SI Review Exam 1

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					     “One important key to success is self-                 do…However, it’s possible that they may incorporate
                confidence.                                 larger, more complex molecules, such as triatomics].
    An important key to self-confidence is
         preparation” - Arthur Ashe                         Instructions for logging into WebMO can be found

 Exam 1 SI Review                                           1. Go to
                                                            2. Then scroll down and follow the instructions under
 Chemistry 332 Fall 2009                                    Tutorial
                                                            3. When you get to point 6 – select calculation for
                                                            Molecular Orbitals
  Emily Allen (, Matt Weber           4. Play around with the different MO’s [for which ever
       (, David Song                   molecule you build] and see if you can match up the
               (                        MO with the WebMO image. This skill will be highly
                                                            valuable for the exam.
1. Practice old exams on ACE. This year the ORGO
department has uploaded a plethora of practice exams        ---So lets start with the MO diagram of an old exam
onto ACE. Use these to your advantage. The types of         problem from Exam I, Spring ’09, question 12-18.
questions on these practice exams will be very similar in   Remember that MO’s are due to the overlap of AO’s or
content and also language. Getting familiar with the        Hybrid AO’s. These interactions give bonding and anti-
types of questions you will see on a real exam will help    bonding molecular orbitals as a result. The relative
build confidence, so definitely take some time to           energies […placement on the MO diagram] is governed
become familiar with the interface. As Pat Mcintre          by stabilization factors such as energies matched vs.
(Former 332 SI leader and current 232 Merit Leader)         mismatched. One huge point is to count the nodes. The
once told me, “At least on the day of the exam, make        more nodes there are, the greater in energy.
ACE your best friend…Be at the top of your game before
you go into the exam room”.                                 ---To start things off lets draw a Lewis structure of an
                                                            azide (N3(-))…Do not forget about formal charge.
2. Be able to identify the AO orbitals involved in          1. count up total number of valence
building MO’s.                                              electrons…(remember that a negative charge will give
                                                            one more electron, and a positive charge will give one
-Seriously, please do not forget to read the question as    less electron.)
far as what type of orbitals you are to construct the       2. connect everything first with a single bond, and then
MO’s from. They will explicitly tell you which AO’s you
should be using.

                                                            play around with localizations of electrons and bond
-Be able to match up MO images (WebMO) with MO              connectivity to satisy the octet rule.
diagram.                                                    3. HONC – (Hydrogen has 1 bond, Oxygen prefers 2
                                                            bonds, Nitrogen prefers 3 bonds, and Carbon prefers 4
-It would be a wise to use WebMO to really see how          bonds)
these MO’s are constructed. So please take some time
before the exam to construct some of your own MO’s
[CO and F2 are nice example calculations that you can

For problems A – G: These questions relate to the LCAO
for azide MO’s. Construct these MO’s from the
combinations of 2s and 2p atomic orbitals following this
LCAO convention. The AO’s from N1, N2, and N3, are
used to create the diagram without rotating the AO
coordinates (all three atoms have their +z axis pointing
in the same direction, and this direction is the same as
the molecules +Z direction). N1 is the central molecule!

Complete this sentence:

Molecular orbital “A” is mainly the contribution of the
N1 {2s, 2px, 2py, 2pz} AO, the N2 {2s, 2px, 2py, 2pz} AO,
and the N3 {2s, 2px, 2py, 2pz}. The AOs for N2 and N3
are of the {same, opposite} sign.

Molecular orbital “B” is mainly the contribution of the
N1 {2s, 2px, 2py, 2pz} AO, the N2 {2s, 2px, 2py, 2pz} AO,
and the N3 {2s, 2px, 2py, 2pz}. The AOs for N2 and N3
are of the {same, opposite} sign.

Molecular orbital “C” is mainly the contribution of the
N1 {2s, 2px, 2py, 2pz} AO, the N2 {2s, 2px, 2py, 2pz} AO,
and the N3 {2s, 2px, 2py, 2pz}. The AOs for N2 and N3
are of the {same, opposite} sign.

