Download Answers, Problem Set 12 (full)... “2,4

Chemistry 122
Mines, Spring 2012
Answers, Problem Set 12 (full)
1. NT1.
(a) What are hydrocarbons? (b) What are unsaturated hydrocarbons and what are saturated hydrocarbons?
Answers: (a) Hydrocarbons are substances that contain only C atoms and H atoms.
(b) A saturated hydrocarbon is one that contains as many H’s per carbon as possible, and so you
could say that it is saturated “with H’s”. It contains no double or triple bonds. Once a double
bond or triple bond is formed, there will be less H’s because C only forms 4 bonds before its
valence shell becomes full. So unsaturated hydrocarbons are those that have at least one
double or triple bond in them--they do not have the maximum number of H's per C atom.
2. 20.44
Name each alkane (provide some reasoning!) [Structures not shown in this key.]
(a) 2-methylpropane The longest chain of carbons is three—hence “propane”. No matter which
way you number the carbons, a methyl group is attached to position “2”—hence “2-methyl”.
(b) 2,4-dimethylhexane Note: This is not 1,2,4-trimethylpentane or 2,4,5-trimethylpentane, because
the longest chain of carbons is six, not five (remember that there can be no “1-methyls”!)—hence
hexane. The closest methyl group to an end is the one on the left (two away from the end,
whereas the one on the right is three away)—hence, the numbering starts from the left. Thus, a
methyl at position 2 and another at position 4—hence “2,4-dimethyl”.
(c) 2,2,3,3-tetramethylbutane The longest chain is four carbons—hence “butane”. There are four
distinct methyl groups, two at position 2 and two at position 3—hence “2,2,3,3-tetramethyl”.
(d) 5-ethyl-2,4-dimethyloctane The longest chain is eight carbons—hence “octane”. The methyl at
the left is two away from the left end, whereas the ethyl is four away from the right end—hence
the numbering starts from the left. Two methyls at positions 2 and 4  “2,4-dimethyl” and one
ethyl at position 5  5-ethyl. since E is earlier in the alphabet than M, the 5-ethyl comes first in
the name (remember, the “di” or “tri” prefixes are not considered in this alphabetization ordering).
(e) 5-methylnonane Don't be fooled! The longest chain of carbons is nine, and there is only ONE
group attached to the chain--a methyl at position 5. That is, this is not 1,4,7-trimethylheptane
(remember, there can be no “1-methyls”!)
(f) 3-ethyl-3-methyloctane You can't have any 2-ethyls! That just extends the longest unbroken
chain of carbons. So here the longest chain is actually 8, and there is a methyl group and an
ethyl group both at position 3. E before M as in (d).
3. NT2.
The following names are incorrect. What is wrong with each, and what are the correct names?
CH3
(a)
CH3CCH2CH2CH3
Answer: The longest unbroken chain of C's is six, not five! What
"appears" to be C1 in this structure is actually a methyl
group attached to position 3! This is 3,3-dimethylhexane.
CH2CH3
4-ethyl-4-methlypentane
(b)
CH2CH3
CH3CHCH2CHCH2CH3
CH3
5-ethyl-3-methylhexane
Answer: The longest unbroken string of C's is seven, not six! What
"appears" to be C1 in this structure is actually a methyl group
attached to position 3 (or 5)! This is 3,5-dimethylheptane.
(Side note: Even if this could somehow be named as an "ethyl-methyl"
compound, it would 2-ethyl-4-methylhexane! That is, the given name in
this problem is also incorrect in how it numbers the carbons. The lowest
number possible for the first branch point carbon is desired--here, "2" (if
counting from the left) instead of "3" (if counting from the right).
PS12-1
Answer Key, Problem Set 12
4. NT3.
Draw all the structural isomers for C8H18 that have the root name "hexane" (longest carbon chain is 6). Name
the structural isomers.
Start by drawing a chain of 6 carbons, and then realize that you have to add "two more" to end up
with an isomer of octane. See if you can generate them in a methodical way, making sure not to
duplicate!
NOTE: I will leave out the H's on these structures just to save time generating this key.
Structure
Name
C
C
C
C
C
C
C
C
C
C
C
C
2,2-dimethylhexane
C
C
C
C
C
C
C
C
C
2,3-dimethylhexane
C
C
C
C
C
C
2,4-dimethylhexane
C
C
2,5-dimethylhexane
C
C
C
C
C
C
C
C
C
C
C
C
C
C
3,3-dimethylhexane
C
C
C
5. 20.37
C
C
C
C
C
C
C
C
C
C
3,4-dimethylhexane
3-ethylhexane
Write structural formulas for each of the nine structural isomers of heptane. Also give IUPAC names for each.
