
a) They are not all diatomic and form univalent ions
b) They are all diatomic and do not form univalent ions
c) They are all diatomic and form univalent ions
d) They are all diatomic and form bivalent ions
Answer: c
Clarification: Halogens are all diatomic and form univalent ions. All halogens can act as both oxidizing and reducing agents and are capable of exhibiting more than one stable oxidation state except Fluorine. Hence, option c is the correct statement.

6. Xenon fluorides are reactive.
a) True
b) False
Answer: b
Clarification: Fluorine is the most electronegative element and is a very powerful oxidising agent. Although generally unreactive, xenon reacts with fluorine and gets oxidised forming fluorides of the type XeF₂, XeF₄ and XeF₆. These fluorides not reactive.

7. What is the hybridization of interhalogen compounds of the type XX’₃ (Bent T-shaped)?
a) sp³d²
b) sp
c) sp³d
d) sp²
Answer: c
Clarification: The structure of all interhalogen compounds of the type XX’₃ involves sp³d hybridization of the central atom X and hence they have trigonal bipyramidal geometries or T-shaped molecules (also called as see-saw).

8. What are interhalogen compounds used as?
a) Reducing agents
b) Aqueous solvents
c) To decreasing reactivity
d) Non-aqueous solvents
Answer: d
Clarification: If we come to the uses of the interhalogen compounds, then one of its main uses is to act as the halogenating agent, i.e., they are used to form other halogen compounds like BrF₅. Interhalogen compounds of fluorine are very useful fluorinating agents.

9. Why is helium added to oxygen which is used by deep-sea divers?
a) It is immiscible with oxygen
b) It is miscible with oxygen
c) It is less soluble in blood
d) It is less poisonous
Answer: c
Clarification: Air contains a large amount of nitrogen. According to Henry’s law, the solubility of gases increases with increase in pressure. Due to the increased pressure, a large amount of nitrogen dissolves in the blood of a deep-sea diver. To prevent this, helium is added to the oxygen cylinders as it is sparingly less soluble in blood.

10. What is the hybridization of interhalogen compounds of the type XX’₅(Square pyramidal)?
a) sp³d²
b) sp²d²
c) sp⁴d³
d) sp³d³
View Answer

Answer: a
Clarification: The structure of all interhalogen compounds of the type XX’₅ involves sp³d² hybridization of the central halogen atom X and hence have octahedral (also called square pyramidal) geometry and with one position occupied by a lone pair.

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

1. Which of the following does not explain the nature of bonding in coordination compounds?
a) Crystal Field Theory
b) Molecular Orbital Theory
c) Valence Bond Theory
d) VSEPR Theory
Answer: d
Clarification: The VSEPR Theory explains the structure of individual molecules based on the electron pairs in their atoms. VBT, CFT, LFT and MOT are theories that explain the nature of bonding in coordination compounds.

2. How many types of hybridisation are possible for complexes with a coordination number of 4?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: According to VBT, a complex with CN=4 can have two possible types of hybridisation and hence geometries. The sp³ hybridisation results in tetrahedral geometry and dsp² hybridisation results in square planar geometry.

3. How many empty orbitals are available in the central metal ion of a complex that has a trigonal bipyramidal geometry?
a) 2
b) 3
c) 4
d) 5
Answer: d
Clarification: One of the assumptions of VBT is that the number of empty orbitals in the metal ion for making bonds with the ligands is equal to its CN. A trigonal bipyramidal geometry has a hybridisation of sp³d and hence a CN=5.

4. A complex having _________ geometry can have more than one type of hybridisation.
a) tetrahedral
b) square planar
c) trigonal bipyramidal
d) octahedral
Answer: d
Clarification: Complexes having octahedral geometry can have either sp³d² or d²sp³ hybridisation depending on whether the outer or inner d orbitals are involved in hybridisation respectively.

5. Which statement regarding [Cr(NH₃)₆]³⁺ is incorrect?
a) It has octahedral geometry
b) It has d²sp³ hybridisation
c) It is diamagnetic
d) It is a low spin complex
Answer: c
Clarification: Cr³⁺ has six empty orbitals- two 3d, one 4s and three 4p orbitals, which are occupied by six electron pairs from six NH₃ molecules to form six d²sp³ hybridised orbitals. This leaves, three unpaired 3d electrons, making it paramagnetic. Also, since the inner d orbitals are involved, it is a low spin complex.

