Select all major signals that would show on an IR spectrum of the molecule below. broad moderate absorption centered around 3300cm1 Obroad, strong absorption around 2900 cm1 sharp absorption at 3300 cm1 Ostrong absorption near 1740 cm 1 Odual peaks centered around 1600 cm1 and 1500 cm1 Osharp absorption at 2150 cm 1 sharp absorption at 3100 cm 1

1Compound 1 has molecular formula C7H16. It shows three signals in the 1H-NMR spectrum, one at 0.85 ppm, one at 1.02 ppm, and one at 1.62 ppm. The relative integrals of these three signals are 6, 1, and 1, respectively. Compound 2 has molecular formula C7H14. It shows three signals in the 1H-NMR spectrum, one at 0.98 ppm, one at 1.36 ppm, and one at 1.55 ppm. The relative integrals of these three signals are 3, 2, and 2, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.

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10. Label three peaks in the following spectrum with the bond that gives rise to them, and identify the functional group that is present in the compound (the compound has a formula of C5H1002)? 100 Transmittance (%) 88&RSRRRRS. 90 70 3200 2800 2400 2000 1800 Micrometers um 1400 1300 Wavenumber (cm-¹) 1000 800 کہہ 600 400

3. Draw a molecule with the molecular formula C6H12O2 that would have the following peak(s) in its IR spectrum. (all major non-fingerprint absorbances are listed) a. 3100-3400 cm-1 (broad) 2900-3000 cm-1

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CH3 c=C H3C H.

Assign peaks to the following RI scan according to the molecule below. 100 %T 90- 80 70- IR52 3500 3000 2500 2000 1750 1500 1250 1000 750 cm-1 H₂N CH3

Q2. This compound has the molecular formula C,H₁,NO₂. Included in this problem are the infrared spectrum, 'H NMR with expansions, and ¹3C NMR spectra data. этотури H Normal Carbon 14 ppm 61 116 SA SA 119 120 129 131 147 167 117 1700 cm DEPT-135 Positive Negative Positive Positive Positive Positive No peak No peak No peak DEPT-90 No peak No peak Positive Positive Positive Positive No peak No peak No peak quartet 10 till l 4 · triplet