Properties of atoms Beauchamp Atomic Structure nucleus ...psbeauchamp/pdf/314_rev_01_atom_props.pdfAufbau Principle - orbitals are filled with electrons in order of increasing energy

Properties of atoms Beauchamp

y:\files\classes\0 Organic Topics - latest\315 topics\20 315 lecture notes, 7-9-15\314 Review Problems\01 review, Zeff of atoms 19p, answers, 6p.doc

Atomic Structure Since molecules are constructed from atoms, we will quickly review the structure of atoms.

Volume of electron cloud compared to volume of nucleus

VeVn

43 (1.33)(3.14)(100,000)3 = 4 x 1015= =

electron clouds determine the overall volumeof atoms

mass protonsmass electrons

18001

protons and neutronsdetermine an atom's mass=

ecore evalence

p,n

1

100,000p = protons = constant # that defines the elementn = neutrons = varies = defines the isotopee = electrons = varies, depending on bonding patterns associated termif electrons = protons (same # of e's and p's) atomif electrons < protons (deficiency of e's) cationif electrons > protons (excess of e's) anion

(eval) = valence electrons = The outermost layer of electrons, which determines the bonding patterns. The usual goal is to attain a noble gas configuration. This is accomplished by losing e's (forming cations) or gaining e's (forming anions) or sharing e's (covalent bonds)

(ecore) = core electrons = The innermost layer(s) of electrons (usually full shells or subshells). These e's are held too tightly for bonding (sharing) and not usually considered in the bonding picture. These e's cancel a portion of the nuclear charge (called shielding) so that the valence e's only see part of the nuclear charge, called Zeffective.

Zeffective = (# protons) - (core e's) = the effective nuclear charge. This is the net positive charge felt by the valence e's (bonding and lone pairs). Zeffective = same # as the column of the main group elements.

nucleus

rern

3

mpme

=

What do atomic orbitals look like?

a. s orbitals – are spherical in shape. The wave nature of electrons (phase) may change as the distance increases from the nucleus but at any fixed distance from the nucleus the phase nature of s electrons will be the same (think of layers on an onion). Nodes are regions in space where the probability of finding an electron is zero (the phase is zero).

b. p orbitals –have a dumbbell shape with two lobes at 180o to one another with opposite phase nature. All p lobes intersect at the nucleus with zero probability of finding any electron density (= node).



a single 2p orbital,has node at nucleus

all three 2p orbitals (2px, 2py, 2pz)are oriented at 90o to one another,represented as px, py and pz.

px

pz

py

c. d and f orbitals have more complicated shapes and are not important to beginning organic chemistry. However they are important for understanding the inorganic world and many biochemical molecules involve the chemistry of d orbitals (like iron in hemoglobin, cobalt in vitamin b12, and zinc, copper, manganese and others in proteins.). We will not discuss these in our course.

lobes in the yz plane

lobes in the xz plane

lobes in the xy plane

lobes in the xy planealong axes

lobes along the z axis

x

y

zdyz dxz dxy dx2-y2 dz2

x x x x

y y y y

z z z z

We rarely have occasion to discuss d orbitals. They are helpful in discussions of sulfur, phosphorous and a few transition metals. Even in those discussions you canmainly look at them as biggerversions of p-like orbitals.

Bonds represent an overlap of orbitals, which are a region where negative electron density is present. Electrons allow positive nuclei to remain in close proximity (come together). In our course covalent bonds will represent two shared electrons between two bonded atoms. We propose two types of covalent bonds: sigma, where electron density is directly between the two bonded atoms and pi, where electron density is above and below (or in front and in back) of a line connecting the two bonded atoms. Sigma bonds are always the first bond and pi bonds are the second (and third) bonds overlapping a sigma bond. Pi bonds are represented by sideways overlap of two p orbitals.

possible sigma bonds - the electron density is directly between the two bonded atoms,

always the first bond

ABA Hpi bonds - the electron density is above and

below the two bonded atoms(or in front and in back), none is directly between,

always second or third bonds

A B bond bond

bond

Problem 1 – What kinds of bonds are between the nonhydrogen atoms below?

H C N H C C H C O C C

H

H

C N

H

H

C O

H

H

H

H

H

C C

H

H

H

H

H

H

a b c d e f g

Atomic Configuration provides a representation of the electrons about a single isolated atom. Atoms represented are NOT in molecules. The usual atomic orbitals (s,p,d,f) are used and electrons fill into these atomic orbitals according to a few simple rules. The electrons closest to the nucleus are held the tightest and called the core electrons (full shells), while those in the incompletely filled outermost shell are called valence electrons. The valence electrons largely determine the bonding patterns and chemistry of the atom in order to gain a Noble gas configuration.



+z

nucleus

1s

2s 2p 2p 2p

3s 3p 3p 3p 3d 3d 3d 3d 3d

n=1 shelln=2 shell

n=3 shell

Electrons fill into atomic orbitals according to the following rules.

1. Pauli Exclusion Principle - only two electrons may occupy any orbital and those electrons must have opposite spins2. Hund's Rule - electrons entering a subshell containing more than one orbital (p,d,f) will spread themselves out over all of the available orbitals with their spins in the same direction until the final subshell is over half filled.3. Aufbau Principle - orbitals are filled with electrons in order of increasing energy (lowest PE to highest PE)

Periodic Table (another view) – Atomic orbitals (and electrons) are found in the following locations of the periodic table.

Elements in the upper right corner have the tightest hold on electronsbecause they have the largest Zeff (towards the right) and are closest to the effective nuclear charge (towards the top).

Elements in the lower left corner have the weakest hold on electronsbecause they have the smallest Zeff (towards the left) and are farthest fromthe effective nuclear charge(towards the bottom).

1s 1s

2s3s

4s5s

6s

7s

3d4d

5d

6d

2p

3p

4p5p

6p

7p

4f5f

Zeffective = +1 +2 +3 +4 +5 +6 +7 +8

n = 1n = 2n = 3

n = 4

n = 5

n = 6

n = 7

Table 1 – First ionization potentials of atoms (equals the energy to remove an electron from a neutral atom)

starting energy

final energy

H +314 (He) +568Li +124 Be +215 B +192 C +261 N +335 O +315 F +402 (Ne) +499Na +118 Mg +177 Al +138 Si +189 P +242 S +239 Cl +300 (Ar) +363K +99 Ca +141 Ga +138 Ge +182 As +226 Se +225 Br +273 (Kr) +323

Group 1A Group 2A Group 3A Group 4A Group5A Group 6A Group 7A Group 8A

Energy to ionize an electron from neutral atoms = IP1 (units are kcal/mole). Compare rows and compare columns.

Zeff = +1 Zeff = +2 Zeff = +3 Zeff = +4 Zeff = +5 Zeff = +6 Zeff = +7 Zeff = +8

Energy = Potential Energy

greater PE (less stable)

lower PE (more stable)

Ionization always has an energy cost to strip an electron from an atom.

ionization potential(kcal/mole)

Atom

electron is lost

Atom + electron

Atom

Atom e-

The general trend, across a row, is that the ionization potential gets larger and the hold on electrons is stronger. Why? The answer is found in the size of Zeffective. As we move from Li (IP1 = +124) to carbon (IP1 = +261) to fluorine (IP1 = +402), the valence shell stays the same (n = 2), but the effective nuclear charge holding those electrons keeps increasing, from +1 to +4 to +7. Because each extra electron goes into the same shell



(n = 2), there is essentially no shielding of the nucleus by any of the additional electrons (the electrons are doing their best to avoid one another). It’s a lot harder to pull an electron away from a +7 charge than a +1 charge. It seems reasonable that fluorine has a stronger attraction for electrons than the other second row elements, even when those electrons are shared in a chemical bond. There are a few deviations that we will ignore.

In a column, all of the atoms have the same Zeffective. What changes in a column is how far away the valence electrons are from that same effective nuclear charge (Zeff). Valence electrons in inner quantum shells are held tighter because they are closer to same Zeff.

Problem 2 - Explain the atomic trends in ionization potential in a row (Na vs Si vs Cl) and in a column (F vs Cl vs Br).

Atomic radii and ionic radii also support these periodic table trends in electron attracting power. Smaller radii, below, indicate 1. a stronger contraction in the same row (shell) of the electron clouds due to higher Zeffective attraction for the electrons and 2. in a column, closer distance of electrons to a particular Zeffective nuclear charge, due to electrons being in a lower principle quantum shell (fewer shells).