Reasoning / Strategy: I find it much easier to draw the carbon frameworks first when generating
isomers, and then I add in the H's, as needed, afterwards (if asked to). So
I generated all nine isomers in the middle first with just frameworks, and
then wrote the semi-condensed structural formulas to the right.
Tip #1: Start with the longest unbroken chain first, then methodically move to smaller "longest" chain
derivatives, generating as many as you can with each of those before moving on. There is one
heptane derivative, two hexane derivatives, five pentane derivatives, and one butane derivative (see
below). Add the names to the after you generate the structures.
Tip #2: Even if you are not asked for the names, if you have time it is always a good idea to name them,
because if you end up with the same name (and you’ve named them correctly!), then you have a
“duplicate” structure (not an isomer)!
Tip #3: You cannot have “1-methyl” or “2-ethyl” type of compounds! If you think that you do, you have not
found the longest unbroken string of C atoms!!
PS12-2
Answer Key, Problem Set 12
Tip #4: In larger molecules, you can have an ethyl and a methyl group attached to the same carbon, or two
ethyl groups attached to the same carbon, but here that doesn’t “work” because you don’t have
enough carbons (i.e., your longest chain wouldn’t be what you thought it was).
n-heptane
2-methylhexane
3-methylhexane
2,2-dimethylpentane
2,3-dimethylpentane
2,4-dimethylpentane
3,3-dimethylpentane
3-ethylpentane
2,2,3-trimethylbutane
6. NT4.
(a)
Draw structures corresponding to the following names:
OH
CH3
CH3 CCH 2CHCH 3
CH2CH3
(b)
HOCH2CH2CH2CH2CCH2CH3
CH 3
2,4-dimethyl-2-pentanol
CH2CH3
5,5-diethyl-1-heptanol
PS12-3
(c)
CH2CH3
CH3CH2CCH2CH2CH3
OH
3-ethyl-3-hexanol
Answer Key, Problem Set 12
7. NT5.
What is wrong with the following names? Give the correct name for each compound.
(a) 2-ethylpropane
(b) 5-iodo-5,6-dimethylhexane
(c) 2-bromo-3-butanol
Answers:
(a) The longest chain of C’s is 4, not 3. The correct name is: 2-methylbutane
(b) The longest chain of C’s is 7, not 6. The numbering starts from right, not left. The correct
name is:
3-iodo-3-methylheptane
(c) The numbering of the carbons is incorrect. The end to which the –OH group is closest is
where the numbering should be begin. The correct name is: 3-bromo-2-butanol
Strategy / Reasoning:
1) Draw the compound the way the (incorrect) name would suggest.
2) Analyze the structure to see the error(s) in the provided name (and describe).
3) Determine the correct name
Execution of Strategy:
(a) propane  3-carbon chain; 2-ethyl  ethyl group at position 2:
C—C—C
C
You can see that by adding an ethyl at position 2, the chain has been
elongated by one carbon. That is, the longest chain of carbons is actually
four and that left-hand carbon is actually a methyl group attached to position
2. Hence “2-methylbutane” is the correct name.
C
(b) hexane  6-carbon chain; 5-iodo-5,6-dimethyl  I and CH3 at position 5, and CH3 at
position 6:
I
You can see that the longest chain of carbons is actually seven,
C—C—C—C—C—C
and that the iodo and methyl groups are on position 3. I comes
before M in the alphabet. Hence “3-iodo-3-methylheptane”.
C C
(c) 3-butanol  4-carbon chain with an –OH at position 3. 2bromo  Br at position 2. Thus (see right):
Br
An alcohol determines the numbering of the carbon chain, with the end nearest
C—C—C—C
the –OH group being C1. Thus, the actual name is “3-bromo-2-butanol”.
OH
8. NT6.
Copy this structural formula on your paper and circle and identify the functional groups. Be
careful not to count a functional group “twice” (i.e., even though a carboxylic acid has an –OH group in it, you do not
refer to that –OH group as an alcohol).
ketone
aldehyde
O
carboxylic
acid
HO
O
C
CH2
C
O
CH
C
H
OH
alcohol
9. NT7.
Using the same molecule as in the previous problem, note which hydrogen is the “acidic hydrogen” and then
+
draw the structural formula of the ion that would be formed if this proton (H ) were to dissociate from the molecule
(show the formal charge on the structure).
acidic
hydrogen
-1 charge is formed on O when
+
an H is released
O
O
O
HO C
CH2 C
CH C
O
H
O
O
O
C
CH2 C
CH C
OH
OH
PS12-4
H
+
H+
Answer Key, Problem Set 12
10. NT8.
Copy this structural formula (which is the drug oleandrin, a cardiac glycoside) on your paper and circle and
identify the functional groups.
ester
alcohol
ether
ester
ether
alcohol
11. NT9
Write equations showing how you could prepare each of the following esters from the appropriate alcohols and
carboxylic acids.
