Chapter-6 Haloalkanes and Haloarenes — Online MCQ Test
CHEMISTRY · Grade 12 · CBSE(NCERT)
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.
The formula of a haloalkane with 3 carbon atoms and one chlorine is:
- A. C₃H₇Cl
- B. C₃H₆Cl
- C. C₃H₈Cl
- D. C₃H₅Cl
Answer: A. C₃H₇Cl
For alkane CₙH₂ₙ₊₂, a 3-carbon alkane is C₃H₈. Replacing one H with Cl gives C₃H₇Cl.
For alkane CₙH₂ₙ₊₂, a 3-carbon alkane is C₃H₈. Replacing one H with Cl gives C₃H₇Cl.
What is the correct IUPAC name for (CH₃)₃C-Br?
- A. 3-bromo-3-methylpropane
- B. 2-bromo-2-methylpropane
- C. 1-bromo-1-methylpropane
- D. tert-butyl bromide
Answer: B. 2-bromo-2-methylpropane
The structure is a 3-carbon chain with Br and CH₃ both at position 2 (the central carbon), giving 2-bromo-2-methylpropane.
The structure is a 3-carbon chain with Br and CH₃ both at position 2 (the central carbon), giving 2-bromo-2-methylpropane.
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.
In aqueous solution, (CH₃)₃C-Br undergoes hydrolysis to form (CH₃)₃C-OH. Which mechanism operates here?
- A. SN2
- B. SN1
- C. E1
- D. E2
Answer: B. SN1
Tertiary haloalkane undergoes hydrolysis via SN1 mechanism because tertiary carbocation is relatively stable.
Tertiary haloalkane undergoes hydrolysis via SN1 mechanism because tertiary carbocation is relatively stable.
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.
Consider the reaction: R-Br + NaI → R-I + NaBr (Finkelstein reaction). Why is this reaction driven forward in acetone?
- A. NaI is more reactive than NaBr
- B. I⁻ is a better nucleophile than Br⁻ in acetone
- C. R-I is less soluble/NaBr is more soluble in acetone, shifting equilibrium
- D. Both a and b
Answer: C. R-I is less soluble/NaBr is more soluble in acetone, shifting equilibrium
NaBr is more soluble in acetone and precipitates out, while R-I remains soluble, driving the SN2 reaction forward.
NaBr is more soluble in acetone and precipitates out, while R-I remains soluble, driving the SN2 reaction forward.
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.
Compound A (sec-butyl chloride) is dissolved in water containing ¹⁸OH. After the reaction, the ¹⁸O label is found in the product. What mechanism predominantly operates, and what is the evidence?
- A. SN2, because ¹⁸O directly attacks from backside
- B. SN1, because ¹⁸O can attack both faces of carbocation intermediate equally
- C. E1, because labeled water attacks the carbocation π-system
- D. SN2, because water solvates the leaving group
Answer: B. SN1, because ¹⁸O can attack both faces of carbocation intermediate equally
Presence of ¹⁸O in product indicates nucleophilic water attack. In aqueous solution with secondary halide, SN1 carbocation pathway operates with water attacking from both faces.
Presence of ¹⁸O in product indicates nucleophilic water attack. In aqueous solution with secondary halide, SN1 carbocation pathway operates with water attacking from both faces.