Quantum Numbers/Electron Configurations

Quantum Numbers/Electron Configurations The Schrodinger equation, which was shown in class, is a wave function and the solutions to this function provide information about the probability of finding electrons in a particular space. Each electron in an atom can be uniquely described using four quantum numbers: Principle quantum number (n): This number describes the most probable distance from the nucleus and thus the size of the orbital. Because electrons farther from the nucleus also have higher energy, this number describes the overall energy of an electron. Each value of n (integers 1,2,…) refers to a shell of electrons. Angular momentum quantum number (l): This number describes the shape of an orbital and the value ranges from 0,1,…(n-1). These are referred to as subshells.

l value Subshell Number of

Orbitals

0 s 1 1 p 3 2 d 5 3 f 7

Magnetic quantum number (ml): This number describes the orientation of an orbital in space. It ranges from –l to l. For s orbitals, a sphere only has one orientation, but for p orbitals, there are three orbitals and they have three orientations px, py, pz which are oriented along the 3 perpendicular axes. Spin quantum number (ms): It represents the spin of an electron and can only take one of two values: up or down. Valence Electrons/Octets/Bonding You’ve learned how to assign electrons to the different subshells to obtain the atomic configuration for a particular atom. We are particularly interested in the electrons contained in the outermost shell. These outermost electrons are called valence electrons and they are the only electrons that participate in bonding.

Examples:

Element Configuration Valence Electrons Lithium 1s22s1 1 Magnesium 1s22s22p63s2 2 Carbon 1s22s22p2 4 Silicon 1s22s22p63s23p2 4 Nitrogen 1s22s22p3 5 Oxygen 1s22s22p4 6 Sulfur 1s22s22p63s23p4 6 Fluorine 1s22s22p5 7

You can notice from the table below that for main group elements (groups 1,2, 13-18) the number of valence electrons corresponds to the group (column number or (column number – 10) for columns 13-18) on the periodic table. Transition metals are harder to predict so often you must actually write out the electron configuration. Since valence electrons are the only electrons involved in bonding, they are largely responsible for the different properties of atoms and thus you can use the columns of the periodic table to predict properties.

Elements tend to form bonds in order to form a closed shell, meaning all orbitals in that particular shell have the maximum two electrons. So for main group elements, this corresponds to achieving s2p6 in the outermost shell. This is often called the octet rule because the bonded atoms prefer to have eight valence electrons either through sharing or giving/taking.

Group 1 and Group 2 metals have only one or two valence electrons respectively, thus they are most likely to lose electrons and form cations.

Na Na+ e-

Mg Mg2+ 2e-

Group 7 elements have seven valence electrons and thus only need one to fill their octet. They can gain this electron either through removing it from another atom to form an ionic bond or through sharing with another atom to form covalent bonds.

F e- F-

F H H F

Lewis Structures/Formal Charge The number of valence electrons tells you the number of bonds an atom prefers to form. For example, carbon has 4 valence electrons and will form 4 bonds while nitrogen has 5 valence electrons and prefers to form 3 bonds. This does not mean that nitrogen cannot form more than three bonds, but if it does, it will be charged. When nitrogen has four bonds, it has 8 valence electrons which is good, but it is sharing all eight in bonds, so it really only has four to itself. Since it normally has 5, it is one electron short and thus has a positive charge. This can be calculated through finding the formal charge which is basically finding the difference between the expected number of valence electrons in an isolated atom and the number of electrons it actually has when it is within a compound either through lone pairs or covalent bonds. Each covalent bond contains two shared electrons, but for formal charge calculations, an atom is considered to have “possession” of one per bond.

