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ORGANIC CHEMISTRY I (CHEM 323)
EXAM 1
September 20, 2016
There are five pages, NINE questions, and a total of 110 points in this exam, which gives an extra 10 points on
top of that required for a “perfect” paper. Please read each question carefully and possibly more than once.
Good luck…
(11)
1.
Give the molecular formula(s) for each of the alkanes given below.
1.1.
Explain how the alkanes above can be differentiated on the basis of their 13C NMR
characteristics (i.e. how many signals)
(12)
2.
Show the acid-base equilibrium observed below by providing structures for the right-hand-side of
the equation. Use curly arrows to designate bond breaking and/or bond formation. Indicate formal
charges (if any). Note that the reactants are in equimolar quantities.
2.1.
The equilibrium above favors (circle one of the answers)
(a) the right-hand-side of the equation
(b) the left-hand-side of the equation
2.2.
Give a narrative reason for your choice of answer in 2.1.
Water, on the right-hand-side of the equation is less acidic, therefore more stable than lactic acid (on the left).
Equilibria are driven by thermodynamics. Right side more stable, which means right side of the equilibrium is favored
(6)
3.
Give the IUPAC name of the following branched alkane:
4,6-diethyl-2,2-dimethyloctane
(15)
4.
Circle the best possible correct answer (only one) for each of the following multiple-choice
questions:
Inspired by questions from Melissa and Ann-Janin
4.1.
Among the pairs of structures listed below, one pair consists of structures which are NOT
constitutional isomers of each other. Identify the correct pair.
(1) A and B
4.2.
(2) C and D
(3) D and E
(4) A and F
(5) B and F
In which one of the molecules below there is both an sp2-sp2 and
an sp2-s overlap?
CH3CH CHCH3
CH3
CH3C CCH3
CH3CH2C
CH
CH3CH2 CH2CH3
CH3
(1)
4.3.
(2)
(3)
(4)
Which of the following statements provides a clue to the fact that in predicting boiling
points, it is necessary to consider the extent of inter-molecular forces.
(1)
(2)
(3)
(4)
Highly volatile substances experience strong inter-molecular forces.
Substances with strong inter-molecular forces are unreactive.
In the gas phase molecules do not experience inter-molecular interactions.
No chemical reactions are observed in the gas phase.
Question submitted by Sarah
4.4.
C NMR utilizes 13C rather than 12C. What is the difference between 13C and 12C?
(1) number of neutrons (2) number of protons
(4) 13C is radioactive; whereas 12C is stable
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(3) number of electrons
4.5.
Which of the following transformations requires the breaking of bonds?
(13)
5.
Nitric acid, HNO3, is a strong acid. Provide a structure for the conjugate base of HNO3. Identify
formal charges (if any)
5.1.
Write resonance structures for the conjugate base of nitric acid. How does this confirm
that nitric acid is a strong acid?
The resonance structures illustrate the fact that there is extensive electron delocalization
within the conjugate base, which makes the conjugate base exceptionally stable and a
“weak base”; therefore, the undissociated acid would have considerable tendency to
dissociate, making it a strong acid.
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(12)
6.
Draw the chair form of tert-butylcyclohexane (shown at the end of the line)
Inspired by questions from DexterM and Liana
6.1.
In its more stable conformation
6.3.
6.2.
In its unstable conformation
Point out the major destabilizing factors on your drawing for 6.2.
(17)
7.
Complete the reaction below by providing a structure for the product formed .
Identify formal charges (if any).
7.1.
You have seen this reaction (or a very similar one) in the last four weeks. However, the
expertise we are expecting to acquire in organic chemistry will allow us to “predict” the
products of such reactions. Our starting point for the reaction is that both reactants are
reactive.
7.1.1.
Explain, on the basis of Lewis Model/Theory, which reactant is reactive and
why. Also comment on the type of reactivity.
BF3 has an incomplete octet; therefore it is unstable/reactive.
Sharing an electron pair with another atom would complete boron’s octet and
allow it to adopt a noble gas electron configuration. Thus boron is susceptible to
attacks by electron pair donors, i.e. nucleophiles/Lewis bases; as a result BF 3
can be classified as an electrophile/Lewis acid.
7.1.2
Explain, on the basis of Orbital-overlap Model/Theory, which reactant is
reactive and why. Also comment on the type of reactivity.
The double bond in cyclopentene consists of a -bond (face-to-face overlap) and a -bond (sideways overlap of p-orbitals).
The extent of overlap for the latter is significantly less compared to a -overlap, which is an indication that the -bond is
weaker and the bonding -electron pair is likely to be attracted by strongly electrophilic species such as BF 3. This will result
in the breaking of the -bond and formation of a new -bond. The alkene will be functioning as a nucleophile.
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(10)
8.
The molecule below, illustrated with “partial Newman-Projection Representation” , cannot
undergo free-rotation.
8.1.
Draw the molecule in a more conventional form (without the “partial Newman-Projection
Representation”). Your drawing should indicate the spatial orientations of the
substituents.
8.2.
Give the IUPAC name of the molecule above (and in 8.1). Make sure the name specifies
the structure of the specific stereoisomer you have drawn
cis-1,2-dimethylcyclohexane
(14)
9.
The complete combustion products for the molecules below are the same.
9.1.
Write a balanced equation for the complete combustion of either one of the molecules
showing the products formed.
9.2.
Do you think the heat of combustion values for the molecules above will be the same or
different? (circle the correct answer)
(a) the same
(b) different
9.3.
Explain your choice of answer for 9.2 using an energy diagram.
Although the two compounds are constitutional isomers
and the stoichiometry of the complete combustion
products are identical, propylcyclopropane is less stable
than cyclohexane. The former is subject to torsional as
well as angle strain. As illustrated by the energy
diagram, the heat of combustion of the less stable
isomer is greater than that of the more stable isomer.
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