Questions & explanations
1. The condensed formula of a compound is CH₃CH(CH₃)CH₂CH₃. Which of the following bond-line representations correctly corresponds to this compound?
- A five-carbon straight chain with a branch at the second carbon: a zigzag line with a methyl group attached to the second carbon from the left.
- A four-carbon straight chain with no branches.
- A five-carbon chain with a methyl group on the third carbon.
- A six-carbon straight chain.
Answer: A five-carbon straight chain with a branch at the second carbon: a zigzag line with a methyl group attached to the second carbon from the left.
The condensed formula CH₃CH(CH₃)CH₂CH₃ represents 2-methylbutane: a five-carbon chain (butane backbone with a methyl substituent on carbon 2). In bond-line notation, a zigzag line of four segments (representing four carbons) has a methyl group (a single line) attached to the second carbon (the second vertex). Options B (four-carbon straight chain) would be butane, C (methyl on third carbon) would be 3-methylbutane (which is same as 2-methylbutane but numbering is wrong; actually 3-methylbutane is not preferred but it's the same structure? but here the description says 'on the third carbon' which would give the same actual structure if the chain is oriented differently, but the question expects the correct representation. However, in standard IUPAC numbering, the substituent is on carbon 2, not 3. The correct bond-line representation shows the methyl branch on the second carbon from any end, but the description 'third carbon' could cause confusion. To be precise, the only option that matches the five-carbon skeleton with one branch is option A. Option D is six carbons. Thus option A i
2. Which of the following pairs correctly matches a common name with its IUPAC name?
- Acetone – Propanal
- Formalin – Methanal
- Acetic acid – Ethanoic acid
- Ethylene glycol – 1,2-Ethanediol
Answer: Acetic acid – Ethanoic acid
The common name 'acetic acid' corresponds to the IUPAC name ethanoic acid (CH₃COOH). Acetone is propan-2-one, not propanal. Formalin is an aqueous solution of methanal (formaldehyde), but 'formalin' itself is not a pure compound name; the IUPAC name of formaldehyde is methanal. Ethylene glycol is indeed 1,2-ethanediol, but the IUPAC name is ethane-1,2-diol (the given '1,2-Ethanediol' is acceptable but less standard; however, the correct pair here is Acetic acid–Ethanoic acid, which is a direct and unambiguous match. Therefore, the correct answer is option C.
3. Considering electronic effects, which of the following represents the correct order of increasing acidity for the given phenols?
- p-Methylphenol < Phenol < p-Chlorophenol < p-Nitrophenol
- Phenol < p-Methylphenol < p-Chlorophenol < p-Nitrophenol
- p-Chlorophenol < Phenol < p-Methylphenol < p-Nitrophenol
- p-Nitrophenol < p-Chlorophenol < Phenol < p-Methylphenol
Answer: p-Methylphenol < Phenol < p-Chlorophenol < p-Nitrophenol
The acidity of phenols increases with electron-withdrawing substituents and decreases with electron-donating groups. The nitro group is strongly electron-withdrawing (both –I and –R), making p-nitrophenol the most acidic. Chlorine exerts a –I effect (electron-withdrawing) but no resonance donation due to its lone pairs being less effective in the benzene ring, so p-chlorophenol is more acidic than phenol. The methyl group is electron-donating (+I), so p-methylphenol is the least acidic.
4. Which purification technique is most suitable for separating a mixture of two volatile liquids with different boiling points?
- Simple distillation
- Fractional distillation
- Crystallisation
- Column chromatography
Answer: Fractional distillation
Fractional distillation is used when the boiling points of the two liquids are close to each other; simple distillation suffices when the boiling point difference is large (as noted correctly later in the same reveal). Simple distillation is effective only when the boiling point difference is large. Crystallisation is for solids, and column chromatography separates based on adsorption. Therefore, fractional distillation is the most suitable method.
5. Which of the following tests is used for the detection of nitrogen in an organic compound?
- Beilstein test
- Lassaigne's test
- Test with sodium nitroprusside
- Dumas test
Answer: Lassaigne's test
Lassaigne's test (sodium fusion test) is a qualitative test for nitrogen, sulfur, and halogens in organic compounds. Nitrogen is detected by the formation of Prussian blue. The Beilstein test detects halogens (copper wire test), sodium nitroprusside test detects sulfur, and Dumas method is a quantitative estimation of nitrogen. Hence, Lassaigne's test is correct for qualitative detection of nitrogen.
6. The neopentyl cation, (CH₃)₃C–CH₂⁺ (a primary carbocation), undergoes a 1,2-methyl shift. Which more stable carbocation is formed?
- (CH₃)₂C⁺–CH₂CH₃ (a tertiary carbocation)
- (CH₃)₃C–CH₂⁺ (neopentyl cation, unchanged)
- CH₃CH₂CH₂CH₂CH₂⁺ (n-pentyl cation, primary)
- (CH₃)₃C⁺ (tert-butyl cation)
Answer: (CH₃)₂C⁺–CH₂CH₃ (a tertiary carbocation)
The neopentyl cation (CH3)3C–CH2+ is primary; since the adjacent carbon bears no hydrogen (only three methyl groups), a 1,2-methyl shift occurs instead of a hydride shift, moving a methyl group to the cationic carbon and generating the more stable tertiary carbocation (CH3)2C+–CH2CH3. Rearrangements of this kind always move a carbocation towards a more stable (tertiary > secondary > primary) form.
7. Which type of reaction involves the replacement of an atom or group in an organic molecule by another atom or group?
- Addition reaction
- Substitution reaction
- Elimination reaction
- Rearrangement reaction
Answer: Substitution reaction
A substitution reaction is one in which an atom or a group of atoms in a molecule is replaced by another atom or group. Addition reactions add atoms across a multiple bond, elimination reactions remove atoms to form a multiple bond, and rearrangement reactions involve a change in connectivity without change in molecular formula. Thus, the correct answer is substitution.
8. Which of the following reactions involves homolytic fission of a covalent bond?
- Hydrolysis of methyl chloride by aqueous NaOH
- Chlorination of methane in the presence of sunlight
- Dehydration of ethanol with concentrated H₂SO₄
- Acid-catalyzed hydration of ethene
Answer: Chlorination of methane in the presence of sunlight
Homolytic fission of a bond produces free radicals (species with unpaired electrons). In the chlorination of methane, the Cl–Cl bond undergoes homolytic cleavage under sunlight to form two chlorine radicals (Cl•), which initiate the chain reaction. The other options involve heterolytic bond fission, where the bond breaks unevenly to form ions (e.g., H⁺, OH⁻, etc.).
9. Which of the following statements correctly distinguishes nucleophilic and electrophilic reagents?
- Nucleophiles are electron-rich species that attack electron-deficient centers, while electrophiles are electron-deficient species that attack electron-rich centers.
- Nucleophiles are electron-deficient species that attack electron-rich centers, while electrophiles are electron-rich species that attack electron-deficient centers.
- Both nucleophiles and electrophiles are electron-rich species that attack electron-rich centers.
- Both nucleophiles and electrophiles are electron-deficient species that attack electron-deficient centers.
Answer: Nucleophiles are electron-rich species that attack electron-deficient centers, while electrophiles are electron-deficient species that attack electron-rich centers.
Nucleophiles possess a lone pair or negative charge and are attracted to electron-deficient (positively polarized) carbon atoms. Electrophiles are electron-deficient (positive charge or empty orbital) and seek out electron-rich centers. The other options reverse these definitions or incorrectly state both categories as the same. Hence the first option is correct.
10. Which of the following is NOT a reaction intermediate?
- Carbocation
- Carbanion
- Free radical
- Transition state
Answer: Transition state
Reaction intermediates are stable species with a finite lifetime that can be isolated or detected, whereas a transition state is a fleeting arrangement of atoms at the energy maximum of a reaction coordinate and cannot be isolated.
11. Which of the following correctly represents the order of stability for alkyl free radicals?
- CH₃• > CH₃CH₂• > (CH₃)₂CH• > (CH₃)₃C•
- (CH₃)₃C• > (CH₃)₂CH• > CH₃CH₂• > CH₃•
- CH₃CH₂• > (CH₃)₃C• > (CH₃)₂CH• > CH₃•
- CH₃• > (CH₃)₃C• > (CH₃)₂CH• > CH₃CH₂•
Answer: (CH₃)₃C• > (CH₃)₂CH• > CH₃CH₂• > CH₃•
Alkyl radicals, like carbocations, are stabilised by hyperconjugation and the inductive effect; tertiary radicals are more stable than secondary, which are more stable than primary, and methyl radicals are the least stable.
12. Which of the following carbocations is the most stable?
- CH₃⁺
- CH₃CH₂⁺
- (CH₃)₂CH⁺
- (CH₃)₃C⁺
Answer: (CH₃)₃C⁺
The stability of carbocations increases with the number of alkyl substituents due to hyperconjugation and the inductive effect; tertiary carbocations are the most stable, followed by secondary, primary, and finally methyl.