8. When (1R,2R)-2-bromocyclohexanol is treated with NaOH, the product does not exhibit a broad IR stretch at 3000-3500 cm³¹ and does not rotate plane polarized light. However, when (1R,2S)-2-bromocyclohexanol is treated with NaOH, the product exhibits a strong, broad IR signal at 3000-3500 cm³¹, and is found to be optically active. Draw the two starting materials and propose a mechanism for their reactions with NaOH that is consistent with these observations. (16 pts)

An unknown compound C of molecular formula C6H12O32 exhibits a strong absorption in its IR spectrum at 1718 cm−1 and the given 1H NMR spectrum. What is the structure of 1H NMR spectrum. What is the structure of C?

An unknown compound C of molecular formula C6H12O3 exhibits a strong absorption in its IR spectrum at 1718 cm−1 and the given 1H NMR spectrum. What is the structure of C?

Propose 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 Air

4. 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)

When compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet, a 3Hdoublet, a 6H doublet, and two 1H multiplets. The 1H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identifycompounds A and B.

Treatment of W with CH3Li, followed by CH3I, affords compound Y (C7H14O) as the major product. Y shows a strong absorption in its IR spectrum at 1713 cm−1, and its 1H NMR spectrum is given below, (a) Propose a structure for Y. (b) Draw a stepwise mechanism for the conversion of W to Y.

When methyl 5-oxopentanoate is treated with vinyl Grignard, a compound is produced whose formula is C,H1002. In the IR spectrum of C7H1002, an intense absorption appears at 1740 cm-1 and a weaker absorption appears at 1650 cm-1. Seven signals appear in the 13C NMR spectrum of C,H1002. The DEPT spectrum shows that there is one carbon that is bonded to no hydrogens, two CH carbons, and four CH2 carbons. Draw the structure of C,H1002. 1.AMGB., THF → C7H10O2 `OCH3 2. NH4CI, H2O Methyl 5-oxopentanoate o=/

Treatment of W with CH3Li, followed by CH3I, affords compound Y (C7H14O) as the major product. Y shows a strong absorption in its IR spectrum at 1713 cm−1, and its 1H NMR spectrum is given below. (a) Propose a structure for Y. (b) Draw a stepwise mechanism for the conversion of W to Y.

When acetonitrile is treated with concentrated sulfuric acid and tert-butanol, followed by water, a product is formed whose 'H NMR spectrum exhibits the following three signals: singlet, 1.3 ppm, 9 H; singlet, 2.0 ppm, 3 H; and broad singlet, 8.2 ppm, 1 H. Its IR spectrum exhibits one broad absorption of medium intensity between 3300 and 3500 cm-1, and a narrow, intense absorption near 1650 cm¯1. A key intermediate is shown. Draw the structure of the product, and draw the complete, detailed mechanism for the reaction. CH3 CH3 H2SO4 (conc) H2O ? + H3C H3C H3C-C=Ñ-c `CH3 CH3 HO