Organic Chemistry Exam 4 Study Guide Important Information:
Professor’s Email: email@example.com
Class Website: organic.utep.edu/courses/2324
For this I am going to examples from the homework and Practice Exam 2014:
Draw the C6H14O isomer with 1H NMR (ppm) 3.4 (4H, t), 1.6 (4H, sex), 0.9 (6H, t). 1st step: We see there is NO O-H Bonding that would show up as (1H, broad s)
This lets us know we have an ether.
So we know we have a O-C-O now its best to work out. We know we have 6H in a triple bond, these number can be broke up so its best to do a methyl to a methylene x2. We also discuss several other topics like What is hammurabi’s code?
So as you can see there is two 3H to 2H which equal the 6H triple bond.
Now let’s do another example:
Draw the C6H14O isomer with 1H NMR (ppm) 1.53 (1H, broad s), 1.49 (4H, q), 1.1 (3H, s), 0.9 (6H, t).
We know there is is an alcohol, but look closely there is NOTHING about 1.53 meaning there is no hydrogens on the carbon that the oxygen is attached to.
We see that were is 3H, s. So lets put it on the carbon that will have NO hydrogen, that will make it show up in a singlet. Now as above with the 6H, t we can break that up to get:
We also discuss several other topics like Who is bryan stevenson?
From 2014 Exam
Match each 1H NMR listing in ppm to a compound on the right.
1. 2.6 (1 H, heptet), 2.1 (3H, s), 1.1 (6H, d) B a.
2. 2.4 (4H, q), 1.1 (6H, t) E b.
3. 9.5 (1H, s), 1.1 (9H, s) D c.
4. 2.4 (2H, t), 2.1 (3H, s), 1.6 (2H, sextet), 0.9 (3H, t) C d. 5. 9.8 (1H, t), 2.4 (2H, quartet), 1.6 (2H, quintet), 1.4 (2H, sextet), 0.9 (3H, t) A e.
Let’s start with 1 and go down the list. (These are confirmed correct answers):
1: There is 1H in a heptet meaning a carbon with one hydrogen has six hydrogen neighbors, this can only be B because there is no others with a carbon with only 1 hydrogen and two methyl groups. Don't forget about the age old question of What is the difference between aggregate and individual opinion?
2: There is 6H in a triplet, this can be broken up into 3H next to 2H x2. If you remember the homework this will be like an ether structure, but we can see from the options there is no ether, but E has the branching out.
3: We know we have 9H in a singlet this means they are bonded to a carbon with NO hydrogens, so we can concluded D every quickly. We also discuss several other topics like In what country has the highest frequency of thunderstorms?
4: We have 3H in a singlet so we need a carbon with no hydrogens, that narrows it down to B,C, and E but we have used B and E already. That means it is C.
5: From the 9.8(1H,t) we know we have 1 hydrogen VERY close to an oxygen or electronegative atom, that has two hydrogen neighbors, well we have already figured out the rest so that only leaves A left. Which is conforimed because there is a carbon with a single hydrogen, two hydrogen neighbors and is very close to an oxygen.
For this I am going to do an example from 2014 and 2015
(Attached is the flow chart, what salvador gives, and a nice chart from mastering organic) *At the end of this packet*
6. Match the following IR to the compound that it corresponds to.
OH N N
a. b. c. d. e.
First look for an alcohol which printing area is 3500-3300 (s,broad). There is a peak around that area that is sharp, but it is NOT broad. So we can rule out A. Now let’s look for an alkyne, that is 2260-2100(v) v means varies, there is a small weak peak in that area so we can conclude we may have an alkyene so that gives us B and E, and possibly C or D. Now lets look at aromatic which is 1600-1450 (v) we do have many peaks around this area. So we have a ring with double bonds so know we are down to C and E. Now let’s look for a nitrile which is 2260-2100(v) well this is the same as alkyne so this doesn’t help us. However at 3300 from an alkyne there will be a sharp line at 3300 which does occur so we know it is now E. Yes alkyne can have two different areas you can look at if you do not look at the full table you may have missed that.
Match each infrared listing (cm-1) to a compound on the right. An IR table is on the second sheet.
1. 2964 (s), 1119 (s) a.
2. 2934 (m), 1730 (s), 1126 (s) b. 3. 2986 (s, broad), 1715 (s) c.
4. 2983 (m), 2813 (m), 2716 (m), 1739 (s) d. 5. 3346 (s, broad), 2972 (s) e.
Let’s pick out the ones with an alcohol first alcohols show up as a broad s, so we know 3 and 5 are alcohols. (Going to start with those two)
3: 2986(s, broad), 1715(s) So we know this is either B or D, D is a carboxylic acid. So if we look at Salvador’s table there is no carboxylic acid listed (it is listed as acid), but there is alkane which B has. That says there should be a sharp peak between 2970-2850 that doesn’t happen in this IR spectra so we know it is NOT B leaving us with only D. The listing for acid is 3000-2500 (s,broad) however we have to go back to the alkane signals.
