2. Fill in the boxes in the following problem. (Hints: The difference between the # of carbons in the product and the # in the starting material tells us the # in the carbonyl. The fact that we form only one product tells us the identity of the carbonyl.) C8H9Br 1H NMR: 7.3-7.1 (m, 5H) 3.5 (t, 2H) 3.1 (t, 2H) 1) Ph3P 2) nBuLi phosphonium ylide carbonyl C1114 (only one product formed)
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- Provide the correct product for the following series of transformation. Hint: The product has a H NMR spectrum with only 2 resonances with relative H integrations of 3:1. 1. Br2 2. 3 eq. NaNH2, NH3 3. CH3I ? Br. Br H3C CH3 A) CH3 C) CH3 B) D) =CH3 A B DPropose a structure for a C,H150,N compound that is unstable in aqueous acid and has the given NMR spectra. 'H NMR: 8 2.30 (6H, s); 8 2.45 (2H, d, J = 6 Hz); 8 3.27 (6H, s); 8 4.50 (1H, t, J = 6 Hz) 13C NMR: 8 46.3, 8 53.2, 8 68.8, 8 102.4 Draw C,H1502N. Select Draw Rings More C N APR étv N Aa MacBook Air4. The 1H NMR spectrum shown below corresponds to one of the molecules A-D. Identify the molecule and then provide the complete IUPAC name for the molecule. Don't forget about stereochemistry. он ÕH ÕH B D 11 10 8 6 5 1 'H NMR: 3.35 (1H, m), 1.66 (1H, m), 1.36 (9H, m, overlapping signals), 0.91 (6H, d), 0.90 (3H, t)
- Identify products A and B from the given 1H NMR data.a. Treatment of CH2 = CHCOCH3 with one equivalent of HCl forms compound A. A exhibits the following absorptions in its 1H NMR spectrum: 2.2 (singlet, 3 H), 3.05 (triplet, 2 H), and 3.6 (triplet, 2 H) ppm. What is the structure of A?b. Treatment of acetone [(CH3)2C = O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1 H) ppm. What is the structure of B?Reaction of p-cresol with two equivalents of 2-methylprop-1-ene affords BHT, a preservative with molecular formula C15H24O. BHT gives the following 1H NMR spectral data: 1.4 (singlet, 18 H), 2.27 (singlet, 3 H), 5.0 (singlet, 1 H), and 7.0 (singlet, 2 H) ppm. What is the structure of BHT? Draw a stepwise mechanism illustrating how it is formed.Predict the product based on the following data: H-NMR: 0.93 ppm (3H, triplet); 1.57 ppm (2H, multiplet – looks like “pentet”); 2.14 ppm (3H,singlet), 2.26ppm (2H, triplet); 3.42 ppm (1H, triplet); 4.83 ppm (1H, multiplet) Weight of product: 156 g/mol IR: Two sharp peaks at 1784 cm-1 and 1742 cm-1 C-NMR: 9.9 ppm; 29.0 ppm; 29.8 ppm; 31.8 ppm; 52.9 ppm; 76.1 ppm; 181.1 ppm; 200.2 ppm
- 1. What spectral features allow you to differentiate the product from the starting material? Mass Spec: IR: H NMR: C NMR:Reaction between this aldehyde and ketone in base gives a compound A with the proton NMR spectrum: ô 1.10 (9H, s), 1.17 (9H, s), 6.4 (1H, d, J 15), and 7.0 (1H, d, J 15). What is its structure? (Don't forget stereochemistry!). When this compound reacts with HBr it gives compound B with this NMR spectrum: õu 1.08 (9H, s), 1.13 (9H, s), 2.71 (1H, dd, J 1.9, 17.7), 3.25 (dd, J10.0, 17.7), and 4.38 (1H, dd, J 1.9, 10.0). Suggest a structure, assign the spectrum, and give a mechanism for the formation of B. H base C11H₂00 HBr B C11H₂1 BroThe compound that exhibits the following spectral data is IR: 1740 cm. 1H NMR: 6 0.9(t, 3H), 1.6(sext, 2H), 2.3(1,2H), 4.6(d, 2H), 5.2(d, 1H), 5.4(d, 1H), 5.9(m, 1H) ppm. El-MS(m/z): 71 (100%) O О
- Identify the following compound from its IR and proton NMR spectra. C,H1,0: 'H NMR 8 3.31 (3H, s); 8 2.41 (1H, s); & 1.43 (6H, s) IR: 2110, 3300 cm- (sharp)2. Provide the structure of E-2,5-dimethyloct-5-enoic acid. 3. a. Name the structure with the following 'H NMR spectrum. 8 (ppm) = 12.71 (1H, s), 8.04 (2H, d), 7.30 (2H, d), 2.41 (3H, s)A compound (C3H,NO) gives the following NMR data. In the box below please draw the structure of the compound. 'H-NMR: 2.06 ppm, s(3H); 7.01 ppm, t(1H); 7.30 ppm, m(2H); 7.59 ppm, d(2H); 9.90 ppm, s(1H) 13C-NMR: 168.14; 139.24; 128.511; 122.834; 118.90; 23.93 • You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading.