Organic Chemistry

Jurgen Torres

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Contents

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Structure and Properties of Organic Compounds

1.1 

Carbon Compounds

Organic chemistry is the study of carbon (C) compounds, all of which have covalent bonds. Carbon atoms can bond to each other to form open-chaincompounds, Fig. 1.1(a), or cyclic(ring)compounds, Fig. 1.1(c). Both types can also have branches of C atoms, Fig. 1.1(b) and (d).Saturatedcompounds have C atoms bonded to each other by single bonds,C— C; unsaturatedcompounds have C’s joined by multiple bonds. Examples withdouble bondsandtriple bondsare shown in Fig. 1.1(e). Cyclic compounds having at least one atom in the ring other than C (a heteroatom) are called heterocyclics, Fig. 1.1(f). The heteroatoms are usually oxygen (O), nitrogen (N), or sulfur (S).

Problem 1.1Why are there so many compounds that contain carbon?

Bonds between C’s are covalent and strong, so that C’s can form long chains and rings, both of which may have branches. C’s can bond to almost every element in the periodic table. Also, the number of isomers increases as the organic molecules become more complex.

Problem 1.2Compare and contrast the properties of ionic and covalent compounds.

Ionic compounds are generally inorganic, have high melting and boiling points due to the strong electro- static forces attracting the oppositely charged ions, are soluble in water and insoluble in organic solvents, are hard to burn, and involve reactions that are rapid and simple. Also, bonds between like elements are rare, with isomerism being unusual.

Covalent compounds, on the other hand, are commonly organic; have relatively low melting and boiling points because of weak intermolecular forces; are soluble in organic solvents and insoluble in water; burn read- ily and are thus susceptible to oxidation because they are less stable to heat, usually decomposing at tempera- tures above 700ºC; and involve reactions that are slow and complex, often needing higher temperatures and/ or catalysts, yielding mixturesof products. Also, bonds between carbon atoms are typical, with isomerism being common.

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CHAPTER 1Structure and Properties of Organic Compounds

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H  H

H  C  C

H  H

H  H

C  C H

H  H

H  H  H

H C C C H H H H

H H

H H H C

H C H H H

H H H

C  C C  C

H H H H

n-Butane unbranched, open-chain (a)

Isobutane branched, open-chain (b)

Cyclopropane unbranched, cyclic (c)

Methylcyclopropane

branched, cyclic

(d)

H  H

H H H  C  C  H

C C

H H H C C H H C

H

H C C H H H

Ethene (Ethylene)

Cyclopentene

Ethyne (Acetylene)

Ethylene oxide

have double bonds

(e)

has a triple bond

Figure 1.1

heterocyclic

(f)

1.2 

Lewis Structural Formulas

Molecular formulasmerely include the kinds of atoms and the number of each in a molecule (as C 4H10 for butane). Structural formulasshow the arrangement of atoms in a molecule (see Fig. 1.1). When unshared electrons are included, the latter are called Lewis(electron-dot)structures[see Fig. 1-1(f)].Covalencesof the common elements—the numbers of covalent bonds they usually form—are given in Table 1.1; these help us to write Lewis structures. Multicovalent elements such as C, O, and N may have multiple bonds, as shown in Table 1.2. In condensedstructural formulas, all H’s and branched groups are written immediately after the C atom to which they are attached. Thus, the condensed formula for isobutane [Fig. 1-1(b)] is CH3CH(CH3)2.

Problem 1.3(a) Are the covalences and group numbers(numbers of valence electrons) of the elements in Table 1.1 related? (b) Do all the elements in Table 1.1 attain an octet of valence electrons in their bonded states?

(c) Why aren’t Group 1 elements included in Table 1.1?

Yes. For the elements in Groups 4 through 7, Covalence  8 (Group number).

No. The elements in Groups 4 through 7 do attain the octet, but the elements in Groups 2 and 3 have less than an octet. (The elements in the third and higher periods, such as Si, S, and P, may achieve more than an octet of valence electrons.)

They form ionic rather than covalent bonds. (The heavier elements in Groups 2 and 3 also form mainly ionic bonds. In general, as one proceeds down a group in the periodic table, ionic bonding is preferred.)

Most carbon-containing molecules are three-dimensional. In methane, the bonds of C make equal angles of 109.5º with each other, and each of the four H’s is at a vertex of a regular tetrahedron whose center is occupied by the C atom. The spatial relationship is indicated as in Fig. 1.2(a) (Newman projection) or in Fig. 1.2(b) (wedge projection). Except for ethene, which is planar, and ethyne, which is linear, the structures in Fig. 1.1 are all three-dimensional.

Organic compounds show a widespread occurrence of isomers, which are compounds having the same molec- ular formula but different structural formulas, and therefore possessing different properties. This phenomenon ofisomerismis exemplified by isobutane and n-butane [Fig. 1.1(a) and (b)]. The number of isomers increases as the number of atoms in the organic molecule increases.

CHAPTER 1Structure and Properties of Organic Compounds

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TABLE 1.1 Covalences of H and Second-Period Elements in Groups 2 through 7

* Exists as B2H6.

TABLE 1.2 Normal Covalent Bonding

Problem 1.4Write structural and condensed formulas for (a) three isomers with molecular formula C  5H12 and

(b) two isomers with molecular formula C3H6.

Carbon forms four covalent bonds; hydrogen forms one. The carbons can bond to each other in a chain:

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or there can be branches (shown circled in Fig. 1.3) on the linear backbone (shown in a rectangle).

We can have a double bond or a ring.

CHAPTER 1Structure and Properties of Organic Compounds

Figure 1.2

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Figure 1.3

Problem 1.5Write Lewis structures for (a) hydrazine, N  2H4; (b) phosgene, COCl2; and (c) nitrous acid, HNO2.

In general, first bond the multicovalent atoms to each other and then, to achieve their normal covalences, bond them to the univalent atoms (H, Cl, Br, I, and F). If the number of univalent atoms is insufficient for this purpose, use multiple bonds or form rings. In their bonded state, the second-period elements (C, N, O, and F) should have eight (an octet) electrons but not more. Furthermore, the number of electrons shown in the Lewis structure should equal the sum of all the valence electrons of the individual atoms in the molecule. Each bond represents a shared pair of electrons.

N needs three covalent bonds, and H needs one. Each N is bonded to the other N and to two H’s.

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  C is bonded to O and to each Cl. To satisfy the tetravalence of C and the divalence of O, a double bond is placed between C and O.

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The atom with the higher covalence, in this case the N, is usually the more central atom. Therefore, each O is bonded to the N. The H is bonded to one of the O atoms, and a double bond is placed between the N and the other O. (Convince yourself that bonding the H to the N would not lead to a viable structure.)

CHAPTER 1Structure and Properties of Organic Compounds

Problem 1.6Why are none of the following Lewis structures for COCl 2 correct?

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The total number of valence electrons that must appear in the Lewis structure is 24, from [2  7](2Cl’s) 4(C) 6(O). Structures (b) and (c) can be rejected because they each show only 22 electrons. Furthermore, in (b), O has 4 rather than 2 bonds, and in (c), one Cl has 2 bonds. In (a), C and O do not have their normal covalences. In (d), O has 10 electrons, though it cannot have more than an octet.

Problem 1.7Use the Lewis-Langmuir octet rule to write Lewis electron-dot structures for: (a) HCN, (b) CO 2,

(c) CCl4, and (d) C2H6O.

  Attach the H to the C, because C has a higher covalence than N. The normal covalences of N and C are met with a triple bond. Thus, H— C ― N: is the correct Lewis structure.

  The C is bonded to each O by double bonds to achieve the normal covalences.

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Each of the four Cl’s is singly bonded to the tetravalent C to give:

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The three multicovalent atoms can be bonded as C— C— O or as C— O— C. If the six H’s are placed so that C and O acquire their usual covalences of 4 and 2, respectively, we get two correct Lewis structures (isomers):

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Problem 1.8Determine the positive or negative charge, if any, on:

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The charge on a species is numerically equal to the total number of valence electrons of the unbonded atoms minus the total number of electrons shown (as bonds or dots) in the Lewis structure.

