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UPSC Combined Geo-Scientist

Chemist

  • 3 descriptive papers
  • 200 marks each
  • 3 h per paper

Mains Paper III covers analytical and organic chemistry.

UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper III

Q10(a). Structure from IR, ¹H-NMR, ¹³C-NMR and MS

  • 10 marks
  • Topic: Spectroscopy
  • Answered
Question

An organic compound having molecular formula C₁₀H₁₆O₄ gave the following spectral data. IR: significant absorption bands at 2904, 1725, 1641, 1244 cm⁻¹. ¹H-NMR (400 MHz, CDCl₃): δ 4.23 (q, J=7.2 Hz, 4H), 2.06 (s, 6H) and 1.29 (t, J=7.2 Hz, 6H). ¹³C-NMR (100 MHz, CDCl₃): δ 165.7, 154.9, 124.8, 60.8, 23.1 and 14.1. EI-MS: prominent peaks at m/z 99 and 155. Deduce the structure of the compound and explain the data.

Answer in brief

Diethyl isopropylidenemalonate, (CH₃)₂C=C(CO₂CH₂CH₃)₂ (diethyl 2-(propan-2-ylidene)propanedioate), M 200.

Key points

  • C₁₀H₁₆O₄: DBE=(2(10)+2-16)/(2)=3: two C=O and one C=C (no rings; IR 1641 C=C, 1725 C=O).
  • 2904 (sp³ C–H); 1725 (conjugated ester C=O, lower than the saturated 1740); 1641 (C=C, tetrasubstituted, conjugated); 1244 (C–O of the ester).
  • ¹H: 4.23 (q, 4H) and 1.29 (t, 6H): two equivalent OCH₂CH₃ groups (ethyl esters); 2.06 (s, 6H): two equivalent vinylic CH₃ groups on a C=C with no neighbouring H. No other H: 4 + 6 + 6 = 16.
  • ¹³C (six signals for ten carbons: molecule symmetric): 165.7 (C=O ester, 2 C); 154.9 ((CH₃)₂C=, the β-carbon, deshielded by conjugation); 124.8 (=C(CO₂Et)₂, the α-carbon); 60.8 (OCH₂); 23.1 (vinylic CH₃); 14.1 (OCH₂CH₃).

The full model answer

Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.

  1. Unsaturation: C₁₀H₁₆O₄: DBE=(2(10)+2-16)/(2)=3: two C=O and one C=C (no rings; IR 1641 C=C, 1725 C=O).
  2. IR: 2904 (sp³ C–H); 1725 (conjugated ester C=O, lower than the saturated 1740); 1641 (C=C, tetrasubstituted, conjugated); 1244 (C–O of the ester).
  3. ¹H: 4.23 (q, 4H) and 1.29 (t, 6H): two equivalent OCH₂CH₃ groups (ethyl esters); 2.06 (s, 6H): two equivalent vinylic CH₃ groups on a C=C with no neighbouring H. No other H: 4 + 6 + 6 = 16.
  4. ¹³C (six signals for ten carbons: molecule symmetric): 165.7 (C=O ester, 2 C); 154.9 ((CH₃)₂C=, the β-carbon, deshielded by conjugation); 124.8 (=C(CO₂Et)₂, the α-carbon); 60.8 (OCH₂); 23.1 (vinylic CH₃); 14.1 (OCH₂CH₃).
  5. Structure: (CH₃)₂C=C(CO₂Et)₂. Fragment count: (CH₃)₂C (3 C), =C (1 C), 2 CO₂Et (6 C) = 10 C.
  6. MS: M^(+·) 200; m/z 155 =M-OC₂H₅ (45), the acylium ion; m/z 99: from 155 by loss of CO (127) and then C₂H₄ (28), 155-28-28=99.

ethyl esters of alkylidene malonates are Knoevenagel products (acetone + diethyl malonate) and are good Michael acceptors for building quaternary centres.

Model answer to Q10(a), 2026 Mains Paper III: solution page 27
Solution page 27 of 30
Read the whole 2026 solution2026 question paper

Coversstructure determinationStructure from IR¹H-NMR

All parts of Question 10