Students solutions manual to accompany organic chemistry

To the Student This solutions manual is designed for use as a supplement to Organic Chemistry by Stephen J.. 3 Student Solutions Manual for Organic Chemistry Rounding off the nonintegra

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

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

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

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

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

chromatography

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

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,

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%

3 Student Solutions Manual for Organic Chemistry

Rounding off the nonintegral ratios gives CaC 0 , which is wrong The substance is CaC20 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 It

is simply the number of covalent bonds the atom typically makes The elements of the second period are most important

Group Element Normal covalence Nonbonding electron pairs

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 I

group 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).] Get into the habit of thinking

in terms of structure

Condensed formulas can be confusing 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 H«N-is poor form

Real charge and formal charge

Ions have real charge The charge is due to different numbers of protons and electrons Simple ions can be formed in several ways:

+

-a electron loss or gain H > H + e

b ionizing reaction H—Cl: > H + :Cl:

c reaction of a molecule with an ion 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:

I

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

5 Student Solutions Manual for Organic Chemistry

Formal charges can be calculated from the formula G - N - iB G i s the

group number, the number of valence shell electrons in the atom N is the number

of nonbonding electrons B is the number of bonding electrons

' Γ

have unusual numbers of bonds

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

cation radical anion 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

For an atom to be hidden is intolerable, so tL H

poor

of plane must be close together Any drawing which

places them far apart either destroys the sense of *K XI

perspective or is totally meaningless / \

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 Next, try doubly

branched structures, if necessary, and so on 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 It is helpful to remember

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

pH is lower for more concentrated solutions of acids

pK is lower for more strongly ionizing acids a

7 Student Solutions Manual for Organic Chemistry

aromatic cyclic acyclic alicyclic substituent functional group heteroatom

hydrogen bond conjugate acid and base

K and pK

—a ^—a 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

group.) Example: Eq 1.4

Variations: Other oxy acids are similar to this

Basicity of amines in water

H R- -N-

H

-H + H 2 0 -? R N - — H + : OH"

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

Student Solutions Manual for Organic Chemistry

H Ή

H :0: H H:N:C:N:H

H 0 H H—C-

H justify negative charge

1-5 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

11 Student Solutions Manual for Organic Chemistry

OH 1-8 a CH3CH CH2OHr CH3CHCH3, CH OCH2CH3

CH CHBrCHBrCH CH^

Doubling the

Student Solutions Manual for Organic Chemistry

c -NH-, amino,~\v // aryl (specifically phenyl)

d -0Hf hydroxyl, plus amino and phenyl

When -OH is directly on phenyl, the compound is phenolic

e phenolic and -COOH, carboxyl

CH3-C-CH2 and CH3CH2C+ are possible, less stable

Since the chlorine atom is less polarizable, 1-chlorobutane should (and does) boil at an intermediate temperature

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)

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 Only the σ bonds are counted for determining bond orientations; the Π 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

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

of movement of this water 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

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) Orbitals of the same type do not all have the same energy A C-C (T bond is not as strong as a C-H

tfbond, but both are stronger than a C-C Π 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 + H20 (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 Determine which elements these are and remember them

19 Student Solutions Manual for Organic Chemistry

Π orbital photochemi stry

3 sp_

S£

bond dissociation energy homolysis

polar bond P^a

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 (< >),

so the angle between the bonding pairs decreases

2-3 a and b tetrahedral, 109.5°

H

a

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

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

i linear, 180°

j trigonal, 120°

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

2-4 a He:—> He + e~" (from lss orbital)

Li —> 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

21 Student Solutions Manual for Organic Chemistry

9 9 9 9 9

g C s_p , 0 s_p ; C-Cl σ s_p - 3 p , C=0 σ S£ -s>£ and Π 2 p - 2 £

3 3 3

h C SJD , Hg SJD; C-H σ s_p - l s _ , C-Hg σ SJD - s p

2-9 Dimethyl ether total bonds = 6 C-H + 2 C-0

total bond energy = (6 x 97 kcal/mol) + (2 x 86 kcal/mol)

