Lecture Notes on UV Spectroscopy Problems

Introduction

• Focus on problems from previous year papers related to UV spectroscopy.
• Aim: Address all types of UV spectroscopy questions.

Key Problems Discussed

Problem 1: UV Spectrum of Acetone

• Peaks at 279 nm and 189 nm.
• Identify electronic transitions:
  • **n→π">*</sup> transition at 279 nm
  • **π→π">*</sup> transition at 189 nm.
• Molecular orbitals energy levels: σ < π < n (non-bonding) < π"></sup> < σ"></sup>
• Energy inversely proportional to wavelength.

Problem 2: Compounds Absorbing UV Radiation

• UV active compounds: Peaks from 200 nm to 800 nm.
• Heptane: Only σ→σ">*</sup> transition, not UV active.
• Ethanol, Cyclohexane: Similar to heptane, not UV active.
• Conjugated compounds (e.g., Benzene, Butadiene) show π→π">*</sup> transitions and absorb UV.

Problem 3: Aniline Absorption

• Aniline absorbs at 280 nm; shifts to 203 nm in acidic medium.
• Reason: Resonance of lone pair with benzene ring is lost in acidic medium (anilinium ion formation).
• Conjugation increases absorption wavelength.

Problem 4: Solvent Choice in UV Spectroscopy

• Solvent criteria:
  • Transparent in UV region (no peaks between 200-800 nm).
  • Less polar (minimal interaction with solute).
• Common solvents: 95% ethanol, water, hexane, cyclohexane.
• Avoid: Benzene, chloroform, carbon tetrachloride (interfering peaks).

Problem 5: Identifying Dienes

• Given values: 176 nm, 211 nm, 215 nm.
• Conjugation and isomer effects on absorption wavelength:
  • Isolated dienes absorb at lower wavelengths.
  • Cis and trans isomers differ in absorption due to conjugation length.

Problem 6: Isomeric Dienes

• Homoannular vs. Heteroannular vs. Isolated dienes.
• Compare lambda max values to distinguish.

Problem 7: Compound Pair Comparison

• Ethyl benzene vs. Styrene: Styrene absorbs at longer wavelength (extended conjugation).
• Cis vs. Trans 1,3-pentadiene: Trans absorbs longer due to more effective conjugation.

Problem 8: Organic Compound C5H8

• Yields n-pentane upon hydrogenation.
• UV spectrum at 176 nm suggests isolated diene or simple alkyne.

Problem 9: Wavelength Order

• Based on conjugation length and number of pi bonds:
  • C (non-conjugated), B (conjugated), A (most conjugated).

Problem 10: Methyl Halides

• **n→σ">*</sup> transition shifts to longer wavelengths from methyl chloride to iodide.
• Reason: Electronegativity affects binding and energy required.*

Problem 11: Lambda Max Calculation

• Use Woodward-Fieser rules:
  • Acyclic conjugated dienes.
  • Alpha, beta unsaturated ketones.
  • Adjust for ring residues, exocyclic bonds, and conjugation.

Conclusion

• Comprehensive coverage of different UV spectroscopy problems.
• Emphasizes understanding transitions, conjugation effects, and solvent choices.
• Encouragement to review both the theory and problem-solving aspects for exams.