Question

A student is given two unknown samples continuing two different compounds with the same molecular formula, C3H7NO, he annaylzes them by IR and 1H NMR

Compound A:

IR: 3366, 3189, 2976, 1662 cm-1

1H NMR: 1.15 ppm, 3H, t

2.24 ppm, 2H, q

6.21 ppm, 2H, broad

Compound B:

IR: 3294, 3099, 2946, 1655 cm-1

1H NMR: 1.98 ppm, 3H, s

2.79 ppm, 3H, s

6.40 ppm, 1H, broad

Identify the structure of each compound and justify your answer.

Answer 1

Compoune A

IR spectrum

3366 cm^-1        - N - H steching frequency

2976 cm^-1      - C - H saturated ssreching frequncy

1662 cm^-1      - C = O Amide sreching frequncy

NMR:

1.15 ppm, 3H, t   - CH₃  attached to - CH₂ -

2.24 ppm, 2H, q       - CH₂ - attached to  - CH₃ and shift is 2.24 indicate that attached to electron withdrawing group ie . -C = O

6.21 ppm, 2H, broad     - NH₂ attached to attached to electron withdrawing group ie . -C = O

thus structure of molecule is CH₃  - CH₂ - CO - NH₂

Compound B

IR spectrum

3294 cm^-1        - N - H steching frequency

2946 cm^-1      - C - H saturated ssreching frequncy

1655 cm^-1      - C = O Amide sreching frequncy

NMR:

1.98 ppm, 3H, s   - CH₃  attached to electron withdrawing group ie . -C = O

2.79 ppm, 3H, s      - CH₃  attached to -NH

6.40 ppm, 1H, broad     - NH attached to attached to electron withdrawing group ie . -C = O

thus structure of molecule is CH₃ - CO - NH  - CH₃