CHEM 2OA3-2009 ASSIGNMENT 1
NAME:_______________________
ID#:_____________
Lab Station(2-72, NOT L01, L02 etc):_____ (On ELM in grade book)
Please circle your lab section below (Chem Bio students circle L10)
Group Monday Tuesday Wednesday Thursday Friday
I L01 L02 L03 L04 L05
II L06 L07 L08 L09 L10
DUE: Tue. Oct. 6th, 2009 AT 2:30 PM.
Students’ Responsibility:
1. Answers must be filled in, in the space provided.
2. Completed work MUST be placed in the drop-off slots, which are in the corridor opposite to the
chemistry office, close to ABB-119. You MUST place your work in the slots designated “CHEM 2OA3/2OB3”, in the CORRECT slot, FOR THE CORRECT LAB DAY.
3. Any assignment not submitted in the correct place and/or after the due date and time will be
graded zero.
4. Students who cannot complete the assignment for medical or other reasons MUST seek exemption
1. [8+2] Nitrogen is a major component of our atmosphere, in part due to the stability of the triple bond between the two atoms. This triple bond has a bond dissociation energy of 945 kJ/mol, one of the highest known, and requires very special mechanisms for activation (nitrogen converting bacteria using metal catalysts, lightning).
a. Draw the valence shell molecular orbital diagrams for molecular nitrogen. b. Include an energy scale, and indicate the relative energies of AOs and MOs. c. For 2 bonus points, try to indicate the correct relative energies of the MO’s.
d. Draw the appropriate number of electrons in the AOs and in the formed MOs, using single headed arrows.
e. Label the MOs according to the conventional nomenclature for MO diagrams.
MO diagram for Nitrogen (N2):
Nitrogen uses three 2p AO from each nitrogen to form the triple bond in N2. The σ bond here is formed by overlap of one lobe each of the 2px orbitals, like in Fluorine. This σ bond has lower stability and hence higher energy (2 bonus points if drawn) than the two π bonds formed by side-ways overlap of the other two p orbital pairs. The 2S2 orbitals are full, lower in energy, do not hybridize with the p orbitals, and do not engage in binding.
When nitrogen has substituents, as in NH3or NR3, it does hybridize in a manner similar to carbon, forming sp3 hybrid orbitals that then form tetrahedral AO arrangements (with one being the lone pair). Note: a reasonable solution using a hybridized approach will be accepted in this assignment, too.) 2. [10] Ethylene is a flat molecule that cannot rotate about its carbon-carbon axis. Explain these two facts,
based on molecular orbital theory. Sketch and label the two carbon-carbon bonding molecular orbitals into the template of the molecule below. Use this sketch to support your (brief) explanation.
Explanation:
Flat: sp2 hybridization establishes a trigonal planar structure for the three sp2 AO’s of each carbon, with angles of 120 degrees between each two sp2 orbitals. Each carbon additionally has one 2p orbital that is vertical to the sp2 plane in that carbon. In-phase overlap of these two singly occupied p orbitals can only happen for a flat arrangement of the molecule.
C
C
H
H
Cannot rotate: the p bond formed y overlap of the two p orbitals consists of electron clouds both above and below the molecular plane. Rotation around the sigma single bond between the two carbons would require breaking the p bond, which would require prohibitive amounts of energy
3. [12] Draw all possible structural isomers for C4H10N (note: there are not necessarily 12 isomers).
NH
2N
H
N
H
NH
2NH
2NH
2N
N
H
4. [10] Add all lone pairs and formal charges to the three molecules shown below. Clearly label all sp and sp2 hybridized atoms, including heteroatoms.
5. [12] Draw bond-line structures of the following three acids in the boxes below, write their pKa values underneath each molecule, and indicate polar bonds with δ+ (delta plus) and δ- (delta minus). Indicate in two words (inductive stabilization, resonance stabilization), why benzoic acid and 2-fluoroacetic acid are more acidic than propionic acid.
propionic acid benzoic acid 2-fluoroacetic acid
pKa = 4.87 pKa = 4.2
why lower than propionic acid?
Please see explanation below:
pKa = 2.59
why lower than propionic acid?
Inductive Stabilization
O
OH δ−
δ− δ+
δ+ OH
O
δ− δ−
δ+
δ+
O
OH F
δ−
δ− δ−
δ+
Why does benzoic acid have a slightly lower pKa then propinoic acid?
O O
Benzoic acid anion
There was a slight error with this question, therefore both answers will be accepted, but here is the rational:
Resonance stabilization is not technically correct. This is because it is not possible to draw a good resonance contributor where the charge from an O- could be moved into the ring system (try this for yourself). So although there is resonance in the benzene ring itself THAT DOES NOT stabilize a negative charge which would be present in the deprotenated benzoic acid. And although you CAN move the negative charge between the two oxygen atoms, you can also do this with propanoic acid, therefore it is NOT the reason why it is BETTER than propanoic acid.
However, inductive stabilization is not entirely correct either. A benzene ring is not like a fluorine atom, it does not exert strong electron withdrawing properties. So although it is located close to the negative charge it does not help share in the negative charge in a simple inductive way.
The real reason that the benzene ring helps stabilize the negative charge, is similar to resonance, but is a concept called electron delocalization (or conjugation). See the image below, although you cannot
technically draw a resonance structure where you can move the negative charge into the ring, there still is a full pi system (each atom has a p orbital). So the electrons can actually move quite freely around the
molecule so the charge is not as localized on the oxygen atoms. This concept is fully covered in Ch13 and in CHEM 2OB3. You are not responsible for this level of detail. Focus on the other questions on this topic from the tutorial.
O O
6. [12] Indicate (by circling A,B,C,D or E) the correct answers to the four questions below:
i. Homolysis is the cleavage of a bond between the same type of atoms, such as Cl2 or Br2. ii. Heterolysis is the cleavage of a bond between a carbon atom and a heteroatom.
iii. Homolysis involves breaking a covalent bond between two atoms such that one electron of this bond ends up on each of the two atoms.
iv. Heterolysis involves breaking a covalent bond between two atoms such that both electrons making up that bond end up on one of the two atoms.
7. [12] Complete the two proton transfer reactions below:
draw the correct products on the right hand side,
add all lone pairs and formal charges on both sides of the reaction arrows,
add the appropriate double headed curly arrows (four in total) indicating electron movements,
N H N
O H
N H
N
O
OH
: :
:
: :
: : O K
..
+
+
+
(potassium ethoxide)
+
+
-..
..
..
.. :-
+
+
K
-..
8. [12] Indicate the correct statements regarding the unimolecular nucleophilic substitution reaction, SN1.
i. SN1 reactions are characterized by only having one molecule involved in the rate-determining step. ii. SN1 reactions proceed with homolytic dissociation of a covalent bond.
iii. SN1 reactions can only occur on tertiary alkyl halides iv. SN1Reactions involve a short-lived ionic intermediate
9. [12] You are tasked with analyzing river water downstream of a suspected spill of small amounts of benzyl alcohol and phenyl methyl ketone. You extract one liter of river water three times with about 10 mL dichloromethane, spot the extract onto a silica TLC plate and elute using hexane as mobile phase.
i.Two dots appeared when visualizing the TLC plate with a UV lamp. Label the TLC plate to indicate which dot likely corresponds to benzyl alcohol and which to phenyl methyl ketone.
PMK
BA
ii. Which of the two compounds would have a lower Rf value (circle the correct answer):
benzyl alcohol phenyl methyl ketone
iii. If you were to load the mixture onto an alumina chromatography column and eluted with hexane, which compound would elute off the column first? (circle the correct answer):
benzyl alcohol phenyl methyl ketone
iv. Briefly explain why:
Phenyl methyl ketone (dipolar interactions) binds less strongly to the polar alumina than benzyl alcohol(hydrogen bonding plus dipolar interactions), is hence less strongly retained on the stationary phase and elutes first.
v. How could you use infrared spectroscopy to confirm which of these two compounds is which?
Phenyl methyl ketone has a strong carbonyl absorption in the 1700 cm-1 range, while benzyl alcohol has a strong absorption at about 3500 cm-1. Either can be used to distinguish the two compounds.
vi. BRIEFLY explain how the order of elution (which one moves fastest) would be affected if the eluting solvent was switched from hexane to methanol?