Examples

+ - H2O H O HO 3

H o H H o H H o H

+ - NH3 NH NH 4 2 H

H N H N H N H H H H H

Formal charge = #valence electrons – (number of lone pair electrons + #bonds

Electronegativity

Above, we have been considering bonding as either purely ionic (electrons lost or gained) or purely covalent (electrons shared), but in reality bonding falls on a spectrum with many bonds having unequal sharing of electrons. This is due to differences in electronegativity which can be thought of as a measure of the atom’s ability to pull electrons towards it. When one atom in a bond is much more electronegative than the other, it can pull the electrons closer to its nucleus, thus concentrating a bit more of the negative charge on itself. These bonds are considered polar and they result in one atom having a partial negative charge and the other (the less electronegative) having a partial positive charge (indicated by a δ- or δ+. Bonding Spectrum

Ionic Polar Covalent Nonpolar Covalent

+ δ- Na+ F- δH F F F Electronegativity 0.9 4.0 2.1 4.0 4.0 4.0 Values

Molecular Orbital Theory

When two atoms come together to form a bond, the orbitals containing the electrons involved in bonding must begin to overlap to form an orbital that is shared by both atoms. This combined orbital is called a molecular orbital because it belongs to the entire molecule rather than a specific atom and just like an atomic orbital, it describes the volume of space where an electron is likely to be found. Bonding must be energetically favorable, otherwise it would not occur. Thus bonding molecular orbitals are lower in energy than the individual atomic orbitals that form them. Intuitively this makes sense as when two atoms are far apart there is no interaction between their nuclei, but as they approach, the positively charged nuclei attract the negatively charged electrons of the other atom. At some distance though, the repulsion of the nuclei of the two atoms will begin to outweigh the attraction to the electrons and moving the nuclei any closer together is energetically unfavorable. This is what determines the atomic distances in bond length. The diagram below shows this effect.

Remember that electrons behave like a standing wave (see image of sinusoidal wave), so atomic orbitals have phases and nodes. For each shell, there is a single s orbital which is spherical in shape. Shells with number two and above have three identical p orbitals oriented at 90 angles to each other. With a standing wave, you have both upward and downward displacement which are the two phases. The the region with no displacement is called the node. Atomic orbitals also have phases and nodes and it is easiest to visualize this with p orbitals.

In the image above, the colors represent the two phases and there is a nodal plane between the two lobes where there is zero probability of finding an electron.

When two waves combine, they can combine in a constructive, additive manner or a destructive manner and atomic orbitals are the same way. When two atomic orbitals overlap to form a bond, you get two separate molecular orbitals because the atomic orbitals combine either in an additive or destructive manner. If they combine such that the in-phase ends overlap (additive), this generates a bonding molecular orbital. If they combine in the opposite way (destructive), you get an antibonding molecular orbital. The following diagram demonstrates this with two p orbitals.

Orbitals can overlap to form two different types of bonds: sigma and pi

Sigma Bonds

Sigma bonds are cylindrically symmetrical which just means that electrons are symmetrically distributed around the bond, or around an imaginary line connecting the centers of the two atoms. The overlap of two s orbitals always results in a sigma bond while two p orbitals can give sigma bonds when they overlap end-to-end (as shown in the figure above). As was mentioned above, you always generate two molecular orbitals when you combine two atomic orbitals which in this case are denoted as σ and σ*. • The σ orbital is the bonding molecular orbital and is lower in energy than either of the p or s orbitals that combined to form it. • The σ* is the antibonding orbital and it is higher in energy than either of the atomic orbitals. The σ* orbital has a node between the two nuclei meaning the electrons are more likely to be found somewhere other than between the two nuclei. This means that the two nuclei can repel each other because the positive nuclear charge isn’t mediated by the negative charge of the electrons. This repulsion explains why this is a higher energy orbital. Much like with atomic orbitals, you always fill the lowest energy molecular orbital first which in this case is always σ. Any electrons in the σ* orbital actually detract from bonding because they are in a higher energy state than they would be if they were in the original s or p atomic orbital. The diagram below just shows the molecular orbital diagram for bonding two H atoms. The two s orbitals from H overlap to give a σ and σ* where the σ is lower in energy and σ* is higher. The two electrons from the two H atoms each fill the σ orbital leaving the σ* empty.

Pi Bonds For p orbitals forming σ bonds, they overlap end-to-end, but the two orbitals can also overlap side-to-side forming what is known as a π bond. This type of overlap results in electron density above and below the axis of the bond (shown below). Just like with σ bonds, you form two molecular orbitals when making a π bond, π and π*. The amount of orbital overlap in π bonds is less than that with σ bonds, thus π bonds are weaker. This image is representing the different phases with +/- signs instead of the colors that had been used in previous diagrams.