When an atom acquires a negative charge (becomes an anion), it expands its size due to the greater electron-electron repulsion. Remember, most of an atom’s mass is in the nucleus, but most of its size is due to the electron cloud.

H = 53 He = 31Li = 167 Be = 112 B = 87 C = 67 N = 56 O = 48 F = 42 Ne = 38Na = 190 Mg = 145 Al = 118 Si = 111 P = 98 S = 88 Cl = 79 Ar = 71K = 243 Ca = 194 Ga = 136 Ge = 125 As = 114 Se = 103 Br = 94 Kr = 88Rb = 265 Sr = 219 In = 156 Sn = 145 Sb = 133 Te = 123 I = 115 Xe = 108

Table 2: Neutral atomic radii in picometers (pm) = 10-12 m [100 pm = 1 angstrom]

3d elements4d elements

Problem 3 - Explain the atomic trends in atomic radii in a row (Li vs C vs F) and in a column (C vs Si vs Ge).

Li = 90 Be = 59 B = 41 C = N = 132 O = 126 F = 119Na = 116 Mg = 86 Al = 68 Si = P = S = 170 Cl = 167 K = 152 Ca = 114 Ga = 76 Ge = As = Se = 184 Br = 182 Rb = 166 Sr = 132 In = 94 Sn = Sb = Te = 207 I = 206

Table 3: Cations (on the left) and anions (on the right) radii in picometers (pm) = 10-12 m [100 pm = 1 angstrom]


+1+1

+1+1

+2

+2

+2

+2

+3

+3+3

+3

-3 -2

-2

-2-2

-1-1

-1-1

as cations as anions Problem 4 - In the tables above:

a. Explain the cation distances compared to the atomic distances (Li, Be, B). (Tables 2 and 3) b. Explain the anion distances compared to the atomic distances (N, O, F). (Tables 2 and 3) c. Explain the cation distances in a row (Li, Be, B). (Table 3) d. Explain the anion distances in a row (N, O, F). (Table 3)

Atoms behave differently in bonding situations when attempting to acquire a Noble gas configuration (ionic, polar covalent and pure covalent). This is often described as the “octet rule,” (8 valence electrons) for main group elements (1A through 8A). Almost always true for second row elements (exceptions possible for higher rows because of the d orbitals).

Electronegativity measures the attraction an atom has for electrons in chemical bonds. There are many proposed scales to do this, we will use the Pauling scale. The larger the number the greater is the attracting power of an atom for the electrons in chemical bonds. Electronegativity will determine nonpolar, polar and ionic characteristics of bonds, and when shapes are included it determines the same attributes in molecules. Compare rows and compare columns in the following table. The largest numbers are found in the upper right corner (strongest attraction for electrons) and the smallest numbers are found in the lower left corner (weakest attraction



for electrons). Group 8A elements don’t have a value listed because the ones toward the top don’t make bonds.

(chi) is the symbol used for electronegativity, which is the property the indicates an atoms attraction for electrons in chemical bonds with other atoms.

H = 2.2Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O = 3.5 F = 4.0 He = noneNa = 0.9 Mg = 1.2 Al = 1.5 Si = 1.9 P = 2.2 S = 2.6 Cl = 3.2 Ne = noneK = 0.8 Ca = 1.0 Ga = 1.6 Ge = 1.9 As = 2.2 Se = 2.5 Br = 3.0 Ar = noneRb = 0.8 Sr = 0.9 I = 2.7 Kr = 3.0

Approximate electronegativity values for some main group elements.

Simplistic estimate of bond polarities using differences in electronegativity between two bonded atoms.

0.4 0.4 < < (1.4 - 2.0)(1.4 - 2.0) <

considered to be a pure covalent bond (non-polar)considered to be a polar covalent bond (permanent charge imbalance)considered to be an ionic bond (cations and anions)


Group 1AZeff = +1

Group 2AZeff = +2

Group 3AZeff = +3

Group 4AZeff = +4

Group 5AZeff = +5

Group 6AZeff = +6

Group 7AZeff = +7

Group 8AZeff = +8

BA A B =bond polarity based on

Elemental hydrogen, fluorine, oxygen and nitrogen are examples of atoms sharing electrons evenly because the

two atoms competing for the bonded electrons are the same. Each line drawn between two atoms symbolizes a two electron bond. These examples also illustrate that pure covalent bonding can occur with single, double and triple bonds and shows that some elements have lone pairs of electrons.

H H OO NNhydrogen molecules with a

single bond, each H has a duet of electrons (like helium)

oxygen molecules with a double bond, each O has an octet

of electrons (like neon)

nitrogen molecules with a triple bond, each N has an octet

of electrons (like neon)

F Ffluorine molecules with a

single bond, each F has a octetof electrons (like neon)

If the two bonded atoms are not the same, then there will be different attractions for the electrons. One atom will have a greater pull for the electrons and will claim a greater portion of the shared electron density. This will make that atom polarized partially negative, while the atom on the other side of the bond will be polarized partially positive by a similar amount. Because there are two opposite charges separated along a bond, the term “dipole moment” () is used to indicate the magnitude of charge separation. Bond dipoles depend, not just on the amount of charge separated, but also the distance by which the charges are separated, as indicated by their bond lengths.

The symbols + and -- represent qualitative charge separation forming a bond dipole. Alternatively, an arrow can be drawn pointing towards the negative end of the dipole and a positive charge written at the positive end of the dipole.

A B A B

d d

+ -

or...........

Two qualitative pictures of a bond dipole are represented below. B is assumed to be more electronegative than A.

bond =amount of charge separated

x distance between charges in cm

= units of Debye (D = 10-18 esu-cm)

bond = ( e ) x ( d ) = bond dipole moment

e = electrostatic charge (sometimes written as q)

d = distance between the opposite charges

The absolute value of a unit charge on an electron or proton is 4.8x10-10 esu

This is often given in angstrums (A = angstrum), but converted to cm for use in calculations (1A = 10-8 cm)

A>B



Because there are two factors that make up bond dipole moments (charge and distance), bonds that appear to be more polar based on electronegativity differences might have similar dipole moments if they are shorter than less polar bonds. The methyl halides provide an example of this aspect. Fluoromethane has a more polar bond, but shorter bond length, while iodomethane has a less polar bond, but longer bond length. The opposing trends of the halomethanes produce similar dipole moments in all of the molecules. Polarity, polarizability and shape have a large influence on physical properties (mp, bp, solubility, etc). We will briefly review these below.

CH3-F 1.86 D139 pm

1.87 D178 pm

1.81 D193 pm

1.62 D214 pm

mp (oC) bp (oC)

-97 -24

-78-142

-94 +3

-66

3.97 g/dm3 (gas)

density

0.53 g/dm3 (gas)

0.92 g/dm3 (gas)

+42

CH3-Cl

CH3-Br

CH3-I 2.28 g/mL (liquid)*

CF = 1.5

CF = 0.7

CF = 0.5

CF = 0.2

dC-X molecule

= dipole mement (measure of polarity)

d = bond distance in picometers (10-12 m)

H = 2.2 C = 2.5 N = 3.0

Cl = 3.2 Br = 3.0 I = 2.7

electronegativity

F = 4.0

O = 3.5

* Notice there is a different phase.

Ion-Ion interactions – There are very strong forces of attraction between oppositely charged ions. The melting points of ionic salts are typically very high. The solubility of many salts in water is often high because water has a concentrated partial positive end (the protons) that can interact strongly with anions and a concentrated partial negative end (the oxygen) that can interact strongly with cations, leading to very high overall solvation energies. However, solvation energy has to be larger than the very large lattice energy (energy of the cations interacting with the anions) or the salt will not dissolve.

Each type of ion is typically surrounded on all sides by oppositely charge ions in some sort of lattice structure. Before a solid can melt the whole network of ions has to break down.

K

Li

Na

O-2

846 / 1717sol. = 1.3g= 3.0

610 / 1383sol. = 550g= 2.2

550 / 1289sol. = 1670g=2.0

467 / 1178sol. = 1510g=1.7

ClF Br I

996 / 1787sol. = 404g= 3.1

858 / 1517sol. = 920g= 3.2

801 / 1465sol. = 359g= 2.3

747 / 1447sol. = 905g= 2.1

660 / 1304sol. = 1840g=1.8

996 / 1787sol. = 344g= 2.4

734 / 1435sol. = 678g= 2.2

681 / 1345sol. = 1400g=1.9

1570 / 2563sol. = reacts=2.5

1132 / 1950dec

sol. = reacts=2.6

>300dec / -sol. = reacts=2.7

The first number is the melting point and the second number is the boiling point (in oC). "sol." = solubility in 100 g water. X is the difference in electronegativity.



Covalent Bonding (Molecules)

A single neutral hydrogen atom would have a single valence electron. This is not a common occurrence in our world because such a hydrogen atom would be too reactive. However, if one were able to generate a source of hydrogen atoms, they would quickly join together in simple diatomic molecules having a single covalent bond with the two hydrogen atoms sharing the two electrons and attaining the helium Noble gas configuration (duet rule). Such a reaction would be very exothermic.

= a b_ = 2.2 - 2.2 = 0

H H HH

The line symbolizes a two-electron, pure-covalent bond based on the calculation below.

If two hydrogen atoms should find one another, they wouldfrom a diatomic molecule with a tremendous release of energy,about 104 kcal/mole.

H = -104 kcal/mole

The atoms of organic chemistry, H, C, N, O, S and halogens, tend to attain a Noble gas configuration by sharing electrons in covalent bonds of molecules. Simple formulas often have only one choice for joining the atoms in a molecule (CH4, NH3, H2O, HF). As the number of atoms increase, however, there are many more possibilities, especially for carbon structures (sometimes incredible numbers of possibilities, C40H82 > 62 trillion isomers!). These possibilities may require single, double and/or triple bonds (sigma and pi bonds will be discussed soon), in chains or rings of atoms. Carbon, nitrogen, oxygen and fluorine can be bonded in all combinations, according to their valencies. We study most of the bond types below. Notice that each atom below only makes as many bonds as electrons needed to complete its octet (C = 4, N = 3, O = 2 and F = 1). N, O and F have lone pairs.

C

H

H

H

H NH

H

H OH H FH

= a b_

= 2.5 - 2.2 = 0.3

pure covalent bond based on the

calculation below

= a b_

= 3.0 - 2.2 = 0.8

polar covalent bond based on the

calculation below

= a b_

= 3.5 - 2.2 = 1.3


calculation below

= a b_

= 4.0 - 2.2 = 1.8


calculation below

2a. Dipole-dipole interactions (partial polarity) – Since the partial charges present in molecules having dipole moments are less than full charges and the bonds are very directional (not “omni” like ions), and attractions for neighbor molecules are weaker than is found in ionic salts. However, polar molecules usually have stronger attractions than nonpolar molecules of similar size and shape. Many polar molecules below are compared with a similar size and shape nonpolar molecules to show how polarity can affect boiling points (an indication of the strength of attractions among neighbor molecules, when other factors are similar). A higher boiling point indicates stronger attractions. Dipole-dipole interactions represent moderate forces of attraction between partially polarized molecules. The molecular dipole moments are indicators of charge imbalance due to a difference in electronegativity , bond length and molecular shape. Fluorine is unusual in that it makes very polar bonds but is not very polarizable and holds on very tightly to its lone pairs, so it sometimes surprises us with lower than expected boiling points. Sometimes atoms that hold onto their electrons very tightly are called ‘hard’ and atoms that have polarizable electrons are called ‘soft’.

Melting points are better indicators of packing efficiency and boiling points are better indicators of forces of attraction. Miscellaneous compounds are compiled below, representing a variety of factors discussed in this topic.



A Few Useful Comparisons

CC

H

CN

H

Hpolarnonpolar

= 0.0 Dbp = -81oCmp = -84oCH2O sol. = insolublepKa = 25

= 2.98 Dbp = +26oCmp = -13oCH2O sol. = misciblepKa = 9.2

CN

H

more polar

C

O

H H

C

C

H H

H H

= 0.0 Dbp = -104oCmp = -169oCH2O sol. = 2.9mg/LpKa = 44

= 2.3 Dbp = -20oCmp = -92oCH2O sol. = 400g/LpKa = NA

nonpolar

polar

C

O

H H

more polar

resonanceresonance

Tbp = 107oCTbp = 84oC

CC

H2C

CN

H3C

H polarnonpolar

= 0.78 Dbp = -23oCmp = -102oCH2O sol. = insoluble

= 3.92 Dbp = +81oCmp = -46oCH2O sol. = miscible

CN

H3C

more polar

resonance

Tbp = 104oC

Problem 5 – What is the effect on the dipole moment of switching a hydrogen for a methyl in each pair below? Propose a possible explanation.

C

O

H2N H

= 3.71 Dbp = +210oCmp = +2oC

CHH

C

O

F CH3

= 2.96 Dbp = +21oCmp = -84oC

C

O

HO H

= 1.41 Dbp = +118oCmp = +17oC

C

O

H2N CH3

= 3.76 Dbp = +222oCmp = +80oC

= 1.74 Dbp = +118oCmp = +17oC

C

O

F H

= 2.02 Dbp = -29oCmp = -142oC

C

O

HO CH3

O

CCH3H

O

= 2.68 Dbp = +20oCmp = -123oC

= 2.3 Dbp = -20oCmp = -92oC

What is the effect of "R" groups? Do they make the molecule more polar, less polar or no different? Electron donation or electron withdrawal through sigma bonds is called an inductive effect.

C

O

H H = 2.3 Dbp = -20oCmp = -92oCH2O sol. = 400g/L

polar

C

O

H H

resonanceC

O

H3C H

= 2.68 Dbp = +20oCmp = -123oCH2O sol. = very sol.

C

O

H3C CH3 = 2.91 Dbp = +56oCmp = -94oCH2O sol. = very sol.

C

O

H3C H

resonance

C

O

H3C CH3

resonance

polar polar

CC

C

= 0.0 Dbp = -34oCmp = -134oC

nonpolar

H3C

H2C

CH3

= 0.08 Dbp = -42oCmp = -188oC

nonpolar

HC

CH2H3C

= 0.37 Dbp = -47oCmp = -185oC

nonpolar

CH

CH3C = 0.75 D

bp = -23oCmp = -103oC

nonpolar

H CH3 NH2 OH F Cl

N

H

= 0.0 Dbp = 81oCmp = 5C

= 0.31 Dbp = 111oCmp = -95oC

= 1.53 Dbp = 184oCmp = -6oC

= 1.55 Dbp = +182oCmp = +40oC

= 1.66 Dbp = +84oCmp = -44oC

= 1.54 Dbp = +131oCmp = -45oC

= 2.37 Dbp = +115oCmp = -42oC

Br

= 1.73 Dbp = +156oCmp = -31oC

I

= 1.91 Dbp = +188oCmp = -31oC

H

H

H

H



2b. Hydrogen bonds – Hydrogen bonds represent a very special dipole-dipole interaction. Molecules that have this feature have even stronger attractions for neighbor molecules than normal polar bonds would suggest. Solvents that have an O-H or an N-H bond are called “protic solvents” and can both donate and accept hydrogen bonds (because they also have lone pairs of electrons). They generally have higher boiling points than similar sized structures without any “polarized hydrogen atoms”. The reasons for a polarized hydrogen atom’s strong attraction for electron density is that there are no other layers of electrons around a hydrogen atom (hydrogen is the only atom to use the n=1 shell in bonding). When a hydrogen atom’s electron cloud is polarized away from the hydrogen atom in a bond with an electronegative atom, usually oxygen or nitrogen in organic chemistry, an especially strong polarization results, resulting in a strong attraction for a lone pair on a neighbor molecule, or even a pi bond. We call such interactions "hydrogen bonds". A molecule that has such a polarized hydrogen is classified as a hydrogen bond donor. A molecule that has a partial negatively charged region that can associate with such a hydrogen is classified as a hydrogen bond acceptor. Quite often the hydrogen bond acceptor is a lone pair of electrons on another oxygen or nitrogen atom, but fluorine, chlorine bromine or sulfur may provide lone pair acceptor sites as well. Weak hydrogen bonds can even form with C=C pi bonds.

O

HH

O

H

H

OH

H

donates hydrogen bond

accepts hydrogen bond

Hydrogen bonding holds the molecules more tightly to one another. This can be seen in higher boiling points among similar structures where hydrogen bonding is possible versus not possible. Many examples below show this property.

Cl

H

CH3C

O

O H

C CH3

O

OH

CCH3C

H

H

H





C

O

H3C CH3

OH

H

H



N

N

N

N

N H

H

O

HDNA

N

N

NH

H

O DNA

guanine cytosine

three hydrogen bonds in G-C base pair

N

N

N

N

DNAN

N

O

O DNAadenine thymine

two hydrogen bonds in A-T base pair

H

N H

H

Problem 6 – Which base pair binds more tightly in DNA, GC or AT, or are they about the same?



