Consider the protons that are in the position ortho- to the amine group in the structure shown below. Will these protons be shifted upfield or downfield relative to protons on an unsubstituted benzene ring?
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- Answer and explain in detail how you arrived at that answer.Step 1: Brief Description The given compound is 4-aminobenzoic acid. The IR and NMR spectrum of this compound also given. Determine the spectrum. Step 2: Determination of IR data The peaks are identified as following: Two bands at 3300 & 3230 cm. This is due to asymmetrical and symmetrical N-H stretching of NH₂ group. A broad stretching between $200-2500 cm** - This broad stretching is due to O-H in carboxylic acid group. Band at 1680-1700 cm¹. C=O Stretching 1600 cm C-C-C (aromatic carbon) Step 3: Determination of PMR spectrum Splitting patterns: It should be noted that spin-spin splitting is observed only between nonequivalent neighboring protons. Equivalent protons do spin-spin couple with one another but splitting is not observed. Chemical shift: A highly shielded proton has a low chemical shift and a highly deshielded proton has a high value of chemical shift. The given peaks are of following: Proton Level Chemical shift (ppm) Relative integration Multiplicity ABC -11.9 7.8 8.3…Determine the hybridization around the N atom in each amine, andexplain why cyclohexanamine is 106 times more basic than aniline.
- How would you explain why phenol is the o-p driver, while nitro benzene is m-directional? Resonance Describe by showing their structures.C) a nitrile D) an amine nitrite salt DIRECTIONS: Each of the following questions or incomplete statements below is followed by four suggested answers lettered A - D. Select or choose the letters of the best answer to the question or incomplete statement. 7. An organic nitrogen compound, X, gives ammonia on warming with dilute aqueous sodium hydroxide, X could be A) ethanamide B) ethylamine C) phenylamine D) amino ethanoic acid Which one of the following reagents reacts 1. in a similar fashion with both phenylamine (C H,NH,) and ethylamine. A) Br,(aq) C) conc. HNO, D) cold HNO (aq) 2. Phenylamine (aniline) can be prepared by reducing nitrobenzene with tin and concerntrated hydrochloric acid followed by the addition of alkali and finally steam distillation. The alkali is added to; A) prevent oxidation of phenylamine. B) liberate free phenylamine from solution C) dissolve excess nitro benzene D) dissolve the phenylamine 8. Arrange the following molecules in their increasing order of base…1. The ¹H-NMR of the product shows signals (two doublets) in the aromatic region and a very broad signal (the signal is almost flat from ~5.0 ppm to -6.5 ppm and may actually extend into the H doublet at ~ 6.7 ppm) which may indicate a "combination" of the amine and acid peaks. (The small peak at ~ 2 ppm should be ignored.) H "HA H d a b C
- Draw the structure of the compound whose data is shown below. Upload your answer to receive full credit. C9H,NO3 There are 7 C-NMR signals. The compound is significantly deactivated toward electrophilic aromatic substitution. H-NMR: 12.7 ppm, s, 1H 10.3 ppm, s 1H 7.91 ppm, d, 2H 7.72 ppm, d, 2H 2.10 ppm, s 3HAmines are weak bases. Table 10.2 in the course textbook provides a list of pKb values for several amines. The pK values for three amines and their structures are shown below. Explain why increased substitution on nitrogen results in an increase of the pKb value. H3C-NH2 PK = 3.36 H3C N-H pKb = 3.27 H₂C' H3C N-CH3 PK = 4.19 H₂CDraw the structure of N,N-diethylethanolamine and label the unique proton environments
- b) 4. Rank the indicated protons in each of the given structures in order of decreasing acidity. Show the details for how you arrived at the answer. SHa HbN & OHCkindly answer 14 and 15. thank you.n ane arther. Charge repulsion: the ethylenediamine and its salts The preferred configuration for ethylenediamine are shown below. As you can see, the preferred conformation in each case depends on the protonation state of the amine groups. NH3 NH2 NH2 H. NH3 H. H. ZHN' H. H. H. H H NH3 Build the molecules and explain how the interplay between charge repulsion and hydrogen bonding explain this behavior. 1