O CH3
(a) CH 3CH 2CH2 CH 2 COCH 3
O
CH 3CH 2CHCOCH
O
CH3
(b)
CH3
CH3CO
(c)
Reasoning / Approach: I find it easiest to find the bond that is broken when an ester is hydrolyzed,
the C-O bond of the ester group (which I have shown with a small arrow), and then go from there
to "see" which acid and alcohol would form--these will be the same two molecules that would need
to be condensed to make the ester. After imagining "cleaving" the appropriate C-O bond (so that
you get two "fragment" structures), an -OH group is added to the C=O (from OH2 "attack" on the
C=O group), resulting in an acid group, and an H is added to the -O-R, resulting in an alcohol
group. MOST OF THE ATOMS DO NOT CHANGE! Remember, no C-C bonds are broken during
condensation or hydrolysis!
To save time, I'm not going to draw the answers to this question myself (I’m using the McMurry and
Fay solution manual “answers”, with my notes and rectangles around the alcohol-derived part of the
ester added). See the next question to see "my" work (showing the actual hydrolysis reaction).
(*I would have written:
HOCH3 )
HOCH
PS12-5
Answer Key, Problem Set 12
12. NT10.
(a) Is the following statement true or false? Comment by explaining what the statement means in a bit more
detail: “Whenever there is a dehydration (condensation) reaction, one can imagine a corresponding hydrolysis
reaction.”
The statement is TRUE because hydrolysis is simply the reverse of condensation. So if you know
any specific hydrolysis or condensation reaction, you know the other. Condensation refers to the
formation of one molecule from two, with the release of water; hydrolysis refers to the breaking up of
one molecule into two, by “using up” one molecule of water.
(b) With your answer to (a) in mind, write the equation representing the hydrolysis reaction that corresponds to part (c) of
the prior problem (NT9).
Hydrolysis of the ester formed in NT5(c) (this is just the reverse of the reaction shown above in the
the answer to NT5, see above!):
bond broken
O
CH3
C
O
+
13. NT11.
O
OH +
C
CH3
HO
H2O
(a) Draw a structural formula for, and state the molecular formula for the compound having exactly 5 carbon
atoms and exactly two double bonds and 1 triple bond.
Answer: In fiddling around with structures, hopefully you figured out that you CAN put two double
bonds next to one another, but you CANNOT put a double bond next to a triple bond (5 bonds
to C!). This basically "forces" the triple bond to be at the end (because with exactly 5 carbons,
there are only four "bonds" to work with--meaning there are only two "positions"--an "end"
position and an "inner" position. That gives:
C
C
C
C
C
Using the "4 bonds to carbon" rule, you should then get:
H
H
C
C
C
C
C
H
H
So the molecular formula for this compound is C5H4.
(b) For the compound with the following condensed structural formula, how many double bonds and
triple bonds are in a molecule? CHCCH2CHCHCCH (Hint: draw the full structural formula, and
use the “4 bonds to carbon” idea.
Strategy: Start by writing the structural formula, realizing that each carbon is bonded to the next
carbon in the formula (this is a hydrocarbon), and the H's after each C "belong" to it. You
should get something like this:
H
C
C
H
H
H
C
C
C
H
PS12-6
H
C
C
Answer Key, Problem Set 12
Then make double or triple bonds wherever necessary to satisfy the "four bonds to C" rule.
You should get:
H
C
C
H
H
H
C
C
C
H
C
C
H
Thus, there are TWO triple bonds and ONE double bond.
14. NT12.
(a) Which two functional groups are essentially always present in any amino acid (Hint: think about the name
for a moment)? (b) Draw the general structure of an amino acid. (c) Which part of an amino acid varies in
different amino acids? (d) What molecule results from the condensation of two amino acids? (e) A polymer of
amino acids that folds up into a specific shape and has a specific biological function is called a ___________.
Answers:
(a) an “amine” and a “carboxylic acid” are present in essentially any amino acid.
(b) Remember to connect the –NH2 group to the –COOH group by putting a single carbon atom in
between the two:
H2N
H
O
C
C
OH
R
(c) R = the group that varies from amino acid to amino acid. It is called the “side chain” or “Rgroup”.
(d) A “dipeptide”
(e) “protein”
15. NT13. (a) What functional group is found in all triglycerides? (b) How many of these groups are
there? (Hint: think about the name for a moment.) (c) What are the common names for the
two basic types of triglycerides, and what is the difference between the two types? (d) What
are the products of hydrolysis of a triglyceride?
