Name_______KEY Name_______ KEY_____________________ _____________________ Date___________________________
CHEM 213 Exam I Summer 2010 Time limit = 60 minutes Part 1: Multiple Choice: 4 pt. @ Directions: Answer ALL questions from this part (#1 – 15) on the answer sheet below. Responses marked elsewhere will not be evaluated. Circle or darken the letter that corresponds to the BEST answer. There is only one BEST answer for each question.
1
a
b
c
d
9
a
b
c
d
2
a
b
c
d
10
a
b
c
d
3
a
b
c
d
11
a
b
c
d
4
a
b
c
d
12
a
b
c
d
5
a
b
c
d
13
a
b
c
d
6
a
b
c
d
14
a
b
c
d
7
a
b
c
d
15
a
b
c
d
8 a b c d __________________________________________________________________________ Part 1
No. correct ______ x 4
=
__________ / 60
Part 2
16
=
__________ / 20
17
=
__________ / 10
18
=
__________ / 10
19
=
__________ / 10
TOTAL
__________ / 110
Name_______KEY_____________________ Date___________________________
Name_______KEY_____________________ Date___________________________ : 1
Which compound is (2Z,4E)-3,4-dimethyl-2,4-octadiene? All structures are 3,4-dimethyl-2,4-octadiene (a) (2E,4E) (c) (2Z,4Z) H3C H
CH2CH2CH3 C
C
C
C
H3C
H3C H3C
H
C C
CH3
H C
C
H
CH2CH2CH3 CH3
(d) (2E,4Z)
(b) 6 H3C
1 H3C
2
3
C
C
4
5
C
C
7
H3C
8
CH2CH2CH3
H
C C
C
C CH2CH2CH3
H H3C
H
H
CH3
CH3
For the 2,3 double bond, on C2 the methyl group ( dark blue) has a higher priority than H (cyan) and on C3 the alkene chain C4-C8 ( C4 is blue) has a higher priority than methyl (cyan). The high priority groups are on the same side so the C2-C3 bond is labeled as “Z” (German: zusammen = together). For the 4,5 double bond, on C4 the alkene chain C1-C3 (C3 is green) has a higher priority than methyl ( lime) and on C5 the propyl group (green) has a higher priority than H ( lime). The high priority groups are on opposite sides so the C4-C5 bond is labeled as “E” (German entgegen = against). 2
In the reaction below, water serves as a(n) CH3
H3C C H3C
CH2
+ + H , H2O
H3C
C
CH3
OH
(Hint: Think of the reaction mechanism.) (c) both base and nucleophile (a) base only (b) nucleophile only (d) neither base nor nucleophile When water serves as a nucleophile, a base is required to remove a proton to make the oxygen electrically neutral. 3
The Cl-C-Cl bond angle of dichlorocarbene (:CCl 2) is most nearly (a) 90° (b) 109° (d) 180° (c) 120° Three groups (lone pair, Cl, Cl) around a central atom = trigonal = 120°
Name_______KEY_____________________
4
Date___________________________ In the BH3-THF molecule, the hydrogen atoms on boron behave chemically as H
H
B
H
O
BH3-THF +
-
(a) H (c) H• (d) H2 (b) H In BH3, B is in group 3 so its cation has a charge of +3. In order to balance the charge, H needs a charge of -1. It will act, therefore as H . 5
Which carbocation will not undergo rearrangement? (Which is the most stable cation?) (c) 2° (a) 3°
(b) 1°
(d) 2°
Name_______KEY_____________________
6
Date___________________________ Which of the compounds below cannot serve as a Diels-Alder diene? Which diene cannot assume an s-cis conformation? (a) s-cis (c) rotate
s-trans – but the single bond can rotate to make it s-cis (d) stuck as s-trans by the ring
(b) s-cis
7
Which cation is more stable? (a) 3°
(c) 1° H
H
H
C
C
CH3
CH3
O
resonance stabilized by lone pair of electrons on O (b) 1°
H
H
H
H
C
C
O
CH2
CH3
H
H
H
H
C
C
O
CH3
CH2
(d) 2°
H
H
H
C
C
O
CH3
CH3
Name_______KEY_____________________
8
Date___________________________ Ozonolysis of an alkene gives the following products. O
O
+
What is the structure of the original alkene? Make the C=O’s face each other and connect them as a C=C
O
9
O
(a)
(c)
(b)
(d)
Which of the following reactions does NOT involve a “three-centered bond” or a three membered ring? (a) Addition of Br2, H2O to an alkene (bromonium ion = 3-centered bond)) (b) Hydrolysis of an epoxide (epoxide = 3 membered ring) (c) Hydroboration-oxidation of an alkene (1. BH 3-THF, 2. H2O2, OH) (4 membered ring) (d) Oxymercuration-reduction of an alkene (1. Hg(OAc) 2 THF, H2O, 2. NaBH4) (3-centered ring w/ Hg)
Name_______KEY_____________________
10
Date___________________________ Which of the following represents the curved arrow notation for the first step in the addition of H-Cl to 1-propene? -
(a) has H as the leaving group H
H
Cl
H3C
CH
CH2
(b) forms 1° cation
H3C
11
(c) forms 2° cation
H
Cl
CH
CH2
H3C
CH
Cl
CH2
-
(d) has H as the leaving group
H3C
H
Cl
CH
CH2
Which of the following represents the highest occupied molecular orbital (HOMO) of 1,3-butadiene? The HOMO looks like the double bond – single bond pattern in the molecule. There are 4 p electrons in 1,3-butadiene (2 e- per p bond x 2 p bonds) (a) Ψ2 2 e- = HOMO
(c) Ψ3 (no electrons)
-
(d) Ψ4 (no electrons)
(b) Ψ1 2 e (lowest energy)
Name_______KEY_____________________
12
Date___________________________ In the reaction below, the thermodynamically favored product is +
H
(a)
products
Cl
(c)
Cl Cl
(b)
(d)
Cl Cl
Find the more stable alkene = more highly substituted C=C 13
Which of the following is the best dienophile in a Diels-Alder reaction? Generally, dienophiles should be electron poor. Alkyl groups are electron donating and C=O groups are electron withdrawing. (a) (c) H
CH3
O
C H C H
CH3 CH3
H
CH3
(d)
(b) O
H
H
H
H
H CH3
CH3 H
O
Name_______KEY_____________________
14
Date___________________________ What is the enol of the ketone shown below? O
(a)
(c) OH
OH
(d)
(b)
OH
OH
2
The ketone sp carbon is also the enol sp2 carbon. 15
What alkene would be best as starting material for the synthesis of 2-bromopentane? Br
(a)
(c)
Markovnikov addition of H-Br gives only 2-bromopentane. Protonation of C1 gives a more stable 2° cation on C2 than protonation on C2 giving a 1° cation. (b)
There is no obvious help from Markovnikov’s rule. Protonation of either C2 or C3 gives 2° cations. Both 2bromopentane and 3-promopentane will be produced
There is no obvious help from Markovnikov’s rule. Protonation of either C2 or C3 gives 2° cations. Both 2bromopentane and 3promopentane will be produced.