Molecular orbital “D” is mainly the contribution of the
N1 {2s, 2px, 2py, 2pz} AO, the N2 {2s, 2px, 2py, 2pz} AO,
and the N3 {2s, 2px, 2py, 2pz}. The AOs for N2 and N3
are of the {same, opposite} sign.

                                                            An energy level diagram for the valence molecular
Molecular orbital “E” is mainly the contribution of the     orbitals of N3(-) is shown here. Each number in the
N1 {2s, 2px, 2py, 2pz} AO, the N2 {2s, 2px, 2py, 2pz} AO,   diagram can be matched to one of the seven MO
and the N3 {2s, 2px, 2py, 2pz}. The AOs for N2 and N3       pictures, A-G. Given the number of valence electrons in
are of the {same, opposite} sign.                           N3(-), the highest occupied molecular orbital
                                                            correspsonds to image {A, B, C, D, E, F, G} and the
Molecular orbital “F” is mainly the contribution of the     lowest unoccupied molecular orbital corresponds to
N1 {2s, 2px, 2py, 2pz} AO, the N2 {2s, 2px, 2py, 2pz} AO,   image {A, B, C, D, E, F, G}. Hint: in matching the energy
and the N3 {2s, 2px, 2py, 2pz}. The AOs for N2 and N3       levels to the images, remember that the MO energy
are of the {same, opposite} sign.                           increases with increasing number of nodes.

This is nice schematic depicting how the molecular orbitals are from the 2s, 2px, 2py, and 2pz AO’s. This helped me
answer a sick nasty crazy problem when I took the Exam I, Sp09.

3. Energy level diagrams for valence molecular orbitals of isomers.

-isomers have the same formula, but different connectivity. Understand how this could ultimately change the MO
diagram and relative placement of energy levels. Keep in mind 2nd row orbital energies of neutral atoms and ions. Page
13 and 21 show the relative energy levels of the valence orbitals for H, C, N, O, and F.

---They might ask you to select the correct energy level diagram for the valence molecular orbitals of different isomers.
Make sure when you select the diagram that it reflects the relative energy levels of the different atoms involved, and to
make sure that these are valid, symmetry allowed combinations.

4. Hydrogen Bonding (Refer to page 70-71 of your lecture notes) and pKa Trends (page 93)

***Essentially, you need a H-bond donor [Usually N and O connected to a H] and a H-bond acceptor [Usually a
nucleophilic N or O]...But really, the frontier orbital interactions between the donor and acceptor are causing this
interaction to occur.

Non-covalent interactions such as hydrogen bonds, ionic bonds, and hydrophobic interactions are weaker than covalent
bonds but can have a great effect on the stability as well as overall reactivity of the molecules.

***Remember, that H-bonds are attractive. Meaning, the closer they are, the “better likelihood” that these orbitals
are going to interact. If the acceptor and donor are not in close proximity with each other, then there is very little
chance for the orbitals to interact and thus no H-bond can form.***
------> Let’s further our analysis by examining how H-bonding affects reactivity…Essentially, H-bonding comes down to
acid-base chemistry. So to truly understand the nature of how H-bonding affects reactivity, we shall direct our attention
to webcast 11.4 [what Professor Moore calls, “Chemical Intuition”].

Let’s consider a problem that deals with H-bonding, pka trends, reactivity, and MO orbital diagrams.

                                                                                         Imagine that these functional
                                                                                         groups were locked in an
                                                                                         enzyme’s active site. Consider
                                                                                         how     the     ambient    pH
                                                                                         determines ionization states
                                                                                         and overall reactivity.

                                                                                         The AO of each of the atoms
                                                                                         has been drawn in for you.
                                                                                         Consider the hybridization
                                                                                         states of the imine and then
                                                                                         construct a MO diagram [all we

really care about is relative positions of the MO’s]

5. Catalysis/Transition States/Rate Determining Step [Don’t pay too much attention to the mechanistic details within
the actual catalysis…What we actually care about is how catalysts speed up reactions. Your understanding for RXN
coordinate diagrams will be immaculate after this section!!!]

