Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Surface Chemistry – Emulsions”.

1. An emulsion is a type of colloid.
a) True
b) False
Answer: a
Clarification: An emulsion is a colloidal dispersion in which both the dispersed phase and the dispersion medium are liquids. It is a suspension of two liquids that usually do not mix. These liquids that do not mix are said to be immiscible.

2. Which of the following is an example of an emulsifier?
a) NaCl
b) CaCO₃
c) C₁₅H₃₁COONa
d) CH₃COOH
Answer: c
Clarification: To obtain a stable emulsion, sometimes, small quantities of certain other substances are added during its preparation. The substances thus added to stabilize the emulsions are called emulsifiers or emulsifying agents.

3. Which of the following is not an example of a water-in-oil emulsion?
a) Cod liver oil
b) Butter
c) Cold cream
d) Milk
Answer: d
Clarification: In an emulsion of water-in-oil (w/o), water is the dispersed phase and oil is the dispersion medium. Cod liver oil, butter and cold cream are all examples of the water-in-oil type of emulsions.

4. Soaps are emulsifying agents.
a) True
b) False
Answer: a
Clarification: Soaps are sodium or potassium salts of higher fatty acids, for example, sodium palmitate (C₁₅H₃₁COONa), sodium stearate (C₁₇H₃₅COONa), etc. A molecule of soap consists of two parts, the hydrophobic part (soluble in oil) and the hydrophilic part (soluble in water).

5. What is the difference between vanishing cream and cold cream?
a) Both are examples of oil-in-water emulsions
b) Vanishing cream is an oil-in-water emulsion whereas cold cream is a water-in-oil emulsion
c) Vanishing cream is a water-in-oil emulsion whereas cold cream is anoil-in-water emulsion
d) Both are examples of water-in-oil emulsions
Answer: b
Clarification: Vanishing cream is an oil-in-water emulsion, that is, oil is the dispersed phase and water is the dispersion medium. Cold cream is a water-in-oil emulsion, that is, water is the dispersed phase and oil is the dispersion medium.

6. Which of the following is not a method to test the type of emulsion?
a) Microscopic method
b) Conductance method
c) Coagulation method
d) Dye method
Answer: c
Clarification: Coagulation or precipitation is a process of aggregating together the colloidal particles to change them into large-sized particles which ultimately settle as a precipitate. The different methods to test the type of emulsion are-microscopic, conductance and dye method.

7. Which of the following statements regarding emulsions is false?
a) Emulsions cannot be separated into their constituent liquids
b) Emulsions show Brownian motion
c) Emulsions show Tyndall effect
d) Emulsions exhibit properties like Electrophoresis and Coagulation
Answer: a
Clarification: Emulsions can be separated into their constituent liquids by boiling, freezing, centrifuging, electrostatic precipitation by adding large amounts of the electrolyte to precipitate out the dispersed phase or by chemical destruction of the emulsifying agent.

8. Which of the following statement about emulsions is true?
a) Oily drugs cannot be prepared in the form of emulsions
b) Digestion does not involve the process of emulsification
c) Disinfectants like Dettol and Lysol give emulsions of water-in-oil type on mixing with water
d) The cleansing action of soap is based upon the formation of water-in-oil emulsion
Answer: b
Clarification: The digestion of fats in the intestines takes place by the process of emulsification. A small amount of the fat reacts with the alkaline solution present in the intestines to form a sodium soap.

9. What is the emulsifier present in milk that makes it stable?
a) Maltose
b) Lactose
c) Lactic acid bacillus
d) Casein
Answer: d
Clarification: The emulsifier or emulsifying agent present in dairy emulsions is Casein. Casein is a slow-digesting dairy protein that people often take as a supplement.The most common form of casein is sodium caseinate.

10. What is the dispersion of a liquid in another liquid called?
a) Gel
b) Foam
c) Emulsion
d) Aerosol
Answer: c
Clarification: The dispersion of a liquid in a solid is called gel. The dispersion of a gas in a liquid medium is called foam. The dispersion of a liquid in another liquid is called emulsion. The dispersion of a solid or liquid in a gaseous medium is called aerosol.