5: 3346 (s,broad), 2972(s) (Explained above B)
1: So looking at the table we can see esters, ethers, and alcohols (alcohols show up in two areas, but the (s,broad) is the best way to know if you have one) there is a singal 1300-1000 (s), so it can be A or E. A is an ester, so the second way to know you have an ester is a signal at 1750-1735 (s), we do not have this, so it is E, the ether. The 2964(s) can be concluded from alkane bonding.
2: We have 1126(s) as above we know this is in the range for ethers,esters and alcohols, there is only 1 option left for this type which is A. Also for esters there is a sharp signal between 1750-1735 which does occur in this IR spectra given.
4: Is C, an aldehyde. An aldehyde(The carbon oxygen double bond) shows up as 1740-1720(s) we have a signal at 1739 and it is sharp. Then we have three medium singals, the table shows for an aldehyde there is 2900 and 2700 that medium signals can show up, we have three singals around those two numbers.
Carbon NMR can be very easy once you get the hang of it so I will show you from 2014 and 2015. 2014(only doing 2)
Match each compound to the number of 13C NMR signals it should have? Answers may be repeated.
7. A a. 2
8. C b. 3
9. B c. 4
10. C d. 5
11. E e. not a.-d.
How I like to look at this is what parts of the molecule are equal to each other. In the first one there is two singals, the four methyls would be equal because they are both on a carbon with no hydrogens, the two carbons with no hydrogesnare equal as well giving 2 signals.
8 (or second one) is 4 look at it like the two methyls are “Equal”, the two methylenes are “equal”, but the bond bonded area is not because the molecule doesn’t have symmetry, so the double bond will have two singals, making 4.
Match each compound on the left to the number of 13C NMR signals it has. Answers may be repeated. 1. A a. 1
2. C b. 2
3. C c. 3
4. D d. 4
5. B e. not a.-d.
#1: All the signals are equal this is like with cyclohexane only has 1 signals.
The second one I am going to explain for this is number 4, as you may think there should only be 2 singals since three methyls, but look at the position, two are in front, one is in back, this will effect the NMR. The two in the front are equal, so that’s one signal. The two carbons that they are attached to are equal, so there is a second signals. The one in the back gets its on signal, and the carbon it is attached to gets its own as well.
Chiral Centers – Achiral, Chiral, Meso
Chiral centers are tetrahedral atoms (usually carbons) that have four different substituents. Each chiral center in a molecule will be either R or S.
Any carbons with a double bond or two/three hydrogens will not be a chiral center. Any carbons with two of the same group (two methyls) will not have a chiral center.
O NNN ON
We know from the practice exam this has 3 chiral centers, the arrow being one. I am going to show you why that is. Below is an Image I have highlighted the areas that cannot be chiral centers, areas that can I have found the priority
and drawn in any hydrogens for viewing. O
Meso, Achiral, Achiral
1. trans-1,4-dimethylcyclohexane b a. chiral 2. tran-1,3-dimethylcyclohexane a b. meso
3. cis-1,2-dimethylcyclohexane b c. achiral but not meso d. not a.-c.
A molecule is achiral if it is superimposable on its mirror image. Most achiral molecules do have a plane of symmetry or a center of symmetry. Achiral molecules that contain a stereocenter are called meso.
This is trans-1,4-dimethylcyclohexane.
This kind of molecule would be achiral because you can create
a mirror image of it, however there is a plane of symmetry, so
this is MESO!
This is trans-1,3-dimethylcyclohexane. As you can see
there are two forms of this molecule to be in the 1 and 3 position.
This molecule is chiral because there is two centers at the
methyls, the first one is S, and the second appears to be R but is really S because the methyl is in the back and the hydrogen is in the front. Trans-(1S,3S)-1,3-dimethylcyclohexane
This is cis-1,2-dimethylcycloehexane. First there is a plane of
symmetry in the molecule, meaning even if there is chiral centers
we know that it is meso right away.
Pink means there will not have the same biological or IR/NMR spectra.
Yellow means they will have the same.
Below is a flow chart to help you understand this:
(On Next Page)
Connectivity -- representation of the attachments of atoms in a molecule Examples of each one below:
Finding the Number of Steroisomers:
This is the 2n – meso structure rule meaning 2 to the number of isomers minus the number of meso structures (I call these fake chirals).
1,3-dibromopentane has how many stereoisomers in its family?
a. 1 b. 2 c. 3 d. 4 e. not a.-d.
First draw it out:
Well, how many isomers can be drawn from this? You could move the Br in the 1 position to the 3, but wait that would make it the 1 position. So there is 1 isomer of this structure. Now, does this have plane of symmetry? If it was butane it would but its pentane so no. Now plug it into the equation: 21-0=2 so the answer is B.
Optical Rotation Calculations- Taken from the Moodle
Given a 10 to 94 mixture of enantiomers (also called optical isomers), what is the enantiomeric excess (e.e.) of this mixture? Give two significant digits in your answer. No units are required. ��. �� =���������� − ����������
���������� + ����������
94 − 10
94 + 10 = .81
Given the specific rotation [α] of a mixture of enantiomers is -45° and the specific rotation of the pure enantiomers is ±75°, what is the percent enantiomeric excess (100 % x e.e.) of this mixture? Give two significant digits in your answer. No units are required.
