4-methoxy benzoic Acid 13C NMR chemical shift (delta ppm). 68 118 124 Cg Hq 03. Interpretation C atom of -OCHZ C atoms of aromatic ring attached to -OCH3 C atoms of benzene ring attached to hydrogens land 4 C atoms of aromatic ring attached to -Cool group -COOH group 164 170 C atoms of 240 220 200 180 160 140 120 100 80 60 40 20 0 -20
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- values : Experimental values of IR Peaks compared to literature values for Benzoic Acid (ATR method) Bond Stretch Functional group Experimental Peak(cm-1) Literature Peak(cm-1) O-H Carboxylic acid 2800 3200-2200 C=O Carboxylic Acid 1674.9 1677.5 C=C (ring) Aromatic stretch 1579.5 1581-1418.5 C-H Aromatic Sp2 3100 C-O stretch 1285.5 O-H Alcohol 929.89 table 3: Experimental values of IR peaks compared to literature values for 2-Naphthol Bond Stretch Functional group Experimental Peak(cm-1) Literature Peak(cm-1) O-H Alcohol 3219.1 3400-3080 C-C Aromatic (ring) stretch 1507.6 C=C Aromatic stretch 1597.7 1627.3-1377.8 C-H Aromatic Sp2 3050 C-O Secondary alcohol stretch 1168.9 IR results used to prove that the compounds are effectively separated. Discuss on bothdiagnostic peaks and fingerprint region.CH3 2. An unknown isomer of C4H9Cl has the ¹³C NMR and DEPT-135 spectra shown below: 13C NMR: 90 70 |||| A 50 8 (ppm) CH3CH₂CH₂ CH₂ CI 30 10 DEPT-135: CH3 CH3-CHCH₂-CI B 90 70 Circle the compound that is represented by these spectra. Clearly explain your reasoning. you will need to explain how each of the other three isomers was eliminated from consideration. 50 8 (ppm) CH3CH₂ CH-CI CH3 C 30 10 CH3 CH3-C-CI CH3 DRank the four groups (a-d) from highest to lowest priority using Cahn-Ingold-Prelog rules. a b -F O A B C D O BDCA O BCDA O CBDA
- Match the spectra to the structures: 3-bromotoluene and methyl 4-methoxybenzoate. CH 3 3-bromotoluene 200 Spectrum B 200 Br (ignore the CDCI3 signal) CH3O methyl 4-methoxybenzoate OCH3 Spectrum A 100 100Match the 13C NMR data to the appropriate structure.The H NMR spectra below is representative of what compound? Chem. shift Rel. area 2.26 1.50 7,04 1.00 1.00 7.37 TMS 10 4 O ppm Chemical shift (8) o-bromotoluene toluene p-bromotoluene m-bromotoluene 2. 6, Intensity
- A couple questions using H-NMR and IR spectroscopy.Times New Roma 12 A. A (II) Study this aromatic molecule and answer the following questions: Br H2 Br H3 NO2 a) How many 'H-NMR signals this compound will show? B What will be the splitting (coupling) of H1, H2, and H3? H1 .. H2 .... H3 .. c) Which of the two hydrogens, H1 or H3 will show larger coupling, and which will show smaller coupling? 2. BWhich carbon would have a signal farthest downfield in the 13C NMR spectrum for this molecule? d. CI H. C. a O A. a B. b OD.d E. e
- Propose a structure consistent with each set of data. a.Compound J: molecular ion at 72; IR peak at 1710 cm−1; 1H NMR data J: (ppm) at 1.0 (triplet, 3 H), 2.1 (singlet, 3 H), and 2.4 (quartet, 2 H)b. Compound K: molecular ion at 88; IR peak at 3600-3200 cm−1; 1H NMR K: data (ppm) at 0.9 (triplet, 3 H), 1.2 (singlet, 6 H), 1.5 (quartet, 2 H), and 1.6 (singlet, 1 H)Construct a structure of the given 1H NMR data signals.A. C5H12O 0.91 δ (3H, triplet) 1.19 δ (6H, singlet) 1.50 δ (2H, quartet) 2.24 δ (1H, singlet) B. C4H10O 0.90 δ (6H, doublet) 1.76 δ (1H, multiplet) 3.38 δ (2H, doublet) 3.92 δ (1H, singlet) C. C5H10O 1.09 δ (6H, doublet) 2.12 δ (3H, singlet) 2.58 δ (1H, septet)ASSIGNED QUESTIONS· - 1. Compound A is converted into compo:ind B with LIAIH4. Treatment of comoound A with strong minerat acid at elevated temperatures yields compound f. Re- đuction of çompound f with LIAIHA gives compound p. Spectra for the compounds are shown below. Write structures for compounds . R. £ and R. Compound A LiAIHy B Cg Hn N Ht LiAl Hy, D cg HioO WAVNO CO IR - Spectrum 4=? IR = ? NMR Spectrum 5 TMS uV Spectrum Absorbance X 100