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

1. Which of the following is not true about phosphine?
a) Phosphine is a colorless gas
b) Phosphine has a rotten fish smell
c) Phosphine is inflammable
d) Phosphine is highly poisonous
Answer: c
Clarification: Phosphine is a colorless gar with rotten fish smell and is highly poisonous. In pure state, phosphine is non-flammable but becomes inflammable owing to the presence of P₂H₄ or White phosphorous (P₄) vapours.

2. Phosphine acts as a Lewis acid.
a) True
b) False
Answer: b
Clarification: Phosphine acts as a Lewis base and reacts with acidic species to form phosphinium salts.
PH₃ + HBr → PH₄Br
Phosphine reacts with hydrogen bromide to form phosphinium salts which is a property of a base.

3. Phosphine is prepared from which of the following methods?
a) By reacting calcium phosphide with water or dilute HCl
b) By directly reacting phosphorous with hydrogen
c) By passing dry chlorine over P₂H₄
d) Heating while phosphorous with halides
Answer: a
Clarification: Phosphine is prepared by the reaction of calcium phosphide with water or dilute HCl.
With water: Ca₃P₂ + 6H₂O → 2PH₃ + 3Ca(OH) ₂
With dilute HCl: Ca₃P₂ + 6HCl → 2PH₃ + 3CaCl₂.

4. Which of the following is a property of phosphine?
a) It is insoluble in water
b) It is inflammable
c) The solution of phosphine in water decomposes in the presence of light
d) It acts as a Lewis acid
Answer: c
Clarification: Phosphine is slightly soluble in water and is non-flammable in pure form. It is a Lewis base and the solution of phosphine in water decomposes in the presence of light giving red phosphorous and hydrogen gas.

5. Which of the following gas is used as Holmes signal?
a) Hydrogen per oxide
b) Nitrogen
c) Acetylene
d) Phosphine
Answer: d
Clarification: When a mixture of calcium carbide and calcium phosphide placed in a container is made to react with water, it produces the gases phosphine and acetylene. This signal produced due to the burning gases is called Holmes signal.

6. What is the reaction involved in Holmes signal?
a) Ca₃P₂ + 6H₂O → 2PH₃ + 3Ca(OH) ₂
b) Ca₃P₂ + 6H₂O → 2PH₃ → P₂H₄ + P₄O₆ and CaC₂ + H₂O → C₂H₂
c) Ca₃P₂ + 6HCl → 2PH₃ + 3CaCl₂
d) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
Answer: b
Clarification: The reactions involved in Holmes signal are:
Ca₃P₂ + 6H₂O → 2PH₃ → P₂H₄ + P₄O₆ and CaC₂ + H₂O → C₂H₂
Calcium carbide and calcium phosphide both react with water to give acetylene and phosphine respectively. These gases burn to give the signal.

7. What is the hybridization of phosphine?
a) sp² hybridized
b) sp³ hybridized
c) sp hybridized
d) No hybridization
Answer: d
Clarification: The hybridization of phosphine seems like sp² but in reality the molecule has no hybridization as it forms all bonds using its pure p orbitals. This can be proved from its bong angle data which shows that its bond angles are 93.5°.

8. What is the structure of phosphine?
a) Trigonal pyramidal
b) Trigonal bi-pyramidal
c) Rhombohedral
d) Pyramidal
Answer: a
Clarification: Phosphine has a trigonal pyramidal structure with molecular symmetry. The length of the bond between phosphorous and hydrogen is 1.42 × 10^-10 m and the bond angles are equal and are known to be 93.5°.

9. Which of the following compounds react with water to give phosphine?
a) Phosphorous trichloride
b) Phosphorous pentachloride
c) Black phosphorous
d) Aluminium phosphide
Answer: d
Clarification: Metal phosphides such as aluminium phosphide reacts with water to from phosphine and metal hydroxide.
Reaction: AlP + 3H₂O → PH₃ + Al(OH) ₃.

10. Phosphine like ammonia has very high affinity for water.
a) True
b) False
Answer: b
Clarification: The ammonia molecule has the capability to form a dative bond due to the lone pair of electrons. There is a formation of H-bond due to strong dipole-dipole attraction between ammonia and water molecules where as in phosphine the H-bond is weak and the P-H bond is non-polar so, it is only slightly soluble in water but highly soluble in non-polar solvents.

Chemistry Multiple Choice Questions on “P-Block – Group 18 Elements”.

