Q60. Zymase NaOI−−C6H12O6 → A → B + CHI3 Δ The number of carbon atoms present in the product B is JEE Main 2022 (29 Jun Shift 1) JEE Main Previous Year Paper
What This Question Tests
This question involves identifying the disproportionation product of manganate, determining the oxidation state and number of unpaired electrons in the higher oxidation state product, and calculating its spin-only magnetic moment.
Concepts Tested
Formulas Used
μ = √n(n+2) BM
📚 NCERT Sections This Tests
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.
9.9 — Give The Structures Of A, B And C In The Following Reactions:
Chemistry Class 12 · Chapter 9
9.9 Give the structures of A, B and C in the following reactions: B NaOH Br 2 C (i) CH 3 CH 2 I NaCN A Partial OHhydrolysis CuCN H 2 O/H NH 3 (ii) C 6 H 5 N 2 Cl A B C 4 B HNO 2 C (iii) CH 3 CH 2 Br KCN A LiAlH 0 C 2 2 O/H HCl B H C (iv) C6 H 5 NO 2 Fe/HCl A NaNO273 K 3 A NaOBr B NaNO 2 /HCl C (v) CH 3 COOH NH 2 B C 6 H 5 OH C (vi) C6 H 5 NO 2 Fe/HCl A HNO273K 279 Amines Reprint 2025-26
6.22 — What Happens When
Chemistry Class 12 · Chapter 6
6.22 What happens when (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride is treated with aqueous KOH, (v) methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN? Answers to Some Intext Questions 6.1 6.2 (i) H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding acid, HI which is then oxidised by it to I2. 6.3 (i) ClCH2CH2CH2Cl (ii) ClCH2CHClCH3 (iii) Cl2CHCH2CH3 (iv) CH3CCl2CH3 191 Haloalkanes and Haloarenes Reprint 2025-26
📋 Question Details
- Chapter
- d-block & f-block Elements
- Topic
- Manganese Compounds
- Year
- 2022
- Shift
- 29 Jun Shift 1
- Q Number
- Q60
- Type
- Numerical
- NCERT Ref
- Class 12 Chemistry Ch 8: d- and f-Block Elements
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