Jones M., Fleming S.A. Organic Chemistry

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1.10 Additional Problems 49

PROBLEM 1.65

Make the molecular orbitals for square H

allowing the molecular orbitals of H

to interact as shown below:

Order the new molecular orbitals by counting nodes, and add

the proper number of electrons. You might check your answer

by deriving the same orbitals. Do this by bending the molecular

orbitals for linear HHHH developed in Problem 1.64.

PROBLEM 1.66 Generate the molecular orbitals for planar

ammonia, NH

. Do this by taking combinations of the molecu-

lar orbitals for triangular H

(see Problem 1.62 for these

orbitals) and the atomic orbitals of nitrogen.

(a) Show clear pictures of the molecular and atomic orbitals you

are using to make the molecular orbitals of planar ammonia.

(b) How many molecular orbitals will ammonia have?

Show clearly how these are generated.

(d) Order the bonding and nonbonding molecular orbitals in

terms of energy. Place them on a scale relative to the energy

of a lone, nonbonding 2p orbital. You do not have to order

the antibonding orbitals.

(e) Place the appropriate number of electrons in the orbitals. Be

careful to indicate the spin quantum number for each elec-

tron (use an up or down arrow to show spin).

PROBLEM 1.67 Generate the molecular orbitals for linear

methylene, by combining the atomic orbitals of

carbon with the molecular orbitals of hydrogen, H

(a) Show clear pictures of the molecular and atomic orbitals

you are using.

(b) How many molecular orbitals will linear methylene have?

–

(d) Order these molecular orbitals in terms of energy. Place

them on a scale relative to the energy of a lone, nonbonding

carbon 2p orbital.

(e) Place the appropriate number of electrons in the orbitals,

being careful to indicate the spin quantum number for each

electron (use an up or down arrow to show spin).

Use Organic Reaction Animations (ORA) to answer the

following questions:

PROBLEM 1.68 Choose the reaction titled “Unimolecular

nucleophilic substitution” and click on the Play button. Do you

suppose the first step of this reaction is a homolytic or a het-

erolytic cleavage? Observe the Highest Occupied Molecular

Orbital (HOMO) track by clicking on the HOMO button.

The location (orbital) of the most available electrons will be

shown throughout the reaction. Notice that the electron density

goes with the bromine as it comes off. That should help you

answer this question.

PROBLEM 1.69 Choose the “Introduction” on the

bottom left of the Table of Contents page. Read this short

document. Under the “Technical Issues” heading there is a

discussion of solvent effects. After reading this section, how

do you think using a polar solvent in the “Unimolecular

nucleophilic substitution” reaction might affect the answer

to the previous question? That is, would a polar solvent

have more impact on a homolytic or a heterolytic cleavage?

PROBLEM 1.70 Choose the reaction “Alkene hydrohalo-

genation” and observe the molecule that initially comes

to the screen. It has a carbon–carbon double bond.

Click on the HOMO button. Observe the calculated area

for the π bond that is shown to answer the following

questions. Is the electron density of a π bond constrained

to the space between the carbons? Do you suppose the π

bond electrons are held more or less tightly than σ bond

electrons?

Alkanes

2.1 Preview

2.2 Hybrid Orbitals: Making a

Model for Methane

2.3 The Methyl Group (CH

) and

Methyl Compounds (CH

2.4 The Methyl Cation (

Anion (



), and Radical

(CH

)

2.5 Ethane (C

), Ethyl

Compounds (C

X), and

Newman Projections

2.6 Structure Drawings

2.7 Propane (C

) and Propyl

Compounds (C

2.8 Butanes (C

), Butyl

Compounds (C

X), and

Conformational Analysis

2.9 Pentanes (C

) and Pentyl

Compounds (C

2.10 The Naming Conventions for

Alkanes

2.11 Drawing Isomers

2.12 Rings

2.13 Physical Properties of Alkanes

and Cycloalkanes

2.14 Nuclear Magnetic Resonance

Spectroscopy

2.15 Acids and Bases Revisited:

More Chemical Reactions

2.16 Special Topic: Alkanes as

Biomolecules

2.17 Summary

2.18 Additional Problems

A CELESTIAL VIEW Does it include molecular orbitals?