6. The ferricyanide complex ion is _________
a) paramagnetic
b) outer orbital
c) spin free
d) tetrahedral
Answer: a
Clarification: The oxidation state of Fe in [Fe(CN)₆]^3- ion is +3, which leaves it with five unpaired electrons in 3d orbitals. Two of these pair up leaving one unpaired electron in 3d and six vacant orbitals to form d²sp³ hybridised octahedral inner orbital complex.

7. What is the number of unpaired electrons in [Fe(H₂O)₆]²⁺ if it is known to be a high spin complex?
a) 0
b) 2
c) 3
d) 4
Answer: d
Clarification: The configuration of Fe²⁺ is 3d⁶. The complex has a CN=6 has should have sp³d² hybridisation. This leaves the 3d orbitals as it is with 4 unpaired electrons.

8. Identify the magnetic nature of the complex from its electronic configuration as shown.

a) Strongly paramagnetic
b) Weakly paramagnetic
c) Diamagnetic
d) Ferromagnetic
Answer: b
Clarification: There is one unpaired electron in the 3d orbitals. The presence of unpaired electrons makes a complex paramagnetic. In this case, since only 1 unpaired electron is present, it is a weak paramagnet.

9. Coordination entities with only three ligand groups attached to the metal ion can have d²sp³ hybridisation.
a) True
b) False
Answer: a
Clarification: This happens in the case of didentate ligands which have six pairs of electrons from three molecules. For example, [Co(ox)₃]^3- has d²sp³ hybridisation and is diamagnetic.

10. Identify the correct geometry and magnetic behaviour of the complex from the configuration shown.

a) Square planar, diamagnetic
b) Square planar, paramagnetic
c) Tetrahedral, diamagnetic
d) Tetrahedral, paramagnetic
Answer: a
Clarification: The hybridisation type of the complex is dsp² which corresponds to square planar geometry. Because there are no unpaired electrons, it is diamagnetic.

11. Predict the geometry of the complex from the electronic configuration shown.

a) Linear
b) Square planar
c) Tetrahedral
d) Octahedral
Answer: a
Clarification: The hybridisation type is sp which involves only two orbitals, so the geometry has to be linear and the coordination number is 2.

12. [Co(en)₂(NH₃)Cl]²⁺ is known to be a diamagnetic complex. What is the type of hybridisation it shows?
a) dsp²
b) sp³
c) d²sp³
d) sp³d²
Answer: c
Clarification: Co³⁺ has 3d⁶ configuration. For a complex to be diamagnetic, it should have no unpaired electrons. So, the single electrons in the 3d orbital will have to pair up and form 3 pairs leaving two 3d orbitals vacant. This results in d²sp³ hybridisation.

13. The compound K₄[Mn(CN)₆] is diamagnetic.
a) True
b) False
Answer: b
Clarification: The atomic number of Mn is 25 and the outer shell configuration of Mn²⁺ is 3d⁵. Four out of these five electrons pair up leaving 1 unpaired electron in 3d orbital and forming d²sp³ hybridisation. Thus, it is paramagnetic.

14. Identify the incorrect statement regarding VBT.
a) It does not explain the colour of coordination compounds
b) It is unreliable in the prediction of geometries of 4-coordinate complexes
c) It does not explain the kinetic stabilities of coordination compounds
d) It can distinguish between strong and weak ligands
Answer: d
Clarification: One of the limitations of VBT is that it cannot differentiate between weak and strong ligands. The other options are also its limitations.

Chemistry Multiple Choice Questions on “Alcohols, Phenols and Ethers Nomenclature – 1”.

1. According to IUPAC, what is the suffix used for naming alcohols?
a) -al
b) -ol
c) -one
d) -OH
Answer: b
Clarification: According to IUPAC, the naming of alcohols is done by substituting the ‘e’ in the name of the alkane with the prefix ‘ol’.

2. What is the common name of Butan-2-ol?
a) n-Butyl alcohol
b) sec-Butyl alcohol
c) Isobutyl alcohol
d) tert-Butyl alcohol
Answer: b
Clarification: Butan-2-ol has a parent chain of 4 carbon atoms and the OH group at the second carbon. Such configurations have the prefix sec- in the common name.

3. How many carbon atoms are present in the parent chain of tert-Butyl alcohol?
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: The IUPAC name of tert-Butyl alcohol is 2-methylpropan-2-ol, which has two substituent groups at the C-2 position, namely one methyl and one hydroxyl group.