  The sum of the valence electrons (6 for O, 4 for C, and 3 for three H’s) is 13. The electron-dot formula shows 14 electrons. The net charge is 13  14  1, and the species is the methoxide anion,

CHAPTER 1Structure and Properties of Organic Compounds

There is no charge on the formaldehyde molecule, because the 12 electrons in the structure equals the number of valence electrons—that is, 6 for O, 4 for C, and 2 for two H’s.

This species is neutral, because there are 13 electrons shown in the formula and 13 valence electrons: 8 from two C’s and 5 from five H’s.

There are 15 valence electrons: 6 from O, 5 from N, and 4 from four H’s. The Lewis dot structure shows 14 electrons. It has a charge of 15  14  1 and is the hydroxylammonium cation, [H 3NOH] .

  There are 25 valence electrons, 21 from three Cl’s and 4 from C. The Lewis dot formula shows 26 electrons. It has a charge of 25  26  1 and is the trichloromethide anion, :CCl –.

1.3 

Types of Bonds

Covalent bonds, the mainstays of organic compounds, are formed by the sharing of pairs of electrons. Sharing can occur in two ways:

(1) A⋅+⋅B→A : B

(2)  A+:B→A : B coordinate covalen t

acceptor  donor

In method (1), each atom brings an electron for the sharing. In method (2), the donor atom (B:) brings both electrons to the marriage with the acceptor atom (A); in this case, the covalent bond is termed a coordinate covalentbond.

Problem 1.9Each of the following molecules and ions can be thought to arise by coordinate covalent bond- ing. Write an equation for the formation of each one and indicate the donor and acceptor molecule or ion:

(a) NH ; (b) BF ; (c) (CH ) OMgCl ; and (d) Fe(CO) .

4 4 3 2 2 5

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(a)

(b)

(c)

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(d)

Notice that in each of the products there is at least one element that does not have its usual covalence—this is typical of coordinate covalent bonding.

Recall that an ionic bond results from a transferof electrons (M· A· � M :A ). Although C usually forms covalent bonds, it sometimes forms an ionic bond (see Section 3.2). Other organic ions, such as CH3COO (acetate ion), have charges on heteroatoms.

Problem 1.10Show how the ionic compound Li  F  forms from atoms of Li and F.

These elements react to achieve a stable noble-gas electron configuration (NGEC). Li(3) has one electron more than He and loses it. F(9) has one electron less than Ne and therefore accepts the electron from Li.

1.4 

Functional Groups

Hydrocarbonscontain only C and hydrogen (H). H’s in hydrocarbons can be replaced by other atoms or groups of atoms. These replacements, called functional groups, are the reactive sites in molecules. The C-to-C double and triple bonds are considered to be functional groups. Some common functional groups are given in Table 1.3.

CHAPTER 1Structure and Properties of Organic Compounds

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Compounds with the same functional group form a homologous serieshaving similar characteristic chemical properties and often exhibiting a regular gradation in physical properties with increasing molecular weight.

Problem 1.11Methane, CH 4; ethane, C2H6; and propane, C3H8, are the first three members of the alkane homologous series. By what structural unit does each member differ from its predecessor?

These members differ by a C and two H’s; the unit is — CH2— (a methylene group).

Problem 1.12(a) Write possible Lewis structural formulas for (1) CH 4O; (2) CH2O; (3) CH2O2; (4) CH5N; and (5) CH3SH. (b) Indicate and name the functional group in each case.

The atom with the higher valence is usually the one to which most of the other atoms are bonded.

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(a)

(b)

1.5 Formal Charge

The formal charge on a covalently bonded atom equals the number of valence electrons of the unbonded atom (the group number) minus the number of electrons assigned to the atom in its bonded state. The assigned number is one half the number of shared electrons plus the total number of unshared electrons. The sum of all formal charges in a molecule equals the charge on the species. In this outline, formal charges and actual ionic charges (e.g., Na ) are both indicated by the signs and .

Problem 1.13Determine the formal charge on each atom in the following species: (a) H  3NBF3; (b) CH3NH ;

and (c) SO2 .

