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CBSE · Class 12 · Chemistry

Haloalkanes and Haloarenes

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The C–X bond in haloalkanes is polar because:
  1. both atoms have equal electronegativity
  2. the bond is ionic
  3. carbon is more electronegative
  4. the halogen is more electronegative than carbon
2.
The IUPAC name of (CH3)2CHI is:
  1. 2-iodopropane
  2. 2-iodobutane
  3. 2-iodo-2-methylpropane
  4. 1-iodobutane
3.
2-chloropropane ((CH3)2CHCl) is a:
  1. tertiary (3°) alkyl halide
  2. primary (1°) alkyl halide
  3. vinylic halide
  4. secondary (2°) alkyl halide
4.
Of chloroethane and 2-chloropropane, which reacts faster by the SN2 mechanism?
  1. chloroethane
  2. 2-chloropropane
  3. both react at the same rate
  4. neither reacts
5.
2-bromopropane reacts with AgCN. The main product is:
  1. the isocyanide (CH3)2CHNC
  2. 2,3-dimethylbutane
  3. propan-2-ol
  4. the nitrile (CH3)2CHCN
6.
The correct order of boiling points is:
  1. 2-chlorobutane < 1-bromopropane < 1-chlorobutane
  2. 1-chlorobutane < 1-bromopropane < 2-chlorobutane
  3. 2-chlorobutane < 1-chlorobutane < 1-bromopropane
  4. 1-bromopropane < 1-chlorobutane < 2-chlorobutane
7.
The percentage of iodine by mass in CH3CH2CH2I is: (C = 12, H = 1, Cl = 35.5, Br = 80, I = 127)
  1. 127%
  2. 25.29%
  3. 74.71%
  4. 42.76%
8.
Assertion (A): 1-bromopropane reacts faster than 2-bromopropane by the SN1 mechanism.
Reason (R): SN1 reactions go through a carbocation, and more substituted carbocations are more stable.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Why are haloalkanes only slightly soluble in water, though they are polar?
10.
Give the IUPAC name and the common name of CH3CH2Br, and classify it as primary, secondary or tertiary.
11.
Name the major alkene formed when 2-chlorobutane is heated with alcoholic KOH. Which rule decides it?
12.
Find the molar mass of 2-chloro-2-methylpropane ((CH3)3CCl) and the percentage of halogen in it. (C = 12, H = 1, Cl = 35.5, Br = 80, I = 127)
13.
Arrange in increasing order of boiling point, and give a reason: 1-chloropropane, 1-bromopropane, 2-bromopropane.
14.
What is formed when 1-bromobutane is treated with sodium in dry ether? Name the reaction.
15.
What is formed when propene reacts with HBr with a peroxide? Name the reaction.
16.
165.6 g of 2-iodo-2-methylpropane is completely hydrolysed by aqueous KOH. What mass of 2-methylpropan-2-ol is formed? (C = 12, H = 1, O = 16, Cl = 35.5, Br = 80, I = 127)
Section C — Word problems4 × 4 = 16 marks
17.
Explain the SN2 mechanism with an example, and say how the structure of the halide affects its rate.
18.
Identify the compounds in: propene —HBr→ A —alc. KOH→ B —HBr, peroxide→ C.
19.
Arrange 2-iodopropane, 1-iodo-2-methylpropane, iodoethane in order of decreasing reactivity by (a) SN2 and (b) SN1, and explain each order.
20.
63.9 g of iodomethane is heated with sodium in dry ether. Write the equation, name the product, and find its mass if the yield is 60%. (C = 12, H = 1, Cl = 35.5, Br = 80, I = 127)
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Chemistry

Answer key

Haloalkanes and Haloarenes · AW-12C-HAL-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) the halogen is more electronegative than carbon—
2(a) 2-iodopropanecommon name: isopropyl iodide
3(d) secondary (2°) alkyl halidecount the carbons attached to the carbon bearing the halogen
4(a) chloroethaneSN2 needs backside attack: less crowding is faster, 1° > 2° > 3°
5(a) the isocyanide (CH3)2CHNC—
6(a) 2-chlorobutane < 1-bromopropane < 1-chlorobutaneboiling points: 2-chlorobutane 68.3 °C, 1-bromopropane 71 °C, 1-chlorobutane 78.5 °C
7(c) 74.71%M = 170; 127 ÷ 170 × 100
8(d) A is false, but R is true.A is false: 2-bromopropane gives the more substituted, more stable carbocation. R is true.
9To dissolve, energy is needed to break the hydrogen bonds between water molecules, and haloalkanes cannot form hydrogen bonds with water to release as much energy.—
10bromoethane; ethyl bromide; primary (1°)count the carbons on the C–X carbon
11but-2-ene; Saytzeff (Zaitsev) rule: the more substituted alkene is the major productdehydrohalogenation (β-elimination)
1292.5 g/mol; 38.38%C4H9Cl
131-chloropropane < 2-bromopropane < 1-bromopropaneboiling point rises with molar mass and falls with branching; data: 46.6 °C, 71 °C, 59.4 °C
14octane; Wurtz reaction2R–X + 2Na → R–R + 2NaX
151-bromopropane; anti-Markovnikov (peroxide) addition—
1666.6 g0.9 mol of halide gives 0.9 mol of alcohol (M = 74)
17In SN2 the nucleophile attacks the carbon from the side opposite the leaving group in one step through a five-coordinate transition state; the configuration is inverted. E.g. CH3Cl + OH− → CH3OH + Cl−, rate = k[CH3Cl][OH−]. Bulky groups block the backside, so the order is CH3X > 1° > 2° > 3°.—
18A = 2-bromopropane, B = propene, C = 1-bromopropanefollow each reagent in turn
19(a) SN2: iodoethane > 1-iodo-2-methylpropane > 2-iodopropane, since backside attack is hindered by bulky groups; (b) SN1: 2-iodopropane > 1-iodo-2-methylpropane > iodoethane, since tertiary carbocations are most stable (halides of the same class are ranked by size)1° > 2° > 3° for SN2; 3° > 2° > 1° for SN1
202CH3I + 2Na → C2H6 + 2NaI; ethane; 4.05 g2 mol of halide give 1 mol of alkane (M = 30)