1. Helium is the second most abundant gas in the universe.
a) True
b) False
View Answer

Answer: a
Clarification: Helium is the second element in the periodic table and has an atomic number of 2. Helium is formed as the result of the fusion of two hydrogen atoms. Helium is the most abundant gas in the universe (23% as compared to 76% of hydrogen).

2. Which of the following noble gases do not occur in the elemental state in the atmosphere?
a) Helium
b) Neon
c) Argon
d) Radon
Answer: d
Clarification: Radon is the last-named member of the 18^th group of elements in the periodic table. It is radioactive in natureand as a result, all other noble gases except radon occur in the elemental state in the atmosphere.

3. Which of the following is the second most abundant noble gas in the atmosphere?
a) Helium
b) Argon
c) Neon
d) Krypton
Answer: c
Clarification: Argon is the most abundant noble gas in the atmosphere, accounting for about 0.934% by volume. Neon is the second most abundant noble gas in the atmosphere (about 0.00182%) followed by helium as the third most abundant noble gas (about 0.000524%).

4. The main commercial source of helium is natural gas.
a) True
b) False
Answer: a
Clarification: The main commercial source of helium is natural gas which mainly contains hydrocarbons along with varying amounts of carbon dioxide, nitrogen, hydrogen sulphide and helium (2-7%).

5. Which of the following scientists first discovered helium in the solar spectrum during a total solar eclipse?
a) Lord Rayleigh
b) J N Lockyer
c) William Ramsay
d) Antoine Lavoisier
Answer: b
Clarification: J N Lockyer first discovered helium in the solar spectrum during a total solar eclipse on 18^th August, 1868. Later on, in 1895, it was discovered on the planet Earth by a scientist named William Ramsay.

6. What is the half-life of radon?
a) 10 days
b) 4.56 days
c) 3.82 days
d) 5.46 days
Answer: c
Clarification: Half-life of a radioactive substance refers to the amount of time that it takes for half of the radioactive atoms in a sample to decay into a more stable form. Radon has a half-life of 3.82 days.

7. What are the decay products of radium?
a) Radon and oxygen
b) Radon and nitrogen
c) Lanthanum and oxygen
d) Radon and Helium
Answer: d
Clarification: Radium is a radioactive element with the atomic number 88. It is the 6^th element in group 2 of the periodic table. Radium (Ra) on radioactive decay, gives Radon (Rn) and Helium (He) as the products.

8. Which of the following noble gases is not obtained via fractional distillation?
a) Krypton
b) Argon
c) Neon
d) Helium
Answer: d
Clarification: Neon, argon, krypton and xenon are obtained by fractional distillation of air. Fractional distillation of liquid air gives oxygen, nitrogen and a mixture of noble gases. The individual noble gases are then separated by adsorption over coconut charcoal which adsorbs different gases at different temperatures.

9. Which of the following statements is incorrect about noble gases?
a) They are monoatomic
b) They are colourless
c) They are odourless
d) They all have an outer electronic configuration of ns²np⁶
Answer: d
Clarification: The general outer electronic configuration for noble gases is ns²np⁶. But, for helium, the outer electronic configuration is 1s² as it its atomic number is only 2. All noble gases are monoatomic, odourless and colourless.

10. What is the ratio of the molar heat at constant pressure to the molar heat at constant volume for a noble gas?
a) 1.5
b) 1.67
c) 1.73
d) 1.37
Answer: b
Clarification: All noble gases are monoatomic in nature due to their stable outer electronic configuration. The ratio of their specific or molar heat at constant pressure to the specific or molar heat at constant volume is equal to 1.67, which is the same for all monoatomic gases.

11. What is the electron gain enthalpy for noble gases?
a) > 0
b) c) = 0
d) It is not defined for noble gases
Answer: a
Clarification: Noble gases have completely filled subshells. As a result, there is no vacant room in their valence shell and hence the additional electron has to be placed in an orbital of the next higher shell. Therefore, energy has to be supplied in order to add an electron and so, the electron gain enthalpy of noble gases is positive (> 0).

12. What is the boiling point of Helium?
a) 7.8 K
b) 0 K
c) 4.2 K
d) 3.7 K
Answer: c
Clarification: The melting and boiling points of noble gases are very low. This is because the atoms are held together by very weak Van der Waals forces of attraction in both liquid and solid states. Helium has the lowest boiling point, equal to 4.2 K, among all known substances.