The image below shows the relative energies of the different σ and π bonds from the 2 shell. Since the π bonds are weaker than σ, they are slightly higher in energy. Also, σ bonds resulting from s orbitals are lower in energy than σ from p orbitals because the s orbitals were lower in energy to begin with. Antibonding orbitals can contain electrons as long as the overall energy of the filled molecular orbitals is less than that of the atomic orbitals. If you have so many electrons to put into the molecular orbitals that this is not the case, then the compound you are considering is unlikely to form.

Hybridization The idea of hybridized orbitals helps to explain known properties of molecules. For example, methane (CH4) has four hydrogen atoms bonded to a central carbon atom and all of the bonds angles are the same (109.5°). When you think about carbon, it only has two unpaired electrons in its outer shell (both in p orbitals), so you might think that it would only form two bonds. To explain why carbon can form four bonds, you could promote one of the electrons from the s orbital to the empty p orbital which results in four unpaired electrons. This would allow for four covalent bonds to form, but there is still one problem. p orbitals are higher in energy than s orbitals so the four bonds formed would not be identical, but we know that that they are in fact identical in methane. Orbital hybridization helps to address this inconsistency. If you combine the single s orbital with the three p orbitals, you can create four degenerate (equal energy) orbitals. These are H known as sp3 hybridized orbitals. These four orbitals adopt a spatial H arrangement that minimizes repulsion, thus keeping the four orbitals as far H H apart as possible. This explains the tetrahedral geometry of methane and the 109.5° bond angles.

p p p p p p sp3 sp3 sp3 sp3 s s

In methane, the sp3 orbitals of carbon are overlapping with the s orbitals of H to form a σ bond.

In ethylene (C2H4), each carbon molecule is forming four bonds, but only to three different atoms. In this case you want three identical hybrid orbitals to form three sigma bonds (C-C and two C-H) which occurs through the hybridization of the s orbital and two H H of the three p orbitals. This results in 3 degenerate sp2 orbitals and one p C C orbital. The C-C bond is a result of sp2-sp2 orbital overlap while the C-H bonds H H are from sp2-s orbital overlap since the H only has an s orbital. The remaining p orbital can form a π bond which gives the second bond between the two carbon atoms. The geometry of ethene can also be inferred from the orbitals. In this case the 3 sp2 orbitals want to be as far apart as possible which results in trigonal planar geometry with bond angles of 120°. Also, the π bond cannot rotate because of the electron density above and below the C-C bond thus that bond is rigid.

p p p p sp2 sp2 sp2 s

Molecules with triple bonds are much the same except that you form 2 sp orbitals which leaves 2 p orbitals free to form two separate π bonds.

Resonance Sometimes molecules have several structures that contribute equally to the overall structure of the molecule. Resonance structures provide a way of representing delocalized electrons, meaning the electrons are spread across more than one bond. It is important to realize that the individual resonance structures themselves do not exist in equilibrium. There is actually just one structure that is a hybrid of all of the individual resonance structures.

O O O C O O O O O O

S C N S C N

Isomers The term isomer refers to compounds that have the same molecular formula (meaning the same atoms) but they do not have identical structures. There are two types of isomers: constitutional isomers and stereoisomers Constitutional Isomers: differ in the way their atoms are connected

OH OH O C3H8O propan-2-ol propan-1-ol methoxyethane isopropanol propanol Stereoisomers: atoms are connected the same way but they differ in their arrangement in space. This is a result of the chirality of the molecule. A chiral object is one that has a nonsuperimposable mirror image. Chiral molecules have an asymmetric center which simply means there is a tetrahedral atom bonded to four different groups. Since a tetrahedral atom is usually carbon, one can identify chiral centers by finding carbon atoms bonded to four different groups. We call the two mirror image molecules enantiomers.

*Remember that the solid wedge represents a bond that is coming out of the plane of the paper and the hashed wedge represents a bond that is going into the paper. Normal lines indicate bonds in the plane of the paper.

