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Exam Explanations

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Chemistry
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Santa Barbara City College
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Exam 1 answers
Q1. Monochlorination can occur in as many as non-equivalent positions there are in
ethylcyclohexane. The six positions that can be chlorinated are shown below
1
2
3
4
5
6
4
5
Thus, there can be as many as six monochlorination products at any of this positions.
Q3. In this case, it’s always useful to identify the monomer within the polymer chain. The
monomer is the single unit that repeats over and over within the chain. It is highlighted in red in
the structure below.
Thus, considering the electronic movement, the structure that resembles the most this monomer
is B. It is not D because the linkage between aromatic rings would contain 4 carbons instead of 2.
Q4. Indeed the product shown is one of the product due to the formation of a free radical in that
position. However, free radicals have resonance structures. One of these resonance structures
leads to a free radical in the terminal carbon, and thus, two products are formed. The two free
radicals are shown below
Each one of them leads to a different bromination product.

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Q5.
The dissociation energy for F-F bond is 155 kJ/mol. It might seem high, but the main problem is
to control the chain reaction started by fluorine free radical. Given its small size it’s extremely
reactive and can initiate a great number of propagations steps making it difficult to form specific
products.
On the other hand, bond dissociation energy for I-I is about 152.5 kJ/mol. However, even when
it’s about the same energy, iodine is a bigger atom capable of stabilizing the recently formed free
radical and thus, poorly reactive in the propagation step making it useless in these reactions.
Q6. This is a hydrobromination reaction promote by peroxides, which induces the addition to occur
via free radical. Thus, the addition of bromine occurs in the less substituted carbon of the double
bond, instead of the typical product (which would be the selected answer B). The correct answer
is C. Neither A nor D are formed, since no free radicals are formed, or could have resonance
structures, in this positions.
Q7. With respect to the resonance theory, in the case of 3-chloroprop-1-ene an allylic carbocation
is formed, which is stabilized by resonance making it far more stable than in the case of 1-
chloropropane in which a primary carbocation is formed.
As for the Hammond postulate, given that the carbocation formed by the alkene is more stable,
and therefore lower in energy, it resembles the most the structure of the starting material and the
energy needed for the reaction to complete is considerably lower than in the case of 1-
chloropropane.
Q11. The kinetic product is the one that forms the fastest but it’s less stable when compared to the
thermodynamic product. In the case of alkenes, the kinetic product is associated with the less
substituted alkene, which is structure B. Structure A is the thermodynamic product.

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Exam 1 answers Q1. Monochlorination can occur in as many as non-equivalent positions there are in ethylcyclohexane. The six positions that can be chlorinated are shown below 2 4 5 6 3 1 4 5 Thus, there can be as many as six monochlorination products at any of this positions. Q3. In this case, it’s always useful to identify the monomer within the polymer chain. The monomer is the single unit that repeats over and over within the chain. It is highlighted in red in the structure below. Thus, considering the electronic movement, the structure that resembles the most this monomer is B. It is not D because the linkage between aromatic rings would contain 4 carbons instead of 2. Q4. Indeed the product shown is one of the product due to the formation of a free radical in that position. However, free radicals have resonance structures. One of these resonance structures leads to a free radical in the terminal carbon, and thus, two products are formed. The two free radicals are shown below Each one of them leads to a different bromination product. Q5. The dissociation energy for F-F bond is 155 kJ/mol. It might seem high, but the main problem is to control the chain reaction started by fluorine free radical. Given its small size it’s extremely reactive and can initiate a great number of propagations steps making it difficult to form specific products. On the other hand, bond dissociation energy for I-I is about 152.5 kJ/mol. However, even when it’s about the same ener ...
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