100 ∗ (rotational
purity ) = percent e. e
100 ∗ (−45
−75) = 60%
Given a 61 % enantiomeric excess for a mixture of enantiomers and that the specific rotation of the pure enantiomers is ±979°. what is the absolute value of the specific rotation that should be observed for this mixture? Give two significant digits in your answer. The units are degrees but you do not have to put the units in your answer.
100 = ������ ������������ ∗ �������� ���������������������� = ���������������� ����������������
100 = .61 ∗ +979 = 597.17
That number would be rounded to 600 to be correct number of significant digits. Given a 21 to 33 mixture of enantiomers and that the specific rotation of the pure enantiomers is ±814°, what is the absolute value of the specific rotation that should be observed for this mixture? Give two
significant digits in your answer. The units are degrees but you do not have to put the units in your answer.
���������� − ����������
���������� + ����������∗ �������� ���������������������� = ���������������� ����������������
33 − 21
33 + 21 ∗ (���������� ���� ���� �������� ���� ����������? )814 = 180.8
Given the specific rotation [α] of a mixture of enantiomers is -5° and the specific rotation of the pure enantiomers is ±84°, what is the percentage of the major enantiomer of this mixture? Give two significant digits in your answer. No units are required.
100 ∗ (rotational
purity ) =percent e. e + 1
100 ∗ (−5
−84) =. 059 + 1
2∗ 100 = 52.9% ���� �������������� ���� 53%
Given a 11 to 88 mixture of enantiomers by a method called Chiral HPLC (High Performance, Pressure or Price Liquid Chromatography) with an observed specific rotation of 43°, what would be the specific rotation of the pure (+)-enantiomer? Give two significant digits in your answer. The units are degrees but you do not have to put the units in your answer.
���������� − ����������
���������� + ����������= ��. ��
��. ��= ���������������� ���� �������� (+)�������������������� 88 − 11
88 + 11 = .78 = (43
. 78) = 55
H3 H6 H7
H1 and H11 a a. equivalent
H3 and H10 a b. enantiotopic
H4 and H8 b c. diastereotopic
H4 and H9 a d. constitutional heterotopic
H5 and H6 d e. not a.-d.
Let’s take H1 and H11- First to note they are in the same plane. So this makes it a little easier. Pretend to replace H1 with say a Br and then H11 with Br is the name still the same? Yes it is. The molecule if the Br was at H1 or H11 would be the same so they are equivalent.
H3 and H10 – are equivalent as well, same plane, same name.
H4 and H8- First the symmetry is equivalent so they could be homotopic (meaning the same or equivalent) or enantiotopic. Now picture putting a Br in the H4 and H8 position if you “mirrored” the image would they still be the same? No they wouldn’t because they Br in H4 that is in front, would be behind for H8 if we mirrored the image, so they are enantiotopic.
H4 and H9- the symmetry is equivalent because you can cut it in half and have equal parts. Now we know we have homotopic (equivalent) or enantiotopic. So we mirrored the image would H4 turn into H9? Yes it would the mirror of the molecule would make it the same, so they are homotopic.
H5 and H6- If you cut this in half it would be equal so we know we are dealing with diastereotopic or constitutional heterotopic. Now do they have the same bonding connectivity? If you recall the definition of connectivity it means do they have the same attachment? No they don’t one is in the front and one is in the back, so this means them constitutional heterotopic.
D and L
The convention that is used to designate the configurations of chiral carbons of naturally occurring compounds is called the D and L convention or system. To use it, we view the molecule of interest according to the following rules:
1. The main carbon chain is oriented vertically with the lowest numbered carbon at the top. The numbering used for this purpose must follow the IUPAC rules:
2. Next, the structure must be arranged at the particular chiral carbon whose configuration is to be assigned so the horizontal bonds to that carbon extend toward you and the vertical bonds extend away from you.
3. Now the relative positions of the substituents on the horizontal bonds at the chiral centers are examined. If the main substituent is on the left of the main chain, the l configuration is assigned; if this substituent is on the right, the d configuration is assigned.
For example, the two configurations of the amino acid, alanine, would be represented in perspective or projection as 15 and 16. The carboxyl carbon is Cl and is placed at the top. The substituents at the chiral carbon connected to the horizontal bonds are amino (—NH2) and hydrogen. The amino substituent is taken to be the main substituent; when this is on the left the acid has the l configuration, and when it is on the right, the d configuration. All of the amino acids that occur in natural proteins have been shown to have the l configuration.
Glyceraldehyde, CH2OHCHOHCHO, which has one chiral carbon bonded to an aldehyde function, hydrogen, hydroxyl, and hydroxymethyl (CH2OH), is of special interest as the simplest chiral prototype of sugars (carbohydrates). Perspective views and Fischer projections of the d and l forms correspond to 17 and 18, respectively, where the carbon of the aldehyde function (—CH=0) is Cl: