Student's Solutions Manual
to Accompany

Organic Chemistry

by Weininger and Stermitz
Thomas J. Cogdell

University of Texas at Arlington

Academic Press, Inc.

(Harcourt Brace Jovanovich, Publishers)
Orlando San Diego San Francisco New York
London Toronto Montreal Sydney Tokyo Säo Paulo


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Copyright ® 1984 by Academic Press, Inc.
All rights reserved.
No part of this publication may be reproduced or transmitted in
any form or by any means, electronic or mechanical, including
photocopy, recording, or any information storage and retrieval
system, without permission in writing from the publisher.
Academic Press, Inc.
Orlando, Florida 32887
United Kingdom Edition Published by Academic Press, Inc.
(London) Ltd.
24/28 Oval Road, London NW1 7DX
ISBN: 0-12-742362-1

Printed in the United States of America


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To the Student
This solutions manual is designed for use as a supplement to Organic Chemistry
by Stephen J. Weininger and Frank R. Stermitz.

In addition to providing complete

answers to all the problems in the text, it also contains several study features to
help broaden and strengthen your knowledge of the material presented in each
chapter.

These features are utilized in the organization of the manual as follows:

Study Hints
Problem-solving Skills
Important Terms
New Mechani sms
Reactions
For Synthesis
Special Purpose
Answers to Problems
You will no doubt need to memorize much of the text material in order to
master essential chemical concepts.

However, your study of organic chemistry can


be facilitated in two other important ways.

By learning to recognize the various

types of mechanisms involved in organic reactions, you will eventually be able to
understand and predict many of their characteristics such as rate, products, and
effects of reaction conditions.

Similarly, by learning the typical reactions for a

class of compounds, such as alcohols, you can eliminate the necessity of memorizing
the reactions of hundreds of individual compounds. Mastering these two valuable
concepts will help you better organize your study of organic chemistry, and will
reduce the amount of rote memorization required.


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Introduction

The Nature of Organic Chemistry

The purpose of this chapter is to acquaint you with the impact of organic
chemistry on our technological society and with the kinds of things organic
chemists do.
Note the definition of organic chemistry and how its emphasis has changed with
time.
alone.

Early ideas were somewhat mystical; the modern concept is based on structure

But a profound idea never entirely disappears.

For instance, the faith

that some people now place in the virtues of "organic" food owes something
to the thoroughly discredited vital force theory.
Important Terms
hydrocarbon

biosynthesis

natural product

reaction mechanism

synthesis

reaction intermediate

covalent bond

spectroscopy

tetravalent

chromatography

biochemistry



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1

A Survey of Organic Structures
Carbon Skeletons and Functional Groups

Chapter 1 is a condensed review of a course in general chemistry.

Included

are those principles of chemistry that govern structures of organic compounds;
structures, in turn, determine the way compounds may react.
For more thorough explanations of topics in this chapter, the following
general chemistry textbooks are recommended (earlier editions would also be
satisfactory).
Chemical Principles, 5th ed., W. L. Masterton, Emil J. Slowinski, and C. L.
Stanitski, Saunders College Publishing, 1981.
General Chemistry, 3rd ed., R. H. Petrucci, MacMillan Publishing Co., 1982.
Chemistry, 2nd ed., T. L. Brown and H. E. LeMay, Prentice-Hall, Inc., 1981.
General Chemistry, 3rd ed., J. E. Brady and G. E. Humiston, John Wiley and Sons,
1983.
Chemistry, 5th ed., C. E. Mortimer, Wadsworth Publishing Company, 1983.
Chemical Principles, 2nd ed., R. S. Boikess and E. Edelson, Harper and Row, 1981.
Fundamentals of Chemistry, 4th ed., F. Brescia, J. Arents, H. Meislich, and A.
Turk, Academic Press, 1980.
Additional textbooks are listed under Supplementary Reading at the end of
Chapter 1.



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Chapter One

2

Study Hints
Problem-solving Skills
After completing this chapter you should be able to do the following (text
problems that develop or test understanding of each skill are given in
parentheses):
1.

Obtain formulas from analysis percentages (Problems 10, 11, 12)

2.

Draw and use Lewis structures (1, 2, 3)

3.

Recognize structures that are ions, free radicals, or have formal charges (4,
5, 6, 16)

4.

Use molecular models, preferably your own set (7)

5.


Draw structures with three-dimensional projection (13)

6.

Draw structures of all isomers (8, 14)

7.

Distinguish functional groups and carbon skeletons (9, 15)

8.

Relate physical properties, acidity, and basicity to functional groups (17,
18, 19, 20)

Obtaining formulas
To convert from mass percent to proportions of atoms in figuring empirical
formulas, remember to divide by atomic numbers.

You may have had more experience

in making the conversion in the opposite direction in order to calculate molecular
weights from molecular formulas.
From formula to mass, multiply by atomic weights.
From mass to formula, divide by atomic weights.
Use four significant figures in your calculations.

Do not be tempted to round

off any number until after you have tried to find the lowest common denominator.

For example, a certain compound is 31.29% calcium, 18.74% carbon, and

49.97%

oxygen.
Mass

Nonintegral

Least Number

Percent

Ratio

of Atoms

.7807

1.000

Ca

1.998

C

4.000

0


Ca

31.29 —40.08 ->

C

18.74 4-12.01 -> 1.560

0

49.97 -f 16.00 -» 3.123

-» Divide by .7807 ->


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3

Student Solutions Manual for Organic Chemistry

Rounding off the nonintegral ratios gives CaC 0 , which is wrong.
CaC20 .

The substance is

Any set of coefficients that is not very close to whole numbers after

division by the least common denominator is probably not correct.


Try higher

multiples of coefficients.
There was no problem converting empirical formulas to molecular formulas
given in the text, so study the example discussed in the chapter (pages 12-13).
Drawing Lewis structures
Each element has a normal covalence that will become very familiar to you.
is simply the number of covalent bonds the atom typically makes.

It

The elements of

the second period are most important.
Group

1

2

3

4

5

6

7


Li

Be

B

C

N

0

F

Normal covalence

1

2

3

4

3

2

1


Nonbonding electron
pairs

0

0

0

0

1

2

3

Element

Covalent compounds of Li, Be, and B will not conform to the octet rule, while
those of C, N, 0, and F normally will.
and F.

Remember the nonbonding electrons of N, 0,

Donft apply these concepts to ionic compounds.

For instance, C a = 0 for


calcium oxide is a mistake.
Carbon atoms will be tetravalent in neutral compounds.

Unless ionic charge or

an odd electron is indicated, there is an error in any structure (occurring at this
point in the course) with three bonds to C.

Five bonds are never correct, since

carbon has just four orbitals in its valence shell.
Elements from the third period and beyond are not limited to an octet of
electrons, but their simplest compounds do follow the octet rule.

They have the

same normal covalence and nonbonding electrons as the element at the top of their

I
group.

I

That is, — S i — is like — C — , — S — is like — 0 — ,

and so on.

Most simple molecular formulas have a unique Lewis structure.

Work problems


like 1-1, 1-2, and 1-3 to develop skill in expanding molecular formulas to
structural formulas.

[Note:

Some of the relatively simple formulas in those

problems have two possible structures (isomers).]
in terms of structure.

Get into the habit of thinking


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Chapter One

Condensed formulas can be confusing.

4

The objective is to represent as

compactly as possible the carbon skeleton and the identity and placement of
functional groups.

Although monovalent atoms (H, F, Cl, Br, I) are generally given

after the atom to which they bond, the sequence can be inverted when it will help

to show the overall connections within the molecule.

For example, in ethanol, if

it is desired to show the hydroxyl group at the beginning of a molecule instead of
at the end, it is better to give it as HOCH^CKL· than as CH^OHCH- because this shows
clearly that the 0 bonds to one C, not both of them.
Other groups also may be inverted when placed at the left.

H«N-

is good for ami no, though H-C- for methyl is poor form.
Real charge and formal charge
Ions have real charge.
electrons.