Problem 7 – Provide an explanation for the different boiling points in each column. What is there such a large deviation for NH3, H2O and HF?

HFmp = -84oCbp = +20oC = 1.86 D = 1.8

mp = -114oCbp = -85oC = 1.0 D = 1.0

mp = -87oCbp = -67oC = 0.8 D = 0.8

H2Omp = 0oCbp = +100oC = 1.80 D = 1.2

mp = -82oCbp = -60oC = 1.0 D = 0.4

mp = -66oCbp = -41oC = ? D = 0.4

NH3

mp = -78oCbp = -33oC = 1.42 D = 0.8

mp = -132oCbp = -88oC = 0.0 D = 0.0

mp = -111oCbp = -62oC = 0.0 D = 0.0

CH4

mp = -182oCbp = -164oC = 0.0 D = 0.3

mp = -185oCbp = -112oC = 0.0 D = 0.3

mp = -165oCbp = -88oC = 0.0 D = 0.2

HClH2SPH3

SiH4HBrH2SeAsH3

GeH4

4B 5B 6B 7B

Hemp = -272oCbp = -269oC = 0.0 D = NA

mp = -249oCbp = -246oC = 0.0 D = NA

mp = -189oCbp = -185oC = 0.0 D = NA

Ne

Ar

8B

Temp(oC)

100

50

0

-50

-100

-150

boiling points (oC)

CH4

GeH4

SiH4

SnH4

SbH3

PH3

NH3

row2

row3

row4

row5

-200

-250

-300

H2O

HF

He

H2S

H2Se

H2Te

HCl HBr

HI

Ne

Ar

Kr

AsH3

Answer: Each dotted line represents the boiling points in the hydrides in a column in the periodic table, plus the Noble gases. In the Noble gases there is a continuning trend towards higher boiling point as the main atom gets larger with more and more polarizable electron clouds. Where bonds to hydrogen become polar, hydrogen bonding becomes important, leading to stronger attractions for neighbor molecules and higher boiling points deviate from the trends observed due to dispersion forces. Deviations are clearly seen with NH3, H2O and HF. Melting points are not plotted.

mp = -51oCbp = -35oC = 0.4 D = 0.5

mp = -49oCbp = -2oC = ? D = 0.1

mp = -88oCbp = -17oC = 0.0 D = 0.2

mp = -146oCbp = -52oC = 0.0 D = 0.2

HIH2TeSbH3SnH4mp = -157oCbp = -153oC = 0.0 D = NA

Kr

Problem 8 – Explain the difference in boiling points of the following molecules.

H3C

H2C

CH3 H3CO

CH3

= 0.08 Dbp = -42oCmp = -188oCH2O sol. = 40mg/L

= 1.30 Dbp = -22oCmp = -141oCH2O sol. = 71g/L

OHH3C

OH



2. Dispersion forces / van der Waals interactions / London forces (nonpolar attractions)

Most of the examples in the table below are gases at room temperature, an indication that there are no strong forces of interaction between the molecules. However, clearly there is some attraction between the molecules, because at room temperature bromine is a liquid and iodine is a solid, and ultimately each gaseous substance in the table does condense to a liquid and solidify to a solid. Also there is quite a range of differences in melting and boiling points among the different compounds.

Substance 2 -259 -253 0.0 gas 28 -210 -196 0.0 gas 32 -218 -183 0.0 gas 38 -219 -188 0.0 gas 71 -101 -35 0.0 gas 160 -7 + 59 0.0 liquid 254 +114 +184 0.0 solid

H HN N

O OF F

Cl ClBr Br

I I

MW (g/mol)

mp (oC)

bp (oC)

DipoleMoment ()

Phase at room temperature

puzzle?



A plot of the trends in melting points and boiling points in the halogen family and Nobel Gas family helps us to see the differences more clearly (below). The difference melting and boiling points show there are differences in the forces of attraction between the halogen molecules even though all of the molecules are nonpolar. Bromine is even a liquid and iodine a solid at room temperature. The higher boiling points (I2 > Br2 > Cl2 > F2) are due to greater polarizability of the larger atoms, where the electrons are held less tightly.

Temp. (oC)

200

150

100

50

0

-50

-100

-150

-200

-250

250

F2

Cl2

Br2

I2

bp

mp

The halogen molecules are similar in shape and nonpolar. There is a smooth, increasing trend in both melting and boiling points. This is suggestive of some factor increasing the forces of attraction between molecles of the halogen family as they get larger. The smooth trend in melting point is not typical, because it can vary so much with differences in shapes. The even change in melting points is observed here because the halogen molecules all have a similar, rigid shape.

Phase at room temperatureF2 = gas He = gasCl2 = gas Ne = gasBr2 = liquid Ar = gasI2 = solid Kr = gas

-273 absolute zero = 0 K

He

Ne

Ar

Kr

The Noble gas molecules are similar in shape and nonpolar. There is a smooth, increasing trend in both melting and boiling points. This is suggestive of some factor increasing the forces of attraction between molecles of the halogen family as they get larger. The smooth trend in melting point is not typical, because it can vary so much with differences in shapes. The even change in melting points is observed here because the halogen molecules all have a similar, spherical shape.

bpmp

room temp 25oC

Dispersion forces are temporary fluctuations of negative electron clouds from one direction to another, relative to the less mobile and more massive positive nuclear charge. These fluctuations of electron density induce fleeting, weak dipole moments. Polarizability is the property that indicates how well this fluctuation of electron density can occur about an atom. Within a column (same Zeff), larger atoms are more polarizable, because they do not hold as tightly to their valence electrons as smaller atoms, since the electrons are farther away from Zeff. Thus, atoms lower in a column are more polarizable than atoms higher up. In a row it’s harder to predict. Atoms to the right have a larger Zeff which should make them less polarizable, but it might seem that lone pairs are held less tightly by only one atom, instead of two atoms in a bond. However, fluorine is not very polarizable (it holds on very hard to even its lone pairs) and appears to indicate that larger Zeff is more important. Picture a cotton ball (polarizable electron clouds = sticky) and a marble (nonpolarizable electron cloud = not sticky).

+Z

In a nonpolar molecule the and are centered, on average. This would seem to indicate that in nonpolar molecules there is no polarity or attraction between molecules. So why do such substances liquify and solidify? Why aren't they always gases?

+Z +Z +Z

+ - + -

and are not centered creating tempory polarity.

Weak, fluctuating polar forces of attraction between molecules.

Fast moving electrons shift position relative to slow moving nuclei, creating a temporary imbalance of charge,

which induces a similar distortion of the electron clouds in neighbor structures and a weak attraction for neighbor molecules.

+Z = nuclear protons = electron cloud

Dispersion Forces



While the examples in the table above are not organic molecules, they are simple and we can use this simplicity to learn important ideas that apply to organic molecules. Simple diatomic molecules must have a linear geometry, since two points determine a straight line. When both atoms are the same, the bonding electrons must be shared evenly. There can be no permanent distortion of the electron clouds toward either atom, so there are no polar bonds.

Zeff = +4 Zeff = +5 Zeff = +6 Zeff = +7

C N O F

Cl

Br

I

Zeff = +7

Zeff = +7

Zeff = +7

Polarizability is greater because there is a weaker hold on the electrons because they are farther away from the same effective nuclear charge, so they are more easily distortable.

Periodic trends in polarizability, .

Features that increase polarizability:

1. smaller Zeff, favors C > N > O > F

2. valence electrons farther from the nucleus when Zeff is similar I > Br > Cl > F.

Polarizability is larger with smaller Zeff because the electrons are not held as tightly, so they are more easily distortable.

Dispersion forces are cumulative, so when the contact surface area is larger, the interactions are stronger

(because there are more of them). Higher molecular weight alkanes have more carbon atoms to interact than lower molecular weight alkanes (even though only similar weak dispersion forces are present in both).



Alkane boiling pointmethane, CH4 -162ethane, C2H6 -89propane, C3H8 -42butane, C4H10 0pentane, C5H12 36hexane, C6H14 69heptane, C7H16 98octane, C8H18 126nonane, C9H20 151decane, C10H22 174undecane, C11H24 196dodecane, C12H26 216

Alkane boiling pointtridecane, C13H28 235tetradecane, C14H30 254pentadecane, C15H32 271hexadecane, C16H34 287heptadecane, C17H36 302octadecane, C18H38 316nonadecane, C19H40 330icosane, C20H42 343henicosane, C21H44 356doicosane, C22H46 369tricosane, C23H48 369 triacotane, C30H62 450tetracotane, C40H82 563

CH4

CH3CH2CH2CH2CH2CH3

CH3CH2CH2CH2CH2CH3

CH4

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3

Larger molecules have more contact surface area with neighbor molecules. Greater dispersion forces mean a higher boiling point.

