12-01 - 12/1/2010 Which of the following reactions will not...

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Unformatted text preview: 12/1/2010 Which of the following reactions will not give the alcohol shown below? A Which of the following reactions will not give the alcohol shown below? H O H H B O H O H H C Which of the following reactions will not give the alcohol shown below? A Which of the following reactions will not give the alcohol shown below? H H3C O Questions about Exam 3? What is the major product of the following reaction (show stereochemistry)? D Cl 1 12/1/2010 What is the major product of the following reaction (show stereochemistry)? What What is the major product of the following reaction (show stereochemistry)? OH CH3 H Br CH3 H OH CH3 Br Br OH H A D Cl B C D How could compound D be prepared? OH An unknown compound A has the following properties: CH3 Br 1. The formula for A is C6H14O and the infrared spectrum shows a strong absorption at 3367cm-1. alkane OH 2. Treatment of A with acid results in the formation of an alkene B (C6H12) which upon treatment with BH3 followed by HOOH/ OH OH on least substituded carbon gives back compound A. 3. Alkene B reacts with H-Br to give an alkyl bromide C whose 1H NMR spectrum is shown below. 2 equivalent CH3 groups CH2-CH3 CH2-CH2 H A CH3 Br H D 3. Treatment of alkyl bromide C with potassium t-butoxide gave a new alkene D that, upon treatment with O3 followed by zinc produced formaldehyde and 2-pentanone. Alkene D is 3-methyl-1-pentene. A. True B. False Alkyl bromide C must be 2-bromo-2-methylpentane. A. True B. False H H C O O 2-methyl-1-pentene D 3. Treatment of alkyl bromide C with potassium t-butoxide gave a new alkene D that, upon treatment with O3 followed by zinc produced formaldehyde and 2-pentanone. 3. Treatment of alkyl bromide C with potassium t-butoxide gave a new alkene D that, upon treatment with O3 followed by zinc produced formaldehyde and 2-pentanone. 2 12/1/2010 Alkyl bromide C must be 2-bromo-2-methylpentane. A. A. True B. False 3. Alkene B reacts with H-Br to give an alkyl bromide C whose 1H NMR spectrum is shown below. Alkene B must be 2-methyl-1-pentene. A. A. True B. False 3. Alkene B reacts with H-Br to give an alkyl bromide C whose 1H NMR spectrum is shown below. D D Alcohol A must be 2-methyl-3-pentanol. A. True B. False 2. Treatment of A with acid results in the formation of an alkene B (C6H12) which upon treatment with BH3 followed by HOOH/ OH gives back compound A. How could the following alcohol be prepared? H Na :H Br OH OH B D What is wrong about the following reaction sequence? A. The hydride base (pKa H2 = 35) is not able to remove the acetylenic hydrogen (pKa = 25). B. The alkyl bromide does not undergo SN2 reactions. C. The acetylide does not undergo SN2 reactions. D. The acetylide will react more rapidly with the acidic The alcohol (pKa = 16). H Na :H Br OH OH What strategy could be successful for accomplishing the following transformation? H Na :H H3C OTMS Si CH3 CH 3 H N + Br O C H3 H3C Si Cl = TMS-Cl CH3 F N imazole H Na :H Br O H + TMS-F OH 3 12/1/2010 What spin-spin coupling pattern would be predicted for HB? Assume the following spin-spin coupling constants JAB = 12 hz JBC = 5 hz JAC = 0 hz Sketch the spin-spin coupling pattern predicted for HB based upon the following coupling constants. JAB = 12 hz JBC = 5 hz JAC = 0 hz What is the spin-spin coupling pattern that would be observed for HB? A. doublet of triplets B. triplet of doublets JAB = 12 hz JBC = 5 hz JAC = 0 hz What is the spin-spin coupling pattern that would be observed for HB? 3.2 ppm C. doublet of quartets B. quartet of doublets JAB = 12 hz JBC = 5 hz JAC = 0 hz 5 12 What is the spin-spin coupling pattern that would be observed for HB? 3.2 ppm JAB = 12 hz JBC = 5 hz JAC = 0 hz 4 ...
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