The charge is due to different numbers of protons and

Simple ions can be formed in several ways:
+
>

a. electron loss or gain

H

b. ionizing reaction

H—Cl:


c. reaction of a molecule with an ion

H
>

+
H

e
+

:Cl:

H90 + H

—>

H~0

Path (c) has the widest application in organic chemistry.
You can recognize ions from structure by realizing that an element does not
have its normal covalence:
H

0+

I

H


or

H

0: "

H
Three bonds to oxygen

One bond to oxygen

Formal charge occurs when two different atoms bonded to each other in a
covalent structure have opposite charges.

Each of these atoms will have an unusual

covalence number, one with an extra bond, the other with one less than usual.
Ozone is an example:

V


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5

Student Solutions Manual for Organic Chemistry

Formal charges can be calculated from the formula G - N


- iB.

group number, the number of valence shell electrons in the atom.
of nonbonding electrons.

G i s the

N is the number

B is the number of bonding electrons.
/'
'Γ

For example, the structure of dimethylsulfoxide
(DMSO) is given at the right.

Sulfur and oxygen

ChL—S—ChL

have unusual numbers of bonds.
For S:

G = 6, N = 2, B = 6

For 0:

G - N - *B = 1

G=6, N=6,


B=2

G - N - iB = -1
.. —
:0:
Q^ _S-i-£H
3 ·· 3

Therefore, the formal charges can be added as at
right.

You can infer that DMSO is very polar from

this structure.
The formula G - N - iB also works to calculate real charge, as illustrated in
the structures below, which represent the prototypes of the reactive species carbon
can form.
C
HI

H—C—H

C
H

4 - 0 - 3 = 1

4 - 1 - 3 = 0


methyl
cation

methyl
radical

H—C—H

H
4 - 2 - 3 =
methide
anion

-1

4 - 2 - 2 = 0
methylene, an
unstable molecule

Drawing molecular formulas with perspective
o

The bond angles at a tetrahedral carbon are 109 .

If a line drawing is made

from an accurate model, it is convenient to place two of the bonds on the plane of
o

the paper.


Those two should be 109

apart for realism.

H

fc
From the model, you can see that this places
one of the other two bonds directly behind the other.

/ \
Cl^-^Cl
109°


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Chapter One

tL H
C

For an atom to be hidden is intolerable, so
/

the representation is twisted to one side slightly

—


—

\

ci

ci

900
to that all four atoms are clearly shown.

V

Ul
r

When doing t h i s , the two atoms which are out
of plane must be close together.

6

/

\Lli

poor

Any drawing which

places them far apart either destroys the sense of

*K

perspective or is totally meaningless.

XI

/ \
%
CI
H
impossible

Drawing isomers
No simple scheme can be devised to follow when drawing all of the isomers of a
given formula, but a systematic approach will help.

Start with the longest

straight chain possible and place the substituent atoms in all possible positions
(avoiding duplications).

Then try one carbon less in the main chain, placing it in

various positions to generate all singly-branched structures.
branched structures, if necessary, and so on.

Next, try doubly

Problem l-8b illustrates this

approach.
Oxygen atoms in the formula may be placed either between any C and H (forming OH)
or between any two C atoms (forming ethers; see Prob. l-8a).

Any formula which may

have a double bond in it (except C~H.) will have isomeric ring structures (see
Prob. l-8d).
Acidity and basicity
Your laboratory work will involve use of acids and bases in some reactions and
even more often in product isolation and purification.
that pK

a

It is helpful to remember

is like pH in more ways than the formulas that define them.

pH is lower for more concentrated solutions of acids.
pK

a

is lower for more strongly ionizing acids.


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7

Student Solutions Manual for Organic Chemistry

Important Terms
empirical formula

normal chain

molecular formula

branched chain

combustion analysis

constitutional isomer

molecular ion (M )

aliphatic

Lewis structural formula

aromatic

nonbonding electrons

cyclic

octet rule

acyclic

condensed structural formula

alicyclic

multiple bond

substituent

cation

functional group

anion

heteroatom

radical

hydrogen bond

formal charge

conjugate acid and base

molecular model

K and pK
—a
^—a

isomer

all functional group names in
Table 1-6 (p. 27)

Reactions
It is the tremendous number of different chemical reactions that has given
organic chemistry the reputation of being all memorization.

