Q7(c)(i). Compound A (M = 102) and its cleavage products B and C
A compound (A) containing C, H and O has molecular weight 102 and displays two signals in the ¹H NMR spectrum at δ 1.1 (d) and 3.55 (septet) in the integral ratio of 6 : 1. Treatment of A with 1 mole of HI gives rise to B and C. In the IR spectrum, B gives a strong absorption band at 3330 cm⁻¹, whereas its ¹H NMR spectrum shows signals at δ 1.05 (d, 6H), 3.6 (septet, 1H) and 4.4 (s, 1H, disappeared with D₂O). The ¹H NMR spectrum of C gives signals at δ 1.9 (d, 6H) and 4.25 (septet, 1H). Reaction of A with excess of HI gives only C. Identify A, B and C. Write all the reactions involved.
Model answer
The free solution for 2022 answers questions 1, 4, 5, 6 and 8, the ones picked in its time plan. Question 7 is answered in full in the Chemistry Paper II, 2021 to 2025 book, with every part of every paper.
Part of VKC Basic
Read the key points and the full model answer with VKC Basic
VKC Basic (₹999 for 6 months) opens every solved answer, the notes, the handbook and the daily answer archive. Every Mains test series includes it.
UPSC Civil Services (Main) Chemistry Paper II: Model Answers 2021 to 2025
- All eight questions of each paper, with the question printed above its answer
- Mechanisms drawn out and labelled on the scheme
- Topic-wise weightage and a topic-wise index
CoversNMR structure determinationisopropyl septetether cleavage with HIdiisopropyl etherD₂O exchange
All parts of Question 7
- 7(a) Secondary structures of proteins compared with tertiary structure15 marks
- 7(b)(i) Butanoyl chloride with 2 NaBH₄, then H₃O⁺10 marks
- 7(b)(ii) But-2-yne with sodium in liquid ammonia5 marks
- 7(c)(i) Compound A (M = 102) and its cleavage products B and C10 marks
- 7(c)(ii) Three isomeric carbonyl compounds C₅H₁₀O10 marks