Question

This is all i was given. No mass just the percentages of the atoms

3) Unknown 311 (56.25% C, 3.91% H, 12.5% 0) 4000 3600 3200 2800 2400 2000 1800 1600 1400 1200 1000 0 ppm 100 Relative Intensity 25 50 75 m/z 100 125

Answer 1

The mass percentage of the atoms present in the compound are given, which doesn't make upto 100%. Thus the missing atom there with unique atomic mass was first to be discovered as can be seen in the answering images. MS and IR spectra leads us to the chemical formula and NMR data provides the data to construct the exact molecule.

Given C = 56.25% NMR shows 5-H H = 3.91 % atoms. O = 12.5% . Mass spectra shows base peak at 128 . 5 008 p 794/ Molecular ion (18x30 googles, peak. O IR data: broad peak at 3300-3400 cm? → H-bonded O-H. NMR: Peaks around 8-(6.5-7.0) ppm. → Aromatic protons. Mass pen of oxygen in the compound = 128 x 1201 16 number of o-atoms = 1 of Mass carbon = 72 - = 128x S 56.25 160 I coatoms = 12 number of Mass of Hydrogen = 128* 3.91 25. = number of H- atoms = 5 There should molecule to be a fulfil i the atom present in the molas molecular mass at 128. Hence Cherrical formula = CG H goce Degree of unsaturation = 146 - 14-6 2 = 4

O-H So with one exchangeble of proton and rest aromatic protons, the compound is a mono-chlorophenol. Now, the NMR shifts are - 8~6.1ppm Hydroxyl group proton. 8~(6.5-7.2 ppm 4 - different aromatic H- protons . can only be when or ca is a meta position writ. –OH group. Hence the COM compound is

Given C = 56.25% NMR shows 5-H = 3.91 % . O = 12.5% .

Mass spectra shows base peak at m/z=128 =>Molecular ion peak.

IR data: broad peak at 3300-3400 cm → H-bonded O-H.

NMR: Peaks around delta-(6.5-7.0) ppm. → Aromatic protons.

Mass of oxygen in the compound = 128 x 0.125 = 16 => number of O-atoms = 1.

Mass of carbon = 128 x 56.25/100 = 72 160 => number of C-atoms = 6.

Mass of Hydrogen = 128* 3.91/100 = 5 => number of H- atoms = 5.

There should be a Cl-atom present in the molecule to fulfil the molecular mass of 128.

Hence Chemical formula = C₆H₅OCl. Degree of unsaturation = (14-6)/2 = 4

So with one exchangeable O-H proton and rest aromatic protons, the compound is a mono-chlorophenol. Now, the NMR shifts are -> delta~6.1ppm = Hydroxyl group proton. delta~(6.5-7.2) ppm = 4 - different aromatic protons -- can only be when, Cl is at meta position with respect to –OH group. Hence the compound is --> meta-chlorophenol.