That reputation is at

least partially deserved, because memorizing reactions enables you to recognize
them more easily.

One way to memorize reactions is to use index cards.

Write the

reactants on one side and products on the other, adding any other
appropriate facts, such as the name (if it is a name reaction) and the mechanistic
type to which it belongs, or alternative reagents which could be used to accomplish
the change.

Sets of such cards can also be bought, but the problem with them is

how to concentrate on the most important reactions.

To help in this regard, brief

lists of reactions of widest application will be provided throughout this manual.
Reactions of no utility in synthesis, such as 1.1, 1.2, and 1.3 in
this chapter, will be omitted.


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Chapter One

Special Purpose

cid i o n i z a t i o n equilibrium
:0:

:0:

II

R

C

0

?

H + H„0

R—

(R represents any

II

'S

—0:

—c—

+ H-0

carbon structural
group.)

Example:

Eq. 1.4

Variations:

Other oxy acids

are similar to this.

Basicity of amines in water
H

R-

-N-

-? R

-H + H 2 0

N-—H

: OH"

+

H

H
Example: Eq. 1.7
Answers to Problems
1-1

In electron dot structures, all valence shell electrons of every atom should

be shown.

For line bond structures, the nonbonding electrons will be left off

except when they are involved in subsequent reaction.
H
a.

:C1 : C Cl

:C1
:C1
:C1 : C Cl

:C1

H

H

Cl—C—Cl

c.

I

H : C : C1:

:CJ:

Cl

H

Cl

d. H:B:H

Cl—C—Cl
I
Cl

e.

H:N:0:H
H

H H
1-2

a.

H:C:C:Br:
H H
H HH
H:C:C:C:H
H HH

H

H

K—ί—C—Br
1

H

CH 3 CH 2 Br

H

H H H
HC1

H

1

1

1

1

1

1

H H

CHoCrl^CHô

H
HCCl
I
Cl
H

I

HòH
1
H

8


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Student Solutions Manual for Organic

H H

HH
c

·

Chemistry

I

I

H—C—-C-

H:C:C:F:

CH 3 CH 2 F

HH

HH

H H

:Br:C:C:Br:

Br—C—C—Br
I I
H H

HH

i

I

BrCH 2 CH 2 Br

An isomer CH^CHBr^ i s a l s o p o s s i b l e .
H H H
e.

H
I

H:C:C:N:H

H
H

I

I

CH 3 CH 2 NH 2

I

H H

/Ö.

H:C:C

I

H—C—£—N—Ή

HH

f.

H H

,0

H

V

" Ö:H

H—C—£
I
\
H
0—Ή

CH3C00H

Other isomers a r e p o s s i b l e .

H Cl

H :C1
g. H : C : C
H :C1

Cl:

i

l

K—C—Cl I
H Cl

CH3CCI3

An isomer CH^ClCHCl« is possible.

3 a. H:C:::C:H

b.

H:C:::N:

H C=C Ή

H—C=N

H H
c. C::N
H
H
d. H:C:N::0
H


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Chapter One

HH

1 1

H:C:C::0

1

=0

H

H

C::C::0

C= =C=0

H

H

H Ή
H :0: H

j

H

0

I II

H:N:C:N:H

I1

H—c — H

H 0 H
I It I
H—CI
I
H
H

H:C:C:C:H

H

H

1-4

H :θ: Η

10

H

I

H—N—C—N—H

The nonbonding electrons are necessary to

I
H
1-5

justify negative charge.

Two approaches may be used.

The easy one is to recall that a balanced

equation has the same net charge on each side.

—>

Thus for (a), NH~ + H

NH. .

Alternatively, since tetracovalent N has shared an electron pair, its charge
must have increased, from neutral to +1.

The formula G - N - iB also gives +1.

Pick a system you like and stick to it.
W
c.

e.

1-6

d.

hh—C—H (neutral)

In each pair, the structure with formal charges is less probable,

a. C::0 or H—C=(A-H
M
i

H
b.

-

c.

"

H—0—€=N:

or

H—N=C=0

approximately equal

H — 0 — N = 0 or

H—N=0
I _"
:0:

d.

