Questions & explanations
1. Which of the following aromatic hydrocarbons is most associated with carcinogenicity due to its metabolic activation to diol epoxides?
- A) Benzene
- B) Toluene
- C) Benzo[a]pyrene
- D) Naphthalene
Answer: C) Benzo[a]pyrene
Benzo[a]pyrene is a polycyclic aromatic hydrocarbon (PAH) that is a known carcinogen. It undergoes metabolic activation by cytochrome P450 enzymes to form highly reactive diol epoxides (e.g., benzo[a]pyrene-7,8-diol-9,10-epoxide) that can bind covalently to DNA, causing mutations and cancer. Benzene is also carcinogenic but via different mechanisms (e.g., inducing leukemia), not through diol epoxide formation. Toluene and naphthalene are less potent carcinogens and do not primarily act via diol epoxide intermediates.
2. Which of the following pairs of compounds are chain isomers?
- A. n-Butane and 2-methylpropane
- B. Ethane and propane
- C. Propane and cyclopropane
- D. Methane and ethane
Answer: A. n-Butane and 2-methylpropane
Chain isomers have the same molecular formula but different carbon skeletons. n-Butane (C4H10) and 2-methylpropane (C4H10) both have four carbon atoms but differ in the arrangement of the carbon chain. Ethane and propane, methane and ethane have different numbers of carbon atoms and are not isomers. Propane and cyclopropane have the same formula but cyclopropane is cyclic, so it is a ring-chain isomer, not a chain isomer.
3. Which of the following substituents is an activating group for electrophilic aromatic substitution?
- A) –NO₂
- B) –CN
- C) –NHCOCH₃
- D) –SO₃H
Answer: C) –NHCOCH₃
In electrophilic aromatic substitution, activating groups donate electron density to the benzene ring, increasing its reactivity. The acetamido group (-NHCOCH₃) is an ortho/para director and activates the ring due to resonance donation from the nitrogen lone pair. In contrast, –NO₂, –CN, and –SO₃H are strongly electron-withdrawing (deactivating) groups that remove electron density and lower ring reactivity.
4. Which type of isomerism is exhibited by alkanes having the same molecular formula but different carbon skeletons?
- A. Position isomerism
- B. Functional isomerism
- C. Chain isomerism
- D. Metamerism
Answer: C. Chain isomerism
Alkanes with the same molecular formula but different arrangements of the carbon chain are called chain isomers. Position isomerism requires a functional group at different positions, which alkanes lack. Functional isomerism involves different functional groups; metamerism involves different alkyl groups on either side of a functional group. Neither applies to alkanes.
5. The acidic character of a terminal alkyne is primarily attributed to:
- sp² hybridization of the carbon bearing the hydrogen.
- sp hybridization of the carbon bearing the hydrogen.
- the presence of a π bond adjacent to the hydrogen.
- the high electronegativity of the hydrogen atom.
Answer: sp hybridization of the carbon bearing the hydrogen.
In terminal alkynes, the hydrogen is attached to a carbon with sp hybridization. The greater s-character (50%) of the sp orbital makes the C–H bond more polar and the hydrogen more easily removed as H⁺, imparting acidity. sp² (33% s) and sp³ (25% s) carbons are less acidic, and the presence of π bonds alone does not account for the acidity. Hence option B is correct.
6. Which of the following reactions is used to prepare an alkane with double the number of carbon atoms present in the alkyl halide?
- A. Kolbe's electrolysis
- B. Wurtz reaction
- C. Decarboxylation
- D. Hydrogenation
Answer: B. Wurtz reaction
In the Wurtz reaction, two alkyl halides react with sodium metal to give an alkane with twice the number of carbon atoms of the alkyl halide. Kolbe's electrolysis yields an alkane from the carboxylate ion, not from an alkyl halide. Decarboxylation reduces the number of carbon atoms. Hydrogenation adds hydrogen to an alkene, not doubling the carbon count.
7. Hydration of propyne in the presence of HgSO₄ and dilute H₂SO₄ gives which product?
- Propanal
- Propanone (acetone)
- Propan-1-ol
- Propanoic acid
Answer: Propanone (acetone)
Hydration of alkynes follows Markovnikov's rule. For propyne (CH₃C≡CH), water adds such that the OH group goes to the more substituted carbon (the middle carbon) and the H to the terminal carbon, forming an enol that tautomerizes to a ketone. The product is propanone (acetone), not an aldehyde or alcohol. Thus option B is correct.
8. Which of the following compounds can exhibit geometrical (cis-trans) isomerism?
- 1-Butene
- 2-Butene
- 2-Methyl-2-butene
- Propene
Answer: 2-Butene
Geometrical isomerism requires each carbon of the double bond to be attached to two different groups. In 2-butene each double-bonded carbon has one –H and one –CH₃, so cis and trans forms exist. In the other options at least one carbon carries two identical groups (e.g., two H or two CH₃), preventing cis–trans isomerism.
9. The conversion of n-hexane into benzene in the presence of a catalyst at high temperature and pressure is known as:
- A. Isomerisation
- B. Aromatisation
- C. Pyrolysis
- D. Cracking
Answer: B. Aromatisation
Aromatisation is the process of converting aliphatic hydrocarbons (e.g., n-hexane) into aromatic hydrocarbons (e.g., benzene) under heat and a catalyst. Isomerisation changes the carbon skeleton without cyclisation. Pyrolysis and cracking break larger molecules into smaller ones, not forming aromatic rings directly.
10. When ethyne is treated with excess bromine water, the major product is:
- 1,2-dibromoethene
- 1,1,2,2-tetrabromoethane
- 1,1-dibromoethene
- 1,2-dibromoethane
Answer: 1,1,2,2-tetrabromoethane
Alkynes undergo two successive electrophilic additions with bromine. The first addition yields a dibromoalkene (e.g., 1,2-dibromoethene), and with excess bromine, a second addition occurs to give the tetrahaloalkane (1,1,2,2-tetrabromoethane). Thus the final product with excess bromine is the tetrabromo compound.
11. Which reagent is commonly used for the dehydrohalogenation of an alkyl halide to form an alkene?
- Aqueous KOH
- Alcoholic KOH
- Conc. H₂SO₄
- Zn / HCl
Answer: Alcoholic KOH
Dehydrohalogenation (elimination of HX) is favoured by a strong base in an alcoholic medium. Alcoholic KOH abstracts a β-hydrogen and eliminates the halogen, yielding an alkene. Aqueous KOH promotes substitution (alcohol formation). Conc. H₂SO₄ is used for dehydration of alcohols, and Zn/HCl is a reducing agent.
12. According to Markovnikov's rule, the addition of HBr to 2-methylpropene will predominantly form:
- 1-Bromo-2-methylpropane
- 2-Bromo-2-methylpropane
- 1-Bromo-1-methylpropane
- 2-Bromo-1-methylpropane
Answer: 2-Bromo-2-methylpropane
Markovnikov's rule states that the hydrogen atom of HX adds to the carbon of the double bond that already has more hydrogen atoms. In 2-methylpropene, the terminal carbon carries two hydrogens, so H adds there, and Br adds to the more substituted carbon (tertiary), giving 2-bromo-2-methylpropane.