Br Br

C H C HO HO H

Fisher Projections: Sometimes Fisher projections are used to show the 3D arrangement of molecules in 2D space. The asymmetric center is located at the intersection of two perpendicular lines. The horizontal line represents bonds that project out of the plane of the paper towards you and vertical line represents bonds extending back behind the plane of the paper. You can start by putting the four groups wherever you want and then to draw the enantiomer, you exchange two of the groups. While it doesn’t matter which two you interchange, typically it is the two on the horizontal line.

CH CH 3 3 H OH HO H

Br Br Enantiomers share all of their physical properties such as density, melting point, etc. The one property that they do not share is the way they interact with polarized light. Polarized light has light waves oscillating only in a single plane. If you pass polarized light through a solution of achiral molecules, the light that emerges with the same plane of polarization. Chiral compounds however rotate the polarized light in either a clockwise or counterclockwise direction and this rotation can be measured. If one enantiomer rotates light in the clockwise direction, the other enantiomer will rotate it in the counterclockwise direction to the same degree. A compound that rotates light clockwise is dextrorotatory (+) and one that rotates counterclockwise is levorotatory (-). The image below shows a diagram of how a chiral compound rotates light.

One result of this is that a mixture of a chiral compound that is 50% one enantiomer and 50% of the other enantiomer will not rotate the light. This type of mixture is called racemic.

Since chiral compounds have almost identical physical properties, the enantiomers cannot be distinguished by an achiral compound. Many of the fundamental biological molecules such as amino acids and sugars are chiral and many enzymes are chiral and thus can tell the difference between two enantiomers.

Multiple Asymmetric Centers If you have more than one asymmetric center in a molecule, there are going to be 2n stereoisomers of that molecule (where n is the number of asymmetric centers). If the molecules have opposite configurations at each of the stereocenters, then they will be mirror images and thus enantiomers. But what if the two molecules have the same configuration at one center and opposite configuration at the other? They will no longer be mirror images. They are called diastereomers. These relationships are easiest to see with Fisher projections as shown below.

CH3 CH3 CH3 CH3 Cl H OH HO H H OH HO H H Cl Cl H Cl H H Cl OH CH3 CH3 CH3 CH3

1 2 3 4

On the left is the compound with two asymmetric carbons and on the right are the Fisher projections. The pairs of compounds 1&2 and then 3&4 are enantiomers of each other because at each of the horizontal lines, they are completely opposite of each other. 1&3 and 1&4 are diastereomers of each other because they are the same at one stereocenter and opposite at the other. There are several other pairs of diastereomers.

Unlike enantiomers, diastereomers do not have the same physical and chemical properties and thus can react differently. This can be explained by the fact that the OH and Cl groups are closer together in space in 1&2 and farther apart in 3&4. This spatial difference can affect the types of interactions the molecules can have and thus their physical properties.

Intermolecular Forces These are forces (attraction/repulsion) that act between molecules. They are much weaker than intramolecular forces (bonds). There are several classes of intermolecular forces that vary in strength.

Ionic Forces: This is just the interaction between two ions – it results in an ionic bond.

Na+ Cl-

Dipole-Dipole: Bonds have polarity and often these polar bonds result in a molecule that has an overall dipole. The partial positive charge of one molecule can be attracted the partial negative charge of another molecule. δ- O

δ+ δ- + - + - O δ δ δ δ H Cl H Cl δ+

Hydrogen Bond: A hydrogen bond is a specific example of a dipole-dipole interaction. It involves a hydrogen bonded to a strongly electronegative atom (F, N, O). This hydrogen has a significant partial positive charge because of the large electronegativity difference and is the donor. A different electronegative atom (F, N, O) will be attracted to the H, creating an intermolecular force. This atom is the hydrogen bond acceptor. Hydrogen bonds are extremely relevant in biological systems because water, which can be both an acceptor and a donor, is ubiquitous. They also play a major role in protein structure.

donor O H H O H acceptor H

Induced Dipole (van der waals): Since electrons are constantly moving, even nonpolar bonds can have instantaneous polarity where there is a slight unbalance of the electron distribution towards one atom. This induced dipole can then trigger an instantaneous partial charge in a neighboring atom creating an attraction.