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(a)

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The sum of all formal charges equals the charge on the species. In this case, the  1 on N and the  1 on B cancel

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(b)

CHAPTER 1Structure and Properties of Organic Compounds

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and the species is an unchanged molecule:

(c)

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These examples reveal that formal charges appear on an atom that does not have its usual covalence and does not have more than an octet of valence electrons. Formal charges always occur in a molecule or ion that can be conceived to be formed as a result of coordinate covalent bonding.

Problem 1.14Show how (a) H 3NBF3 and (b) CH3NH can be formed from coordinate covalent bonding. Indi- cate the donor and acceptor, and show the formal charges.

(a)

(b)

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SUPPLEMENTARY  PROBLEMS

Problem 1.15Why are the compounds of carbon covalent rather than ionic?

With four valence electrons, it would take too much energy for C to give up or accept four electrons. There- fore, carbon shares electrons and forms covalent bonds.

Problem 1.16Classify the following as (i) branched chain, (ii) unbranched chain, (iii) cyclic, (iv) multiple bonded, or (v) heterocyclic:

(a) (iii) and (iv); (b) (i); (c) (ii); (d) (v); (e) (iv) and (ii).

TABLE 1.3 Some Common Functional Groups

TABLE 1.3(continued)

1 The italicized portion indicates the group.

2 A primary (1º) amine; there are also secondary (2º) R2NH, and tertiary (3º) R3N amines.

3 Another name is propanamine.

CHAPTER 1Structure and Properties of Organic Compounds

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Problem 1.17Refer to a periodic chart and predict the covalences of the following in their hydrogen compounds: (a) O; (b) S; (c) Cl; (d) C; (e) Si; (f) P; (g) Ge; (h) Br; (i) N; (j) Se.

The number of covalent bonds typically formed by an element is 8 minus the group number. Thus: (a) 2; (b) 2; (c) 1; (d) 4; (e) 4; (f) 3; (g) 4; (h) 1; (i) 3; (j) 2.

Problem 1.18Which of the following are isomers of 2-hexene, CH 3CH  CHCH2CH2CH3?

CH3CH2 CH  CHCH2CH3 (b) CH2  CHCH2CH2 CH2CH3

(c) CH3 CH2CH2CH  CHCH3

All but (c), which is 2-hexene itself.

Problem 1.19Find the formal charge on each element of

and the net charge on the species (BF3Ar).

Problem 1.20Write Lewis structures for the nine isomers having the molecular formula C 3H6O, in which C, H, and O have their usual covalences; name the functional group(s) present in each isomer.

One cannot predict the number of isomers by mere inspection of the molecular formula. A logical method runs as follows. First write the different bonding skeletons for the multivalent atoms, in this case the three C’s and the O. There are three such skeletons:

To attain the covalences of 4 for C and 2 for O, eight H’s are needed. Since the molecular formula has only six H’s, a double bond or ring must be introduced onto the skeleton. In (i), the double bond can be situated three ways, between either pair of C’s or between the C and O. If the H’s are then added, we get three isomers: (1), (2), and (3). In (ii), a double bond can be placed only between adjacent C’s to give (4). In (iii), a double bond can be placed between a pair of C’s or C and O, giving (5) and (6), respectively.

CHAPTER 1Structure and Properties of Organic Compounds

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In addition, three ring compounds are possible:

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Bonding and Molecular Structure

2.1 

Atomic Orbitals

An atomic orbital(AO) is a region of space about the nucleus in which there is a high probability of finding an electron. An electron has a given energy, as designated by (a) the principal energy level (quantum num- ber) n, related to the size of the orbital; (b) the sublevel s,p, d,f, org, related to the shape of the orbital; (c) except for the s, each sublevel having some number of equal-energy (degenerate) orbitals differing in their spatial orientation; and, (d) the electron spin, designated ↑ or ↓. Table 2.1 shows the distribution and desig- nation of orbitals.

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TABLE 2.1

CHAPTER 2Bonding and Molecular Structure

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The sorbital is a sphere around the nucleus, as shown in cross section in Fig. 2.1(a). A porbital is two spherical lobes touching on opposite sides of the nucleus. The three porbitals are labeled p x, py, and pz because they are oriented along the x-, y-, and z-axes, respectively [Fig. 2.1(b)]. In a porbital, there is no chance of find- ing an electron at the nucleus—the nucleus is called a node point. Regions of an orbital separated by a node are assigned  and  signs. These signs are not associated with electrical or ionic charges. The sorbital has no node and is usually assigned a .

