CH223 001 Exam 2 - CH223 001 Exam 2...

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CH223 001 Exam 2 (Test) SHAW ROBERT H CH 223, section 001, Fall 2008 Instructor: Kay Sandberg Web Assign Current Score: 0 out of 100 Due: Monday, November 10, 2008 02:15 PM EST Description Before opening this exam, please put away all notes, textbook, flashcards, periodic tables or any other aids except for your model kit. Also, close all browser windows except for the exam browser window and only open links within the exam. You may have blank paper for scratch work. You are on your honor. Once you access the exam, you will have 75 minutes. You cannot stop the clock once it has started. Instructions Periodic Table Reagent Table The due date for this assignment is past. Your work can be viewed below, but no changes can be made. Request Extension 1. --/2 points Given that this ring is aromatic, draw the protonated species in greatest abundance, that is, draw the conjugate acid. You should NOT use the X tool to add hydrogens. You will need to use the +/- tool on the toolbar to add charge and a hydrogen. Note - you only get one submission for this one. c1c[nH+]c[nH]1 Solution or Explanation The pair of electrons on the nitrogen bonded to the hydrogen is needed to give the requisite number of pi electrons for aromaticity by making that nitrogen sp2 hybridized with the pair of electrons in a p-orbital. Since that nitrogen's pair of electrons is needed for aromaticity, they are not free to be the basic lone pair (without destroying aromaticity.) CH223 001 Exam 2 http://www.webassign.net/[email protected]/student.pl?q=2008120718... 1 of 14 12/7/2008 1:53 PM
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2. --/2 points Which compound has the highest p K a value for the indicated proton? A B C Solution or Explanation The higher the pKa, the harder it is to remove the proton. Removal of the proton bonded to C leaves behind a negative charge to be resonance delocalized on the C and O; removal of the proton bonded to N leaves behind a negative charge to be resonance delocalized on the N and O; removal of the proton bonded to O leaves behind a negative charge to be resonance delocalized on both O's. Since all 3 of the atoms (C,N,O) are in the same period (thus being close to the same size) the more electronegative the atom, the more stable it is carrying the negative charge. Deprotonation of the O leads to the most stable anion, therefore it will be the easiest to remove thereby resulting in the lowest pKa. 3. --/3 points This question deals with tautomers. A) Draw the skeletal structure of the tautomer. CH223 001 Exam 2 http://www.webassign.net/[email protected]/student.pl?q=2008120718... 2 of 14 12/7/2008 1:53 PM
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C/C=N/C C/C=N\C B) Select the more stable species structure given tautomer you drew Solution or Explanation The imine is more stable than the enamine because pi electrons are more stable when they are closer to electronegative atoms. 4. --/5 points A) Rank the following in order from the most electrophilc carbonyl carbon to the least electrophilic carbonyl carbon by inputting the corresponding letter A - D.
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