-150

-100 -50

0 50

100

150

200 250

300

350

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18 C19 C20

Boiling PointTemp (oC)

Straight chain alkanes

From the examples above, you can see that even the weak dispersion forces of attraction become significant when a large number of them are present.

bp of water

mp of water

In alkane isomers (having the same number of atoms, CnH2n+2), more branching reduces contact with neighbor molecules

and weakens the intermolecular forces of attraction. Linear alkanes have stronger forces of attraction than their branched isomers because they have a greater contact surface area with their neighbor molecules. Branches tend to push neighbor molecules away. This is very evident in boiling points, where all of the forces of attraction are completely overcome and the linear alkane isomers have higher boiling points (stronger attractions) than the branched isomers. The strength of these interactions falls off as the 6th power of distance. A structure twice as far away will only have 1/64 the attraction for its neighbor.

More atoms increase the contact surface area with neighbor molecules (not isomers).

Less branching increases contact surface area with neighbor molecules in these isomers.

H2CH3C

H2CH3C

H2C CH3

CH

CH3

H3C CH3bp = -42oC bp = -0.5oCbp = -12oC

CH311

1

12

164

=

=

6

6

H2CH3C

H2C CH3

bp = -0.5oC

Problem 9 – Provide an explanation for the different boiling points in each series.



e. H2CH3C

HCH3C

H2C CH3

C

CH3

H3C CH3

bp = +30 oC

H2C

CH3

H2C CH3

bp = +36 oCCH3

bp = +10 oC

CH3H3C

bp = -89 oC

bp = -0.5 oC

bp = -23 oC

F Cl Br I

bp = +85 oC bp = +130 oC bp = +155 oC bp = +188 oC

Obp = +5 oC

OH

bp = +78 oC

bp = -47 oC

O

bp = +20 oC

HO O

bp = +101 oC

H2N O

bp = +210 oC

bp = -42 oC

N

bp = +82 oC

CO O

bp = -78 oC(sublimes)

a.

b.

c.

d.

e. CH4 CH3Cl CCl4

bp = -164 oC bp = -24 oC bp = +40 oC bp = +61 oC bp = +77 oC

CH2Cl2 CHCl3

H3C

H2C

CH3 H3C

H2C

OH

H3C

H2C

NH

Hbp = -42 oC bp = +17 oC bp = +78 oC

f.



Problem 10 – Match the given boiling points with the structures below and give a short reason for your answers. (-7oC, +31 oC, +80 oC, +141oC, 1420oC)

O O

OH

2-methylpropene2-methyl-1-butene2-butanone propanoic acid

MW = 56 g/molMW = 70 g/molMW = 72 g/mol MW = 74 g/mol

KCl

MW = 74.5 g/mol

potassium chloride

Solutes, Solvents and Solutions

Mixing occurs easily when different substances, having similar forces of interaction, are combined. This observation is summarized in the general rule of “Like Dissolves Like”. To discuss the solution process, there are a few common terms that we need be familiar with.

Solutes are substances that are dissolved in a solvent. There may be as few as 1 solute (ethanol in water) or as many as 1000’s of dissolved solutes (blood).

The solvent is the liquid in which the solute(s) is(are) dissolved. The solvent is usually the major component of the mixture. Sometimes when water is present, it is considered to be the solvent, even when present in only 1% (as is often the case in sulfuric acid/water mixtures).

A solution is the combination of the solvent and the solute(s).

To dissolve a solute in a solvent, the forces of interaction among solute molecules must be overcome and the forces of interaction among solvent molecules must be overcome. These are energy expenses. In return, new interactions develop among solute and solvent molecules. These are energy gains. The balance between these energy expenses and energy gains determines whether a solute will dissolve. A solution is formed when the energy gains are greater than the energy expenses. The important interactions are listed below in decreasing order of energy importance.

Possible energy expenses due to (solute / solute) and (solvent / solvent) interactions

1. ion-ion (lattice energy)2. hydrogen bonds3. dipole-dipole4. dispersion forces

increasingenergy

Possible energy gains from (solute / solvent) interactions

1. ion-dipole (solvation energy)2. hydrogen bonds3. dipole-dipole4. dispersion forces

Solvents are classified as polar or nonpolar based on their dielectric constant (see table below). If the dielectric constant is greater than 15, then the solvent is considered polar. If less than 15, then the solvent is considered nonpolar. This is example of an arbitrary division, created for our convenience.

Water and alcohols (= “organic water”) are in the special class of solvents that have a polarized hydrogen atom in a covalent bond (O-H > N-H). Solvents in this class tend to be especially good at interacting with both types of charge. Such solvents are called “protic solvents”, because they can both donate and accept hydrogen bonds. In our course, protic solvents have an O-H bond or an N-H bond and tend to be better at solvating (thus, dissolving) polar and ionic solutes than solvents without this feature.



O

H

RO

R

HO

H

R

OR

HO

R

H

O

R

H

OR

H

OR

H

Protic solvents are good at solvating both cations and anions. Many partial negative solvent dipoles (lone pairs on oxygen or nitrogen atoms) can take the place of an anion, while many partial positive solvent dipoles (polarized hydrogen atoms) can take the place of the cation.

Protic solvents usually have an O-H bond.

- -

-

-

++

++

Solvents that are classified as polar ( > 15, see below), but do not have a polarized hydrogen are called polar

aprotic solvents. They tend to be very good at solvating positive charge, but not so good at solvating negative charge. This turns out to be very helpful when anions need to react in chemical reactions, and the use of such solvents is very common in organic chemistry when that is the goal.

Aprotic solvents do NOT have an O-H (or N-H) bond.

Polar aprotic solvents are good at solvating cations and not so good at solvating anions.

C

N

CH3

C

N

CH3

C

N

H3C

C

N

H3C

C

N

CH3

Nitrogen and oxygen lone pairs can interact strongly with cationic charge.

C

N

CH3

Polar aprotic solvents interact poorly with anionic charge, so they tend to be available to react.

acetonitrile(ethanenitrile)

- -

--

+ +

Solvents with e < 15 and no polarized hydrogen atom are called nonpolar solvents and generally do not mix well with polar and ionic substances. In biochemistry such substances are referred to as hydrophobic (don’t mix well with water) or lipophilic (mix well with fatty substances). This is a good feature because they can keep different aqueous solutions (blood vs. cytosol) apart and prevent mixing (cell membranes).

Three different kinds of solvents that chemists tend to talk about are: nonpolar solvents, polar aprotic solvents and polar protic solvents. Examples are shown below.



water H2O 78 proticmethanol (MeOH) CH3OH 33 proticethanol (EtOH) CH3CH2OH 24 protic2-propanol (i-PrOH) (CH3)2CHOH 20 proticmethanoic acid (formic acid) HCO2H 50 protic*ethanoic acid (acetic acid, HOAc) CH3CO2H 6 protic

2-propanone (acetone) CH3COCH3 21 polar aproticdimethyl sulfoxide (DMSO) CH3SOCH3 46 polar aproticdimethyl formamide (DMF) HCON(CH3) 37 polar aproticacetonitrile (AN) CH3CN 36 polar aproticnitromethane CH3NO2 36 polar aprotichexamethylphosphoramide (HMPA) O=P[N(CH3)2]3 30 polar aprotic

ethyl ethanoate (EtOAc) CH3CO2CH2CH3 6 nonpolarethyl ether (Et2O) CH3CH2OCH2CH3 4 nonpolartetrahydrofuran (THF, cyclic ether) (CH2)4O 8 nonpolardichloromethane (methylene chloride) CH2Cl2 9 nonpolartrichloromethane (chloroform) CHCl3 5 nonpolartetrachloromethane (carbon tetrachloride) CCl4 2 nonpolarhexane CH3(CH2)4CH3 2 nonpolar

dielectricCommon Solvents Formula constant () protic/aprotic

*Ethanoic acid violates our arbitray solvent polarity rules but is a strong hydrogen bonding solvent, so is included in the protic solvent group.

