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Finalreviewpracticekey - Chemistry 14CL Exam Review Key...

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Unformatted text preview: Chemistry 14CL Exam Review Key Winter 2008 1. The stoichiometric equation for the aldol condensation of benzaldehyde with acetone is given below. MW: 102g/mol 58g/mol 234g/mol (a) If a reaction of 1.00 mL of benzaldehyde (density = 1.0415 g/mL) is reacted with 0.360 mL of acetone (density =0.7899 g/mL), what is the theoretical yield (in grams) of product? 1.0415 g 102 g/mol = 1.02Ξ10-2 mol benzaldehyde, which will give 5.1 X 10-3 mol product 0.360 mL X 0.7899 g/mL 58 g/mol = 4.9Ξ10-3 mol acetone Since two moles of benzaldehyde are required for each mole of product, acetone is the limiting reagent and the theoretical yield is (4.9 X 10-3 mol) (234 g/moL) = 1.15 g (b) If the actual experimental yield for the reaction was 89%, how many mL of acetone would be required to produce 4.00 grams of product? If the yield were 89% then in the previous example, 0.360 mL of acetone would give (1.15 g) (0.89) = 1.02 g of product. 1.02 g product 0.360 mL = 4.00γπροδυχτ ΞμΛ X = 1.41 mL O H O O + 2 NaOH/H 2 O C 2 H 5 OH 2. The chromium oxidation of compound I below yields a mixture of products ( II – V ). Identify the order of elution that would occur if you used an alumina stationary phase with diethyl ether to separate the mixture of products (II, III, IV, V) by column chromatography. (Assume there is no starting material in the products.) (a) Write balanced redox reactions for the reaction of K 2 Cr 2 O 7 with I to give each of the products. Starting material I Products: CH 3 CH 3 CH 3-C-OH OH H C 6 H 5 C O O C 6 H 5 C C 6 H 5-C-C(CH 3 ) 3 O II III IV V (a) Depending on conditions, chromium 6 can reduce to chromium 4 as CrO 2 , chromium 3 as the ion, or chromium 2 as the ion. Based on the reaction of chromium 6 reducing to chromium 3 Half reaction for the oxidizing agent for all reactions Cr 2 O 7 2- + 14 H + + 6 e- = 2 Cr +3 + 7 H 2 O Half reactions for reducing agents Let R = C 6 H 5 (phenyl), R ’ = C(CH 3 ) 3 , t-Bu (a) RR ’ CH-OH = RR ’ C=O + 2H + + 2 e- (b) H 2 O + RR ’ CH-OH = R ’ OH + RCOH + 2 H + + 2e- (c) 2 H 2 O + RR ’ CH-OH = R ’ OH + RCOOH + 4 H + + 4 e- Balanced reactions: (a) 3RR ’ CH-OH + Cr 2 O 7 2- + 8 H + = 2 Cr +3 + 7 H 2 O + RR ’ C=O (a) (b) 3RR ’ CH-OH + Cr 2 O 7 2- + 8 H + = 2 Cr +3 + 7 H 2 O + R ’ OH + RCOH (c) 6RR ’ CH-OH + 4 Cr 2 O 7 2- + 7 H + = 8 Cr +3 + 16 H 2 O + 6 R ’ OH + 6 RCOOH (b) First eluted ______ II _______ 2 OH C 6 H 5-CH-C(CH 3 ) 3 Reason: The ketone is the least polar of the oxygen functional groups; in addition there are two nonpolar hydrocarbon group—an alkyl and an aryl group. Second eluted ____ IV _______ Reason: The aldehyde is slightly more polar than the ketone and this compound has only one hydrocarbon group—a polarizable aryl group....
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Finalreviewpracticekey - Chemistry 14CL Exam Review Key...

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