Figure 2.1

Three principles are used to distribute electrons in orbitals:

Aufbau or building-up principle.Orbitals are filled in order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, and so on.

Pauli exclusion principle.No more than two electrons can occupy an orbital and then only if they have opposite spins.

Hund’s rule.One electron is placed in each equal-energy orbital so that the electrons have parallel spins, before pairing occurs. (Substances with unpaired electrons are paramagnetic—they are attracted to a magnetic field.)

Problem 2.1Show the distribution of electrons in the atomic orbitals of (a) carbon and (b) oxygen.

A dash represents an orbital; a horizontal space between dashes indicates an energy difference. Energy increases from left to right.

Atomic number of C is 6.

↑↓↑↓ ↑ ↑ 

1 2 2 x 2 y 2s z s  p p p

The two 2pelectrons are unpaired in each of two porbitals (Hund’s rule).

Atomic number of O is 8.

↑↓↑↓ ↑↓  ↑ ↑

1 2 2 x 2 y 2s z s  p p p

2.2 

Covalent Bond Formation—Molecular Orbital (MO) Method

A covalent bond forms by overlap (fusion) of two AO’s—one from each atom. This overlap produces a new orbital, called a molecular orbital (MO),which embraces both atoms. The interaction of two AO’s can produce two kinds of MO’s. If orbitals with like signs overlap, a bonding MOresults which has a high

CHAPTER 2Bonding and Molecular Structure

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electron density between the atoms and therefore has a lower energy (greater stability) than the individual AO’s. If AO’s of unlike signs overlap, an antiboding MO*results which has a node (site of zero electron density) between the atoms and therefore has a higher energy than the individual AO’s. An asterisk indicates antibonding.

Head-to-head overlap of AO’s gives a sigma( σ)MO—the bonds are called σ bonds, [Fig. 2.2(a)]. The

corresponding antibonding MO* is designated σ* [Fig. 2.2(b)]. The imaginary line joining the nuclei of the bonding atoms is the bond axis, whose length is the bond length.

Figure 2.2

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Two parallel porbitals overlap side by side to form a pi( π) bond [Fig. 2.3(a)] or a π* bond [Fig. 2.3(b)]. The bond axis lies in a nodal plane (plane of zero electronic density) perpendicular to the cross-sectional plane of the π bond.

Single bonds are σbonds. A double bond is one σ and one π bond. A triple bond is one σ and two π bonds (a πz and a πy, if the triple bond is taken along the x-axis).

Although MO’s encompass the entire molecule, it is best to visualize most of them as being localized

between pairs of bonding atoms. This description of bonding is called linear combination of atomic orbitals (LCAO).

Figure 2.3

CHAPTER 2Bonding and Molecular Structure

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Problem 2.2What type of MO results from side-to-side overlap of an sand a porbital?

The overlap is depicted in Fig. 2.4. The bonding strength generated from the overlap between the sAO and the  portion of the porbital is canceled by the antibonding effect generated from overlap between the  s and the  portion of the p. The MO is nonbonding(n); it is no better than two isolated AO’s.

Figure 2.4

Problem 2.3List the differences between a σ bond and a π bond.

σ Bond π Bond

1.  Formed by head-to-head overlap of AO’s. 1. Formed by lateral overlap of porbitals

(or pand dorbitals).

2.  Has cylindrical charge symmetry about bond axis. 2. Has maximum charge density in the

cross-sectional plane of the orbitals.

3.  Has free rotation. 3. Does not have free rotation.

4.  Has lower energy. 4. Has higher energy.

5.  Only one bond can exist between two atoms. 5. One or two bonds can exist between two atoms.

Problem 2.4Show the electron distribution in MO’s of (a) H

, (b) H+, (c) H, (d) He . Predict which are unstable.

2 2 2 2

Fill the lower-energy MO first with no more than two electrons.