---Given the Reaction Coordinate Diagram for an                ---Redraw the RXN Coordinate Diagram for a catalyzed
uncatalyzed reaction what is the RDS?                              Reaction, noting TS energy levels.

Consider this case study of Acetoacetate
Decarboxylase…What is the rate determining step?
How does ambient pH and pKa factor in with the
overall reactivity of this enzyme? Let’s say the reaction
conditions are unfavorable for this reaction to occur,
what might be some factors that could contribute to

6. Tautomerization (For this one, consider how different tautomers contribute to overall reactivity of a reaction)
-You will most likely not have to draw arrows for this exam for any problem. Therefore, I will not spend too much time as
far as the mechanistic considerations of a tautomerization. What is important is that tautomers although just an
oxidation or reduction step away from each other, have considerably different relativities. In your consideration of these
tautomers, check out how the different tautomers would participate in H-bonding. How does tautomerization change
H-bond donors and H-bond acceptors?

Also, think of how these pi-conjugated systems are able to share electrons in a massive alpha, beta unsaturated type of
resonance donations.

---Keto/Enol                                    Exam II Fall 08 #22

---imine                                        Exam II Fall 05 #2


***Really be sure to understand how H-
bonding interactions change as a result
of this tautomerization. Indicate the H-
bond donors and H-bond acceptors at
physiological pH. What are the frontier
orbitals involved in the H-bonding?

8. Frontier Orbitals
                                                                           HOMO: Highest Occupied Molecular
                                                                           Orbital, this is the nucleophilic site,
                                                                           typically either the sigma, pi, or
                                                                           nonbonding orbitals. This is where the
                                                                           most reactive (highest energy)
                                                                           electrons sit.
                                                                           LUMO: Lowest Unoccupied Molecular
                                                                           Orbital, this is the electrophilic site,
                                                                           typically either the empty atomic
                                                                           orbital, pi*, or sigma*. This is the most
                                                                           stable orbital that attacking electrons
                                                                           can go.

How to identify frontier orbitals:
       Draw the reaction arrows in the reactant
       Identify where the electrons are coming from that are doing the attacking [HOMO]
              o Lone pair: n
              o Pi bond: pi
              o Sigma bond: sigma                                                  The usual
       Identify the bond that breaks when the atom is attacked [LUMO]           energy levels
              o No bond breaks: a                                                       σ*
              o Pi bond: pi*                                                            π*
              o Sigma bond: sigma*                                                      a
       Now you know the HOMO and LUMO, but is it sigma or pi type?                     n
              o Draw the MO for the HOMO and for the LUMO                               π
              o How do they line up?                                                    σ
                       Head-on: sigma-type (this is the best interaction)
                          Side-by-side: pi-type (this is weaker than sigma-type)
                          If you do not have sigma or pi-type overlap, there is NO INTERACTION.

Draw the frontier orbitals:

                   HOMO                                           LUMO

You should be able to fill out sheet 1 pretty quickly by the test. Remember that sp hybrid orbitals look and behave
similarly to 2px atomic orbitals

Exam I, Fall 05, Q5

The Syn frontier orbitals   The Anti frontier orbitals

9. Determining Hybridization

          Sp3: tetrahedral, 109.5 degree bond angles, no pi bonds
           o draw four sp3 atomic orbitals in a MO diagram (since sp3 orbitals are 75% p character and 25% s
               character, the sp3 orbitals are ¾ between the s and p atomic orbitals)
           o all three p orbitals are “used” in making the sp3 hybrid orbitals, so you are not left over with any p
        Sp2: trigonal planar, 120 degree bond angles, contains 1 pi bond
           o draw three sp2 atomic orbitals and one p orbital in a MO diagram (since sp2 orbitals are 66.67% p
               character and 33.33% s character, the sp2 orbitals are 2/3 between the s and p atomic orbitals)
           o two p orbitals are “used” in making the sp2 hybrid orbitals, so you are left over with one p orbital
        Sp: linear, 180 degree bond angles, contains 2 pi bonds
           o draw two sp atomic orbitals and two p orbitals in your MO diagram (since sp orbitals are 50% s character
               and 50% p character, the sp orbitals are halfway between the s and p atomic orbitals)
           o one p orbital is “used” in making the sp hybrid orbitals, so you are left over with two p orbitals
         Pay attention to geometry! It will help you figure out what direction the frontier orbitals will point