4. What is the IUPAC name of Isopropyl alcohol?
a) Propan-1-ol
b) Propan-2-ol
c) 2-Methylpropan-1-ol
d) 2-Methylpropan-2-ol
Answer: b
Clarification: Isopropyl alcohol has a parent chain of three carbon atoms with the hydroxyl group at the second carbon, hence propan-2-ol.

5. Glycerol is a _________ alcohol.
a) monohydric
b) dihydric
c) trihydric
d) tetrahydric
View Answer

Answer: c
Clarification: The IUPAC name of glycerol is Propane-1,2,3-triol, signifying that there are three hydroxyl groups attached to three different carbon atoms.

6. Which of the following is not a polyhydric alcohol?
a) Cyclohexanol
b) Ethylene glycol
c) Propylene glycol
d) Benzene-1,2-diol
Answer: a
Clarification: Polyhydric alcohols have two or more hydroxyl groups in their structure and are named as either glycols in the common system or as diols and triols in the IUPAC system.

7. Identify the correct IUPAC name of the given compound.

a) 3-Ethylbutan-2-ol
b) 2-Ethylbutan-3-ol
c) 3-Methylpropan-2-ol
d) 3-Methylpropan-4-ol
Answer: c
Clarification: The longest parent carbon chain include the ethyl group and has 5 carbon atoms. The hydroxyl group is at C-2 and methyl group at C-3 to satisfy the lowest set of locants rule.

8. Identify the correct IUPAC name for the given compound.

a) Butane-diol
b) Butylene glycol
c) Butane-1,3-diol
d) Butane-1,4-diol
Answer: c
Clarification: This is a dihydric alcohol with OH groups at the first and third carbon positions. Butylene glycol is also the correct name but in the common system.

9. Which of the following is an incorrect name for the compound with aromatic ring having hydroxyl groups at para positions?
a) Hydroquinone
b) Benzene-1,4-diol
c) Resorcinol
d) Quinol
Answer: c
Clarification: Resorcinol is the common name for the dihydric phenolic compound with the OH groups at meta positions or 1, 3 positions.

10. What is the common name of the simplest hydroxy derivative of benzene?
a) Phenol
b) Cresol
c) Cyclohexanol
d) Glycerol
Answer: a
Clarification: The simplest hydroxy derivative of benzene is phenol, which is its common as well as IUPAC name. It is a monohydric compound.

11. Which of the following phenols is not dihydric?
a) p-Cresol
b) Catechol
c) Resorcinol
d) Quinol
Answer: a
Clarification: m-Cresol is a monohydric phenol with a methyl group at meta position with respect to OH. Its IUPAC name is 3-Methylphenol.

12. The compound o-Cresol is named as 2-Hydroxymethyl benzene in the IUPAC system.
a) True
b) False
Answer: b
Clarification: The C attached to the hydroxyl group is considered as the first carbon and hence the methyl group is present at the second carbon. So the correct name would be 2-Methylphenol.

Chemistry Multiple Choice Questions on “Nomenclature and Structure of Carboxyl Groups”.

1. Butyric acid was first obtained from _______
a) red ants
b) vinegar
c) rancid butter
d) goats
Answer: c
Clarification: The common names of some carboxylic acids gives an idea of where it was first obtained from in nature. Butyric acid comes from the Latin word butyrum meaning butter.

2. What is the common name of the simplest dicarboxylic acid?
a) Oxalic acid
b) Malonic acid
c) Acetic acid
d) Propionic acid
Answer: a
Clarification: The simplest dicarboxylic acid is ethanedioc acid, which has two carbon atoms both of which are a part of the carboxyl groups. Its formula is (HOOC-COOH).

3. Which of the following acids have a double bond in their structure?
a) Isobutyric acid
b) Succinic acid
c) Acrylic acid
d) Glutaric acid
Answer: c
Clarification: The IUPAC name of acrylic acid is prop-2-enoic acid, which indicates the presence of a double bond at the second carbon, the one next to the carboxyl carbon. Its formula is (CH₂=CHCOOH).

4. What is the correct IUPAC name of the compound CH₃CH=CHCH=CHCOOH?
a) Hexenedioc acid
b) Hexa-2,4-dienoic acid
c) Penta-1,3-dienioc acid
d) Pentenedioc acid
Answer: b
Clarification: The compound consists of six carbon atoms including the carboxyl carbon thus, it is a derivative of hexanoic acid. The only substituents are two double bonds at the second and fourth carbon considering that the carboxyl carbon is the number one.