13. Which of the following noble gases is most soluble in water?
a) Helium
b) Radon
c) Krypton
d) Neon
Answer: b
Clarification: Of all the noble gases, radon is the most soluble in water. Noble gases are non-polar substance, however, when dissolved in water, they are slightly soluble. This is due to dipole induced dipole interactions.

14. Which of the following noble gases can diffuse through rubber?
a) Helium
b) Xenon
c) Argon
d) Krypton
Answer: a
Clarification: Hydrogen is the lightest and the smallest noble gas, having an atomic number of 2. Owing to its small size and inert nature, it has the unusual property of diffusing through most commonly used laboratory materials such as rubber, glass or plastic.

15. Which among the following noble gases does not form clathrates?
a) Argon
b) Xenon
c) Krypton
d) Helium
Answer: d
Clarification: Noble gases can form compounds in which the gases are entrapped in the cavities of crystal lattices. Such compounds are called clathrates. Only Argon, Krypton, Xenon and Radon are known to form clathrates among the noble gases.

Chemistry Problems for IIT JEE Exam on “Bonding in Coordination Compounds – 2”.

1. The crystal field theory considers the metal-ligand bond to be a _______ bond.
a) covalent
b) ionic
c) polar
d) hydrogen
Answer: b
Clarification: The CFT is an electrostatic model which considers the bond between metal ion and the ligand to be ionic arising due to the electrostatic interactions between them.

2. In CFT, which of the following ligands will be treated as point dipoles?
a) Cl
b) NO₂
c) CN
d) NO
Answer: d
Clarification: Anionic ligands are treated as point charges and neutral ligands are treated as point dipoles in crystal field theory.

3. Which of the following orbitals does not belong to t_2g orbitals?
a) d_xy
b) d_yz
c) d_xz
d) d_z²
Answer: d
Clarification: t_2g is a set of three d orbitals of lower energy in octahedral complexes. The two remaining d orbitals of higher energy form the e_g set.

4. Identify the correct order of increasing field strength of the ligands.
a) SCN^–, NH₃, NCS^–, CN^–
b) NCS^–, SCN^–, NH₃, CN^–
c) SCN^–, NCS^–, NH₃, CN^–
d) NCS^–, NH₃, SCN^–, CN^–
Answer: c
Clarification: According to the spectrochemical series, thiocyanate is weaker than isothiocyanate which is in turn weaker than amine and cyanide ligands.

5. By how much percentage does the energy of e_g orbitals increase by from the average field splitting energy in octahedral complexes?
a) 40%
b) 45%
c) 50%
d) 60%
Answer: d
Clarification: The energy of the two e_g orbitals will increase by (3/5)Δ_o, that is 60% of the average energy that separates the field in octahedral complexes.

6. Identify the correct order of decreasing field strength of the neutral ligands.
a) CO, en, NH₃, H₂O
b) CO, en, H₂O, NH₃
c) en, CO, NH₃, H₂O
d) en, CO, H₂O, NH₃
Answer: a
Clarification: According to the spectrochemical series, CO is the strongest ligand. The ligands en, amine and water are weaker than CO and their field strength decreases in that order.

7. Identify the correct relation between Δ_o and Δ_t, where Δ_o denotes crystal field splitting in octahedral complexes and Δ_t denotes crystal field splitting in tetrahedral complexes.
a) Δ_ot
b) Δ_o > Δ_t
c) Δ_o = Δ_t
d) Δ_o ≥ Δ_t
View Answer

Answer: b
Clarification: It has been observed that splitting energy in tetrahedral fields is considerably less than in octahedral complexes. To be precise, it has been found that Δ_t = (4/9) Δ_o.

8. Which of the following ligands is most likely to form high spin complexes in octahedral fields?
a) Cl^–
b) OH^–
c) C₂O₄^–
d) CN^–
Answer: a
Clarification: Cl- is the weakest field ligand in the given options. The splitting field Δ_o is small and the energy separation between the e_g and t_2g orbitals are narrower. Hence, the electrons tend to occupy the eg orbitals easily and are likely to have a greater number of half-filled orbitals.

9. What will be the electronic configuration of d⁵ in terms of t_2g and e_g in an octahedral field when Δ_o

a) t_2g⁵ e_g⁰
b) t_2g² e_g³
c) t_2g³ e_g²
d) t_2g⁰ e_g⁵
Answer: c
Clarification: When the field splitting is less than the pairing energy, the energy gap between the two sets of d orbitals is small. This allows the electrons to occupy the higher energy orbitals once the lower orbitals are singly occupied. Thus, the configuration will be t_2g³ e_g², with one electron in each orbital.

10. Tetrahedral fields can easily form low spin complexes.
a) True
b) False
Answer: b
Clarification: The splitting energy in tetrahedral fields is almost half of that of octahedral fields. Due to this, the orbital splitting energies are not very large, which makes it difficult for electron pairing and therefore low spin complexes are rarely formed.

11. Observe the orbital splitting diagram of a tetrahedral field and find the value of x.

a) 0.178
b) 0.267
c) 0.4
d) 0.44
Answer: a
Clarification: The value of x is asked in terms of Δ_o and not Δ_t. The t₂ orbitals is at an energy of (2/5) Δ_t above the average energy. This is equal to (2/5) times (4/9) Δ_o = 0.178Δ_o.

12. As the crystal field splitting energy in octahedral field increases, the wavelength of light absorbed _________
a) Increases
b) Decreases
c) Remains the same
d) May increase or decrease
Answer: b
Clarification: If Δ_o is less (as in weak field ligands), the excitation energy will be the smallest and the largest wavelength will be absorbed. On the other hand, Δ_o is high (as in strong field ligands), the excitation energy will be the largest and the smallest wavelength will be absorbed.

13. The complex [Co(H₂O)₆]³⁺ absorbs the wavelength of light corresponding to orange colour. Predict the colour of the coordination compound based on this information.
a) Red
b) Yellow
c) Blue
d) Colourless
Answer: c
Clarification: This is because of the concept of complementary colours. The colour of the complex is complementary to that which is absorbed and is generated from the wavelength that is left over.

14. Hexaaquatitanium(III) chloride is a coloured compound that loses its colour on heating.
a) True
b) False
Answer: a
Clarification: Heating the compound removes the water which is the ligand. In the absence of a ligand, crystal field splitting does not occur and hence the compound becomes colourless.

Chemistry Problems for IIT JEE Exam,

Chemistry Written Test Questions for IIT JEE Exam on “Alcohols, Phenols and Ethers Nomenclature – 2”.

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

a) But-3-enol
b) But-3-en-2-ol
c) But-1-en-3-ol
d) But-2-en-3-ol
Answer: b
Clarification: The numbering preference is given to the hydroxyl group over the double bond and hence, the OH lies on the second carbon and double bond on the third.

2. What is the correct IUPAC name of the following phenol?

a) Benzene-1,2,4-triol
b) 1,2,4-Benzenetriol
c) Benzene-1,2,4-triphenol
d) 1,2,4-Triphenol
Answer: a
Clarification: Polyhydric phenols are named as hydroxy derivatives of benzene in the IUPAC system. The positions 1,2,4 comply with the lowest set of locants rule.

3. What is the correct common name of 2-Bromophenol?
a) 2-Bromophenol
b) m-Bromophenol
c) o-Bromophenol
d) 2-Hydroxybromobenzene
Answer: c
Clarification: Phenol is used as the common as well as the IUPAC name and haloarenes with hydroxy groups are named as halo derivatives of phenols.

4. Which of the following is given highest preference while naming when present in the same structure?
a) Hydroxyl group
b) Nitro group
c) Triple bond
d) Halogen group
Answer: a
Clarification: Functional groups are given the highest preference while naming compounds according to the IUPAC system.

5. What is the correct common name for CH₃OC₂H₅?
a) Methyl methyl ether
b) Methyl ethyl ether
c) Ethyl ethyl ether
d) Ethyl methyl ether
Answer: d
Clarification: It is an unsymmetrical ether with one methyl group and one ethyl group. It is named such that the alkyl group are in alphabetical order.

6. Which of the following compounds is ethoxyethane?
a) CH₃OCH₃
b) CH₃OC₂H₅
c) C2H₅OC₂H₅
d) C2H₅OC₃H₇
Answer: c
Clarification: Ethoxyethane is an ether which is a derivative of ethane with one hydrogen atom replaced by ethoxy group (OC₂H₅).