H2 has a total of two electrons, therefore:

Stable(excess of two bonding electrons).

  H, formed from H  and H·, has one electron:

↑↓ 

σ σ *

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↑

*

Stable(excess of one bonding electron). Has less bonding strength than H 2.

  H, formed theoretically from H: and H·, has three electrons:

↑↓ ↑

*

Stable(has net bond strength of one bonding electron). The antibonding electron cancels the bonding strength of one of the bonding electrons.

He2 has four electrons, two from each He atom. The electron distribution is

↑↓ ↑↓

σ σ *

Not stable(antibonding and bonding electrons cancel, and there is no net bonding). Two He atoms are more stable than a He2 molecule.

CHAPTER 2Bonding and Molecular Structure

Problem 2.5Since the σ MO formed from 2sAO’s has a higher energy than the σ * MO formed from 1s AO’s, predict whether (a) Li2, (b) Be2 can exist.

The MO levels are as follows: σ1sσ*1s σ2s σ*2s, with energy increasing from left to right.

Li2 has six electrons, which fill the MO levels to give

↑↓ ↑↓ ↑↓ 

σ1s σ * σ 2s σ *

designated (σ1s)2(σ*1s)2(σ2s)2. Li2 has an excess of two electrons in bonding MO’s and therefore can exist; it is by no means the most stable form of lithium.

Be2 would have eight electrons:

↑↓ ↑↓ ↑↓ ↑↓

σ1s σ * σ 2s σ *s

There are no net bonding electrons, and Be2 does not exist.

Stabilities of molecules can be qualitatively related to the bond order,defined as

Bond order ≡ (Number of valence electrons in MO’s) – (Number of valence electrons in MO*’s)

2

The bond order is usually equal to the number of σ and π bonds between two atoms—in other words, 1 for a single bond, 2 for a double bond, 3 for a triple bond.

Problem 2.6The MO’s formed when the two sets of the three 2porbitals overlap are

π π σ π*  π*  σ

2 py  2 pz  2 px  2 py  2 pz

2 px

(the π and π* pairs are degenerate). (a) Show how MO theory predicts the paramagnetism of O2. (b) What is the bond order in O2?

The valence sequence of MO’s formed from overlap of the n 2 AO’s of diatomic molecules is

σ σ * π π σ π*

π*  σ *

2s  2s  2 py  2 pz  2 px  2 py

2 pz  2 px

O2 has 12 elecrons to be placed in these MO’s, giving

(σ 2s

)² (σ* )² (π

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2 py

)² (π

––––––––

2 pz

)² (σ

––––––––

2 px

)² (π*

y

)¹(π* )¹

z

  The electrons in the two, equal-energy, π* MO*’s are unpaired; therefore, O2 is paramagnetic.

Electrons in the first two molecular orbitals cancel each other’s effect. There are 6 electrons in the next 3 bonding orbitals and 2 electrons in the next 2 antibonding orbitals. There is a net bonding effect due to 4 electrons. The bond order is 1/2 of 4, or 2; the two O’s are joined by a net double bond.

2.3 

Hybridization of Atomic Orbitals

A carbon atom must provide four equal-energy orbitals in order to form four equivalent σ bonds, as in methane, CH4. It is assumed that the four equivalent orbitals are formed by blending the 2sand the three 2pAO’s give four new hybrid orbitals, called sp³ HO’s (Fig. 2.5). The shape of an sp³ HO is shown in Fig. 2.6. The larger lobe, the head, having most of the electron density, overlaps with an orbital of its bonding mate to form the bond. The smaller lobe, the tail, is often omitted when HO’s are depicted (see Fig. 2.11). However, at times the tail plays an important role in an organic reaction.

The AO’s of carbon can hybridize in ways other than sp³, as shown in Fig. 2.7. Repulsion between pairs of elec- trons causes these HO’s to have the maximum bond angles and geometries summarized in Table 2.2. The sp² and sp HO’s induce geometries about the C’s as shown in Fig. 2.8. Only σ bonds, not π bonds, determine molecular shapes.

CHAPTER 2Bonding and Molecular Structure

Figure 2.5

Figure 2.6

Figure 2.7

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TABLE 2.2

* See Fig. 1.2.

