Questions & explanations
1. For (R)-2-bromobutane, what is the correct priority order of substituents according to CIP rules?
- Br > CH2CH3 > CH3 > H
- Br > CH3 > CH2CH3 > H
- CH2CH3 > Br > CH3 > H
- Br > CH2CH3 > H > CH3
Answer: Br > CH2CH3 > CH3 > H
According to CIP rules, priority is based on atomic number of the atom directly attached to the chiral centre. Bromine (atomic number 35) gets highest priority. Next, compare the carbon atoms: CH2CH3 has a carbon attached to two carbons, while CH3 has a carbon attached to one carbon, so CH2CH3 > CH3. Hydrogen (atomic number 1) is lowest. Thus, the correct order is Br > CH2CH3 > CH3 > H.
2. Which pair of compounds are ring-chain isomers?
- Butane and 2-methylpropane
- Benzene and cyclohexane
- Ethanol and dimethyl ether
- Propene and cyclopropane
Answer: Propene and cyclopropane
Ring-chain isomers have the same molecular formula but differ in having a cyclic or open-chain structure. Propene (CH3-CH=CH2) and cyclopropane (C3H6) both have the formula C3H6, with propene being open-chain and cyclopropane cyclic. Benzene and cyclohexane have different formulas; ethanol and dimethyl ether are functional group isomers; butane and 2-methylpropane are chain isomers.
3. How many distinct compounds (including stereoisomers) exist for the molecular formula C4H8?
- 5
- 6
- 7
- 8
Answer: 6
Degree of unsaturation = 1, so isomers have one double bond or one ring. Structural isomers: 1-butene, 2-butene (cis/trans), isobutene, cyclobutane, methylcyclopropane. 2-butene has two stereoisomers (cis and trans), but they are counted as distinct compounds. Total distinct compounds = 6 (1-butene, cis-2-butene, trans-2-butene, isobutene, cyclobutane, methylcyclopropane).
4. Which of the following will show optical activity?
- 2-propanol
- Racemic mixture of 2-butanol
- Meso-tartaric acid
- 2-butanol
Answer: 2-butanol
Optical activity is shown by chiral substances that rotate plane-polarized light. 2-butanol has a chiral carbon (C2 with OH, H, CH3, and C2H5) and no internal compensation, so it is optically active. Achiral molecules (2-propanol), racemic mixtures (equal amounts of enantiomers cancel rotation), and meso compounds (internal compensation) are optically inactive.
5. If the specific rotation of (R)-2-butanol is +13.9°, what is the specific rotation of (S)-2-butanol?
- -13.9°
- 0°
- +13.9°
- Cannot be determined from R/S configuration
Answer: -13.9°
Enantiomers rotate plane-polarized light by equal magnitude but in opposite directions. Since (R)-2-butanol is dextrorotatory (+13.9°), its enantiomer (S)-2-butanol must be levorotatory with the same magnitude, i.e., -13.9°. The sign of rotation is not directly linked to R/S configuration, but here the specific rotation is given, so the opposite sign is known.
6. How many structural isomers of C5H12O are possible, considering only saturated acyclic alcohols and ethers?
- 14
- 12
- 16
- 10
Answer: 14
Degree of unsaturation = 0, so only saturated acyclic structures. Alcohols: 8 (n-pentanol, 2-pentanol, 3-pentanol, 2-methyl-1-butanol, 3-methyl-1-butanol, 2-methyl-2-butanol, 3-methyl-2-butanol, 2,2-dimethyl-1-propanol). Ethers: 6 (methyl n-butyl, methyl isobutyl, methyl sec-butyl, methyl tert-butyl, ethyl n-propyl, ethyl isopropyl). Total = 14.
7. Which of the following alkenes shows cis-trans isomerism?
- 2-butene
- 1-butene
- 2-methyl-2-butene
- ethene
Answer: 2-butene
Cis-trans isomerism requires each sp² carbon to have two different groups. In 2-butene, each sp² carbon has one H and one CH₃, so it shows cis-trans. 1-butene has a terminal CH₂= group (two H on one carbon), 2-methyl-2-butene has a carbon with two methyl groups, and ethene has two H on each carbon — none satisfy the condition.
8. Which of the following Fischer projections represents the same molecule as the wedge-dash structure with CHO at top, CH2OH at bottom, OH on a wedge (towards viewer), and H on a dash (away)?
- Fischer with CHO at top, CH2OH at bottom, OH on left, H on right, but rotated 90° clockwise
- Fischer with CHO at top, CH2OH at bottom, OH on right, H on left
- Fischer with CHO at top, CH2OH at bottom, OH on left, H on right
- Fischer with CHO at top, CH2OH at bottom, OH on right, H on left, but rotated 90° anticlockwise
Answer: Fischer with CHO at top, CH2OH at bottom, OH on left, H on right
The wedge-dash shows OH towards viewer and H away. In Fischer projection, horizontal bonds come towards viewer, vertical bonds go away. So OH (towards) must be on a horizontal bond, and H (away) on a vertical bond. With CHO at top and CH2OH at bottom, the correct Fischer has OH on left and H on right. Option a matches this.
9. Which isomer of 1-tert-butyl-4-methylcyclohexane is more stable?
- cis-1-tert-butyl-4-methylcyclohexane
- Neither is stable; the compound exists only as a chair conformation
- Both are equally stable
- trans-1-tert-butyl-4-methylcyclohexane
Answer: trans-1-tert-butyl-4-methylcyclohexane
In trans-1-tert-butyl-4-methylcyclohexane, both substituents can be equatorial in the most stable chair conformation. The large A-value of tert-butyl (21 kJ/mol) strongly favors equatorial position. In cis isomer, one substituent must be axial, causing significant 1,3-diaxial strain. Hence trans is more stable.
10. Which of the following alkenes shows geometrical isomerism?
- 1-Butene
- Ethene
- 2-Methyl-2-butene
- 2-Butene
Answer: 2-Butene
Geometrical isomerism requires each sp² carbon to have two different groups. In 2-butene, each sp² carbon has one methyl and one hydrogen, so cis and trans isomers exist. 1-Butene has a CH₂ group with two identical H, ethene has two H on each carbon, and 2-methyl-2-butene has one carbon with two methyl groups.
11. Which of the following compounds shows keto-enol tautomerism?
- Benzaldehyde
- Acetic acid
- Acetone
- Chloroform
Answer: Acetone
Keto-enol tautomerism requires a carbonyl group with at least one alpha-hydrogen. Acetone (CH3COCH3) has alpha-hydrogens on both sides of the carbonyl, so it can tautomerize to its enol form (propen-2-ol). Benzaldehyde lacks alpha-hydrogens, acetic acid's enol is destabilized, and chloroform has no carbonyl.
12. Which statement about metamerism is correct?
- Metamers have the same molecular formula but different functional groups.
- Metamers have the same molecular formula and same functional group but differ in alkyl groups attached to a divalent atom.
- Metamers have different molecular formulas but the same functional group.
- Metamers have the same molecular formula and same carbon skeleton but differ in the position of a substituent.
Answer: Metamers have the same molecular formula and same functional group but differ in alkyl groups attached to a divalent atom.
Metamerism is a type of structural isomerism where isomers have the same molecular formula and the same functional group, but differ in the alkyl groups attached to a divalent atom (e.g., –O–, –S–, –NH–). For example, diethyl ether (C2H5–O–C2H5) and methyl propyl ether (CH3–O–C3H7) are metamers.