7. Identify the incorrect name for CH₃-O-C₆H₅.
a) Methyl phenyl ether
b) Methoxybenzene
c) Anisole
d) Phenetole
Answer: d
Clarification: This is an aromatic ether with a methoxy substituent and has the common name anisole or methyl phenyl ether.

8. What is the IUPAC name of CH₃-O-CH₂-CH₂-OCH₃?
a) 1,2-Dimethoxyethane
b) 1,3-Dimethoxybutane
c) 2,3-Dimethoxyethane
d) 1,4-Dimethoxypropane
Answer: a
Clarification: The base hydrocarbon is ethane, in which one hydrogen of each carbon atom is replaced by a methoxy group, making it a dimethoxy ether.

9. What is the IUPAC name of CH₃(CH₂)₆-OC₆H₅?
a) Heptyl phenyl ether
b) 1-Phenoxyheptane
c) 1-Hepoxybenzene
d) Phenyl heptyl ether
Answer: b
Clarification: Since heptane is a larger group than benzene, it is chosen as the parent hydrocarbon and the compound is named as an aryloxy derivative of heptane.

10. The name 1-Ethoxy-2, 2-dimethylcyclohexane is correct according to IUPAC nomenclature.
a) True
b) False
Answer: b
Clarification: The methyl group should be given higher preference over the alkoxy group in ethers. The correct naming will be 2-Ethoxy-1, 1-dimethylcyclohexane.

11. Phenetole is an aromatic ether.
a) True
b) False
Answer: a
Clarification: Phenetole is also known as ethoxybenzene or ethyl phenyl ether and has an aromatic ring on one side.

Chemistry Written Test Questions for IIT JEE Exam,

Chemistry Exam Questions for IIT JEE Exam on “Methods of Carboxylic Acids Preparation”.

1. Which of the following is known as Jones reagent?
a) KMnO₄ in alkaline medium
b) CrO₃ in H₂SO₄
c) K₂Cr₂O₇ in acidic medium
d) KMnO₄ in H₂SO₄
View Answer

Answer: b
Clarification: Chromium trioxide in an aqueous solution with sulphuric acid in known as Jones reagent. It Is an important compound in the preparation of carboxylic acids from alcohols.

2. Benzoic acid is obtained from the oxidation of _______ with alkaline KMnO₄ followed by treatment with mineral acid.
a) phenol
b) benzaldehyde
c) acetophenone
d) benzyl alcohol
Answer: d
Clarification: Only primary alcohols undergo oxidation in the presence of common oxidising agents followed by reaction with H₃O⁺ to give the respective carboxylic acids. This reaction is proceeded by the removal of both hydrogen atoms from the alpha carbon and formation of a double bond between C and O.

3. Identify the most suitable reagent for the conversion of ethanal to acetic acid.
a) Alkaline KMnO₄; H₃O⁺
b) Jones reagent
c) Tollen’s reagent
d) LiAlH⁴
Answer: c
Clarification: Aldehydes are easily oxidized to carboxylic acids having the same number of carbon atoms as the parent aldehyde, when reacted even with mild oxidising agents like Tollen’s reagent.

4. Identify the product B in the reaction chain shown.

a) Ethanoic acid
b) Propanoic acid
c) Butanoic acid
d) 2-Methylpropanoic acid
Answer: b
Clarification: Ethyl bromide on reaction with alc. KCN gives ethyl cyanide by nucleophilic substitution. Nitriles can be hydrolysed to give amides and on further heating give a carboxylic acid which contains one more carbon than the original ethyl bromide. Hence, the acid contains three C atoms, making it propanoic acid.

5. Identify X in the following conversion.

a) Alkaline KMnO₄; H₃O+
b) KOH; heat
c) H₃O⁺; heat
d) KMnO₄-KOH; heat
Answer: c
Clarification: Amides undergo hydrolysis in the presence of heat to give corresponding carboxylic acids along with ammonia gas. In the reaction shown, benzamide undergoes oxidation to form benzoic acid.

6. Benzoic acid can be prepared by oxidation of tert-Butyl benzene.
a) True
b) False
Answer: b
Clarification: tert-Butyl benzene consists of a tertiary group with no benzylic H. It is highly stable and does not undergo oxidation even under drastic conditions, to give aromatic carboxylic acids.

7. Which of the following pairs do not give the same compound on heating with alkaline potassium permanganate?
a) Toluene and propyl benzene
b) Toluene and n-Butyl benzene
c) Propyl benzene and isopropyl benzene
d) o-Xylene and n-Butyl benzene
Answer: d
Clarification: All mono-substituted alkyl benzenes, with primary or secondary alkyl groups, on vigorous oxidation give benzoic acid. The entire side chain is oxidised irrespective of the length. In case of o-xylene, there are two methyl groups on the benzene ring and both of them are oxidised to carboxyl group, resulting in phthalic acid.

8. p-Xylene on reaction with acidified potassium dichromate at high temperature gives ________
a) benzoic acid
b) phthalic acid
c) terephthalic acid
d) no reaction
Answer: c
Clarification: p-Xylene is a dimethyl substituted benzene which on vigorous oxidation, gets oxidised to an aromatic dicarboxylic acid, with the two carboxyl groups at para positions with respect to each other. This compound is called terephthalic acid.

9. Which of the following cannot be converted to benzoic acid on reaction with KMnO₄-KOH followed by H₃O⁺?
a) Ethyl benzene
b) Acetophenone
c) 4-Methylacetophenone
d) Styrene
Answer: c
Clarification: The ethyl group, acetyl group and ethene group are all oxidised to potassium carboxylate groups which are further oxidised to carboxyl groups. In the case of 4-Methylacetophenone, there are two groups that will be oxidised to carboxyl groups, hence forming terephthalic acid instead of benzoic acid.

10. How can methyl magnesium bromide be converted to propanoic acid?
a) Jones reagent
b) KMnO₄-KOH; heat
c) H₃O⁺; heat
d) CO₂-dry ether; H₃o⁺
Answer: d
Clarification: Methyl magnesium bromide (Grignard reagent) on reaction with CO₂ forms an addition product containing an additional C carbon atom. This is decomposed in the presence of mineral acid to form propanoic acid.

11. 3-Chlorophenyl magnesium bromide on reaction with dry ice followed by acidification in mineral acid gives _________
a) 3-Chlorophenol
b) 3-Chlorophenylethanoic acid
c) 3-Chlorobenzaldehyde
d) 3-Chlorobenzoic acid
Answer: d
Clarification: The MgBr group of 3-Chlorophenyl magnesium bromide (Grignard reagent) will be substituted by COOH group in the above reaction, to give a halogen substituted aromatic carboxylic acid.

12. Benzoic ethanoic anhydride on hydrolysis gives __________
a) benzoic acid and methanoic acid
b) benzoic acid and ethanoic acid
c) phenylethanoic acid and methanoic acid
d) no products
Answer: b
Clarification: Benzoic ethanoic anhydride (C₆H₅COOCOCH₃) is easily hydrolysed with water to give its corresponding acids, benzoic acid (by adding H to C₆H₅COO) and ethanoic acid (by adding OH to COCH₃).

13. Ethanoyl chloride on hydrolysis with aqueous NaOH gives _______
a) acetate ion
b) acetic acid
c) propanoic acid
d) no reaction
Answer: a
Clarification: Acid chlorides are readily hydrolysed with aqueous base to give carboxylate ions which further give corresponding carboxylic acids on acidification. It can be directly obtained by direct hydrolysis with water.

14. The final product(s) of basic hydrolysis followed by acidification of ethyl butanoate is _______
a) ethanoic acid
b) butanoic acid
c) ethanoic acid and butanoic acid
d) butanoic acid and ethanol
Answer: b
Clarification: Ethyl butanoate (CH₃CH₂CH₂COOC₂H₅) on basic hydrolysis forms CH₃CH₂CH₂COONa and ethanol. Then this sodium carboxylate compound gets acidified to give butanoic acid.

15. Acidic hydrolysis of ethyl benzoate directly gives benzoic acid.
a) True
b) False
Answer: a
Clarification: Acidic hydrolysis of esters directly gives carboxylic acids. In this case, the ethyl group of the ester is separated out as ethanol along with the main product, benzoic acid.