Answers: (a) ester;
(b) three;
(c) Fats and oils are what I was looking for here. For practical purposes, the difference is that fats
are solids at room temperature and oils are liquids at room temperature. (Note: The word “oil” has
other meanings that I’m not talking about here. For example, crude oil that is pumped from the ground as a natural
resource is not a triglyceride. I’m only talking about oils that are used in cooking for example, corn oil, vegetable oil,
olive oil, etc.)
Someone pointed out to me that another possible answer to this question could be “saturated”
and “unsaturated”. The differences between these two are that saturated fats or oils contain
fatty acids “tails” that have no double bonds in them, whereas unsaturated fats or oils contain
tails with at least one double bond. Typically the more unsaturated the tails, the lower the
melting point of the triglyceride, although one cannot simply say that all fats are saturated and
all oils are unsaturated because fats and oils are not pure substances, but rather are mixtures
of different triglycerides, some of which may be saturated and some of which may be
unsaturated. Oils tend to have a higher percentage of poly-unsaturated tails whereas fats
tend to have a higher percentage of saturated tails.
PS12-7
Answer Key, Problem Set 12
(d) glycerol and three fatty acids (see below)
glycerol
“part”
fatty acid
“parts” (three)
O
CH2
O
C
O
CH
O
C
O
CH2
O
C
[though they are
not actually fatty
acids until the
molecule is
hydrolyzed!]
hydrolysis
O
HO
CH2 OH
glycerol
O
HO
CH OH
CH2 OH
C
O
HO
16. NT14.
Fatty
acids
C
C
Which of the compounds in problems 20.53 and 20.54 (as well as the compound below (part e) exhibit cistrans isomerism? Explain. (Do NOT name these compounds!)
NOTE: The key to this kind of problem is determining which groups are bonded to each of the two
carbon atoms of the double bond. If the molecule can be abbreviated C(A)(B)=C(Y)(Z), then
the molecule with exhibit cis-trans isomerism unless either: A = B OR Y = Z. (It is irrelevant
whether A = Y or Z B = Y or Z!!)
20.53 (a) CH2=CH-CH2-CH3
NO (because two H’s are bonded to the 1st C; i.e., A = B)
(b) CH3-C(CH3)H-C(CH3)=CH-CH3 YES (b/c CH3 ≠ CH3-C(CH3)H
AND
H ≠ CH3)
st
(c) CH2=CH-C(R)H-CH2-CH2-CH3 NO (b/c two H’s are bonded to the 1 C; i.e., A = B)
(d) CH3-CH(CH3)-CH=C(R)CH3
20.54 (a) CH3-CH2-CH=CH-CH2-CH3
(b) CH3-C(CH3)H-CH=CH-CH3
YES (b/c H ≠ CH3-CH(CH3)
YES (b/c CH3-CH2 ≠ H
AND
R ≠ CH3)
AND H ≠ CH2-CH3)
YES (b/c CH3-C(CH3) ≠ H
AND H ≠ CH3)
(c) CH3-C(CH3)H-CH=C(CH2CH3)-C(CH3)H-CH3
YES (b/c CH3-C(CH3)H ≠ H
AND CH2CH3 ≠ C(CH3)H-CH3)
(d) CH3-C(CH3)2-CH=C(CH3)-CH2-CH3 YES (b/c CH3-C(CH3)2 ≠ H
AND CH3 ≠ CH2-CH3)
CH2CH3
(e)
|
CH3CH2CH==C—CH2CH3
NO (b/c on the right C, Y = Z (i.e., CH2CH3 = CH2CH3))
PS12-8
Answer Key, Problem Set 12
17. 20.42.
Determine whether the molecules in each pair are the same or enantiomers.
NOTE: A tetrahedral carbon is chiral if and only if all four groups are different. Note that this issue is fundamentally
different than the issue of cis-trans isomerism because there is free rotation around a single bond. In fact, the way to
solve this kind of enantiomer (optical isomer) problem is to rotate one of the structures until two atoms are in the same
exact positions, and then you can see if the other two “overlap” (i.e., are “superimposable” and thus “the same”) or if
they are in swapped positions (in which case the two structures are enantiomers / optical isomers.
Answers:
(a) Since two of the groups are methyl groups (i.e., the same), the compounds are “the same”.
(There is a plane of symmetry, so the molecule is achiral.)
(b) These two are not superimposable (which can be seen via a rotation), and thus they are
enantiomers.
(c) Although all four groups bonded to C are different here (i.e., the molecule is chiral), these two
structures are superimposable (which can be seen via a rotation), and thus they are “the same”
(not enantiomers).
I will try to put in structures for these compounds at some point (I didn’t make it on April 30th!) in order
to help you see how to “rotate” these structures.
PS12-9