2.1 Preview 51

God never saw an orbital.

—WALTER KAUZMANN, FEBRUARY, 1964

2.1 Preview

The chemical reactions a given compound undergoes are a function of its structure.

No compound reacts exactly like another, no matter how similar their structures.Yet

there are gross generalizations that can be made, and some order can be carved out

of the chaos that would result if compounds of similar structure did not react in

somewhat the same fashion. To a first approximation, the reactivity of a molecule

depends on the functional groups (p. 22) it contains.

The inside front cover of this book shows some of the functional groups that

are important in organic chemistry. This material is not to be memorized, and it is

certainly not complete. You will want to work through it, however, to be certain you

can write correct Lewis structures for each group and to begin to become familiar

with the variety of functional groups important in organic chemistry. These differ-

ent kinds of molecules will all be discussed in detail in later chapters. We start in

this chapter with alkanes, the simplest members of the family of organic molecules

called hydrocarbons—compounds composed entirely of carbon and hydrogen.

Alkanes all have the molecular formula C

2n2

. The simplest alkane is

methane,CH

. Vast numbers of related molecules can be constructed from

methane by replacing one or more of its hydrogens with other carbons and their

attendant hydrogens. Linear arrays can be made,as well as branched structures and

even rings (ring compounds called cycloalkanes, have a slightly different formu-

la: C

). Figure 2.1 shows a few schematic representations for these molecules.

Walter Kauzmann (p. 42) was chairman of the Princeton University chemistry department in 1964 when MJ

interviewed for a job. He made this comment in response to a typically convoluted molecular orbital answer

to some simple question.

The simplest

representations—not much

information here

Here, the connectivity is

shown, but there is no

three-dimensionalit

Three-dimensional structures.

Solid wedges come toward

ou, dashed wed

es retreat

The ultimate schematic

structures—neither carbons,

nor h

dro

ens are shown

Methane

Propane

Cyclopropane

H H

FIGURE 2.1 Several ways of drawing

alkane structures.

52 CHAPTER 2 Alkanes

Before we can go further, we need to make sense of these schematics, and as usual,

we must start with structure. The first task is to describe the shape of methane.

Once we have done this, we can go on to investigate other members of the al-

kane family.

PROBLEM 2.1 Draw Lewis structures for the linear alkanes butane (C

) and

pentane (C

ESSENTIAL SKILLS AND DETAILS

1. Hybridization. Above all, it is important to master the structural model introduced in

this chapter, hybridization.The hybridization model does a good job of allowing us to

predict the general structures of compounds and it is nicely suited for following

electron flow in chemical reactions.Thus, it fits in well with the curved arrow

formalism introduced in Chapter 1 (p. 23).

2. Structures. It is important to be able to use the various structural formulas, which range

from richly detailed three-dimensional representations to the ambiguous condensed

formulas that give no hint of the three-dimensional complexity often present in a

molecule.

3. Difference.The concept of difference is essential in organic chemistry. When are two

atoms the same (in exactly the same environment) and when are they different (not in

the same environment)? This question gets to the heart of structure and is much

tougher to answer than it seems. In this chapter, we will introduce this subject and we

will return to it in Chapter 4.

4. Names. In truth, it is not necessary to know every nuanced detail of the naming

convention for alkanes, but you do need to know some nomenclature.

5. The cis/trans convention. Cyclic molecules (rings) have sides—above the ring and

below the ring—and attached groups can be on the same or opposite sides of the

ring.

6. The Newman projection. Drawing and “seeing”Newman projections is a critical skill in

organic chemistry.

7. Bond strengths.You surely do not have to memorize all possible bond strengths, but by

the end of this chapter you should have a good idea of some of the important ones,

such as , , and .

2.2 Hybrid Orbitals: Making a Model for Methane

2.2a Hybridization Our task is to devise a bonding model for the structure

of the simplest alkane, methane, CH

. Physical chemists tell us the structure:

methane is a tetrahedron, with all carbon–hydrogen bonds the same length

(Fig. 2.2). We had an earlier look at a tetrahedron in Problem 1.5 (p. 16), in

which the molecule carbon tetrachloride appeared. Our job now is to work out

a bonding scheme that leads to the experimentally determined tetrahedral struc-

ture of methane.