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Figure 2.8

CHAPTER 2Bonding and Molecular Structure

Problem 2.7The H 2O molecule has a bond angle of 105º. (a) What type of AO’s does O use to form the two equivalent σ bonds with H? (b) Why is this bond angle less than 109.5º?

(a)

O = ↑↓ ↓ ↑↓ ↑ ↑

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(ground state)

8 1s 2s 2 px 2 py 2 pz

O has two degenerate orbitals, the py and pz, with which to form two equivalent bonds to H. However, if O used these AO’s, the bond angle would be 90º, which is the angle between the y- and z-axes. Since the angle is actually 105º, which is close to 109.5º, O is presumed to use sp³ HO’s.

O = ↑↓ ↑↓ ↑↓ ↑ ↑

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sp(³ HO's )

⁸ 1s

2sp³

(b) Unshared pairs of electrons exert a greater repulsive force than do shared pairs, which causes a contraction of bond angles. The more unshared pairs there are, the greater is the contraction.

Problem 2.8Each H—N—H bond angle in :NH 3 is 107º. What type of AO’s does N use?

7 N = ↑↓↑↓ ↑ ↑ ↑

1s 2s 2 px 2 py 2 pz

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(ground state)

If the ground-state N atom were to use its three equal-energy pAO’s to form three equivalent N—H bonds, each H—N—H bond angle would be 90º. Since the actual bond angle is 107º rather than 90º, N, like O, uses sp³ HO’s:

7N = 1s

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2sp³

(sp³ HO’s)

Apparently, for atoms in the second period forming more than one covalent bond (Be, B, C, N, and O), a hybrid orbital must be provided for each σbond and each unshared pair of electrons. Atoms in higher periods also often use HO’s.

Problem 2.9Predict the shape of (a) the boron trifluoride molecule (BF 3) and (b) the boron tetrafluoride anion (BF ). All bonds are equivalent.

The HO’s used by the central atom, in this case B, determine the shape of the molecule:

5 B = ↑↓↑↓ ↑

1s 2s 2 px 2 py 2 pz

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(ground state)

There are three sigma bonds in BF3 and no unshared pairs; therefore, three HO’s are needed. Hence, B uses

sp² HO’s, and the shape is trigonal planar. Each F—B—F bond angle is 120º.

B = ↑↓ ↑  ↑  ↑

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(sp² hybrid state)

⁵ 1s

2sp²

2 pz

The empty pz orbital is at right angles to the plane of the molecule.

  B in BF has four σ bonds and needs four HO’s. B is now in an sp³ hybrid state:

B = ↑↓ ↑  ↑  ↑

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sp(³ hybrid state )

⁵ 1s

2sp³

used for bonding

The empty sp³ hybrid orbital overlaps with a filled orbital of F , which holds two electrons:

The shape is tetrahedral; the bond angles are 109.5º.

CHAPTER 2Bonding and Molecular Structure

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Problem 2.10Arrange the s, p, and the three sp-type HO’s in order of decreasing energy.

The more scharacter in the orbital, the lower the energy. Therefore, the order of decreasing energy is

p> sp ³ > sp² > sp> s

Problem 2.11What effect does hybridization have on the stability of bonds?

Hybrid orbitals can (a) overlap better and (b) provide greater bond angles, thereby minimizing the repulsion between pairs of electrons and making for great stability.

By use of the generalization that each unshared and σ-bonded pair of electrons needs a hybrid orbital, but

π bonds do not, the number of hybrid orbitals (HON)needed by C or any other central atom can be obtained as HON  (Number of σ bonds) + (Number of unshared pairsof electrons)

The hybridized state of the atom can then be predicted from Table 2.3. If more than four HO’s are needed, dorbitals are hybridized with the sand the three p’s. If five HO’s are needed, as in PCl 5, one dorbital is included to give trigonal-bipyramidalsp ³dHO’s [Fig. 2.9(a)]. For six HO’s, as in SF 6, two dorbitals are included to give octahedralsp ³d² HO’s [Fig. 2.9(b)].

Figure 2.9

TABLE 2.3