Questions & explanations
1. The heat of hydrogenation of ethene is -137 kJ/mol and that of 2,3-dimethylbut-2-ene is -110 kJ/mol. What is the approximate heat of hydrogenation of cis-2-butene?
- -120 kJ/mol
- -126 kJ/mol
- -115 kJ/mol
- -110 kJ/mol
Answer: -126 kJ/mol
Heat of hydrogenation becomes less negative with increasing alkyl substitution due to hyperconjugation. Ethene (0 alkyl) = -137, tetrasubstituted (4 alkyl) = -110. The difference per alkyl group is (137-110)/4 ≈ 6.75 kJ/mol. cis-2-butene is disubstituted (2 alkyl), so heat = -137 + 2×6.75 ≈ -123.5, closest to -126 kJ/mol.
2. Which statement about cis-2-butene and trans-2-butene is correct?
- cis-2-butene has a higher boiling point and a higher melting point than trans-2-butene
- cis-2-butene has a lower boiling point and a lower melting point than trans-2-butene
- cis-2-butene has a higher boiling point but a lower melting point than trans-2-butene
- cis-2-butene has a lower boiling point but a higher melting point than trans-2-butene
Answer: cis-2-butene has a higher boiling point but a lower melting point than trans-2-butene
cis-2-butene has a net dipole due to substituents on the same side, leading to stronger dipole-dipole interactions and a higher boiling point (3.7°C vs 0.9°C). However, trans-2-butene is more symmetric and packs better in the crystal lattice, giving it a higher melting point (-106°C vs -139°C).
3. The major product formed when 3-methylbut-1-ene is treated with HCl is:
- 2-chloro-3-methylbutane
- 1-chloro-2-methylbutane
- 1-chloro-3-methylbutane
- 2-chloro-2-methylbutane
Answer: 2-chloro-2-methylbutane
Addition of HCl to 3-methylbut-1-ene initially gives a secondary carbocation at C2. This carbocation undergoes a 1,2-hydride shift from C3 to form a more stable tertiary carbocation at C3. Chloride then attacks the tertiary carbocation, yielding 2-chloro-2-methylbutane as the major product.
4. What is the product when cis-2-butene is treated with cold dilute alkaline KMnO4?
- meso-2,3-butanediol
- racemic-2,3-butanediol
- 2,3-butanedione
- butane-2,3-diol (no stereochemistry specified)
Answer: meso-2,3-butanediol
Cold dilute alkaline KMnO4 (Baeyer's reagent) adds two OH groups syn to the double bond via a cyclic manganate ester. For cis-2-butene, syn addition gives the meso compound because the two chiral centers have opposite configurations (R,S) and the molecule has a plane of symmetry.
5. Starting from ethyne as the only carbon source, which sequence of reagents gives cis-2-butene?
- NaNH2, CH3Br, NaNH2, CH3Br, Na/liq NH3
- NaNH2, CH3Br, H2/Lindlar, NaNH2, CH3Br
- NaNH2, CH3Br, NaNH2, CH3Br, H2/Lindlar
- NaNH2, CH3Br, NaNH2, CH3Br, H2/Pt
Answer: NaNH2, CH3Br, NaNH2, CH3Br, H2/Lindlar
Ethyne is deprotonated by NaNH2 to form sodium acetylide, which undergoes SN2 with CH3Br to give propyne. Repeating gives but-2-yne. Lindlar's catalyst (H2/Pd poisoned) reduces the triple bond to cis-alkene. Thus, the correct sequence is NaNH2, CH3Br, NaNH2, CH3Br, H2/Lindlar.
6. Bromination of trans-2-butene in CCl4 gives which product?
- trans-2,3-dibromobutane
- racemic-2,3-dibromobutane
- cis-2,3-dibromobutane
- meso-2,3-dibromobutane
Answer: meso-2,3-dibromobutane
Bromination proceeds via anti addition through a bromonium ion. trans-2-Butene has methyl groups on opposite sides. Anti addition places the two Br atoms on opposite sides, giving a molecule with two chiral centres of opposite configuration, which is meso (optically inactive).
7. A hydrocarbon A (C5H10) decolourises Br2/CCl4 and Baeyer's reagent, gives no precipitate with ammoniacal AgNO3, and on reductive ozonolysis yields only acetone and acetaldehyde. Identify A.
- pent-1-ene
- 2-methylbut-2-ene
- pent-2-ene
- 2-methylbut-1-ene
Answer: 2-methylbut-2-ene
Degree of unsaturation = 1 indicates an alkene. Reductive ozonolysis cleaves the double bond, giving carbonyl compounds. Acetone (CH3COCH3) and acetaldehyde (CH3CHO) come from (CH3)2C=CHCH3, i.e., 2-methylbut-2-ene. Joining the carbonyl carbons yields the alkene structure.
8. Which of the following reagents gives anti-Markovnikov addition to propene?
- HCl in the presence of peroxide
- HBr in the presence of peroxide
- HI in the presence of peroxide
- H₂O in the presence of acid
Answer: HBr in the presence of peroxide
Only HBr shows the peroxide effect because the H–Br bond is weak enough to be broken by radicals, and the bromine radical adds to the less substituted carbon, leading to anti-Markovnikov product. HCl bond is too strong, and HI bond is too weak (iodine atoms recombine).
9. What is the stereochemical outcome of adding Br2 to cis-2-butene?
- racemic mixture of (2R,3R)- and (2S,3S)-2,3-dibromobutane
- meso-2,3-dibromobutane
- a single enantiomer of 2,3-dibromobutane
- a mixture of meso and racemic 2,3-dibromobutane
Answer: a single enantiomer of 2,3-dibromobutane
Br2 adds via anti addition through a bromonium ion intermediate. For cis-2-butene, anti addition places the two Br atoms on opposite sides of the original double bond, giving a chiral molecule. The product is a racemic mixture of (2R,3R)- and (2S,3S)-2,3-dibromobutane.
10. Which product is formed when propene reacts with HBr in the presence of benzoyl peroxide?
- 2-bromopropane
- 1-bromopropane
- 1,2-dibromopropane
- 2,2-dibromopropane
Answer: 1-bromopropane
Benzoyl peroxide initiates a radical chain mechanism. A bromine radical adds to the less substituted carbon (C1) of propene, forming a more stable secondary radical on C2. This radical abstracts a hydrogen from HBr, yielding 1-bromopropane (anti-Markovnikov product).
11. How many structural isomers (including cycloalkanes) are possible for the molecular formula C4H8?
- 3
- 4
- 5
- 6
Answer: 4
For C4H8, structural isomers include acyclic alkenes and cycloalkanes. Acyclic alkenes: but-1-ene, but-2-ene (cis/trans are stereoisomers, not structural), and 2-methylpropene. Cycloalkanes: cyclobutane and methylcyclopropane. Total distinct structural isomers = 4.
12. An unknown alkene on reductive ozonolysis gives acetone and propanal. What is the alkene?
- 3-methylpent-1-ene
- 3-methylpent-2-ene
- 2-methylpent-1-ene
- 2-methylpent-2-ene
Answer: 3-methylpent-2-ene
Reductive ozonolysis cleaves the C=C bond to give carbonyl compounds. Acetone (CH3)2C=O indicates a disubstituted carbon, and propanal CH3CH2CHO indicates a terminal carbon. Joining the carbonyl carbons gives (CH3)2C=CH-CH2-CH3, which is 3-methylpent-2-ene.