.. ..- ..+
H—0—€=0—Ή

or

H—C—O—H
II

:0:

"

a,b. The model can be oriented to resemble the drawing.

vv\
H H

H H

HH

HH

.. .

CH 3 C—0:

CH3—(j—H (neutral), CH3—Ư:"

H

1-7

II

H ,+
«

+ H—0—H


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11

Student Solutions Manual for Organic Chemistry

OH
1-8

a.

CH3CH CH2OHr CH 3 CHCH 3 , CH OCH 2 CH 3
CH.

CH-

CH CH CH CH Br, CH CH CHBrCH , CH CHCH Br, CH CCH
Br
CH,
CH.CH.CH_CHJ:H0,
6
Z
Z
Z
ό

CH,

CH0CH_CHCH0,CH0CCH0
ό
Ζ
J
J i J

CH,
CH,

CH 2 =CHCH 2 CH 3f CH 3 CH=CHCH 3f CH 2 =CCH 3f CH 2 CH 2 , CH 2
CH 2

XcH

CH 3 CH 2 CH 2 CHBr 2

C H C H CHBrCH Br

2
CH CHBrCH CH Br

BrCH2CH2CH CH Br

CH CH2CBr2CH

CH CHBrCHBrCH

CH„

CHn
CH3CHCHBr

1-9

CH 2

CH CH 3

CH^

CH CBrCH2Br

a. -I iodo

BrCH CHCH2Br

Br bromo, -COOH carboxy

alkyl iodide

bromocarboxylic acid
Br

-NH2 ami no

0
OH

alkylamine

ether, amine (or alkoxyamine)

Cl,
OH

f. -F fluoro, C=0 oxo

-Cl chloro, -OH hydroxy

fluoroketone

chloroalcohol

1-10
1-11

C

12 H 13 N °2

C

10 H 14°3
The analysis gives an empirical formula Q.r ro H,„ „-ISL Λ Λ .
6.58 10.47 1.00
analysis gives (approximately) C ^ ^ j j ^ ·

Doubling the


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Chapter One

C

13 H 21 N 2

calculates

76.05% C, 10.31% H, 13.64% N.

C

13 H 20 N 2

calculates

76.42% C, 9.86% H, 13.71% N.

C

H

N

calculates 76.38% C, 10.25% Hf 13.36% N.

These are all within 0.30% absolute error.

Others are possible, especially with a

few H atoms more or less.
1-12

a.

1-13 a.

98

b.

98

c. 82

HH

d.

VV"
C

H

..H

/·.

0

H H

H

^ C C^3CH3
H

b. H
.H
\
\ C _ N ^H

πH πH

:P

H

H

e. H- / V - H
.c—c.

H

1-14

a.

I'

H

H

I

I

H

H

H

HH

CH CHF2,CH FCH F
CH 3

b.

CH 3 CH 2 CH 2 CH 3 , CH 3 CHCH 3

C.

CH CH -O-O-H, CH O-OCH , CH OCH OH, HOCH CH OH, CH CH(OH),

d.

CH0C=CH, CH =C=CHn, CH=CH
3
2
2
χ /
CH 2

e.

CH3CH2C=CH

CH C=CCH

CH0—CHn
CH
, 2= C H, 2

CH=C
CH
χ /

(Note:

CH 0
3

CH3CH=C=CH
CH=CH

\ /

CHCH

CH2=CHCH=CH2
CH

CH 0

1 NJ 2
CH —CH
2

Two extra structures are given for good measure.)
NH 0
,2

1-15

12

CH,3

f.

CH C H C H J H , CH CHCH , CH CH NHCH , CH NCH

a.

-Cl, chloro

b.

chloro and -Ff fluoro


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Student Solutions Manual for Organic Chemistry

// aryl (specifically phenyl)

c.

-NH-, amino,~\v

d.

-0Hf hydroxyl, plus amino and phenyl
When -OH is directly on phenyl, the compound is phenolic.

e.

phenolic and -COOH, carboxyl
0

II
f.

hydroxyl, carboxyl, -C-, oxo or carbonyl, and C=C, alkenyl,
The ring is not considered functional because it is
saturated; it has no multiple bond.