= 75 4 kcal/mol Ethanol total bonds = 5 C-H + 1 C-C + 1 C-0 + 1 0-H

total bond energy = [(5 x 97) + (1 x 84) + (1 x 86) + (1 x 104)]

kcal/mol

= 759 kcal/mol Ethanol has the lower energy (total molecular potential energy), since to

decompose it completely to atoms would require 5 kcal/mol more energy than the

decomposition of dimethyl ether to form the same atoms

2-11 HCl is a stronger acid than HF Since these are not of the same period,

is not sufficient to consider only electronegativity Because of its larger si

in which negative charge is dispersed, Cl is more stable than F , so HCl is mo

ionized in water

2-12 CH3NH2 + H20—»CH3NH~ + Η3<3+

Since N is more electronegative than C, an N-H bond is more readily

dissociated ionically than a C-H bond

2-13 a CH , nonbonding electron pair in CH ~ compresses the angle between

bonding pairs

b NH- , nonbonding electron pair in NH^

c H C=0 is trigonal, angle near 120°; H C(OH) tetrahedral

Student Solutions Manual for Organic Chemistry

d CH trigonal; CH2 trigonal but with a compressed angle due to the nonbonding electron pair

e CH^Cl, because two Cl atoms in CH^Cl« compress the HCH angles more

2-17 a The CH groups have an added

repulsion due to nonbonded

interactions of the H-atoms

(< >)

b In H 2 0 r nonbonding electrons on

the 0 compress the H-O-H angle by

"pushing" the 0-H bonding electrons

This force is present in dimethyl

ether also, but the nonbonding

interactions between methyl groups

2-22 Using average bond energies, all C-H bonds are equal, so the only calculation

is to look for the bond broken and the bond made The number of other bonds

remains constant

When only a Π bond of a double bond is broken instead of the entire bond, its bond energy is figured as the difference between the double bond energy and the single bond energy Thus, for a C-C Π bond, 146 kcal/mol - 84 kcal/mol = 62

kcal/mol

a break Π , make σ; (+146 - 84) - 84 = - 22 kcal/mol

The ring structure is more stable, since its formation from the alkene

releases 22 kcal/mol

b break Π C-C and σ 0-Hf make IlcO and σ CH; (+146 84) + 104

-(176 - 86) - 97 = -21 kcal/mol

The form with a carbon-oxygen double bond is more stable, by 21 kcal/mol

c break Π C-C and N-H, make Π N-N and C-H; (146 -84) + 101 - (100 - 60)

3

Alkanes and Cycloalkanes

The Three-Dimensional Structure of Hydrocarbons and Their Derivatives

Study Hints

Problem-solving Skills

After completing this chapter you should be able to do the following:

1 Draw structures of alkanes and cycloalkanes and name them (Problems 1, 5, 22,

23f 24, 25, 26, 27, 41, 42, 44)

2 Classify structural groups or positions as primary, secondary, tertiary, or quaternary (2, 3)

3 Explain trends in the physical properties of alkanes

4 Draw conformations of alkanes with respect to rotation of groups about a σ bond; use Newman and perspective drawings; correlate potential energy of

conformations to forces within the structure (6, 7, 8, 9, 10, 28, 29, 30)

5 Predict sign of entropy change Δ ^ in conformational equilibrium from numbers

of conformers or rigidity of a conformation (11, 12, 15, 31, 34)

6 Apply the angle strain theory correctly (13, 17b, 18, 33, 45)

7 Draw equatorial and axial substituents on chair and boat cyclohexane

conformations and relate them to molecular energy (16, 34, 35, 36, 37, 38,

39, 40)

8 Draw conformations of cyclopropane, cyclobutane, and cyclopentane (17a, 19)

9 Recognize eis and trans stereoisomers of cycloalkane derivatives (20, 21)

10 Draw polycyclic alkanes to indicate molecular shape and substituent projection (46, 47)

Naming compounds

The root names in Table 3-3 (p 69) should be memorized, at least through

decane These are not only the root names, but the basis for the substituent group names as well The alkyl group names in Table 3-2 (p 67) are also essential

Be sure to understand the system of classifying carbons as primary, secondary, tertiary, and quaternary from their bonding to other carbon atoms

The advice about rewriting structures when they are given as condensed

structural formulas is valuable Working from the condensed formula can lead to mistakes such as miscounting chain length, omitting substituents, or failing to recognize the longest chain

Common errors in using IUPAC names:

1 Not recognizing the longest chain, especially when a long group is drawn off-line from the rest of the chain

2 Numbering from the wrong end

3 Choosing the end that gives the lowest sum of locator numbers, contrary to step 3 (p 69 in textbook), which specifies that the lowest possible single locator number is used

4 Alphabetizing by prefixes di-, tri-, and so on, instead of by group names such as ethyl, methyl, and propyl

5 Forgetting to put cyclo- into the names of ring compounds

Physical properties

A frequently quoted rule of thumb about the boiling points of homologs is that the boiling point increases approximately 30° per added -CH~- Comparison to Table 3-3 shows that this is true for those alkanes which are common solvents But, the first member of a family is usually exceptional Also, the difference in boiling point diminishes gradually with chain length

29 Student Solutions Manual for Organic Chemistry

Conformations of alkanes

The use of models that permit rotation around σ bonds is very helpful at making the interconversion between conformers familiar However, models lack any semblance of the barriers caused by bond and group repulsion, so you have to use your imagination

