Partial Periodic Table - Organic Chemistry at CU Boulder ...
Partial Periodic Table - Organic Chemistry at CU Boulder ...
Partial Periodic Table - Organic Chemistry at CU Boulder ...
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CHEMISTRY 3311, Fall 2004<br />
Professor Walba<br />
First Hour Exam, September 23<br />
scores:<br />
1) 26<br />
2) 20<br />
3) 20<br />
4) 18<br />
5) 16<br />
100<br />
<strong>CU</strong> Honor Code Pledge: On my honor, as a University of Colorado <strong>at</strong><br />
<strong>Boulder</strong> Student, I have neither given nor received unauthorized assistance.<br />
Name (printed): _________________________________<br />
Key<br />
Sign<strong>at</strong>ure: ______________________________________<br />
Recit<strong>at</strong>ion TA Name: _____________________________<br />
Recit<strong>at</strong>ion day and time: ___________________________<br />
This is a closed-book exam. The use of notes, models,<br />
calcul<strong>at</strong>ors, scr<strong>at</strong>ch paper, or any other paraphernalia will not be<br />
allowed during the exam. Please put all your answers on the<br />
test. Use the backs of the pages for scr<strong>at</strong>ch.<br />
PLEASE read the questions carefully!<br />
<strong>Partial</strong> <strong>Periodic</strong> <strong>Table</strong><br />
1A<br />
8A<br />
1<br />
H<br />
2A 3A 4A 5A 6A 7A<br />
2<br />
He<br />
3<br />
Li<br />
4<br />
Be<br />
5<br />
B<br />
6<br />
C<br />
7<br />
N<br />
8<br />
O<br />
9<br />
F<br />
10<br />
Ne<br />
11 12 13 14 15 16 17 18<br />
Na Mg Al Si P S Cl Ar<br />
35<br />
Br<br />
53<br />
I
Name:<br />
1) (26 pts) a) Draw a valid valence bond structure for each of the possible isomers (both constitutional<br />
isomers and stereoisomers) with the molecular formula C 3H 5F. Draw each isomer ONLY ONCE.<br />
b) Draw a valid valence bond structure for each of the following molecules. Carefully show all lone pairs,<br />
unpaired electrons, and formal charges in your structures.<br />
H<br />
CH 4 CH 3 (CH 3) + (CH 3) -<br />
H<br />
C<br />
F<br />
H<br />
H<br />
F<br />
c) For 1,2-butadiene, indic<strong>at</strong>e the hybridiz<strong>at</strong>ion of each carbon. Put your answer for each carbon in the box to<br />
the right its number.<br />
H<br />
C C C<br />
H 1 2 3<br />
H<br />
H<br />
CH 3<br />
4<br />
C<br />
H<br />
H<br />
F<br />
1<br />
2<br />
F<br />
H<br />
Page 2 of 7<br />
C<br />
H<br />
H<br />
H<br />
F<br />
C<br />
sp 3<br />
2 sp2 sp 4 sp3 H<br />
H
Name:<br />
2) (20 pts) a) Circle the stronger Brønsted acid in each of the following pairs of structures.<br />
H 2O<br />
CH 3NH 2<br />
H<br />
H<br />
H 3O +<br />
CH 3CH 3<br />
H<br />
H<br />
b) Circle the compound with the more stable carbonyl group in each of the following pairs of compounds.<br />
O<br />
H 3C H<br />
O<br />
H 3C CH 3<br />
O O<br />
O<br />
O<br />
O<br />
Page 3 of 7<br />
NH 3<br />
OH<br />
OH<br />
NH 2<br />
OH<br />
O<br />
O O<br />
O<br />
OH 2<br />
N<br />
H
Name:<br />
3) (20 pts) a) Give the stereochemical descriptor (E or Z) for the following alkenes.<br />
b) Circle the arom<strong>at</strong>ic compounds in the following list.<br />
B<br />
H<br />
H<br />
F<br />
N<br />
H<br />
Z E<br />
Page 4 of 7<br />
H<br />
F OH<br />
N<br />
H H H H<br />
c) I have given a valence bond structure of N-methylacetamide below. Give the two additional major<br />
resonance contributors to the structure. Be sure to show all lone pairs and formal charges in your structures.<br />
O<br />
H 3C N CH 3<br />
H<br />
O<br />
H 3C N CH 3<br />
H<br />
O<br />
H 3C N CH 3<br />