You can have any combination of the above as well. For example a molecule with a permanent dipole such as HCl could induce a dipole in a nonpolar molecule such as CCl4. In terms of relative strengths

Ionic > Ion-Dipole > Dipole-Dipole > Dipole-Induced Dipole > Induced-Induced

Intermolecular forces also have a significant impact on physical properties such as boiling points. The stronger the intermolecular interaction, the more energy you must put into the system to break the interaction.

Carbohydrates Referred to based on the number of carbon atoms – triose, pentose, etc. They are visualized in a variety of ways including Fisher projections which are helpful since monosaccharides have multiple sterocenters. Sugars can exist in either the linear form that is shown by the Fisher projection or a cyclic form which is shown by the Haworth projection as seen below. You don’t need to know the details of the reaction, but the alcohol (OH) on the C5 attacks the aldehyde (CHO) of C1 to form a ring with C6 extraneous to the ring. The red arrows shown in the figure represent electron movement (it is often called arrow pushing). In aqueous solutions, glucose (shown below) exists almost exclusively in the cyclic form. You will also see glucose drawn in a chair conformation. All of these representations provide different information about the structure, the details of which you will learn in organic chemistry. For this course, it is just important to be able to recognize different forms as sugars.

1CHO

2 H OH

3 HO H

4 H OH 5 H OH

6 CH2OH

6 6

CH2OH CH2OH 6 5 5 HOH2C OH H O OH 5 O 4 4 HO 4 OH C OH 1 1 O 1 HO OH OH OH 2 OH 3 2 3 2 3 OH OH

Haworth Projection Chair Conformation

The figure below shows the Fisher projections of glucose and galactose in their two enantiomeric forms (D and L). D-glucose and L-glucose can be differentiated by certain enzymes, but not taste receptors, hence they both taste identically. Otherwise their physical properties are the same because they are enantiomers. Glucose and galactose, however, are daistereomers and even though they only differ at one of four chiral centers, they have distinctly different properties. Galactose tastes less sweet than glucose and is metabolized through a different pathway.

CHO CHO CHO CHO

H OH HO H H OH HO H HO H H OH HO H H OH

H OH HO H HO H H OH

H OH HO H H OH HO H

CH2OH CH2OH CH2OH CH2OH

D-glucose L-glucose D-galactose L-galactose The numbering system for sugars remains consistent regardless of the number of carbons. The carbon immediately adjacent to the ring oxygen is C1 and then you work counterclockwise around the ring. The carbon outside of the ring is always last.

Sugars can form long chains of polysaccharides with each monomer attached with a glycosidic linkage. In this case, the alcohol on C1 bonds with C4 of the next molecule. Some of the most common polysaccharides are cellulose (structural material in plants), starch or glycogen (glucose storage) and chitin (exoskeletons)

Lipids Lipids have great structural variability but they are classified based on their solubility in nonpolar solvents. They are able to dissolve in nonpolar solvents because they have a large hydrocarbon (chain of carbon and hydrogen) component. One of the main classes of lipids are fatty acids which have a hydrocarbon chain attached to a carboxylic acid. The chain length can vary and it can either be saturated (all single bonds) or unsaturated (at least one double bond). Unsaturated fatty acids have kinks due to the rigid double bond that does not allow for rotation and these kinks prevent tight packing.

O O

OH OH Lauric Acid Palmitoleic Acid Saturated Unsaturated

Fatty acids can be combined with a variety of molecules to make more complex lipids. An example is combining glycerol with three fatty acid chains to triacylglycerols (triglycerides) which are found in many oils and fats.

O O

R1 OH H2C OH H C O R1 O 2 O HC OH R2 OH HC O R2 O H2C OH O

R3 OH H2C O R3

Glycerol Fatty Acids Triacylglycerol

Another important class of lipids are phospholipids which contain a phosphate group. Many are similar to triacylglycerol except they have replaced one of the fatty acid chains with a phosphate. O

H C O R 2 O 1

HC O R2

O

H2C O P OH O Lipids perform many critical functions in biological systems ranging from being the fundamental component of membranes to hormones (steroids) to vitamins.

