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Unformatted text preview: Copyright 2010 James K Whitesell
10-1 You might start this exercise by drawing all of the isomers of C7H16 of which
there are nine: Pick one with both secondary and tertiary carbon atoms and simply add an OH group to
OH 10-2 Through this exercise you will hopefully equate light with energy. Here we
need light of wavelength that corresponds to the strength of the Cl—Cl bond (58
kcal/mole). The relationship between wavelength and energy is:
λ = 28,600/E and thus in this case, λ = 28,600/58 = 493 nm. The book is wrong is
assigning this to the ultraviolet/visible range---it is well into the visible which starts at
about 390 and spans to 750 nm for the typical human eye.
It is useful to store the correspondence of 71.5 kcal/mole with 400 nm. Simple math
can then be used to determine the energy of other wavelengths (e.g., 800 nm
corresponds to 35.75 kcal/mole, twice the wavelength half the energy).
10-3 We determine δ by dividing the observed frequency by the frequency of the
machine, as done in the answer in the book to this exercise.
10-4 The primary factor in the frequency for protons is electron density. The presence
of nearby more electronegative atoms reduces electron density and shifts the signal to
lower field. There are two different types of hydrogens in chloromethyl methyl ether
(ClCH2OCH3) as it is normally called, those on the methyl carbon where the carbon
is attached to an oxygen and those on the other carbon which has both an oxygen and
a chlorine attached. The latter are observed at lower field.
This is an interesting molecule to pick as an example. Bis-chloromethyl methyl ether
(ClCH2OCH2Cl) is always a contaminant in chloromethyl methyl ether. The former is a
very highly carcinogenic compound. It is presumed that each of the carbons bearing
chlorine can undergo substitution and in living systems, these substitutions can result in
cross linking between strands of double helix DNA. 10-5 You were not expected to come up with exact numbers, as indicated in the back
of the book, but just ranges. It would be worthwhile for you to look, for example, at
all of the methyl groups in the (a)-(c) and see how they vary depending on to what the
carbon is attached.
10-6 Worked out in the textbook.
10-7 Both of these molecules are symmetrical and thus have only one type of
hydrogen. Each would result in a spectrum with a single signal.
10-8 The answer in the back of the book is adequate.
10-9 This exercise is worked out on page 403 as "Working with Concepts".
10-10 Every carbon atom is unique and therefore, every set of hydrogen atoms on the
carbon atoms is also unique.
10-11 Two chlorine atoms on a cyclopropane ring can either be on the same or
adjacent carbon atoms. In the later case, they can be cis or trans to each other, thus
there are three dichlorocyclopropanes.
Cl Cl Cl Cl For 1,1-dichlorocyclopropane, all of the hydrogen atoms are identical and so we predict a
single signal. The 1,2-isomers can be distinguished from the 1,2-isomer because each has
two different types of hydrogens, those on carbon bearing chlorine and the two of the
methylene group. The cis isomer can be distinguished from the trans as the two hydrogen
atoms of the methylene group are identical in the trans isomer but not in the cis (in the
cis, one hydrogen atom is on the same side and thus cis to the chlorine atoms whereas the
other is trans to the chlorines. 10-12 (a) Diethylether has two identical ethyl groups with the methylene group of
each shifted downfield by the oxygen. The methyl groups appear as a triplet as a
result of the adjacent two hydrogens and the methylene groups are split into identical
quartets by the methyl groups. (b) Here we can see the downfield shift effect of
bromine. The central methylene has four identical adjacent hydrogen atoms and thus
appears as a quintet whereas the other two methylenes are split into triplets by the pair
of hydrogens on the central carbon atom. (c) There are two identical methyl groups
lacking adjacent hydrogens and thus appear as a singlet. There is an ethyl group with
the expected quartet and triplet, and finally a hydrogen on oxygen. Should should be
aware that the position of an OH hydrogen varies greatly with the solution pH.
