1) B. Primary alkyl halides react faster under SN2 reactions than secondary or tertiary alkyl
halides. Vinyl and aryl alkyl halides tend to not react under SN2 reactions, partially due to the
repulsion between the pi system and the attacking nucleophile. These facts eliminate choice C,
D, and E. Between choices A and B, iodine is a better leaving group, making it easier for the
attacking nucleophile to push the iodine off of the alkyl halide than with the bromine.
2) D. The product molecule provided has R stereochemistry (methoxy has the highest priority,
followed by ethyl, methyl, and lastly hydrogen. Remember to read the reverse stereochemistry
because the low priority group is coming out of the page). Because this reaction is using
backside attack (an SN2 mechanism), the stereochemistry was inverted in the course of the
reaction. This means that the reactant had S stereochemistry. Molecule D is the only one that
has S stereochemistry. For the reactant, Br has the high priority, followed by ethyl, methyl, and
finally hydrogen. Recall that in Fisher projections, the horizontal groups are coming of the page
and the vertical groups are going into the page.
3) B. Reaction of an alkene or alkyne with NBS and light creates an allylic halide through a
radical reaction mechanism. The only molecule that is not an allylic halide (a halide on the
carbon adjacent to a double bond) is molecule B. Molecules A and D are formed from the
original reaction. Molecule C is formed from a resonance structure that exists when the radical
is on the terminal primary carbon (the intermediate that formed molecule D). Resonance with
radicals happens in a similar manner to resonance with cations or lone pairs. The main
difference is that only one electron is moving at a time, whereas with lone pairs and cations two
electrons generally move in each step.
4) E. Of the two possible products (2,3-dimethyl-1-butene and 2,3-dimethyl-2-butene), the more
stable product is 2,3-dimethyl-2-butene due to there being more substitution about the double
bond. So, we are trying to increase the yield of the minor product. The major difference
between each of the answer choices is the size of the molecule (all of them have similar basicity
and nucleophilicity). So, we are likely looking at some sort of steric hindrance issue. In order to
form the major product, the base must deprotonate a tertiary carbon. This is difficult to do due to
the degree of substitution of that carbon atom. In order to form the minor product, the base only
has to deprotonate a primary carbon. If a larger molecule were used, it would be difficult to
make the major product due to the large steric hindrance between the tertiary carbon and the
large base. The minor product would be much easier to make due to only having to deprotonate
a primary carbon, where there is much less steric hindrance. Molecule E is the largest of the
answer choices.
5) B
6) B
7) A
8) C
9) C
10) B
11) B
12) A
13) C
14) E
15) D
16) a)
b) The product IR should not contain any alkyne hydrogen bond stretches near 3300 cm-1 or
any alkyne carbon-carbon bonds near 2200 cm-1. It will contain alkene carbon-hydrogen bond
stretching near 3000 cm-1, alkane carbon-hydrogen bond stretching around 2900 cm-1, and a
carbon-nitrogen triple bond stretch near 2200 cm-1 (this may be difficult to differentiate from the
carbon-carbon triple bond stretch in the reactant.
17)
d) Molecule A will react faster. Molecules have to have an anticoplanar geometry in order to
react in SN2 reactions. In the case of a cyclohexane molecule, this means that the leaving group
and proton must both be axial, and must be facing opposite directions (one up and one down). In
molecule A, this geometry is seen in its more stable chair conformation. However, in molecule
B this geometry only exists in its less stable chair conformation. So, molecule B must first go
from its most stable conformation to its less stable conformation, and then is able to react. This
raises the activation energy for its reaction (because it needs energy to perform a ring flip),
which slows the reaction.
18) The first graph corresponds to molecule D. The main differentiating features are the
presence of a low –OH stretch, which is characteristic of carboxylic acids, and the presence of
relatively low C=C and C=O stretches, which suggests that there are conjugated double bonds in
the molecule.
The second graph represents molecule C. The main features are the presence of both
C=O (~1750) and C-O (around 1200) stretches and the lack of aldehyde hydrogen peaks at 2800
and 2700 cm-1.
The third graph is molecule B. This graph contains aldehyde hydrogen stretching peaks
at 2700 and 2800 cm-1. It also shows a C=O stretch near 1700. The main deciding feature
between A and B is the presence of the aldehyde peaks in the third graph and the lack of them in
the fourth graph (which correlates to the ketone molecule A).
19)
The reaction of HBr with the original alkene will create a secondary cation adjacent to a
quaternary carbon. The methyl group can shift to make a tertiary carbocation, creating the
intermediate for the major product (F).
For product E, there will be no reaction. A weak nucleophile (EtOH) was added in a polar protic
solvent (EtOH). These conditions favor and SN1/E1 reaction. However, the alkyl halide is
primary, so it will not react under first order conditions.