It is important to practice drawing structures in perspective, either working problems or just trying to represent the critical portion of a molecule in the most effective way Since molecules of moderate size prohibit showing all of the atoms

in perspective, narrowing the picture to the critical, perhaps reactive, sites

of the molecule requires spatial perception

Effects of energy and entropy on conformations

The natural tendency of molecules to acquire minimum energy and the equally natural trend toward randomness nearly always work against each other Energy is usually minimum in the conformer with the biggest groups opposite each other But, since there is only one such form, and usually many alternatives, entropy favors forming something else The study of energy as a consequence of conformation

is called conformational analysis

Angle strain

Baeyerfs theory as applied to rings larger than cyclopentane seems foolish to

us now, because when we build a model of cyclohexane we can hardly get it to stay flat It spontaneously slips away into some puckered form, either chair or boat Try it and see Baeyer had no models

Our models represent that there i^ strain in cyclopropane and cyclobutane, but

to different extents, depending on the flexibility of the material they were

Errors in drawing cyclohexane derivatives

1 Drawing a group iji between axial and equatorial

positions

Remedy: Draw both bonds at any carbon with a

substituent One of those (the equatorial one)

must be parallel to the ring bonds designated

by arrows

2 Drawing a bond in the axial position in the

opposite direction, even when the equatorial

group is approximately correct

Remedy: Axial groups must always point outward

from the point of the ring that they occupy

3 Failure to recognize trans disubstituted

stereoisomers correctly when the substituents

are both equatorial

Remedy: Add the missing H atoms If they are

trans, the methyl groups are also trans

conformation eclipsed Newman projection anti and gauche conformers Baeyer strain

chair conformer twist boat conformer

31 Student Solutions Manual for Organic Chemistry

bicyclic and polycyclic bridgehead position steroid

2 Conformational equilibria

Since these are equilibria, they may be written in either direction The sign

of the energy change A G depends on the direction A G is positive for the reaction as written above and K would be less than one; if the reaction was

-eq turned around, A G would be negative and K > 1 In most cases, a kind of

intuitive conformational analysis will enable you to decide which form

3-6 ^f an attractive force between bonding electrons existed, the eclipsed

conformer would have lower energy than the staggered conformer It would look like

a sine wave with minima at 0°, 120°, and 240°

3-7 Three staggered, three eclipsed Statistically, these conformations are

equally likely

33 Student Solutions Manual for Organic Chemistry

3-8

120 180 Dihedral Angle

Assumptions: Eclipsed conformations should still be maxima and staggered forms minima If polarizability of Cl atom leads to a stabilization, assumed less than 2.9 kcal/mol, then the magnitude of the stabilizing force would be greatest when they are close together Therefore, the conformer with Cl atoms eclipsed becomes the least rotation barrier and the gauche forms become more stable than the anti form

3

120 c

CH 3 180°

3 T 11

"3 X, CH 3 ^C„ 3 X«. , B3 , CH3

300

'3 240°

The 0° and 240° forms are the most stable, since they have only one

methyl-methyl gauche interaction

3-10 a

c " 3 vrv H

I

Structures II and III are equivalent in potential energy

b 41

3-11 At low temperature, the 0° and 240° forms will predominate since they have the least energy They will still dominate at elevated temperature by two to one even if the temperature is so high that the proportions become determined

statistically

3-12 a Two free radicals are more disordered than a single molecule

b The liquid solution is more disorderly than the crystalline solid NaCl The change in the water is also significant, but less than the change in the salt

c Since there is only one anti conformer, but two gauche forms, they

represent the more probable state among random possibilities The rotation

therefore increases entropy

3-13 Cyclopropane would produce the most energy for a given mass Its combustion would produce a higher temperature from release of the strain along with the

combustion reactions, generating higher pressure, and consequently more thrust

35 Student Solutions Manual for Organic Chemistry

c The axial conformation of isopropylcyclohexane (b at right) has a

preferred orientation of the two methyl groups that avoids the 1,3-diaxial

interaction, and thus is only slightly higher in free energy than methyl- or

ethylcyclohexane This restriction to a single conformation makes a slight

contribution to the free energy due to the cost in entropy The molecule has lost its three-way randomness

3-17 a H b From Figure 3-11 (p 84), the

strain energy per CH«

energy) However, since the molecule is folded, as in Equation 3.8, the two methyl groups interact more like the gauche form of butane (about 0.9 kcal/mol)

CH M CH >o> 3 \— \

3

\

CHn

eis and trans-1,3-dimethylcyclopentane, stereoisomers

(H atoms are omitted for clarity.)