Questions: C6H15N 1H NMR a= 6= 13C NMR

Solution Steps

• The molecular formula is C6H15N.
• Degree of unsaturation formula: \( \text{DU} = \frac{2C + 2 - H + N}{2} \)
• Plugging in the values: \( \text{DU} = \frac{2(6) + 2 - 15 + 1}{2} = \frac{12 + 2 - 15 + 1}{2} = \frac{0}{2} = 0 \)
• Conclusion: The degree of unsaturation is 0, indicating no double bonds, triple bonds, or rings.

• The IR spectrum shows peaks around 3300 cm\(^{-1}\) (N-H stretch), 2900 cm\(^{-1}\) (C-H stretch), and 1600 cm\(^{-1}\) (C=C stretch).
• Conclusion: The presence of N-H and C-H bonds, and possibly an aromatic ring (though DU suggests otherwise).

• The 1H NMR spectrum shows peaks at around 0.9 ppm (triplet), 1.2 ppm (multiplet), and 2.7 ppm (quartet).
• Conclusion: The triplet at 0.9 ppm suggests a CH3 group adjacent to a CH2 group. The quartet at 2.7 ppm suggests a CH2 group adjacent to a CH3 group. The multiplet at 1.2 ppm suggests a CH2 group adjacent to other CH2 groups.

Final Answer