IR spectral region check: In which wavenumber range does a typical carbonyl (C=O) stretching band appear?

Difficulty: Easy

Correct Answer: 1870–1650 cm^-1

Explanation:


Introduction / Context:
The carbonyl stretch is one of the most diagnostic features in an IR spectrum. Recognizing its typical range allows rapid identification of aldehydes, ketones, esters, acids, amides, and acyl derivatives with only minor shifts due to conjugation or ring strain.



Given Data / Assumptions:

  • We are considering fundamental C=O stretching vibrations.
  • Conjugation, hydrogen bonding, and ring strain can shift the exact position but remain within a characteristic window.
  • Units are cm^-1 (wavenumbers), standard for IR.



Concept / Approach:
The C=O stretching band generally lies between about 1870 and 1650 cm^-1. Stronger carbonyls (acid chlorides, anhydrides) appear at higher end; amides and conjugated systems appear at lower end. Specific subranges (e.g., 1740–1720 cm^-1) may describe particular subclasses, but the accepted broad window is 1870–1650 cm^-1.



Step-by-Step Solution:
Recall canonical C=O range: approximately 1870–1650 cm^-1.Compare options: choose the broad, correct window rather than subclass-specific or unrelated regions.Select 1870–1650 cm^-1.



Verification / Alternative check:
Spectral tables list aldehydes/ketones ~1725–1705, esters ~1750–1735, acid chlorides ~1810–1790, amides ~1690–1630—all within the 1870–1650 window.



Why Other Options Are Wrong:

  • 1740–1720 cm^-1: Fits many esters/aldehydes but is too narrow.
  • 3640–3250 cm^-1: O–H/N–H stretching region.
  • 160–110 cm^-1: Far-IR region; not typical for C=O stretch.
  • 2350–2100 cm^-1: Cumulated systems (CO2 asymmetric stretch, C≡C/C≡N areas); not C=O.



Common Pitfalls:
Assuming one fixed value for all carbonyls; always consider substituent effects and conjugation.



Final Answer:
1870–1650 cm^-1.


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