Download Chem 2423-Test 2 - HCC Learning Web

Name: ________________________ Class: ___________________ Date: __________
Orgo1-Test 2
Short Answer
Draw structures corresponding to each of the following names.
1. cis-1-sec-butyl-2-ethylcyclopentane
2. 3,5-dicyclohexylnonane
3. Draw the two stereoisomers of 1,3-dibromocyclobutane.
For the pair of molecules below, circle the more stable one.
CH 3
H
CH 3
H
H
CH 3
H
CH3
and
H
CH 3
CH3
4.
H
Classify each reaction below as a(n):
a.
b.
c.
d.
addition
elimination
substitution
rearrangement
Place the letter corresponding to the correct answer in the blank to the left of the reaction.
5. _____
6. _____
7. _____
1
ID: A
Name: ________________________
ID: A
Identify the functional groups present in the compound below and predict the direction of polarity in each.
8. mustard gas Cl−CH2CH2−S−CH2CH2−Cl
9.
Classify each structure below as a nucleophile or electrophile and briefly explain your choice.
10.
11.
12.
Identify the nuclephile and electrophile in each reaction below and label them.
13.
14.
2
Name: ________________________
ID: A
Consider this reaction when answering the following question(s):
15. This reaction is an example of:
a.
b.
c.
d.
a substitution reaction.
a rearrangement reaction.
an addition reaction.
an elimination reaction.
Use the reaction energy diagram below to answer the following question(s).
16. The reaction depicted in this reaction energy diagram can best be described as:
a.
b.
c.
d.
a slow exothermic reaction
a fast exothermic reaction
a slow endothermic reaction
a fast endothermic reaction
17. The transition state is found at point _____ on the diagram.
18. The products are found at point _____ on the diagram.
19. The free-energy change for the reaction is indicated at point _____ on the diagram.
20. The reactants are found at point _____ on the diagram.
3
Name: ________________________
ID: A
Consider the reaction of 2-bromo-2-methylpropane with water, shown below, to answer the following
question(s).
Diagram 1: The first step of this reaction is shown below.
Diagram 2: The second and third steps of the reaction are shown below.
21. Using the bond dissociations values in the table below, calculate the ∆H° for the reaction in Diagram 2
above. Show your calculations for full credit.
Bond
(CH3)3C−Br
(CH3)3C−OH
HO−H
H−Br
D (kJ/mol)
263
380
498
366
Calculate the degree of unsaturation in each formula below. Show your calculations.
22. pinene, C10H18
23. diazepam (Valium), C16H13N2OCl
Draw structures corresponding to each name below. Either condensed or line structures may be used.
24. 3,6-dimethyl-1,4-cyclohexadiene
4
Name: ________________________
ID: A
Provide names for each structure below. Be sure to include the cis, trans or E, Z designations where
applicable.
25. Name:
Rank each set of substituents using the Cahn-Ingold-Prelog sequence rules by numbering the highest priority
substituent 1 and numbering the lowest priority substituent 4. Place the number in the blank below the
substituent.
26.
Assign E or Z configurations to each alkene below.
27.
28.
Predict the major organic product for each reaction below.
29.
5
Name: ________________________
ID: A
30.
6
ID: A
Orgo1-Test 2
Answer Section
SHORT ANSWER
1. ANS:
CH 3
CH 2 CH3
CHCH 2 CH 3
H
H
PTS: 1
2. ANS:
PTS: 1
3. ANS:
Br
Br
and
Br
Br
cis
trans
PTS: 1
4. ANS:
CH 3
H
CH 3
H
H
CH 3
H
CH3
and
H
CH 3
CH3
H
PTS: 1
5. ANS:
c
PTS: 1
1
ID: A
6. ANS:
d
PTS: 1
7. ANS:
a
PTS: 1
8. ANS:
PTS: 1
9. ANS:
PTS: 1
10. ANS:
Azide is a nucleophile since it has a net negative charge (and lots of electron pairs!).
PTS: 1
11. ANS:
Hydronium ion is an electrophile since it has a positive charge.
PTS: 1
12. ANS:
Phenol can be a nucleophile or an electrophile
PTS: 1
2
ID: A
13. ANS:
PTS: 1
14. ANS:
PTS: 1
15. ANS:
b
PTS: 1
16. ANS:
b
PTS: 1
17. ANS:
B
PTS: 1
18. ANS:
D
PTS: 1
19. ANS:
C
PTS: 1
20. ANS:
A
PTS: 1
3
ID: A
21. ANS:
∆H°
= ∆H° bonds broken − ∆H° bonds formed
= 761 kJ/mol − 746 kJ/mol
= +15 kJ/mol
Bonds Broken
(CH3)3C−Br
HO−H
263 kJ/mol
498 kJ/mol
761 kJ/mol
(CH3)3C−OH
H−Br
Bonds Formed
380 kJ/mol
366 kJ/mol
746 kJ/mol
PTS: 1
22. ANS:
The saturated ten carbon compound would have 22 hydrogens so the number of degrees of unsaturation is:
(22 − 18) ÷ 2 = 4 ÷ 2 = 2.
PTS: 1
23. ANS:
Oxygen does not affect the base formula. A hydrogen is added to the base formula for each halogen and
subtracted for each nitrogen so the base formula for diazepam is C16H12. The saturated 16 carbon compound
would have 34 hydrogens so the number of degrees of unsaturation for diazepam is: (34 − 12) ÷ 2 = 22 ÷ 2 =
11.
PTS: 1
24. ANS:
PTS: 1
25. ANS:
(2E,4E)-5-ethyl-6-methyl-2,4-heptadiene
PTS: 1
26. ANS:
PTS: 1
27. ANS:
E
PTS: 1
4
ID: A
28. ANS:
E
PTS: 1
29. ANS:
PTS: 1
30. ANS:
PTS: 1
5