5. Which of the following is not a dicarboxylic acid?
a) Malonic acid
b) Glutaric acid
c) Adipic acid
d) Carballylic acid
Answer: d
Clarification: Malonic acid, glutaric acid and adipic acid are dicarboxylic acids with 3, 5 and 6 carbon atoms in their structures respectively. Carballylic acid, also known as tricarballylic acid is a tricarboxylic acid having 3 carbon atoms excluding the three carboxyl carbons in its structure. Its IUPAC name is Propane-1,2,3-tricarboxylic acid.

6. Which of the following has two carboxyl groups in its structure?
a) Succinic acid
b) Crotonic acid
c) Cinnamic acid
d) Mandelic acid
Answer: a
Clarification: Succinic acid is a dicarboxylic acid with 4 carbon atoms in its structure. Its IUPAC name is butanedioic acid. Crotonic acid is but-2-enoic acid, whereas cinnamic acid and mandelic acid consists only a single carboxyl group along with a phenyl group.

7. What is the IUPAC name of terephthalic acid?
a) 2-Phenylethanoic acid
b) Benzene-1,2-dicarboxylic acid
c) Benzene-1,3-dicarboxylic acid
d) Benzene-1,4-dicarboxylic acid
Answer: d
Clarification: Terephthalic acid is an aromatic dicarboxylic acid with the two carboxyl groups on the opposite ends (para position) of a benzene ring. 2-Phenylethanoic acid consists of only a single carboxyl group.

8. What is the IUPAC name of (C₆H₅)CH₂CH₂COOH?
a) 2-Phenylethanoic acid
b) 2-Phenylpropanoic acid
c) 3-Phenylethanoic acid
d) 3-Phenylpropanoic acid
Answer: d
Clarification: The compound has the group (-CH₂CH₂COOH) directly attached to a benzene ring. The preference is given to the carboxyl group and the compound is named as a carboxylic acid derivative. There are total three carbons making it propanoic acid with the benzene ring as substituent on the last carbon (C-3) as a phenyl group.

9. What is the IUPAC name of (CH₃)₃CCH₂COOH?
a) Hexanoic acid
b) 2,2,2-Trimethylpropanoic acid
c) 3,3-Dimethylbutanoic acid
d) 4-Methylpentanoic acid
Answer: c
Clarification: The longest parent carbon chain consists of 4 carbon atoms including the carboxyl carbon and one methyl group attached to the tertiary carbon. The other two methyl groups act as substituents on the carbon, which is now the third carbon.

10. What is the correct IUPAC name of the shown compound?

a) 2-Carboxylmethylcyclopentane
b) 2-Methylcyclopentanecarboxylic acid
c) 2-Methylcyclopentanoic acid
d) (2-Methylcyclopentyl)carboxylic acid
Answer: b
Clarification: The compound is a cyclic carboxylic acid with five carbon atoms in the cyclic ring and one carboxyl group. The methyl group is present as a substituent at the adjacent position on the ring. Hence, it is named as a derivative of cyclopentanecarboxylic acid.

11. Identify the correct IUPAC name of the following compound?

a) 4-Oxocyclohexan-1-carboxylic acid
b) 4-Carboxylcyclohexan-1-one
c) 4-Oxocyclohexylmethanoic acid
d) 4-Oxocyclohexylethanoic acid
Answer: a
Clarification: This is a keto-substituted cyclic carboxylic acid and is named considering the CO group as a substituent at the fourth position given that the carbon attached to the carboxyl carbon is C-1.

12. Which of the following is an incorrect name for the following compound?

a) α-Methylpropionic acid
b) β-Methylbutyric acid
c) Isovaleric acid
d) 3-Methylbutanoic acid

Answer: a
Clarification: The longest parent chain of carbon atoms (including the first carboxyl carbon) is four. The methyl group is substituted at the third carbon, also known as the beta carbon. Isovaleric acid is also a common name for this structure.

13. The carboxyl carbon is more electrophilic than the carbonyl carbon.
a) True
b) False
Answer: b
Clarification: The carboxyl group is capable of forming three resonance structures in comparison to the two formed by carbonyl group. This makes the carboxyl carbon less electrophilic than the carbonyl carbon.

14. Salicylic acid consists of an aldehyde group on the benzene ring other than the primary carboxylic group.
a) True
b) False
Answer: b
Clarification: Salicylic acid consists of a hydroxy group on the benzene ring at ortho, meta or para position with respect to the carboxyl group. It is named as hydroxybenzoic acid in IUPAC system.