The method we use to generate our model of methane goes by the name

hybridization. Remember that we are working out a model, which means that no

matter how useful it is,what we produce here is bound to be flawed in some respects.

It is not the “ultimate truth”by any means! Our strategy will be to combine the four

(a)

(b)

A tetrahedron

109.5

Methane, a tetrahedral molecule

1.09

All angles

are 109.5



WEB 3D

FIGURE 2.2 (a) A tetrahedron.

(b) A tetrahedral molecule, methane,

showing the arrangement of the four

hydrogens about the central carbon.

2.2 Hybrid Orbitals: Making a Model for Methane 53

atomic orbitals of a carbon atom (2s,2p

,2p

) to produce four new orbitals for

the atom, called hybrid orbitals, that can overlap with the 1s orbitals of four hydro-

gens to produce methane.Despite the fancy name hybridization, we are really doing

something quite similar to the combining of atomic orbitals we saw often in Chapter

1. In Chapter 1 we combined orbitals from different atoms to make molecular

orbitals and here we will combine orbitals from the same atom to make hybrid atom-

ic orbitals. Remember that we already know that a combination of four atomic

orbitals must produce four new orbitals (p. 34).

It will greatly help our understanding of the hybridization process if we back

up just a bit and look at two simpler hybridization models before we return to

methane.

2.2b sp Hybridization Let’s first combine the 2s and 2p

atomic orbitals from

some arbitrary atom, as shown in Figure 2.3. The two can be combined in a con-

structive (2p  2s) or a destructive (2p  2s) way to create a pair of hybrid orbitals.

–

2ssp Hybrid orbital

2s Another sp hybrid

orbital

2s 2p

Two sp hybrid orbitals

One sp orbital in schematic form…

…and more correctly:

–

FIGURE 2.3 sp Hybridization.

FIGURE 2.4 A comparison between

overlap of a hydrogen 1s orbital with

(a) an sp hybrid orbital and (b) an

unhybridized 2p orbital. With the sp

hybrid orbital, overlap is maximized

because the non-overlapping back

lobe is small. With the unhybridized

2p orbital, all of the non-overlapping

back lobe is “wasted.”

Note in Figure 2.3 that the hybrid orbitals no longer have equal-sized lobes, as in a

2p orbital. In the 2p  2s combination, we get an expansion of one of the original

lobes (the one with the same sign as the 2s orbital) and a shrinking of the other (the

one with the opposite sign). The second combination, (2p  2s), is also lopsided,

but in the opposite sense.These two new atomic orbitals are called sp hybrid

orbitals. The designation sp means that the orbitals are composed of 50% s

orbital (they have 50% s character) and 50% p orbital (they have 50% p

character).

Figure 2.3 also shows the real shape of an sp hybrid orbital. It is tradi-

tional to use the schematic form in drawings, rather than attempt to repro-

duce the detailed structure of the orbital.

Overlap between an sp orbital of one atom and an orbital of some other

atom will be especially good if the fat lobe of the sp orbital is used (Fig. 2.4a).

Contrast this overlap with the overlap when a 1s orbital overlaps with one

lobe of a 2p orbital of another atom (Fig. 2.4b).The better the overlap, the greater

the stabilization and the stronger the bond. In a sense, overlap of a 1s orbital with

an unhybridized atomic 2p orbital “wastes” the back lobe of the 2p orbital.

Hybridization improves overlap!

Hybridization also minimizes electron–electron repulsion. The two new sp

hybrid atomic orbitals formed from 2p  2s and 2p  2s (Fig. 2.3) are aimed 180°

from each other. The electrons in the bonds are as remote from each other as

possible.

(b)

(a)

Wasted!

sp 1s

2p 1s

54 CHAPTER 2 Alkanes

There are many compounds in which the bonding can be nicely described by sp

hybridization. The classic example is , a linear hydride of beryllium

(Fig. 2.5). Beryllium (

Be) has the electronic configuration 1s

and thus brings

only two electrons to the bonding scheme (Remember: The filled shell 1s electrons

are not used for bonding).We construct our picture of beryllium hydride (BeH

) by

making two sp hybrid atomic orbitals from the Be 2s atomic orbital and one of the

three equivalent Be 2p orbitals, arbitrarily taking the 2p

orbital (which exists even

though the configuration 1s

tells us there are no electrons in this atomic orbital).

