Chapter-6 Haloalkanes and Haloarenes — Online MCQ Test
CHEMISTRY · CLASS 12 SECOND PUC · Karnataka State Board
Practice Chapter-6 Haloalkanes and Haloarenes with a free chapter-wise online MCQ test.
This chapter covers: nomenclature - SN1 mechanism - SN2 mechanism - chemical reactions.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter-6 Haloalkanes and Haloarenes — Important Questions & Answers
What is the IUPAC name of CH₃-CHBr-CH₂-CH₃?
- A. 2-bromobutane
- B. 1-bromobutane
- C. 3-bromobutane
- D. 2-bromopropane
Answer: A. 2-bromobutane
The longest carbon chain has 4 atoms, and Br is at position 2 when numbering from the end nearest to the halogen.
The longest carbon chain has 4 atoms, and Br is at position 2 when numbering from the end nearest to the halogen.
Which of the following is a haloalkane?
- A. Chlorobenzene
- B. Chloroethane
- C. Bromobenzene
- D. Iodobenzene
Answer: B. Chloroethane
Haloalkanes contain halogen bonded to sp³ hybridized carbon. Chloroethane (C₂H₅Cl) is a haloalkane, while others are haloarenes.
Haloalkanes contain halogen bonded to sp³ hybridized carbon. Chloroethane (C₂H₅Cl) is a haloalkane, while others are haloarenes.
Which factor does NOT affect the rate of SN2 reaction?
- A. Concentration of nucleophile
- B. Nature of solvent
- C. Temperature
- D. Concentration of leaving group only
Answer: D. Concentration of leaving group only
SN2 rate depends on both substrate and nucleophile concentrations. Leaving group concentration alone doesn't significantly affect rate in typical conditions.
SN2 rate depends on both substrate and nucleophile concentrations. Leaving group concentration alone doesn't significantly affect rate in typical conditions.
Why does 1-bromo-2-methylpropane NOT undergo SN2 reaction readily?
- A. Because the carbon bearing Br is primary
- B. Because the nearby methyl group causes severe steric hindrance (neopentyl system)
- C. Because Br is a poor leaving group
- D. Because primary carbocations are unstable
Answer: B. Because the nearby methyl group causes severe steric hindrance (neopentyl system)
Despite being primary, the neopentyl structure has a bulky tert-butyl-like hindrance from the adjacent isopropyl group, blocking nucleophilic backside attack.
Despite being primary, the neopentyl structure has a bulky tert-butyl-like hindrance from the adjacent isopropyl group, blocking nucleophilic backside attack.
In the reaction of optically active (S)-2-iodooctane with KCN in DMSO via SN2 mechanism, what is the stereochemical outcome?
- A. Retention of S configuration
- B. Inversion to R configuration with high optical purity
- C. Racemic mixture of R and S
- D. S configuration is maintained with loss of optical activity
Answer: B. Inversion to R configuration with high optical purity
SN2 proceeds with Walden inversion. (S)-2-iodooctane inverts to give (R)-2-cyanooctane with high stereochemical purity.
SN2 proceeds with Walden inversion. (S)-2-iodooctane inverts to give (R)-2-cyanooctane with high stereochemical purity.