HO

H H3CO

H

HC

O

OH

H3CC

CH3

O

H3CS

CH3

O

HC

N

O

(H3C)2NP

N(CH3)2

O

CH3

CH3

H3CC

N

N(CH3)2

NH3C

O

O

H3CC

O

O

H2C

CH3 H3C

H2C

O

H2C

CH3 H2C CH2

CH2

O

H2C C

H

HCl

Cl

water methanol

formic acid acetone

DMSO acetonitrile

nitromethan

HMPA ethyl acetate ether methylene chloride

DMF

THF

Problem 11 – Point out the polar hydrogen in methanol. What is it about dimethyl sulfoxide (DMSO) that makes it polar? Draw a simplistic picture showing how methanol interacts with a cation and an anion. Also use DMSO (below) and draw a simplistic picture showing the interaction with cations and anions. Explain the difference from the methanol picture.

Two resonance structures of DMSO a polar, aprotic solvent

S

O

H3C CH3S

O

C C

Obeys the octet rule, but has formal charge.

Violates the octet rule, but does not have formal charge. Sulfur has 3d orbitals, which provides a possible explanation for drawing a double bond.

= cation

= anion

methanol

OH3C H

H

HH H

H

H



Simplistic Sketch of Mixing Solutes and Solvents

ionic substance

Na Cl

A very large energy input is needed to separate ionic charges from one another (lattice energy).

nonpolarsubstance

octadecane(waxy substance)

OH

H

OH

H

polar protic solvent = water

O

H

H

O

H

H

OH

H

O H

HThe necessary energy to dissolve NaCl comes from many, many ion-dipole interactions (hydration = water, solvation = solvent). All of these smaller interactions between ions and solvent balance the very large ion-ion interactions of the lattice structure and hydrogen bonding of water. If the solvation energy is larger than the lattice energy, the salt will dissolve (NaCl), but if the lattice energy is larger than the solvation energy, the salt will not dissolve (AgCl).

There is no way the nonpolar solvent can compensate for the lattice energy of the salt. The salt stays in its crystalline form and sinks or floats based on its density relative to the solvent. Whatever is more dense sinks and less dense floats (think of a stick and a rock in water).

cost = hydrogen bondingcost = lattice energygain = ion-dipole

cost = dispersion forcescost = lattice energygain = dispersion-ion (too weak)

OH

H

OH

H

polar protic solvent = water

nonpolar solvent = octane(like gasoline)

cost = hydrogen bondingcost = dispersion forcesgain = dispersion-dipole

cost = dispersion forcescost = lattice energy (weak dispersion)gain = dispersion-dispersion (weak)

OH

H

OH

H

The weak dispersion forces do not compensate for the broken hydrogen bonds. Also, there is a large entropy cost for water to structure itself around the nonpolar solute molecules. Water stays separate from the nonpolar solute and is on the top or bottom based on the densities of the substances (greater density is on the bottom).

There are no strong forces to overcome in either the solvent or the solute. There are no strong interactions gained in the solution(only dispersion forces). There is a favorable increase in entropy (DS) from the disorder created in the mixing of the solution, so the substances mix.

CH3(CH2)16CH3

Only weak dispersion forces are present in pentane.

CH3(CH2)6CH3

CH3(CH2)6CH3

nonpolar solvent = octane(like gasoline)

Problem 12 – Carbohydrates are very water soluble and fats do not mix well with water. Below, glucose is shown below as a typical hydrophilic carbohydrate, and a triglyceride is used as a typical hydrophobic fat. Point out why each is classified in the manner indicated.

O

HO

HO

HO OH

OHO

OH

HO

HOOH

HO

glucose (carbohydrate)

O

O

O

O

O

O

typical saturated triglyceride (fat)

=



Problem 13 – Bile salts are released from your gall bladder when hydrophobic fats are eaten to allow your body to solubilize the fats, so that they can be absorbed and transported in the aqueous blood. The major bile salt glycolate, shown below, is synthesized from cholesterol. Explain the features of glycolate that makes it a good compromise structure that can mix with both the fat and aqueous blood. Use the ‘rough’ 3D drawings below to help your reasoning, or better yet, build models to see the structures for yourself (though it’s a lot of work).

HO

H

H H

NH

HO

H

H Hsynthesized in many,

many steps in the body

HOH

OHO

O

Ocholesterol

glycolate(bile salt)1. source for steroid syntheses in the body

2. important constituent of cell membranes2. transported in blood to delivery sites via VLDL LDL HDL

VLDL = very low density lipoprotein, has high cholesterol concentrationLDL = low density lipoprotein, has medium cholesterol concentrationHDL = high density lipoprotein, has low cholestero contcentration

All polar groups are on the same face. Which side faces water and which side faces fat molecules? (See structures below.)

HO OH

OH

OH

CO2

representation of cholesterol as a long flat shape

representation of bile acid (glycolate) as a long bent shape havingtwo different faces, one polar and one nonpolar



Simplistic picture of atoms’ Zeffective and other periodic properties

+3

+11 -2

-2

+1

hydrogenZtotal = +1shield = 0Zeffective = +1IP1 = +314 kcal/moleat. radius = 53 pm# bond = 1lone pairs = 0 = 2.2

+2

heliumZtotal = +2shield = 0Zeffective = +2IP1 = +313at. radius = 31 pm# bond = 0lone pairs = 0 = ?

lithiumZtotal = +3shield = -2Zeffective = +1IP1 = +124at. radius = 167# bond = 1 or ioniclone pairs = 0 = 1.1

+4 -2

beryliumZtotal = +4shield = -2Zeffective = +2IP1 = +215at. radius = 112 pm# bond = 2 or ioniclone pairs = 0 = 1.5

+5 -2

boronZtotal = +5shield = -2Zeffective = +3IP1 = +192at. radius = 87 pm# bond = 3 lone pairs = 0 = 2.0

+6 -2

carbonZtotal = +6shield = -2Zeffective = +4IP1 = +261at. radius = 67 pm# bond = 4lone pairs = 0 = 2.5

+7 -2

nitrogenZtotal = +7shield = -2Zeffective = +5IP1 = +335at. radius = 56 pm# bond = 3lone pairs = 1 = 3.0

+8 -2

oxygenZtotal = +8shield = -2Zeffective = +6IP1 = +315at. radius = 48 pm# bond = 2lone pairs = 2 = 3.5

+9 -2

fluorineZtotal = +9shield = -2Zeffective = +7IP1 = +402at. radius = 42# bond = 1lone pairs = 3 = 4.0

+10 -2

neonZtotal = +10shield = -2Zeffective = +8IP1 = +499at. radius = 38# bond = 0lone pairs = 4 = ?

-8 +12 -2 -8 +13 -2 -8 +14 -2 -8 +15 -2 -8 +16 -2 -8 +17 -2 -8 +18 -2 -8

sodiumZtotal = +11shield = -10Zeffective = +1IP1 = +118at. radius = 190 pm# bond = 1 or ioniclone pairs = 0 = 0.9

magnesiumZtotal = +12shield = -10Zeffective = +2IP1 = +177at. radius = 145# bond = 2 or ioniclone pairs = 0 = 1.2

aluminumZtotal = +13shield = -10Zeffective = +3IP1 = +138at. radius = 118 pm# bond = 3 lone pairs = 0 = 1.5

silicaonZtotal = +14shield = -10Zeffective = +4IP1 = +189at. radius = 111 pm# bond = 4lone pairs = 0 = 1.9

phosphorousZtotal = +15shield = -10Zeffective = +5IP1 = +242at. radius = 98 pm# bond = 3lone pairs = 1 = 2.2

sulfurZtotal = +16shield = -10Zeffective = +6IP1 = +239at. radius = 88# bond = 2lone pairs = 2 = 2.6

chlorineZtotal = +17shield = -10Zeffective = +7IP1 = +300at. radius = 79# bond = 1lone pairs = 3 = 3.5

argonZtotal = +18shield = -10Zeffective = +8IP1 = +363at. radius = 71# bond = 0lone pairs = 4 = ?

