IR Spectroscopy and Mass Spectrometry: Quiz

IR Spectroscopy and Mass Spectrometry: Quiz

7 questions

1. Question 1EasyA strong, sharp IR absorption at 1715 cm⁻¹ indicates which bond?

   AO–HBC–HCC=ODC–O
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   Answer: C=O

   C=O bonds absorb strongly at 1680–1750 cm⁻¹ (aldehydes, ketones, acids, esters).

2. Question 2EasyAn IR spectrum shows a broad absorption at 3350 cm⁻¹ and none near 1700 cm⁻¹. The compound is most likely:

   Aa ketoneBan alcoholCa carboxylic acidDan alkane
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   Answer: an alcohol

   A broad band at 3200–3550 cm⁻¹ is an alcohol O–H. No C=O rules out ketones and acids.

3. Question 3EasyIn a mass spectrum, the peak at the highest m/z (ignoring tiny isotope peaks) is the:

   Abase peakBmolecular ionCfragment ionDfingerprint
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   Answer: molecular ion

   The molecular ion M⁺ is the whole molecule minus one electron; its m/z gives the molar mass. The base peak is the tallest peak.

4. Question 4MediumA compound shows peaks at m/z 78 and 80 in a 3 : 1 ratio. What does it contain?

   AOne bromine atomBOne chlorine atomCTwo oxygen atomsDA benzene ring
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   Answer: One chlorine atom

   ³⁵Cl and ³⁷Cl occur in about a 3 : 1 ratio, giving M and M+2 peaks 2 units apart in that ratio. One Br would give 1 : 1.

5. Question 5MediumPropanone (M = 58) shows its base peak at m/z 43. Which fragment is lost?

   ACH₃ (15)BO (16)CC₂H₅ (29)DCO (28)
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   Answer: CH₃ (15)

   58 − 43 = 15: loss of a methyl group leaves CH₃CO⁺ at m/z 43.

6. Question 6HardWhat is the wavelength of IR radiation with a wavenumber of 3300 cm⁻¹?

   A3.03 μmB3300 μmC0.303 μmD30.3 μm
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   Answer: 3.03 μm

   λ = 1/3300 cm⁻¹ = 3.03 × 10⁻⁴ cm = 3.03 μm (1 cm = 10⁴ μm).

7. Question 7MediumWhich pair of IR absorptions would best identify a carboxylic acid?

   ABroad 3200–3550 cm⁻¹ and 1050 cm⁻¹BVery broad 2500–3300 cm⁻¹ and strong 1700–1725 cm⁻¹CSharp 2250 cm⁻¹ onlyD1650 cm⁻¹ and 3050 cm⁻¹
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   Answer: Very broad 2500–3300 cm⁻¹ and strong 1700–1725 cm⁻¹

   A carboxylic acid has both an O–H (very broad, overlapping the C–H region) and a C=O. The first pair is an alcohol; 2250 is C≡N; 1650 + 3050 suggests an alkene.