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Chem 210 KEY Exam 2 Spring 2015
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Chem 210 KEY Exam 2 Spring 2015Chem 210
Exam 1 FORM A February 25, 2015
Printed Name: KEY______________________________ Section: _________
PSU eMail: ______________________________________
Signature: _______________________________________
FILL OUT AND TURN IN THIS FRONT PAGE WITH YOUR SCANTRON FORM. WE
WILL NOT PROCESS YOUR SCANTRON WITHOUT THIS COMPLETED AND SIGNED
FRONT PAGE.
1) There are 25 multiple-choice questions on this exam. Make sure that you have them all. Each
question is worth 4 pts.
2) On the Scantron sheet, make sure that you fill in your name (print), section, your student ID
number, and the test form (white cover = Form A, yellow cover = test form B).
3) You may take the exam booklet with you. Your score will be based solely on the information on
your Scantron form.
4) Needed data will be provided. The data values provided with the exam will be the ones that you
must use to arrive at an answer.
5) There is no penalty for guessing.
6) No electronic devices of any kind are allowed. This proscription includes, but is not limited to,
calculators, cell phones, cameras, etc. Exposing such devices during the exam, even if they are
not in use, is grounds for imposing a "zero" grade on the exam.
7) Use of molecular model kits is encouraged.
8) Blank paper is provided at the end of the exam for your convenience. You may not use any other
source of paper during the exam.
9) An answer key will be posted soon after the end of the exam.
1
Copyright 2015 Ken Feldman & Pshemak Maslak Use of this document by any third-party for-
profit concern (i.e., PSU KnowHow, Nittany Notes) is in violation of US copyright law. All rights
reserved.
, Chem 210 KEY Exam 2 Spring 2015
1. How many stereoisomers are possible for the following molecule?
(a) 1 (b) 2 (c) 3 (d) 4
The molecule has two stereogenic centers, and could have up to 22 stereoisomers. However,
the stereogenic centers are related by a symmetry plane in the meso isomer, reducing the
number of stereoisomers to 3. (c) is the correct answer.
2. Estimate the strain energy (in kcal/mol) of trans-1,2-dimethylcyclopropane (1).
(a) 29.2 (b) 27.4 (c) 26.1 (d) 24.4
The total strain energy of 1 can be (artificially) broken down into two more specific components
of strain: (a) ring strain, and (b) torsional/steric strain from eclipsing interactions. The former
should not be affected significantly by the methyl substitutions—the cyclopropane ring still has
angle strain—but the latter will be responsive to the CH3-for-H substitution. We know that for
unsubstituted cyclopropane, there are six C–H//C–H eclipsing interactions, and each interaction
costs 1 kcal/mol. Furthermore, we know that the overall strain of unsubstituted cyclopropane is
27.6 kcal/mol. Thus, the amount of the total strain attributable to ring strain is 27.6 – 6 = 21.6
kcal/mol. To this value we must now add back all of the strain attributable to eclipsing
interactions in the dimethyl-substituted species 1: 1 C–H//C–H (1 kcal/mol) + 2 C–H//C–
CH3 (1.4 kcal/mol) = 3.8 kcal/mol per ring face, or 7.6 kcal/mol in total. Adding this torsional
strain value to the ring strain value from above (21.6 kcal/mol) gives 29.2 kcal/mol of total
strain. Of course, a “shortcut” would be to simply replace the four C–H//C–H eclipsing
interactions of unsubstituted cyclopropane (= 4 kcal/mol) with four C–H//C–CH3 eclipsing
interactions in 1 (= 5.6 kcal/mol); 27.6 – 4 +5. 6 = 29.2 kcal/mol. (a) is the correct answer.
2
Copyright 2015 Ken Feldman & Pshemak Maslak Use of this document by any third-party for-
profit concern (i.e., PSU KnowHow, Nittany Notes) is in violation of US copyright law. All rights
reserved.
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