The overlap of the 2p

 2s hybrid orbital with a singly occupied hydrogen 1s orbital

and of the 2p

 2s hybrid orbital with a second hydrogen 1s orbital produces two

bonds directed at an angle of 180° with respect to each other. As a result

of the linear shape of this compound we say that the beryllium is sp-hybridized.

Bonds of this type, with cylindrical symmetry, are called sigma bonds (σ bonds).

The orbital is unchanged through rotation around the axis between the Be and H.

The combination of a Be sp orbital and a H 1s orbital produces the sigma bonding

orbital (sp  1s) and must also yield an antibonding orbital (sp  1s), which in this

case is empty, as shown in Figure 2.5. This antibonding orbital also has cylindrical

symmetry and is therefore properly called a sigma antibonding orbital.

In Chapter 1, we used subscripts A and B to indicate antibonding and bonding

orbitals (Φ

, Φ

), but from now on we can use the simpler notation of an asterisk (*)

to indicate an antibonding orbital. Thus, in Figure 2.5, σ is the bonding molecular

orbital sp  1s and σ* is the antibonding molecular orbital sp  1s.

It is no accident that we use two of beryllium’s atomic orbitals (2s and 2p

) in

our description of the bonding in BeH

. Two hybrid orbitals are needed for bond-

ing to two hydrogens, which means we must combine two unhybridized atomic

orbitals to get them.

Energy

180

sp Hybrid

orbital of Be

H 1s

sp –

The antibonding

␴* orbital

Axis of

symmetry:

Rotation around

this axis changes

nothing

sp +

The bonding

␴ orbital

Axis

FIGURE 2.5 The formation of one

bonding (σ) molecular orbital and

one antibonding (σ*) molecular

orbital from the overlap of a Be sp

hybrid orbital with a hydrogen 1s

atomic orbital. Note that the

interaction diagram shows the 1s

atomic orbital is at a lower energy

than the sp hybrid orbital.The figure

shows formation of only one of the

two beryllium–hydrogen bonds in

BeH

.The other bond is

made in identical fashion through

overlap of the second Be sp hybrid

orbital with the 1s atomic orbital of a

second H atom.

Nothing in this discussion so far is specific to the element Be. Even though we are using this element as our

example, we have been constructing hybrid orbitals for any second row linear HXH molecule.

CONVENTION ALERT

2.2 Hybrid Orbitals: Making a Model for Methane 55

PROBLEM 2.2 What happens to the 2p

and 2p

atomic orbitals of Be in BeH

Elaborate the picture in Figure 2.5 to show the unused, empty 2p atomic orbitals

on Be.

WORKED PROBLEM 2.3 Use the hybridization model to build a picture of the

bonding in the linear molecule . (Recall that a molecular orbital

picture of this linear molecule was constructed in Problem 1.67, p. 49). If you did-

n’t do that problem, you might go back, try it now, and contrast your molecular

orbital answer with your hybridization answer.

ANSWER As with BeH

, the central atom, here carbon, is hybridized sp because

the CH

carbon only has two bonds. The 2s and 2p

atomic orbitals of carbon

are combined to form two sp hybrid orbitals (2s  2p

and 2s  2p

). One of

the bonds in CH

is made through overlap of the 2s  2p

sp hybrid orbital

and the 1s atomic orbital of one hydrogen. The other bond is made through

overlap of the 2s  2p

hybrid orbital and the 1s atomic orbital of the other

hydrogen.

The 2p

and 2p

atomic orbitals in the carbon remain unchanged. The atom

uses two of its four bonding electrons in the carbon–hydrogen bonds, leaving two

for the remaining 2p orbitals. Each contains a single electron.

This kind of thing generally drives physical chemists crazy. As an orbital is nothing more than the

electron probability region, how can there be an “empty orbital”? The problem is at least partially

avoided if the empty orbital is thought of as “the place the next electron would go” as mentioned in

Chapter 1.