10. More on rate laws…Understand reaction coordinate diagrams (Webcast 10.2) and the kinetic model constructed in
Webcast 10.3.

      Each step of a reaction mechanism can be profiled in a reaction coordinate diagram, with each step having to
       pass through an energy barrier each time (known as a transition state)
      The step involving the transition state of highest energy is known as the rate-determining step (RDS).
   General Tips About Rxn Coordinate Diagrams and Kinetic Models

      The molecularity of a reaction step is the number of molecules involved in the transition state (i.e. aA + bB  cC,
       would be bimolecular)
      The kinetic order is referring to the order of a reaction as determined by the RDS (rate-determining step).
           o So… if we had 1A + 2B  3C (and this was the RDS), we would have…

                                                rate= k[A][B]^2

                       Our kinetic order would be 3 (1+2=3), just add the exponents.

      The units of rate coefficients (k).
           o This is a good shortcut to determine the kinetic order of the reaction (but this only works if you are given
               the units of the rate coefficients).
                     Zero Order= mol L-1 s-1, First Order= s-1, Second Order= L mol-1 s-1
      Go over Spring05 #3, Spring04 #7 and Fall05 #3 for practice.
      About the kinetic model….
           o A kinetic model is a quantitative representation of a qualitative reaction coordinate diagram, and it
               allows us to find the rate of formation of various reactants/intermediates over time.
           o The rate law is written directly from the RDS, but it is often useful to rewrite this expression in terms of
               the concentrations of reactants because experimentally, we would have a very good idea of what the
               concentrations of our reactants were.
                     In rewriting our rate law, it is common to be able to eliminate the RDS step from the equation
                        since it has such a high energy barrier and can be considered negligibly small.

                       When a compound is in equilibrium, the rate of formation of the forward and backward
                        reactions is equal; thus, the rate of formation of that compound is 0.

11. Bond Energies: Thermodynamics…the Driving Force behind organic reactions.

Using the table, calculate the bond energy changes that take place and decide whether or not this reaction is favorable.

12. More on pKa/pH/ionization sites.

Henderson-Hasselbach Equation:

pKa = pH + log [HA]/[A-]

Below are a few examples of proton transfer and acid-base electron flow:

----Reaction Rates/Steady State Kinetics:

-Keq = [prod1][prod…]/[react1][react2][reac…]

-Rate of product formation = rate of product generation – rate of product consumption

-This is due to the fact that the reaction lies along an equilibrium scale

The rate forwards (product generation) is usually referred to as k1

The rate backwards (product consumption) is usually referred to as k-1

k1 / k-1 = Keq

-The relative pKa’s will be given to you for the exam. It will be your job to understand the different ionization states.
Really, its just acid/base chemistry. The idea behind weak acids is that they are able to exist in different protonation
states, depending on ambient pH and electrostatic repulsion affects.

If this molecule were in an environment with a pH=4, redraw with all pertinent hydrogens and charges.

pKa1         4.76
pKa2         5.3
pKa3         7

If this molecule were in an environment with a pH=10, redraw with all pertinent hydrogens and charges.

Final advice: The test may reference recent research papers in the exam questions. It is not necessary
(and often not a smart use of your time) to read these papers during the exam in order to answer the
questions correctly. However, if you are completely stumped on a problem, use the reference given to you to
look up the paper online and quickly skim the paper over for any useful figures/mechanisms/diagrams, etc.
***Often looking up a paper will not lead to any useful information without actually READING the paper, which
you DON’T want to waste your time doing. Therefore, DO NOT rely on this on your exam!!!

