Question

its a little hard to see but i have it set up to where the NMR and IR analysis for the unknowns are across from each other and then it goes in order from 1-7. Help me to identify which unknown analysis goes to which compound.

ZOOM + Lab 8: Instrumental Analysis You are given seven unknowns, all of which are white or brown powders. Use the provided IR and H NMR spectra to assign Unknowns 1-7 to the given compounds. Label all spectra and briefly explain how you made your determinations. NH2 OH OH O-vanillin Benzoic acid p-toluidine 4-tert-butylphenol OH OH OH H2N 4-aminobenzoic acid 9-fluorenol salicylic acid

Answer 1

Let us see each spectra one by one

Compound 1: The peak near to 12ppm is for a carbocylic acid proton. The two doublets for two hydrogen each is coming in the aromatic region. So it is clear that the compound is 4-substituted benzoic acid. Now the two protons of the broad peak coming near 5.8ppm is for NH2. Hence the compound is 4-aminobenzoic acid OH 1271 8, 49 HN 5 6.57

Compound 2: The compound contain only two aliphatic protons. OH hydrogen comes around 2ppm and OH attached carbon containing proton comes around 6ppm. All other peaks will be in the aromatic region of 7-8ppm OH 15.07 7.55 7.55 7.28 7.38 7.38

Compound 3: The compound have a peak at shielded region for 3 protons which corresponds to a methyl group. Also a peak of singlet is for an NH2. Other two doublets are in the aromatic region. Hence the compound will be p-toluidine 8.40 ( LNH,.91 6.40 2.32 7.04

Compound 4: This contain a peak around 12ppm which corresponds to COOH. All the other peaks are in the aromatic region. One doublet and two triplets. This corresponds to the compound benzoic acid 7.48 OH,271 8.13 7.48

Compound 5: This contain a peak around 12ppm which corresponds to COOH. Also it have a singlet around 6ppm which is for an OH peak. Other peaks are in the range of 7-8ppm. Hence the compound is salicylic acid 7.51 OH,204 7.58 ОН 7.08

Compound 6 The compound have a peak at 1ppm for 9 protons which corresponds to a tert-butyl group. Also a peak of singlet at 6ppm is for an OH. Other two doublets are in the aromatic region. Hence the compound will be 4-tert-butylphenol 7.37 SOH

Compound 7: The singlet peak comes around 10ppm is for an aldehyde hydrogen. Also there is a singlet 3H around 4ppm which is for a CH3 attached to oxygen. One singlet near to the aromatic region is for OH. Other peaks are in the aromatic region. Hence the compound is o-vanillin 7.29 7.28 H10.19 7.05 OH 3.83

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