Question

X | DOSBox 0.74, Cpu speed: 3000 cycles, Frameskip 0, Program: SPECTOC QUIZ: UNKNONN#7 Molecular Formu la: CH15NO 3 3 6 5 4 2 4000 3808 2000 1800 1 C 1 IR 3400,1690 RCONHR NMR s, 1.0 6,9H 3.9 6 quartet and bs t-Butyl for a total of 3 H's Press Any Key

Using NMR and IR spectroscopy data, what is the molecular structure?

Answer 1

Molecular formula : C₇H₁₅NO

IR : -3400 cm-1 : weak and little broad peak (due to -N-H stretching ) indicates presence of  -NH- group.

- 1690 cm-1 : strong peak due to -C=O stretching, it is carbonyl of Amide group.

NMR :

- Singlet at 1.0 ppm for 9H : 9 equivalent protons indicates presence of ^tBu group ( (CH₃)₃-C- ) ).

-  Triplet at 1.3 ppm for 3H : indicates for Methyl group ( -CH₃ ), which is attached with -CH₂- group, thus coupling gives triplet.

- Quartet with broad spectra at 3.9 ppm , it is due to presence of -NH-CH₂- , 2H of -CH₂- couples with protons of -CH₃ to give a quartet (de-shielded due attached N ) to , proton of -NH- appears as broad signal, due rapid exchange and magnetic property of Nitrogen. Both peaks appears in adjacent areas which gives a complex overlapping peak.

Using all these information most probable structure of molecule for this Spectra is :

(CH₃)₃-C-(C=O )-NH-CH₂-CH₃