Proteins The building blocks of all peptides are amino acids. The 20 natural amino acids have a wide range of properties, but all consist of the same core. There is an asymmetric carbon attached to an amine (NH ), a carboxylic acid (CO H), a hydrogen and an R 2 2 R group. It is the R group that gives each amino acid its identity and the R groups have a wide range of properties. OH H2N Two amino acids can be attached via a peptide bond. Due to resonance, O the peptide bond has partial double bond character and this lack of rotation means the bonds are planar.

R O R 1 H 3 N OH H2N N H O R2 O

Secondary Structure Secondary structure refers to local structure and order of the backbone. Secondary structures form in order to maximize the number of hydrogen bonds that can form (between the amide hydrogen (N-H) and carbonyl oxygen) which helps to minimize energy. It is also important to minimize overlap of the R groups (steric strain) in space as that can have a negative effect through repulsion. The fact that the peptide bond is planar limits the types of conformations the backbone can adopt. α-helix: In an α-helix, the backbone coils around an axis with the R groups of the amino acids facing outwards. This minimizes steric interactions between the R groups. This helix is stabilized by hydrogen bonds: each amide hydrogen is hydrogen bonded to a carbonyl oxygen 4 amino acids away. See adjacent image

β-sheet: A β-sheet has the backbone extending in a zigzag type fashion. Again it is held together by hydrogen bonding, but this time, the bonds are between amino acids in adjacent strands. The strands can be oriented in a parallel or antiparallel fashion. The R groups are facing above or below the plane of the sheet.

Tertiary Structure Tertiary structure describes the 3D folded structure of a protein. The final folded structure of a protein represents a lowest energy state, so one can think about the forces that drive folding in relation to ΔG. The enthalpic component of ΔG relates to bond energies and there are several important intermolecular interactions in protein folding that help to stabilize a folded state. Through folding, oftentimes hydrogen bonds are formed as well as salt bridges (interactions between two ions) and disulfide bonds (between two cysteine amino acids). All of these interactions contribute to the enthalpic component of folding. When you take an unfolded polypeptide which can adopt many conformations and fold it into an ordered 3D structure, you are decreasing the entropy. The primary counter to this entropic loss is due to a concept called the hydrophobic effect. Hydrophobic Effect: When water molecules are in bulk water, they have many degrees of freedom for rotation and movement; however when you add a nonpolar molecule, the water molecules become ordered around that molecule forming hydrogen bonds between water molecules in an attempt to minimize unfavorable interactions with the nonpolar molecule. This results in a decrease in entropy; however, if the nonpolar molecules then clump together, they minimize the surface area exposed to the water. This in turn reduces the number of water molecules that is ordered around the surface of the nonpolar molecule. The water molecules that are no longer around the nonpolar surface are now free to rotate again, thus increasing the entropy of the system. Since peptides often have many nonpolar amino acids, folding that would collect these nonpolar sites into the interior of the protein would be entropically favorable. By excluding water from the interior of the peptide, those ordered water molecules are released back into the bulk water.

Nucleic Acids Nucleotides, the subunit of nucleic acids, are composed of a 5-carbon sugar, a phosphate, and a base. The 5-carbon sugars are ribose (RNA) and deoxyribose (DNA), which is missing the OH at the 2’ position.

There are 5 nitrogenous bases, shown below. Uracil is only found in RNA.

NH2 O NH2 O O

N N N NH N NH NH

N N N N NH N O N O N O H H 2 H H H

Adenine Guanine Cytosine Thymine Uracil

A nucleotide is shown below

NH 2 N N

N N HO O

O

O O O O

Nucleotide triphosphates are the starting material for the synthesis of nucleic acids. The 3’ hydroxyl group of the most recently attached nucleotide attacks the α-phosphorous (phosphorous nearest the sugar) to form a bond. This results in the cleavage of pyrophosphate (2 phosphate groups) which provides the energy to make nucleic acid synthesis favorable. The figure on the following page shows this reaction with arrow pushing.

O O O P OH O P OH O O base base O O

3 3 OH O O O O O base base P O P P P O O 5 O O O O 5 O O O O OH OH