An unknown molecule with an elemental composition of 78.65% Carbon and 8.25% Hyd

Question

An unknown molecule with an elemental composition of 78.65% Carbon and 8.25% Hydrogen has the following Mass; Infrared; 'H and BC NMR spectra. Determine the structure of the unknown. Show all work in the space below to justify the structure you draw. Mass Spec: so eo 80 100 110 120 1101@1101 1D11@200 m/e IR Yave limber, c1 4000 3000 2500 2000 500 1300 1 300 1100 1000 800 800 Yavelength, microns H NMR: pm.s 0 C Spectral Data: Quartet, 27.0 ppm Quartet, 56.0 ppm Doublet, 114.0 ppm Doublet, 129.6 ppm Singlet, 129.7 ppm Singlet, 166.4 ppm

Explanation / Answer

We first need to figure out the molecular formula of the unknown from given:

%C = 78.65

%H = 8.25

%O = 100- 78.65-8.25 = 13.1

Assuming that there is 100.00 g of the substance, molar ratios of C, H and O respectively would be:

C = 78.65g/12.01 g/mol = 6.549 moles

H = 8.25/ 1.008 g/mol = 8.185 moles

O = 13.10/15.99 g/mol = 0.819 moles

Now diving with the smallest mole ratio, we get

C = 6.549/0.819 = 7.998 =8

H = 8.185/.819 = 9.994 = 10

O = .819/.819 = 1

So the formula for the unknown would be C8H10O

1H NMR:

2 siglets with 3 protons at 2.4(a) and 3.8(b) respectively corresponding to 2 methyl groups. One of them (at 3.8 ppm) is deshielded due to an oxygen next to it.

The two sets of peaks (e and f) in the 7.00- 8.00 ppm range corresponds to aromatic protons ans since they are in sets of 2 suggests a 1,4 substitution.

So the structure should be 4-methoxy toluene: H3C-C6H4-OCH3 (para substituted: Structure attached separately)

13CNMR:

27.0 ppm (A) and 56.0 ppm (B) ccoreesponding to C from CH3 and OCH3 respectively

doublets at 114(C) and 129.69(D) corresponding to non-substituted aromatic carbons

129.7(E) and 166.4 (F) corresponding to substituted aromatic carbons

IR:

CH streches at around 3000 cm-1, C-O streches around 1000 cm-1, Aromatic C=C around 1400,

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