HNMR Practice 3

Determining the Structure of C₆H₁₄O from ¹H NMR and ¹³C NMR Spectra

The molecular formula is C₆H₁₄O, and the ¹H NMR spectrum shows a triplet at 3.3 ppm, a hextet at 1.5 ppm, and a triplet at 1.0 ppm. The ¹³C NMR spectrum shows three signals. Let’s deduce the structure step by step.

Step 1: Analyze the Molecular Formula

The molecular formula is C₆H₁₄O.

Degree of Unsaturation (DU):

\[ \text{DU} = \frac{2C + 2 - H}{2} = \frac{2(6) + 2 - 14}{2} = 0 \]

A DU of 0 indicates the molecule is fully saturated (no rings or double bonds). This suggests an ether or alcohol functional group.

Step 2: Examine the ¹H NMR Spectrum

The ¹H NMR spectrum shows:

• A triplet at 3.3 ppm.
• A hextet at 1.5 ppm.
• A triplet at 1.0 ppm.

Key Observations:

• Chemical Shifts:
  • The signal at 3.3 ppm is downfield, suggesting protons near an oxygen atom (e.g., -O-CH₂-).
  • The signals at 1.5 ppm and 1.0 ppm are in the aliphatic region, suggesting protons in alkyl chains.
• Splitting Patterns:
  • A triplet indicates coupling with two equivalent neighboring protons (n+1 rule, where n = 2).
  • A hextet indicates coupling with five equivalent neighboring protons (n+1 rule, where n = 5).

Step 3: Examine the ¹³C NMR Spectrum

The ¹³C NMR spectrum shows three signals.

This indicates three distinct carbon environments in the molecule.

Step 4: Deduce the Structure

a) Triplet at 3.3 ppm:

• This signal corresponds to protons on a carbon adjacent to an oxygen atom (e.g., -O-CH₂-).
• The triplet indicates these protons are coupled to two equivalent neighboring protons.

b) Hextet at 1.5 ppm:

• This signal corresponds to protons on a carbon adjacent to a -CH₂- group (e.g., -CH₂-CH₂-).
• The hextet indicates these protons are coupled to five equivalent neighboring protons.

c) Triplet at 1.0 ppm:

• This signal corresponds to protons on a terminal methyl group (e.g., -CH₃).
• The triplet indicates these protons are coupled to two equivalent neighboring protons.

d) Combining the Information:

• The molecular formula (C₆H₁₄O) and the DU of 0 suggest a saturated ether (no rings or double bonds).
• The splitting patterns and chemical shifts are consistent with a symmetrical ether structure.
• The data fits dipropyl ether (CH₃-CH₂-CH₂-O-CH₂-CH₂-CH₃):
  • The -O-CH₂- protons appear as a triplet at 3.3 ppm.
  • The -CH₂- protons adjacent to -O-CH₂- appear as a hextet at 1.5 ppm.
  • The terminal -CH₃ protons appear as a triplet at 1.0 ppm.

Step 5: Verify the Structure

• Molecular Formula: C₆H₁₄O matches dipropyl ether.
• ¹H NMR Spectrum:
  • Triplet at 3.3 ppm: -O-CH₂- protons (4 protons).
  • Hextet at 1.5 ppm: -CH₂- protons adjacent to -O-CH₂- (4 protons).
  • Triplet at 1.0 ppm: Terminal -CH₃ protons (6 protons).
• ¹³C NMR Spectrum:
  • Three signals corresponding to the three distinct carbon environments in dipropyl ether:
    1. -O-CH₂- carbons (2 equivalent carbons).
    2. -CH₂- carbons adjacent to -O-CH₂- (2 equivalent carbons).
    3. Terminal -CH₃ carbons (2 equivalent carbons).

Step 6: Conclusion

The compound is dipropyl ether (CH₃-CH₂-CH₂-O-CH₂-CH₂-CH₃). The NMR spectra are consistent with its structure, and the molecular formula matches.

Summary of Key Points

1. Molecular Formula (C₆H₁₄O): Suggests a saturated ether (DU = 0).
2. ¹H NMR Spectrum:
   • Triplet at 3.3 ppm: -O-CH₂- protons.
   • Hextet at 1.5 ppm: -CH₂- protons adjacent to -O-CH₂-.
   • Triplet at 1.0 ppm: Terminal -CH₃ protons.
3. ¹³C NMR Spectrum:
   • Three signals corresponding to the three distinct carbon environments in dipropyl ether.
4. Structure: Dipropyl ether is the only symmetrical ether with C₆H₁₄O that fits the data.

Final Answer

The compound is dipropyl ether (CH₃-CH₂-CH₂-O-CH₂-CH₂-CH₃).