16

a.

CH3CH2CH2f CH CHCH3

b.

CH3—0+—H

C.

CH0=CH-CH0, CH0-CH=CH, CH--C=CHof CH 0

CH 0
CH
+

CH 3 -C-CH 2
H
d.

CH^CH^N
H

and CH3CH2C+ are possible, less stable.
H
H, CHINCH-,
H

17

Br~, Cl~, HSO~, N0~, F~, C ^ C O O " , CH3COO~,

18

The reaction is between the acid HF and CH^COO , acting as a base.

CN~M

C

6

H

5

°~

HF + CH3COO" — > F~ + CH3COOH
:0:

II 19

H-C-O-H
HCOOH + H 2 0 — > HCOO~ + H 3 0 +
At equilibrium [H 3 0 + ] = [HCOO~] = .013 x 1.0 M = 0.013 M
[HCOOH] = .987 x 1.0 M = .987 M


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Chapter One

[H~0 ][HCOO ]
K =—^
*[HCOOH]
pK

1-20

a

=

(.013)
(.987)

=

14

Δ

1.7 x 10~ 4

= -log K = -log 1.7 x 1 0 - 4 = 3 . 8
a

1-Bromobutane is more polar than butane because the electronegativity of

bromine will polarize the C-Br bond.

Dipole-dipole attractions between molecules

will exist in the liquid, which must be broken to vaporize 1-bromobutane.
will require a higher temperature to provide sufficient kinetic energy.

This

Probably

more important, the bromine atom carries a relatively large shell of 35 electrons,
which are polarizable, increasing van der Waals forces between molecules.
Since the chlorine atom is less polarizable, 1-chlorobutane should (and does)
boil at an intermediate temperature.


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2

Molecular Bonding

Bond Angles, Bond Energies,
and Bond Lengths

Molecular model sets are built with the ideal bond angles predicted by equal
valence shell electron repulsions.

But the distortion from ideal angles for

nonequivalent electron pairs requires a somewhat "softer" concept of what a
molecule is like.

That is why most organic chemists prefer flexible plastic

molecular models to the more rigid stick models.

The softer models give a more

accurate representation of how atoms actually behave in a molecule.
Study Hints
Problem-solving Skills
After completing this chapter you should be able to do the following:
1.

Predict molecular shape and bond angles for ideal molecules and those
distorted by unequal electron pair repulsions (Problems 1, 2, 3, 13, 17)

2.

Assign electrons to atomic orbitals (5)

3.

Relate electron energies to orbitals they occupy (4)

4.

Combine atomic orbitals into molecular orbitals (6, 14, 19)

5.

Correlate hybridization with molecular shape and bond angles (7, 8, 15)

6.

Utilize the ΊΤ bonding concept to explain molecular shape and bond
angles

7.

Solve problems using bond energies (9, 20, 21, 22)

8.

Recognize direction and size of bond polarity from differences in
electronegativity (10, 11, 12, 23, 24, 25, 26, 27)


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Chapter Two

16

Molecular shapes from VSEPR
Remembering the ideal shapes of molecules is easy because they are the most
symmetrical, geometrical shapes—line, triangle, tetrahedron, then trigonal
bipyramid and octahedron.

The most common mistakes in identifying molecular shapes

are:
1.

Using an incomplete Lewis structure.

Remember the nonbonding electrons on

N, 0, and their analogs from later periods, P, S, etc.

If they are omitted, space

will not be allowed for them and the wrong molecular shape will be expected.
2.

Exaggeration of the space required for multiple bonds.

are counted for determining bond orientations; the Π

Only the

σ

bonds

electrons are not counted in

the first estimation because they are out of the plane of the

σ

electrons and

cause much less electron pair repulsion.
3.

Deformation of real molecules from the ideal shapes are determined mainly

by the nonbonding electrons.

Since nonbonding electrons do not have another

nucleus at the end of their orbital, as bonding electron pairs do, they are more
closely drawn to the one nucleus they do have and, consequently, occupy a larger
portion of the space near it.
Electron configurations
Here are the basics for writing electron configurations.