2.2c sp

Hybridization In BH

, we must make three boron–hydrogen bonds

to form the molecule. The electronic configuration of boron (

B) is 1s

, and

thus there are three electrons available to form bonds with the three hydrogen 1s

Energy

HHC

sp Hybrid

orbital of C

constructed

from 2s

H 1s

sp – 1s

The antibonding

σ∗ orbital

1s/sp Bonds

sp + 1s

Bonding

σ orbital

The 2p

and 2p

orbitals

–

56 CHAPTER 2 Alkanes

atomic orbitals.This time we combine the boron 2s,2p

, and 2p

orbitals to produce

the three orbitals we need in our model. These are called, naturally, sp

hybrid

orbitals (constructed from one s orbital and two p orbitals) and have 33% s charac-

ter and 67% p character. They resemble the sp hybrid orbitals, shown in Figure 2.3.

As with the sp orbital, the fat lobe of the sp

orbital allows for efficient overlap with

a hydrogen 1s or with any other orbital.

2s  2p

 2p

Three sp

hybrid orbitals

Top view Side view

120

FIGURE 2.6 Top and side views of the

arrangement in space of the three sp

hybrid orbitals formed from the 2s,

, and 2p

atomic orbitals of a

boron atom.

PROBLEM 2.4 Boron also forms a compound BF

. Create a bonding scheme and

draw a Lewis structure for this molecule.

How are the three sp

hybrid orbitals needed to form BH

arranged in space?

Your intuition may well give you the answer. They are directed so that electrons in

them will be as far as possible from one another. The angle between them is 120°,

and they are aimed toward the corners of a triangle (Fig. 2.6). An sp

carbon and

the three atoms surrounding it lie in the same plane.

PROBLEM 2.5 The 2p

orbital of boron was not used in our construction of the

three sp

hybrid atomic orbitals for the atom. Sketch it in, using the drawing of

Figure 2.6.

2.2d sp

Hybridization: Methane For CH

we need four hybrid atomic

orbitals because there are four hydrogens to be attached to carbon. Fortunately, we

have just what we need in the 2s,2p

,2p

, and 2p

atomic orbitals of carbon. In our

mathematical description, we combine these orbitals to produce four sp

hybrid

orbitals, which also look similar to sp hybrids (Fig. 2.3).

2s  2p

 2p

Four sp

hybrid orbitals

The familiar lopsided shape appears again, and thus we can expect strong bond-

ing to the four hydrogen 1s atomic orbitals. By now you will surely anticipate that

these orbitals will be aimed in space so as to keep them, and any electrons in them,

as far apart as possible. How are four objects arranged in space so as to maximize

the distance between them? The answer is,at the corners of a tetrahedron (Fig.2.2).

The carbon in methane is hybridized sp

(25% s character; 75% p character) and

the four hybrid atomic orbitals are directed toward the corners of a tetrahedron.

The 1s atomic orbital in each of four hydrogen atoms overlaps with one of the four

2.2 Hybrid Orbitals: Making a Model for Methane 57

orbitals to form a bonding molecular orbital and an antibonding molecular

orbital (Fig. 2.7).The resulting angles are each 109.5°, and the length

of a carbon–hydrogen bond in methane is 1.09 Å (Fig. 2.2).

Energy

FIGURE 2.7 The formation of a

carbon–hydrogen bond through the

overlap of an sp

hybrid orbital with a

hydrogen 1s atomic orbital. Note the

formation of both the bonding (σ)

molecular orbital and the

antibonding (σ*) molecular orbital.

energy

⌬H = +105 kcal/molHomolytic cleavage of one

carbon–hydrogen bond in methane

FIGURE 2.8 Two ways of showing

homolytic cleavage of one

carbon–hydrogen bond in methane.

Each single-barbed arrow represents

movement of one electron.

The carbon–hydrogen bond in methane is very strong; its bond dissociation ener-

gy is 105 kcal/mol.This means that 105 kcal/mol must be applied in order to break

one of the carbon–hydrogen bonds in methane homolytically. This reaction is

endothermic by 105 kcal/mol (Fig. 2.8). Compare this value to the 104 kcal/mol

bond strength of the very strong bond in H

(p. 36). The bond in

methane and the bond in H

have very similar strengths.H

2.2e Why Hybridization? Why do we need to develop this new structural model

called hybridization? If we want to form methane, why don’t we just overlap the 1s

atomic orbitals of four hydrogens with the 2s and 2p atomic orbitals of carbon? Well,

we certainly could have done just that, but the results are not very satisfying—this

approach doesn’t give us a decent approximation of the tetrahedral structure of methane.