+19 -2 -8 -8 +20 -2 -8 -8 +31 -2 -8 -18 +32 -2 -8 -18 +33 -2 -8 -18 +34 -2 -8 -18 +35 -2 -8 -18 +36 -2 -8 -18

potassiumZtotal = +19shield = -18Zeffective = +1IP1 = +99at. radius = 243 pm# bond = 1 or ioniclone pairs = 0 = 0.8

calciumZtotal = +20shield = -18Zeffective = +2IP1 = +141at. radius = 194# bond = 2 or ioniclone pairs = 0 = 1.0

galliumZtotal = +31shield = -28Zeffective = +3IP1 = +138at. radius = 136 pm# bond = 3 lone pairs = 0 = 1.6

germaniumZtotal = +32shield = -28Zeffective = +4IP1 = +182at. radius = 125 pm# bond = 4lone pairs = 0 = 1.9

arsinicZtotal = +33shield = -28Zeffective = +5IP1 = +226at. radius = 114 pm# bond = 3lone pairs = 1 = 2.2

seleniumZtotal = +34shield = -28Zeffective = +6IP1 = +225at. radius = 103# bond = 2lone pairs = 2 = 2.5

bromineZtotal = +35shield = -28Zeffective = +7IP1 = +273at. radius = 94# bond = 1lone pairs = 3 = 3.0

kryptonZtotal = +36shield = -28Zeffective = +8IP1 = +363at. radius = 88# bond = 0lone pairs = 4 = 3.0

Hydrogen, lithium and sodium all have a Zeff of +1, but the pull for the electrons is weaker in K than Na than Li than H because the outer most electron is farther from the Zeff in K than Na than Li than H. The result from this trend is that elements higher in a column attract electrons more strongly than elements lower in a column.

The valenece electrons in Li, C, N, O and F are all in the same n=2 shell, and all have the same number of core electrons (2). The nuclear charge keeps getting larger when moving across the row and the pull on the valence electrons keeps getting stronger. The result from this trend is that elements further to the fight in a row attract electron more strongly than elements on the left.

The combination of both trends is that elements in the upper right corner of the periodic table attract electrons strongest. The electron attracting power of atoms in bonds is referred to as electronegativity, where a higher number indicates a higher electronegativity. See values in the table below.

Core electrons are full shells inside the outermost valence electrons. They are considered to shield the outermost electrons from Ztotal, leaving a net Zeff to attract the valence electrons.The positive number in the middle of each group of circles is Ztotal. Each negative number inside a circle represents a full inner shell (core electrons) and shields Ztotal. The dots in the outer circle represent the valence electrons of each neutral atom feeling the pull of Zeff.

10 "3d" electrons between these two elements.



Possible Answers to the problems Problem 1 – What kinds of bonds are between the nonhydrogen atoms below?

H C N H C C H C O C C

H

H

C N

H

H

C O

H

H

H

H

H

C C

H

H

H

H

H

H

a b c d e f g

1 x sigma2 x pi

1 x sigma2 x pi

1 x sigma2 x pi

1 x sigma1 x pi

1 x sigma1 x pi

1 x sigma1 x pi

1 x sigma0 x pi

Problem 2 - Explain the atomic trends in ionization potential in a row (Na vs Si vs Cl) and in a column (F vs Cl vs Br).

Na (+118), Si (+189), Cl (+300): The ionization potential (IP1) increases across a row because Zeff is larger.

F (+402), Cl (+300), Br (+273): The ionization potential decreased down a column because the electrons are farther away from the “same” Zeff (in this case, Zeff = +7 for the halogens)

Problem 3 - Explain the atomic trends in atomic radii in a row (Li vs C vs F) and in a column (C vs Si vs Ge).

Li (167 pm), C (67 pm), F (42 pm): All of the electron clouds in these examples are using the n=2 shell (similar sizes), but the Zeff gets larger as one moves to the right, which shrinks the electron cloud (gets smaller)

C (67 pm), Si (111 pm), Ge (125 pm): Each atom in a column adds an extra shell of electrons, which gives lower atoms a larger radius.

H = 53 He = 31Li = 167 Be = 112 B = 87 C = 67 N = 56 O = 48 F = 42 Ne = 38Na = 190 Mg = 145 Al = 118 Si = 111 P = 98 S = 88 Cl = 79 Ar = 71K = 243 Ca = 194 Ga = 136 Ge = 125 As = 114 Se = 103 Br = 94 Kr = 88Rb = 265 Sr = 219 In = 156 Sn = 145 Sb = 133 Te = 123 I = 115 Xe = 108

Table 2: Neutral atomic radii in picometers (pm) = 10-12 m [100 pm = 1 angstrom]


Li = 90 Be = 59 B = 41 C = N = 132 O = 126 F = 119Na = 116 Mg = 86 Al = 68 Si = P = S = 170 Cl = 167 K = 152 Ca = 114 Ga = 76 Ge = As = Se = 184 Br = 182 Rb = 166 Sr = 132 In = 94 Sn = Sb = Te = 207 I = 206

Table 3: Cations (on the left) and anions (on the right) radii in picometers (pm) = 10-12 m [100 pm = 1 angstrom]


+1+1

+1+1

+2

+2

+2

+2

+3

+3+3

+3

-3 -2

-2

-2-2

-1-1

-1-1

as cations as anions

Problem 4 - In the tables above:

a. Explain the cation distances compared to the atomic distances (Li, Be, B). (Tables 2 and 3)

Li+ (90 pm), Be+2 (59 pm), B+3 (41 pm): All of the cations hold on to their remaining electron tighter, making smaller radii, and the most electron deficient hold the tightest (smallest radius)

Li (167 pm), Be (112 pm), B (87 pm): All of the neutral atoms have a larger radius of their electron clouds than the cations and B with the largest Zeff has the smallest radius of the three in the n=2 row.



Explain the anion distances compared to the atomic distances (N, O, F). (Tables 2 and 3) N 132 pm), O (126 pm), F (119 pm): All of the anions have greater electron-electron repulsion, which expands the electron clouds over the neutral atoms, N with the greatest negative charge expands the most

b. Explain the cation distances in a row (Li, Be, B). (Table 3) Li+ (90 pm), Be+2 (59 pm), B+3 (41 pm): All of the cations hold on to their remaining electron tighter, making smaller radii, and the most electron deficient hold the tightest (smallest radius)

c. Explain the anion distances in a row (N, O, F). (Table 3) N 132 pm), O (126 pm), F (119 pm): All of the anions have greater electron-electron repulsion, which expands the electron clouds, N with the greatest negative charge expands the electron cloud the most Problem 5 – What is the effect on the dipole moment of switching a hydrogen for a methyl in each pair below? Propose a possible explanation.

C

O

H2N H

= 3.71 Dbp = +210oCmp = +2oC

CHH

C

O

F CH3

= 2.96 Dbp = +21oCmp = -84oC

C

O

HO H

= 1.41 Dbp = +118oCmp = +17oC

C

O

H2N CH3

= 3.76 Dbp = +222oCmp = +80oC

= 1.74 Dbp = +118oCmp = +17oC

C

O

F H

= 2.02 Dbp = -29oCmp = -142oC

C

O

HO CH3

O

CCH3H

O

= 2.68 Dbp = +20oCmp = -123oC

= 2.3 Dbp = -20oCmp = -92oC

In every example, the methyl for the hydrogen increases the dipole moment. The methyl must give up more electron density to the oxygen relative to the hydrogen. This methyl (R group in general) is referred to as inductively electron donating. Hyperconjugation is an alternative explanation using molecular orbitals (HOMO and LUMO). The effect is smallest for the amide because the nitrogen is such a good resonance donator that the inductive is not as significant.

Each additional "R" group (methyl) makes the carbonyl bond (C=O) more polar, indicated that the "R" group is inductively donating relative to the hydrogen.

C

O

H H = 2.3 Dbp = -20oCmp = -92oCH2O sol. = 400g/L

polar

C

O

H H

resonanceC

O

H3C H

= 2.68 Dbp = +20oCmp = -123oCH2O sol. = very sol.

C

O

H3C CH3 = 2.91 Dbp = +56oCmp = -94oCH2O sol. = very sol.

C

O

H3C H

resonance

C

O

H3C CH3

resonance

polar polar

CC

C

= 0.0 Dbp = -34oCmp = -134oC

nonpolar

H3C

H2C

CH3

= 0.08 Dbp = -42oCmp = -188oC

nonpolar

HC

CH2H3C

= 0.37 Dbp = -47oCmp = -185oC

nonpolar

CH

CH3C = 0.75 D

bp = -23oCmp = -103oC

nonpolar

H

H

H

H

The sp2 orbital of an alkene is more electronegative (33% s) than an sp3 orbital (25% s). The methyl substituent gives up some of its electron density because of this (inductively donating). The sp orbital (50% s) of an alkyne is even more electronegative because of the tighter hold by s electrons over p electrons so takes even more electron density from the inductively donating methyl group (has a higher dipole moment).