d) N-methylacetamide has a higher boiling point than N,N-dimethylacetamide. In one short sentence, explain<br />
why this is the case.<br />
O<br />
H 3C N CH 3<br />
H<br />
boiling point = 205°C<br />
O<br />
H 3C N CH 3<br />
CH 3<br />
boiling point = 166°C<br />
N-methylacetamide has a higher boiling point because it can particip<strong>at</strong>e in intermolecular hydrogen bonding.<br />
H
Name:<br />
4) (18 pts) a) Circle the compound with the higher he<strong>at</strong> of combustion (th<strong>at</strong> is, the one th<strong>at</strong> gives off the most<br />
he<strong>at</strong> when you burn it in oxygen).<br />
b) On the following energy diagram, very clearly indic<strong>at</strong>e the rel<strong>at</strong>ive energy of the two isomers given in part<br />
4a by drawing a short horizontal line above each structure indic<strong>at</strong>ing its energy. Make sure it’s easy to see your<br />
lines, and th<strong>at</strong> it’s clear which isomer has higher energy.<br />
E<br />
c) For each of the following pairs of alcohols, circle the one th<strong>at</strong> would react faster with Lucas reagent.<br />
OH<br />
OH<br />
Page 5 of 7<br />
OH OH
4) – continued<br />
Name:<br />
d) Propose an arrow-pushing mechanism for the following transform<strong>at</strong>ion. Assume for this purpose th<strong>at</strong> the<br />
reagent [ZnCl 2, HCl, H 2O] = H + . Show only one valence bond structure for each intermedi<strong>at</strong>e in your<br />
mechanism, but be sure to give all the intermedi<strong>at</strong>es.<br />
OH<br />
H +<br />
ZnCl2, HCl, H2O OH Cl<br />
e) [Note – this is a hard question, but hay, it’s only five points. However, it might be a good idea to finish the<br />
rest of the exam before you tackle this question] When the 3-methyl-2-butanol reacts with Lucas reagent, two<br />
isomeric products are formed, as shown below. Using resonance structures, show how this happens.<br />
ZnCl2, HCl, H2O OH Cl<br />
O<br />
Cl or Cl<br />
H<br />
Page 6 of 7<br />
H + H 2O<br />
Cl<br />
+<br />
Cl
Name:<br />
5) (16 pts) Give the single major organic product of each of the following reactions. In both cases, only one<br />
product is formed, in high yield. Assume th<strong>at</strong> there is excess hydrogen gas (as usual in a c<strong>at</strong>alytic<br />
hydrogen<strong>at</strong>ion), but th<strong>at</strong> the reaction is only allowed to proceed for a few hours <strong>at</strong> room temper<strong>at</strong>ure.<br />
a)<br />
b)<br />
H<br />
N O<br />
H<br />
H<br />
H 2, Pt<br />
room temp<br />
H 2, Pt<br />
room temp<br />
Page 7 of 7<br />
H<br />
H<br />
N O<br />
c) Penicillin was the first important antibiotic (it kills bacteria). Unfortun<strong>at</strong>ely, many bacteria have developed<br />
resistance to penicillin by making an enzyme called penicillinase. For our purposes, penicillinase is a giant<br />
molecule with a nucleophile on it, as indic<strong>at</strong>ed below. Penicillin has two amide groups, with the amide<br />
carbonyls labeled 1 and 2 on the structure. The penicillinase nucleophile <strong>at</strong>tacks one of those carbonyls, which<br />
ends up making the penicillin harmless to the bacteria. A key point is th<strong>at</strong> one of the carbonyl groups is much<br />
more reactive than the other.<br />
1<br />
O<br />
H<br />
N<br />
O<br />
N<br />
2<br />
Penicillin<br />
S<br />
CO 2H<br />
Penicillinase—Nu<br />
c) Which carbonyl group is the most reactive one? #2<br />
d) Propose a structure for the intermedi<strong>at</strong>e formed when the penicillinase—Nu: - reacts with penicillin.<br />
H<br />
O<br />
N<br />
Penicillinase—Nu<br />
O<br />
N<br />
S<br />
CO 2H<br />
H