Q66.When sec-butylcyclohexane reacts with bromine in the presence of sunlight, the major product is : (1) (2) (3) (4)
What This Question Tests
This question tests the understanding of free radical bromination, emphasizing the selectivity for the most stable radical intermediate (tertiary over secondary), leading to the major product.
Concepts Tested
📚 NCERT Sections This Tests
7.7 — Predict The Major Product Of Acid Catalysed Dehydration Of
Chemistry Class 12 · Chapter 7
7.7 Predict the major product of acid catalysed dehydration of (i) 1-methylcyclohexanol and (ii) butan-1-ol
7.6 — Give Structures Of The Products You Would Expect When Each Of The
Chemistry Class 12 · Chapter 7
7.6 Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl –ZnCl2 (b) HBr and (c) SOCl2. (i) Butan-1-ol (ii) 2-Methylbutan-2-ol
6.21 — Primary Alkyl Halide C4H9Br (A) Reacted With Alcoholic Koh To Give Compound (B).
Chemistry Class 12 · Chapter 6
6.21 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
📋 Question Details
- Chapter
- Hydrocarbons
- Topic
- Free radical halogenation
- Year
- 2025
- Shift
- 22 Jan Shift 2
- Q Number
- Q66
- Type
- MCQ
- NCERT Ref
- Class 11 Chemistry Ch 13: Hydrocarbons
More from this Chapter
Q62. The IUPAC name of is (1) 1, 1−diethyl −2, 2−dimethylpentane (2) 4, 4−dimethyl −5, 5−diethylpentane (3) 5, 5−diethyl −4, 4−diemthylpentane (4) 3−ethyl −4, 4−dimethylheptane
Q63.The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly (1) benzoyl chloride (2) benzyl chloride (3) o−and p−chlorotoluene (4) m−chlorotoluene
Q76.Which of the following reactions will yield 2, 2−dibromopropane? (1) CH3 −C ≡CH + 2HBr ⟶ (2) CH3CH ≡CHBr + HBr ⟶ (3) CH ≡CH + 2HBr ⟶ (4) CH3 −CH = CH2 + HBr ⟶
Q79.The compound formed as a result of oxidation of ethyl benzene by KMnO4 is (1) benzophenone (2) acetophenone (3) benzoic acid (4) benzyl alcohol