You should know:

1.

the order of increasing energy of atomic orbitals;

2.

the number of orbitals of each type in a shell (one s_, three £, five d,
seven f_) ;

3.

when each type of orbital begins in the Periodic Table;

4.

the Pauli principle limitation on number and spins of electrons in a
single orbital;

5.

Hund's rule for the sequence of filling degenerate orbitals.

There are two kinds of test questions about electron configurations.

One type

deals with ground state electron configurations, in which all the basics above must
be followed.

In a multiple choice test, trick answers to this type of question may

include errors like using 3d orbitals too soon immediately after 3JD, for example,
using nonexistent orbital types such as l£ or 2d, or pairing electrons


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17

Student Solutions Manual for Organic Chemistry

that should remain separate in degenerate orbitals.
The other kind of question is about excited state electron configurations, in
which one electron is in a higher energy orbital but the Pauli principle must
still be obeyed.
Waves
It is difficult to accept the idea that an electron in an atom or a molecule

behaves like a wave.

Think of it as a special kind of wave, a standing wave

confined to a limited space.

Unlike a wave at the seashore, which moves to the

beach and vanishes, an electron wave is like a wave in a bathtub, rebounding off
the ends and setting up a resonant oscillation.
of movement of this water.

Imagine two of the possible modes

One type has the water going to a peak at the center

when it is down at the walls.

This resembles the s_ orbital.

Another attains a

maximum in one half of the tub and a minimum in the other and oscillates back and
forth, resembling a p orbital.
Orbital signs and nodes
Study the way that the positive and negative waves of carbon p orbitals become
extended over the ethylene molecule in forming the Π bond (Fig. 2-14, p. 44). Note
also that the node in the atomic orbitals becomes a node in the molecular orbital.
These ideas recur in all of the more elaborate

Π-bonding systems to come.

Energies of molecular orbitals
You should now be familiar with the relative energies of atomic orbitals and
their impact on ground state electron configuration, ionization energies, and light
absorption or emission.

The energy levels of molecular orbitals have corresponding

effects, in addition to affecting which bonds in a complex molecule break under
stress.


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Chapter Two

18

For two atoms bonded together, the order of increasing orbital energy is
usually as follows:
(Γ , Π , n,

Π*,

cr* (where n designates nonbonding orbitals).

same type do not all have the same energy. A C-C
tfbond, but both are stronger than a C-C

Π

Orbitals of the

(T bond is not as strong as a C-H

bond.

The strength of C-H

cr bonds also depends upon the hybridization of the carbon atom.
Bond dissociation energies
Bond dissociation energies are more frequently used than average bond energies
because of their importance in the energy changes in certain reaction mechanisms.
The difference in C-H bond strengths in Equations 2.6, 2.7, and 2.8 are very
significant in determining which C-H bond will react in complex hydrocarbons where
primary, secondary and tertiary types may all be present.
The most common mistake in using bond dissociation energies is failure to
recognize that the energy change is appropriate if only that bond is broken.

For

example, don't use the C-0 bond dissociation energy for carbon monoxide (it is
multiple bonded) or CH^OH+H ~*CH3

+ H 2 0 (an HO bond has also been

made and the C-0 bond breaking was not homolytic).
Electronegativities
Since relative electronegativity is indicated by position in the Periodic

Table, bond polarities can usually be inferred without use of the numerical values.
However, the position of H is misleading.

It is much more electronegative than

Group 1 metals, almost equal to carbon.
It is also useful to think of carbon as a reference point and to learn those
elements (only five) that are more electronegative than it is.
elements these are and remember them.

Determine which


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19

Student Solutions Manual for Organic Chemistry

Important Terms
tetrahedron

valence shell electron pair

orbital

repulsion (VSEPR)

ground state

node

ionization energy

excited state

levels

degenerate energy

σ orbital

wave function

bonding and antibonding orbitals

Π orbital

hybrid orbital
2
S£
trigonal

photochemi stry
sp_3

average bond energy

bond dissociation energy

electronegativity

homolysis

«a

polar bond

S£

P^a
Answers to Problems
2-1

Planar CH-, 90° bond angles.

