Transcript Chapter 12 quiz - Ventura College

Organic Chemistry, 6th edition
Paula Yurkanis Bruice
Chapter 12
Radicals:
Reactions of Alkanes
Brian L. Groh
Minnesota State University, Mankato
Mankato, MN
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
12.2
What is the first step in radical halogenation
of alkanes?
A.
B.
C.
D.
E.
Abstraction
Initiation
Propagation
Combination
Termination
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
12.2
What is the first step in radical halogenation
of alkanes?
A.
B.
C.
D.
E.
Abstraction
Initiation
Propagation
Combination
Termination
Initiation is the “initial” step in a radical process.
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
12.4
Calculate the percentage yield of the product
below formed by radical bromination.
[Relative reactivities: 1°(1), 2°(82), 3°(1600)]
Br2
ROOR, heat
Br
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
A.
B.
C.
D.
E.
5%
15%
29%
41%
71%
12.4
Calculate the percentage yield of the product
below formed by radical bromination.
[Relative reactivities: 1°(1), 2°(82), 3°(1600)]
Br2
ROOR, heat
Br
% yield = relative reactivity X no. H’s *100
sum of relative amounts
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
A.
B.
C.
D.
E.
5%
15%
29%
41%
71%
12.6
Which compound will predominate in the
radical bromination of 1-methylcyclohexene?
B
A
C
HBr
ROOR, heat
Br
Br
Br
Br
Br
D
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
E
12.6
Which compound will predominate in the
radical bromination of 1-methylcyclohexene?
B
A
C
HBr
ROOR, heat
Br
Br
Br
Br
In radical additions, the bromine
radical adds to the least substituted
carbon of the C=C.
Br
D
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
E
12.8
Rank the radicals in decreasing order of
stability.
CH2CH3
CH3CHCH3
1
A.
B.
C.
D.
E.
2
CH2CH=CH2
(CH3)3C
3
1>2>3>4
4>3>2>1
3>4>2>1
1>2>4>3
4>2>3>1
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
4
12.8
Rank the radicals in decreasing order of
stability.
CH2CH3
CH3CHCH3
1
A.
B.
C.
D.
E.
2
1>2>3>4
4>3>2>1
3>4>2>1
1>2>4>3
4>2>3>1
CH2CH=CH2
(CH3)3C
3
4
The allyl radical is resonance
stabilized. The other radicals
are only weakly stabilized by
hyperconjugation.
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
12.8
Which compound will give one product upon
radical bromination using NBS, ROOR, and
heat?
A
B
C
D
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
E
12.8
Which compound will give one product upon
radical bromination using NBS, ROOR, and
heat?
A
B
C
D
In compound A, abstraction of any allylic
hydrogen leads to only one resonance
stabilized allylic radical.
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.
E
12.10 Which reaction sequence would likely give the
product in reasonable yield?
OH
H3CO
A.
B.
C.
D.
E.
1. Br2/hʋ; 2. (CH3)3COK/(CH3)3COH; 3. CH3CO3H; 4. CH3O-/CH3OH
1. Br2/ROOR, hʋ; 2. HO-/H2O 3. CH3CO3H; 4. H+/CH3OH
1. Br2/hʋ; 2. (CH3)3COK/(CH3)3COH; 3. H2O2/H2O; 4. CH3O-/H2O
1. NBS/hʋ; 2. CH3O-/CH3OH; 3. H2O2/THF; 4. H2O/CH3OH
1. NBS/hʋ; 2. CH3O-/CH3OH; 3. CH3CO3H; 4. H+/CH3OH
Bruice: Organic Chemistry, © 20101Pearson Education, Inc.
12.10 Which reaction sequence would likely give the
product in reasonable yield?
OH
H3CO
A.
B.
C.
D.
E.
1. Br2/hʋ; 2. (CH3)3COK/(CH3)3COH; 3. CH3CO3H; 4. CH3O-/CH3OH
1. Br2/ROOR, hʋ; 2. HO-/H2O 3. CH3CO3H; 4. H+/CH3OH
1. Br2/hʋ; 2. (CH3)3COK/(CH3)3COH; 3. H2O2/H2O; 4. CH3O-/H2O
1. NBS/hʋ; 2. CH3O-/CH3OH; 3. H2O2/THF; 4. H2O/CH3OH
1. NBS/hʋ; 2. CH3O-/CH3OH; 3. CH3CO3H; 4. H+/CH3OH
Acceptable reactions: Step 1 – A, B, or C; Step 2 – A and C are best;
Step 3 – A, B or E; Step 4 – only A
Bruice: Organic Chemistry, © 2011 Pearson Education, Inc.