1 ,1 2
3 .3 8 Br Cl OH 2 .3 4 O Br
0 .9 0 3 .5 3
1 .1 4 1.48 Cl
Cl 5 .5 8 3.71 (d) Here we see the effect of two chlorines versus one. Note that the effect is not
additive and the second chlorine does not result in as much downfield shift as the
first. (Ponder why this might be.)
10-13 This exercise is worked out on page 412 as "Working with Concepts".
10-14 For small molecules of this size, we general see unique signals for each set of
hydrogen atoms. Thus, while it is possible that a signal for 3 H might be the overlap of a
CH and a CH2 group, it is much more likely that it is the result of a CH3 group. Two
solve a problem of this type, we start by listing the groups that we deduce are present
from the spectrum: CH2 CH3 CH3 CH3 OH (the broad singlet). We compare thus
with the molecular formula and that we are missing a carbon atom and thus there must be
one carbon present with no hydrogen atoms. Next we examine the splitting information
and see that we have two methyl groups with no adjacent hydrogens and that the third
methyl is adjacent to a methylene that in turn is split into a quartet by the methyl. Thus,
we have four groups, two methyls, one ethyl, and one OH to attach to our quaternary
carbon atom. Only one structure results from this analysis:
OH 10-15 This exercise is worked out on page 420 as "Working with Concepts". 10-16 (a) There are a total of four adjacent hydrogen atoms but they are not
identical. It is not possible to predict if the selected methylene group will appear as a
triplet of triplets where each of the adjacent methylene groups have different
splitting constants or as a quintet if the splitting constants are not sufficiently
different to be resolved. The pattern on the left, below, results from two quite
different splitting constants, that on the right from identical constants. Note that the
constants on the left were selected so that none of the peaks would overlap.
Increasing the lower splitting constant will result in overlap of some of the peaks.
Beware---complex patterns do always lend themselves to a simple peak count. 10-17 It is best not to over analysis an exercise such as this. We could go through
each structure and predict what we would expect in each case. However, the exercise
asks simple if we can distinguish these structural isomers base upon splitting. We note
that only two of the structures have methyl groups and one will be a doublet and the
other a triplet. We really do not need splitting to distinguish the third but we can base
on it is the only isomer with a methylene group bearing chlorine with two adjacent
Cl Cl Cl Cl Cl 10-18 One of the benefits of C13 spectroscopy is how readily it reveals symmetry.
In 1-bromopropane, we have three unique carbon atoms so we expect three signals
with one significantly downfield as a result of the attached bromine atom.
10-19 (a) Three identical methyl groups, a quaternary carbon, and a methylene
bearing oxygen. (b) There are eight methylene groups four that are identical to each
other and directly connected to the bridgehead carbons and the other four which are
identical to each other. The bridgehead atoms are unique from the methylene groups
but identical to each other. Thus, we predict three signals. (c) This is a nasty
question as the stereochemistry shown requires the central ring to be a boat. There is
a mirrow plane that divides the molecule left to right but otherwise there are seven
unique carbon atoms. (d) All of the six methylene groups are identical as are the two bridgehead carbons:
10-20 This exercise is worked out on page 426 as "Working with Concepts".
10-21 The key here is that reduction of the aldehyde functional group makes the
functional groups on the two ends of these molecules functionally identical. However,
this a very complex stereochemical situation, best view in more typical
representations that with Fischer projections. Below are the two products of
HO H HO H
A HO H
OH R H OH HO OH
R S R B H OH H OH Note that the product derived from A is a meso stereoisomer and is shown in the
conformation that has a mirror plane. (The central carbon atom IS a center of chirality as
it has two different groups attached by virtue of stereochemistry, R on the left and S on
the right---R has higher priority than does S). The diastereomer on the right has no such
mirror symmetry and the two R centers of chirality are different. Note that here the
central carbon is not a center of chirality---if you switch the OH and H, you can rotate
180 degrees left to right to return to where you started. 10-22 Because DEPT spectra specifically identify carbons as primary, secondary and
tertiary, it would be trivial to differentiate these two isomers as only A has methyl
groups and only B has methylene groups. ...
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