You will run out of time if you look up every reference listed on your exam!!!

However, as a LAST RESORT, looking up the paper can sometimes earn you quick points on the exam.

To look up a paper online:

1. Go to the U of I library website:
2. Select the “Journals” tab.
3. Enter the title of the journal you would like to search.
4. Once the search results come up, select the Electronic Journal Start of Title match you want.
5. Click on the archived issues you want to view (organized by year).
6. This takes you to the journal website. Use the journal’s Advanced Article Search to search for the paper
you want using the paper’s issue, year published, and page number.

Example: I am a CHEM 332 student taking the Fall 08 exam, and I can’t figure out #22, which is shown below.

What on earth is this talking about in this question?!?! This is definitely a “WTF” moment!

I have no idea where to begin.

But look, no worries…I see there is a reference to the research paper that this question came from at the end
of the question. Great! All I have to do is find this article online, and maybe the article will have a picture of
this mysterious C3H2 molecule. So I do the following:

1. I go to the U of I library website. In fact, I was so prepared that I already had
the library website pulled up on my computer screen while taking the CHEM 332 exam, since I thought I
might need to reference a paper or two to help me score more points on the exam.
2. I select the “Journals” tab.
3. I type “J. Am. Chem. Soc.” in the search bar and hit “go” (the journal title will be written in italics in the
4. Next to “Electronic Journals Search” I click on “Start of Title Matches: 1”.
5. Sweet! This takes me to the J. Am. Chem. Soc (JACS) website.
6. At the top right-hand corner of the webpage, I see the yellow box that says “Quick Search”. I click on
“Advanced Search” in this yellow box.

7. I scroll down a bit to the bottom left hand corner of the page where I see the blue box titled “Search by
Citation”. This is what I want, since the exact reference for the paper was provided!
8. Where it says “Select a journal”, I scroll down until I select “Journal of the American Chemical Society”.
9. In the box that says “Vol #”, I type “119”, since the volume number will be written in italics just after the
year the article was published in the reference listed.
10. In the box that says “Page #”, I type “5847”, which is listed right after the issue number in the provided
reference. This is the first page of the article I want in this journal.
11. I hit “Go”. This takes me to a page with the article. I hit “High Res PDF”, but I could also have selected
“High Res HTML”; either one is fine.
12. Now I skim the article quickly (I DON’T want to read the whole article…this will take way too much time).
I’m just looking for helpful figures or diagrams that might help me answer this question. #22 on the Fall 08
exam wants me to draw the Lewis structure of the neutral compound C3H2, so I scan the article looking for a
molecule that fits this description. Found it!!!! On the very first page under the diagram “Scheme 1”, the
molecule labeled “2” and the molecule labeled “5” look like they will work. In both molecules, one of the
carbons is a carbene (a neutral carbon atom with a lone pair and incomplete octet of electrons) in a linear
molecule. Molecules “2” and “5” seem to fit the question’s requirements for a neutral C3H2 compound that has
no rings, no triple bonds, no positive or negative charges, and no radicals. Sweet!!! I only need one, so I pick
one of the two and input this Lewis structure into ACE, and I get the question right!!!

This is the diagram on the first page of the article called “Scheme 1”. It looks like compound 2 or compound 5
is what this exam question from Fall 08 was looking for!

NOTE: The information present is our best interpretation of the materials. We are absolutely trying our best to prepare
you so that you can feel confident about the exam. Emily, Matt, and David, spent most of their weekends working on
this material so that you could have an organized review for EXAM I. So please use this study guide and ask many

Seriously, you guys are going to do awesome!!! Just believe in yourself and all of that work is going to pay off!!! Try and
stay calm during the exam, and definitely get plenty of rest.

“Think before you click, don’t just click before you think” [Slight modification from Professor Van der Donk’s insight on
exam taking strategies.

***We will accept Chipotle Gift Certificates***


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