H CH3 NH2 OH

N

H

= 0.0 Dbp = 81oCmp = 5C

= 0.31 Dbp = 111oCmp = -95oC

= 1.53 Dbp = 184oCmp = -6oC

= 1.55 Dbp = +182oCmp = +40oC

= 2.37 Dbp = +115oCmp = -42oC

Inductive donation by the methyl gives toluene a small dipole moment.

Both aniline and phenol inductively withdraw electron density from the aromatic ring, BUT both nitrogen and oxygen are resonance donors (N > O) and add electron density to the ring. Usually resonancedominates over inductive effects, but not always.

The nitrogen atom in pyridine inductively withdraws from the carbon atomsmaking a larger dipolemoment.

F Cl

= 1.66 Dbp = +84oCmp = -44oC

= 1.54 Dbp = +131oCmp = -45oC

Br

= 1.73 Dbp = +156oCmp = -31oC

I

= 1.91 Dbp = +188oCmp = -31oC

The dipole moment depends on both charge separation (larger for F > Cl > Br > I) and bond length (larger for I > Br > Cl > F). The dipole moments are very similar because of opposing effects. The differences in boiling points is due to the greater polarizability of the larger atoms (I > Br > Cl > F).

Problem 6 – Which base pair binds more tightly in DNA, GC or AT, or are they about the same?

N

N

N

N

N H

H

O

HDNA

N

N

NH

H

O DNA

guanine cytosine

three hydrogen bonds in G-C base pairs bind tighter than two hydrogen bonds in A-T pairs

N

N

N

N

DNAN

N

O

O DNAadenine thymine

two hydrogen bonds in A-T base pair

H

N H

H

Problem 7 – Answer given with table. Problem 8 – Explain the difference in boiling points of the following molecules.

H3C

H2C

CH3 H3CO

CH3

= 0.08 Dbp = -42oCmp = -188oCH2O sol. = 40mg/L

= 1.30 Dbp = -22oCmp = -141oCH2O sol. = 71g/L

OHH3C

OH



Structure a, b and c are similar in size. Structure a only has dispersion forces, while structure b has dispersion forces and polar bonds, while structure c has dispersion forces, polar bonds and hydrogen bonds. Intermolecular forces of interaction increase in order of dispersion forces < polar bonds < hydrogen bonds, so c has the highest bp, followed by b and lowest is a. Compound d is like c except it has an extra methyl, thus greater dispersion forces, thus a higher boiling point.

a b c d



Problem 9 – Provide an explanation for the different boiling points in each series.

F Cl Br I

bp = +85 oC bp = +130 oC bp = +155 oC bp = +188 oC

a.

Similar molecules. The highest boiling points go with the largest dispersion forces (greatest polarizability, I > Br > Cl > F)

bp = -0.5 oCO

bp = +5 oC

OH

bp = +78 oC

b.

Structure 3 has H bonding and polarity (strongest attractions), structure 2 has polar bonds and structure 1 only has dispersion forces (weakest attractions)

bp = -47 oC

O

bp = +20 oC

HO O

bp = +101 oC

H2N O

bp = +210 oC

CO O

bp = -78 oC(sublimes)

c.

Structure 1 only has weak dispersion forces, structure 2 had polar bonds (stronger), structure 3 has H bonds too (stronger), structure 4 has the strongest polarity and H bonds with both N-H bonds (highest). Finally CO2 has polar bonds, but its symmetry cancels out its polarity. The oxygen atoms hold so tightly to their electrons that they are less polarizable than structure 1 and sublimes at a lower temperature.

bp = -23 oCbp = -42 oC

N

bp = +82 oC

d.

Structure 1 (propane) only has weak dispersion forces, and has a bent shape that prevents closer contact with neighbor molecules.Structure 2 (propyne) only has weak dispersion forces, but has a linear shape that allows closer contact with neighbor molecules, so stronger attractions. Structure 3 has the shape of propyne, and relatively strong polarity which makes stronger attractions and a higher bp.

e. H2

CH3CHCH3C

H2C CH3

C

CH3

H3C CH3

bp = +30 oC

H2C

CH3

H2C CH3

bp = +36 oCCH3

bp = +10 oC

CH3H3C

bp = -89 oC

More carbons have greater contact surface area so stronger dispersion forces (C2 < C5). Straight chains can approach closer than when branches are present (branches tend to push neighbor molecules away = weaker dispersion forces). More branches weaken those attractions even more (D < C < A).

e. CH4 CH3Cl CCl4

bp = -164 oC bp = -24 oC bp = +40 oC bp = +61 oC bp = +77 oC

CH2Cl2 CHCl3

Chlorine is a larger atom (than C) and has polarizable lone pairs. More chlorines = more dispersion forces = higher bp. B, C and D have polarity too. A and E are not polar, but E has four chlorine atoms and larger dispersion forces dominate here (highest bp).

H3C

H2C

CH3 H3C

H2C

OH

H3C

H2C

NH

Hbp = -42 oC bp = +17 oC bp = +78 oC

f.

Bp's follow H bonding possibilities. Structure 3 has the strongest H bonds because O-H is more polar. Structure 2 has some H bonding possibilitiies so a higher bp than structure 1 which only has dispersion forces. All 3 molecues are similar in size, so comparisons are fair.



Problem 10 – Match the given boiling points with the structures below and give a short reason for your answers. (-7oC, +31 oC, +80 oC, +141oC, 1420oC)

O O

OH2-methylpropene2-methyl-1-butene2-butanone propanoic acid

MW = 56 g/molMW = 70 g/molMW = 72 g/mol MW = 74 g/mol

KCl

MW = 74.5 g/mol

potassium chloride

1420oC+141oC+80 oC +31oC

-7oC

Structure 4 has ionic bonds, by far the strongest forces to overcome and has an extremely high bp. Structure 3 has strong H bonding possiblities and the second high bp. Structure 1 has polar C=O bond so higher than structures 2 and 5, which only have weak dispersion forces. Structure 2 is larger and has a larger surface area and more dispersion forces than structure 5, so has a higher bp.

Problem 11 – Point out the polar hydrogen in methanol. What is it about dimethyl sulfoxide (DMSO) that makes it polar? Draw a simplistic picture showing how methanol interacts with a cation and an anion. Also use DMSO (below) and draw a simplistic picture showing the interaction with cations and anions. Explain the difference from the methanol picture.

S

O

H3C CH3

= cations are smaller because they have lost electrons and hold onto the remaining electrons tighter.

= anions are larger because they have extra electron density that is repulsive and expands the electron clouds

OH3C H

OH3C H

O

H3C

H

O

CH3

H

O

CH3

H

OH3C H

OCH3H Methanol is good at solvating

positive and negative charge.

S

O

H3C CH3

SO

CH3

CH3

S O

H3C

H3C DMSO is very good at solvating positive charge but it is poor at solvaing n egative charge.

S

O

H3C CH3

S

O

H3C CH3

polar H

strong bond dipole



Problem 12– Carbohydrates are very water soluble and fats do not mix well with water. Below, glucose is shown below as a typical hydrophilic carbohydrate, and a triglyceride is used as a typical hydrophobic fat. Point out why each is classified in the manner indicated.

O

HO

HO

HO OH

OHO

OH

HO

HOOH

HO

glucose (carbohydrate)

O

O

O

O

O

O

typical saturated triglyceride (fat)

=

Problem 13 – Bile salts are released from your gall bladder when hydrophobic fats are eaten to allow your body to solubilize the fats, so that they can be absorbed and transported in the aqueous blood. The major bile salt glycolate, shown below, is synthesized from cholesterol. Explain the features of glycolate that makes it a good compromise structure that can mix with both the fat and aqueous blood. Use the ‘rough’ 3D drawings below to help your reasoning, or better yet, build models to see the structures for yourself (though it’s a lot of work).

HO

H

H H

NH

HO

H

H Hsynthesized in many,

many steps in the body

HOH

OHO

O

Ocholesterol

glycolate(bile salt)

OH

OH

OH

CO2

Glycolate has a nonpolar, hydrophobic face that can cover the inside of a fat ball and a hydorphilic face that can point outward toward the aqueous blood, which allows fats to be transported throughout the body to reach fat storage cells and other essential locations.

OH

OH

OH

CO2

HO

HO

HO

O2C

HO

HO

HO

O2C

blood

nonpolarfats

inside

H2O

H2O

blood

H2O

blood

H2O

blood

H2Oblood

H2O

blood

H2O

blood

H2O

blood

H2Oblood

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