Chem14C_SU07_E2_Key

Chem14C_SU07_E2_Key - 1 15 points The structure of a common...

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1. [ 15 points] The structure of a common disaccharide is shown on the left. Examine the structure very carefully and answer the questions below. Be specific in your answers. * anomeric carbons a. Write a common name for this sugar? __ Ans: Sucrose b. This sugar consists of two subunits _ Glucose __ and ___ Fructose ____. c. These two subunits are linked by a bond called ___ Ans: Glycoside. d. Is it a reducing or non reducing sugar ? ___ Ans: non-reducing sugar. e. Explain your answer in question (d) using specific chemical terms. Ans: anomeric carbons on glucose and anomeric carbon of fructose are linked by glycoside bond and blocked. Sucrose, therefore, does not have hemiacetal or hemiketal group. Therefore, non-reducing sugar . It is not in equilibrium with the readily oxidized open-chain aldehyde or ketone from its aqueous solution. f. Does this sugar exhibit mutarotation? Yes or No : Ans: no g. Explain your answer. Ans: A nomeric carbons on glucose and anomeric carbon of fructose are linked by glycoside bond. Sucrose, therefore, does not have hemiacetal or hemiketal group. It is not in equilibrium with the open-chain aldehyde or ketone from its aqueous solution. Place an * on the anomeric carbon(s). Ans: pl see pointers above on the structure T
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2. [15 pts] Propose a structure for the following compound using the given 1H NMR and IR data. Formula: C 6 H 12 O 1 HNMR: δ 2.58 ppm (septet, 1H) δ 2.45 ppm (quartet, 2H) δ 1.07 ppm (doublet, 6H) δ 1.01 ppm (triplet, 3H) IR: Strong peak at 1720 cm -1 Show calculation of index of hydrogen deficiency (IHD): Ans; = C-H/2+N/2-X/2+1=6-12/2+0+1 = 1; one ring or one pi bond Show IR analysis: Ans: 1720 cm-1 , Therefore, C = O present No other peaks given; assume absence of OH. Show 1H NMR analysis: Chemical shift, δ splitting Integration protons Implications 2.58 Septet 1 1H CH 3 CH CH 3 2.45 Quartet 2 2H CCH 2 CH3 CHCH 2 CH 2 1.07 Doublet 6 6H CH 3 CHCH 3 1.01 triplet 3 3H CH 3 CH 2 Put pieces together: C6H12O – CO = C5H12= 100 – 28 = 72 One piece Æ
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This note was uploaded on 04/05/2008 for the course CHEM 14C taught by Professor Hardinger during the Spring '08 term at UCLA.

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Chem14C_SU07_E2_Key - 1 15 points The structure of a common...

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