The electron pairs in bonds 90° apart would

either be closer together than for bonds 109° apart, or the bonds would be
lengthened if the electrons moved farther out on the bonds.

Either way would make

the molecule less stable.
2-2

The ideal angle is 109.5°, since the oxygen has four electron pairs around it.
μ

The repulsion between bonding pairs (<

>) is less than

that between a nonbonding pair and a bond pair (<
H
2-3

>),

so the angle between the bonding pairs decreases.

a.
a.

and b.

tetrahedral, 109.5°

c.

distorted tetrahedron.

Cl-C-Cl greater than 109.5° because Cl has

nonbonding electrons; H-C-Cl about 109.5°; H-C-H less than
109.5°
distorted trigonal, Cl-B-Cl greater than 120°; H-B-Cl less than
120°
tetrahedral at both C and 0; H-C-H less than 109.5°; H-C-0 greater

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Chapter Two

20

than 109.5°; C-O-H less than 109.5°
f.

tetrahedral, 109.5°

g.

linear, 180°

h.

distorted trigonal, H-C-H less than 120°; H-C=C greater than
120°, since the two electrons of the Π bond give a slight added
repulsion

2-4

i.

linear, 180°

j.

trigonal, 120°

k.

distorted tetrahedron, H-C-H less than 109.5°

a.

He:—> He
Li

+ e~" (from lss orbital)

— > Li

+ e_

(from 2ε> orbital)

The energy level of the 2^ orbital is higher, so it takes less added
energy to take the electron away.
b.

This ionization requires more energy to remove one of the electrons from

the ls_ orbital, which produces an excited state electron configurations l£ 2s> .
(Incidentally, this proves that the increase in nuclear charge from 2 for He to 3
for Li causes the electrons in the 1^ orbital of Li to become more stable.)
2-5

Element

Electron Configuration
Filled

Na

[Ne]

Mg
Al
Si
p

[Ne]
[Ne]
[Ne]
[Ne]
[Ne]
[Ne]
[Ne]

S
Cl
Ar

3s

3D

3p

3p

+
tY

-H

+

tr

+

A

ΪΨ

A

•r

A

u
u

A '

A

A

Ή
Ή

H

t

ti

Αψ

tv

2-6
Na )

oo

b.

3s

3p

f Na
3s

o-o
3P


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21

Student Solutions Manual for Organic Chemistry

"Q

ΟΌ

Is

e.

OOOO

4P

o

3s.

5p

'Θ oYo

0Cfc>

2 p

is

- _x,

Z

'3p.

Θ

Θ

Pz

Is

Is

Oxygen atomic orbitals actually hybridize s_p upon bonding (see sec. 2:14)
2-7

a.

:ci:
,

£

δ
.

b.

3

H

,

j."-o

0

sp -3p

,

spJ-ls

£,.
°sp
••tfYSST-·
:Cl:

H-

Cl:

J

-3p

sp

d
=

L· ?sp 3 -2p

?sp3-ls

·

-7

A'
H

SP

e.

u

V

o
n3
, csp
-Is

f

3

3

V*sp3-ls
l l 3

sp

*SD3-1S
H / -t

-

3 3

~>T^O

\

s:—'6

3

3

CH3

H

C—S:

SP^

2-8

^

έ

a.

B s_£ 2 ; B - C l

b.

C S £ 2 , N s p 2 ; C-H

c.

C ε>£, 0 sj> ; C=0

d.

N S£;

N=N

3
e.

C SJD ,

f.

-0-

σ

0"

S£

SP

-3£
σ

sp2-l£,

(Γ s p - s p

sp-sp

and

2
B SJD ; C-B

S£3,

Π 2£-2£

N S£2,

=0

σ

IP "5P

S£

s_£ 2 ;

C=N

and

two

Π

3

2
-sp

0-H

Π

σ

S £ 2 - S £ 2 and

Π

2 p - 2 p ; N-H

σ

S£2-l

2£-2£

2JD-2£
3

; C-H
σ

σ s_p

s_£3-ls,

-Is
0-N

(Γ S £ 3 - s _ £ 2 f

N=0

σ

S£2-S£