Download Chapter 11 Introduction to Organic Chemistry Part 2

Chapter 11 Introduction to Organic Chemistry
1.
Which of these molecules is unsaturated?
A. C3H8
B. CH3OH
C. C5H10
D. CH4
E. C4H10
2.
Which of these species are structural isomers of C6H14?
A.
B.
C.
D.
E.
3.
I and II
I and III
II and III
II and IV
III and IV
Which of these pairs are geometric isomers?
A.
B.
C.
D.
1
4.
The two molecules represented below are examples of
A.
B.
C.
D.
E.
5.
isomers
isotopes
alcohols
carboxylic acids
unsaturated hydrocarbons
The two molecules represented below are examples of
A.
B.
C.
D.
E.
geometric isomers.
structural isomers.
optical isomers.
stereoisomers.
none of these
6.
Which of these species is an aromatic compound?
A. C2H2
B. C6H12
C. C6H4Br2
D. C5H10
E. C2H4Br2
7.
Which of these is the systematic name for the compound represented below?
A.
B.
C.
D.
E.
8.
2-ethylbutane
3-methylpentene
3-methyl-1-pentene
3-methyl-1-hexene
2-methylhexane
The systematic name for the compound represented below is
A.
B.
C.
D.
E.
4,5-diethylheptane.
3-propyl-4-ethylhexane.
3-ethyl-4-propylhexane.
3-methyl-4-propylheptane.
2-ethyl-4-propylhexane.
2
9.
The correct structure for 2,3,3-trimethylpentane is
A.
B.
C.
D.
10. Which type of organic compound does not contain a carbonyl group?
A. ethers
B. carboxylic acids
C. ketones
D. aldehydes
E. esters
11. Which is the product of the reaction of one mole of HCl with one mole of 1-butyne?
A. 1-chloro-1-butene
B. 1-chloro-2-butene
C. 2-chloro-1-butene
D. ethyl chloride + acetylene
12.
Write the formula for the alcohol and the carboxylic acid from which the following ester may be
synthesized.
13.
Name the following compounds:
3
14. Which choice gives the structures of the reaction products when the ester below is hydrolyzed in acid
solution?
A.
B.
C.
D.
15. Which of the following compounds are isomers?
I. pentane
II. 2-methylbutane
III. 2,3-dimethylbutane
IV. 2,2-dimethylpropane
V. 1-hexene
16. The oxidation product of 1-propanol when using Cr2O72- as the oxidizing agent is acetone.
True
False
17. The reaction of hydrogen chloride gas with propene will yield 1-chloropropane as the main product.
True
False
18. Cycloalkanes have the general formula CnH2n.
True
False
4
Answers to Chapter 11
1.C
2.D
3.D
4.A
5.B
6.C
7.B
8.D
9.A
10.A
11.C
12.
13. 2,3-dimethybutane; 4,5-dimethyl-2-hexene; 3,3-dimethylpentane
14.B
15.I, II, IV
16.False
17.False
18.True
1
Chapter 11 Introduction to Organic Chemistry
Part 2
1. Name the compounds in the following Newman projections.
a
b
c
2. (a) Which of the structures represents the most stable conformation of 2-methylbutane? Explain why.
(b) Which of the structures represents the least stable conformation of 2-methylbutane? Explain why.
3. Draw the structures of the following compounds
(a) 4-ethyl-2-methylheptane
(b) methylcyclobutane
(c) trans-6-methyl-2-heptene
4. Which of the following compounds are optically active?
CH3
︱
(a) C2H4 (b) C6H6 (c) C6H4Cl2 (d) CH3CH2CHCH2CH2CH3
2
5. Determine whether the following isomers are enantiomers or diastereomers. What characteristic feature of the
molecule can be used to make the decision?
6. Predict the products (A and B) of the following reactions.
7. Write the mechanism for the following reaction.
3
Answers
1 (a) 2-methylbutane
(b) 2-methylbutanol or 2-methyl-1-butanol or 2-methylbutan-1-ol
(c) 2-bromobutane or sec-butyl bromide
2 (a) Although both structures A and C are the staggered conformations of 2-methylbutane and more stable than
structures B and D, which are eclipsed conformations, structure A will be the most stable conformation
because all the methyl groups and hydrogen atoms are farther apart from each other than those in structure C.
Hence there will be less repulsion among the atoms and less energy in structure A.
(b) Although both structures B and D are the eclipsed conformations of 2-methylbutane, structure B will be the
least stable conformation because the methyl groups are closer to each other, leading to stronger repulsion
among the atoms and higher energy in structure B.
3 (a)
or
(b)
or
or
(c)
4 (d)
4
5. The two isomers, configurations (a) and (b), are enantiomers because each molecule contains two stereogenic
centers (asymmetric carbon atoms) and their mirror images are non-superimposable to each other.
6. A.
O─
│
C─
│
CH3
CH3
OH
│
C─
│
CH3
CH3
B.
7. It will be an unimolecular nucleophilic substitution (SN1) reaction
+ OCH3─
→
5