Question

What is the structure of the unknown molecule based on 1H, 13C NMR and IR

NMR-115 (700MHz) MW: 152.15 Analysis: C, 63.15; H, 5.30; o, 31.55 galdenadte 70 60 ppm 80 160 150 140 130 120 110 100 9 190 180

Answer 1

1st Date: C 63.15%; H 5.30%; O 31.55%

$C=\frac{63.15}{12.01}=5.26\; \; \; \; C=\frac{5.26}{1.97}=2.67\; \; \; \; \; C=2.67x3=8$

$H=\frac{5.30}{1.00}=5.30\; \; \; \;\; H=\frac{5.30}{1.97}=2.69\; \; \; \; \; H=2.69x3=8$

$O=\frac{31.55}{15.99}=1.97\; \; \; \; \; O=\frac{1.97}{1.97}=1\; \; \; \; \; \; \; \; O=1x3=3$

Molecular Formula = C₈H₈O₃

Weight Formula = 12x8(C) + 1x8(H) + 16x3(O) = 152.15

Degree of unsaturation:

$\frac{2(8)+2-8}{2}=5\; unsaturations$

In the IR Spectrum we have two characteristical signals, in 3123 cm^-1 a broad peak (O-H strecht) and 1661 cm^-1 a intense peak (CHO, aldehyde).

In the ¹H NMR spectrum, we have:

1) singlet in 3.9 ppm (3H equivalent to methoxy group -OCH₃)

2) singlet in 6.5 ppm (1H equivalent to hydroxyl group -OH)

3) doublet in 7.05 ppm (aromatic zone, 1H equivalent to an phenyl C-H)

4) multiplet in 7.40 - 7.45 ppm (aromatic zone, 2H equivalent with two phenyl C-H)

5) singlet in 9.8 ppm (1H equivalent to aldehyde group -CHO)

In the ¹³C NMR spectrum we have:

1) singlet in 56 ppm (C equivalent to methoxy group -OCH₃)

2) 6 singlets in aromatic zone (from 108 ppm - to 151 ppm equivalent to substituted benzene ring)

3) singlet in 191 ppm (C equivalent to aldehyde group CHO)

With this data we can propose this compound:

4-hydroxy-3-methoxybenzaldehyde