However, let’s follow this flawed procedure. We can learn a lot from examining what is

wrong with models that are too simple. Further, what we do here approximates what

really happens in science in that we make a series of successive structural approxima-

tions in which we come increasingly close to the real methane. In Section 2.2d,we went

right to a satisfactory model for methane, thus shortcutting the intellectual process.

58 CHAPTER 2 Alkanes

Figure 2.9 shows the overlap of the occupied carbon 2s,2p

,and 2p

atomic orbitals

with four hydrogen 1s orbitals. Two predictions are clear: First, we should form one

bond from the overlap of the carbon 2p

orbital and one hydrogen 1s orbital

and one bond from the overlap of the carbon 2p

orbital and another hydro-

gen 1s orbital; second, these bonds must be at 90° to each other because of the 90°

angle between the carbon 2p

and 2p

orbitals. The third and fourth bonds are diffi-

cult to locate exactly, as they are formed from the overlap of a carbon 2s orbital with

two hydrogen 1s orbitals. Both the 2s and 1s orbitals are spherically symmetric, and

thus no directionality can be induced by the shape of the orbitals. We might guess

that these two new carbon–hydrogen bonds would be aimed so as to keep the elec-

trons in the bonds as far from each other as possible, thus minimizing repulsions.

PROBLEM 2.6 In a quick analysis there appear to be too many electrons in the region

of Figure 2.9 where the carbon 2s orbital and the two hydrogen 1s orbitals overlap.

You might think that an atomic orbital with two electrons can’t make a bond with two

other atoms, each bringing an electron. It is true that no orbital can have more than

two electrons (the Pauli principle). You are used to thinking of two-electron bonds,

and our new system is clearly more complicated than that.But there is no violation of

the Pauli principle in the unhybridized model illustrated in Figure 2.9. Explain.

Hint: Remember that overlap of n orbitals produces n new orbitals. See Problem 1.60.

Using unhybridized carbon atomic orbitals to form methane, a model that yields

the structure shown in Figure 2.9,has a fatal flaw.Modern spectroscopic methods show

that there are no 90° angle in methane and that all carbon–hydrogen

bonds in the molecule are of equal length! Knowing that the Figure 2.9 structure is

wrong, we can begin to construct a model that does a better job of describing Nature.

Don’t be offended by this—science works this way all the time.The goal is to use what

we learn from our wrong guesses to come closer next time. We can already anticipate

one problem in this model—the electrons in some of the bonds are only 90° apart.

They could be farther apart, and thus electron–electron repulsion could be reduced.

PROBLEM 2.7 Consider the ammonium ion (



). Predict the hybridization of

the nitrogen and the shape of the molecule.

A second problem with the model used in Figure 2.9 might be labeled “ineffi-

cient overlap” or “wasted orbitals.” Recall from Figure 2.4 that a 2p atomic orbital

is poorly designed for efficient end-on overlap because overlap can take place only

with one lobe.The back lobes of 2p orbitals are unused and thus wasted (Fig. 2.10).

Our unhybridized model for methane has two such overlapping pairs of orbitals.

This partial picture of methane

contains two such 1s/2p end-on

overla

s; two rear lobes are wasted

Waste not!

FIGURE 2.10 End-on overlap of a 2p

atomic orbital with a 1s atomic

orbital.

1s/2p End-on overlap;

notice the wasted lobe

not involved in bonding

Wasted lobe

(1s is not shown)

Methane

(CH

)

1s/2p

Overlap

1s/2s/1s

Overlap

outlined

in bold

90

(1s

)2s

FIGURE 2.9 A structure for methane,

. Bonds are formed by the

overlap of four hydrogen 1s orbitals

with the 2p and 2s atomic orbitals of

one carbon atom.This model requires

some 90° bond angles and requires

the bonds to be of different lengths.

Carbon’s 1s electrons are shown in

parentheses because they are not used

in bonding.