HNMR Practice 1

Determining the Structure of C₄H₈O from ¹H NMR and IR Spectra

The molecular formula is C₄H₈O, and the ¹H NMR spectrum shown below has a triplet at ~3.6 ppm and a triplet at ~1.7 ppm. Additionally, the IR spectrum does not have a peak at 1700 cm⁻¹, which is indicative of a carbonyl group (C=O). The structure is determined to be tetrahydrofuran (THF). Here's how we arrive at this conclusion:

Step 1: Analyze the Molecular Formula

The molecular formula is C₄H₈O.

Degree of Unsaturation (DU):

\[ \text{DU} = \frac{2C + 2 - H}{2} = \frac{2(4) + 2 - 8}{2} = 1 \]

A DU of 1 indicates either one ring or one double bond (C=C or C=O).

Step 2: Examine the IR Spectrum

The IR spectrum does not have a peak at 1700 cm⁻¹, which rules out the presence of a carbonyl group (C=O). This suggests that the oxygen atom in the molecule is likely part of an ether or alcohol functional group rather than a carbonyl.

Step 3: Examine the ¹H NMR Spectrum

The spectrum shows:

• A triplet at ~3.6 ppm.
• A triplet at ~1.7 ppm.

Key Observations:

• Chemical Shifts:
  • Protons near electronegative atoms (e.g., oxygen) are deshielded and appear downfield (higher ppm).
  • Protons in aliphatic chains are shielded and appear upfield (lower ppm).
• Splitting Patterns:
  • A triplet indicates that each set of protons is coupled to two equivalent neighboring protons (n+1 rule, where n = 2).

Step 4: Deduce the Structure

a) Triplet at ~3.6 ppm:

• This signal is downfield, suggesting the protons are near an oxygen atom (e.g., -O-CH₂-).
• The triplet indicates these protons are coupled to two equivalent neighboring protons.

b) Triplet at ~1.7 ppm:

• This signal is upfield, suggesting the protons are in an aliphatic environment (e.g., -CH₂-).
• The triplet indicates these protons are also coupled to two equivalent neighboring protons.

c) Combining the Information:

• The molecule must have a cyclic structure to satisfy the degree of unsaturation (DU = 1).
• The absence of a carbonyl peak in the IR spectrum and the presence of oxygen suggest an ether functional group.
• The coupling patterns and chemical shifts are consistent with a tetrahydrofuran (THF) structure:

  \[ \text{THF: } \text{O-CH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2 \]

• In THF:
  • The -O-CH₂- protons appear as a triplet at ~3.6 ppm (coupled to the adjacent -CH₂- protons).
  • The -CH₂- protons appear as a triplet at ~1.7 ppm (coupled to the adjacent -CH₂- protons).

Step 5: Verify the Structure

• Molecular Formula: THF (C₄H₈O) matches the given formula.
• NMR Spectrum:
  • The two triplets correspond to the two distinct proton environments in THF.
  • The integration ratio of the signals should be 2:2 (for -O-CH₂- and -CH₂-), which fits the spectrum.
• IR Spectrum:
  • The absence of a peak at 1700 cm⁻¹ confirms the lack of a carbonyl group, supporting the ether structure of THF.

Step 6: Conclusion

The compound is tetrahydrofuran (THF). The NMR spectrum is consistent with its structure, the molecular formula matches, and the IR spectrum rules out the presence of a carbonyl group.

Summary of Key Points

1. Molecular Formula (C₄H₈O): Suggests one degree of unsaturation (ring or double bond).
2. IR Spectrum: No peak at 1700 cm⁻¹, ruling out a carbonyl group and suggesting an ether or alcohol.
3. NMR Spectrum:
   • Triplet at ~3.6 ppm: Protons near oxygen (-O-CH₂-).
   • Triplet at ~1.7 ppm: Aliphatic protons (-CH₂-).
4. Structure: THF is the only cyclic ether with C₄H₈O that fits the NMR and IR data.