(d)All will give 2-bromopentane as the major product.
Name_______KEY_____________________ Date___________________________ Part 2: Short Answer: Directions: Answer the questions below to the best of your ability. Bear in mind that not all questions are of equal point value or of equal difficulty.
Provide a reagent or product as needed for the reactions below: Stereochemistry is important! Show it! (4 pt. @)
16
Anti-Markovnikov addition and syn addition of H-OH
A
CH3 H
+ 1. BH3-THF 2. H2O2, -OH
B H3CH2C
+ CH3CO3H
C
H
OH
Epoxidation (stereospecific)
CH2CH3 C
H
H3CH2C C
H
CH2CH3
O C
H
C H3C
(:CH2 addition) CH2N2 or CH2I2, Zn(Cu) +
H C
C
H
CH3
D H3CH2C
CH3 C
+ 1. OsO4 2. Na2SO3
C
H3C
CH2CH3
H
(stereospecific) CH2 CH3
H
H
CH3
Syn hydroxylation (stereospecific)
H3CH2C
OH
OH
C
C
H3C
E H3C
C
C
CH2 CH3
Syn addition of H 2 (H2, Lindlar’s catalyst) +
CH2CH3
H3C
CH2 C
H
CH3
C H
CH3
Name_______KEY_____________________
17a
Date___________________________ Use curved arrows to indicate electron flow in the stepwise mechanism for the reaction below. Be sure to account for the formation of both products. (6 pt.) H
H
H3C H H3C
H
+
H
H CH3
Cl
Cl
C
C
C
C
H
CH2
H C
C
C
+
Cl
H
CH3
H
H3C
C CH2
C
C
H CH3
H H3C
C C
H
H
H
C C
C
H
C
C
H CH3
Cl
H3C
Cl H
H C
C
C
C
Cl
H H
CH2
H
H C
CH3
H
CH3 H
H3C
C
Cl H3C
H
H
C
H3C
C H
H C
CH2
C
H
Cl H
H
CH3
C
CH3
Name_______KEY_____________________
17b
Date___________________________ One might argue that it is also possible to form the products below. Why are they not formed in appreciable amounts? (4 pt.) Cl
H3C H
H
H3C
C C
CH C
C
C
H
H
H
Cl
+
H3C H
C
H
+
CH3
Does NOT occur to a large extent.
CH3
CH
CH3 C
CH3
C
Cl
CH3
Protonation of the “other” double bond gives a 2° cation with a resonance structure of another 2° cation. The products in 17 a form a more stable 3° cation and its resonance structure a 1° cation. Cl
H
Cl
H
C
C
H3C Cl
C
H C
CH C
CH3
C
H
H H3C
H3C
H
C
H
C
H
H
CH3
CH3 H
C
C
H H
H
Cl
H
C
C
H
CH3 H3C C H
C
C CH
H CH3
H3C
Cl
C CH3
CH3
Name_______KEY_____________________
18a
Date___________________________ The “hard insecticide” heptachlor is (banned because of its toxicity and slow decomposition in the environment), shown below, is made by a Diels-Alder reaction. What diene and dienophile are required for that synthesis? (5 pt.) Cl Cl Cl Cl
H Cl
H Cl Cl
Cl
Cl
Cl
Cl H
Cl
Cl Cl
Cl Cl Cl Cl
Cl H Cl
Cl
Cl Cl Cl Cl
H H
Cl Cl
Cl
Name_______KEY_____________________
18b
Date___________________________ What would be the product of the diene and dienophile below? (5 pt.) O
O O
+
Name_______KEY_____________________
19a
Date___________________________ In the reaction coordinate graph below, which product is the “kinetic” product – and why? (4 pt.)
By definition, the kinetic product is the one formed fastest (has the lowest activation energy). In the reactions above, product 1 has the lower activation energy.
19b
What reaction conditions favor the formation of the kinetic product? (6 pt.)
Kinetic products are favored by short reaction times under cold reaction conditions.
