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Example Questions
Example Question #82 : Stereochemistry
Which of these describes an epimer?
Two or more stereoisomers of a compound which have different configurations at one or more (but not all) of the equivalent stereocenters and are not mirror images of each other
There are no such things as epimers.
Stereoisomers that are not superimposable mirror images
One of a pair of stereoisomers, which differ in configuration at only one stereogenic center
One of a pair of stereoisomers, which differ in configuration at only one stereogenic center
In organic chemistry, an epimer refers to one of a pair of stereoisomers, which differ in configuration at only one stereogenic center. Any other stereogenic centers in the compounds are the same in each one. The sugars glucose and galactose are epimers.
Diastereomerism occurs when two or more stereoisomers of a compound have different configurations at one or more, but not all, of the equivalent stereocenters. These stereoisomers are not mirror images of each other. D-erythrose and D-threose are diastereomers.
Note: Epimers are diastereomers that contain more than one stereocenter but differ from each other in the configuration at ONLY one stereocenter. Diastereomers can differ at more than one stereocenter, but not all of them.
Enantiomers are stereoisomers that are non-superimposable mirror images. This means that the molecules cannot be placed on top of one another and give the same molecule. D-threose and L-threose are enantiomers.
Example Question #1 : Help With Comformers
The molecules shown below are best described as __________.
enantiomers
isomers
epimers
diastereomers
isomers
The molecules in this problem are isomers because they each have unique configurations and do not share the same funcitonal groups at the same carbon positions. Enantiomers are reflections of each other. Diastereomers are stereoisomers that differ at one or more stereocenters, while epimers are stereoisomers that differ at only one stereocenter.
Example Question #2 : Help With Comformers
In the most stable chair conformation of the following Haworth projection, the bromine at carbon 1, the hydroxide at carbon 2, and the hydroxide at carbon 4 are __________, __________, and __________, respectively.
axial up . . . equatorial down . . . equatorial up
equatorial up . . . equatorial down . . . axial up
axial down . . . equatorial up . . . equatorial down
equatorial down . . . equatorial up . . . axial down
equatorial up . . . axial down . . . equatorial up
equatorial up . . . equatorial down . . . axial up
Based on the Haworth projection given, the relative orientations of the three substituents are bromine up, hydroxide: down, hydroxide: up. In the most stable chair conformation, the largest substituents are oriented equatorial. If bromine is oriented equatorial up, the hydroxide at carbon 2 is equatorial down and the hydroxide at carbon 4 is axial up.
Example Question #3 : Help With Comformers
What is the relationship between these compounds?
Conformational isomers
Constitutional isomers
Meso
Enantiomers
Diastereomers
Conformational isomers
The two given molecules are conformational isomers. The molecules are superimposable by horizontally rotating either structure 180 degrees and rotating the bond between carbons 2 and 3. Both enantiomers and diastereomers are non-superimposable and constitutional isomers must have non-configurational differences in structure. There is no internal plane of symmetry, thus neither compound can be meso.
Example Question #4 : Help With Comformers
Determine the most stable chair conformation corresponding to the given Haworth projection.
I
III
IV
II
More than one of these
I
The most favorable chair conformation of a six-member ring is that in which the greatest number of large substituents are oriented equatorially. Conformation I corresponds to the given Haworth projection and orients all three large substituents in equatorial positions, making it the least energetic (most favorable) conformation.
Example Question #5 : Help With Comformers
How many constitutional isomers exist for the general molecular formula ?
Four
Three
Five
Six
Four
There are four unique constitutional isomers possible for the given formula; meaning that they may not be made identical by conformational changes. They are illustrated as follows:
Example Question #6 : Help With Comformers
In general, what is the most stable orientation in a Newman projection?
Eclipsed
Totally eclipsed
Anti
They are all equally stable
Gauche
Anti
The anti conformation is most stable in a Newman projection because this is the form where the two largest functional groups are facing opposite directions and are furthest away from each other. However, note that in rare cases, such as in 1,2-ethandiol, the gauche conformation is more stable due to intramolecular hydrogen bonding. All other conformations have the functional groups closer to each other causing some repulsion and instability. The least stable of them all is the eclipsed conformation because this is the form where the functional groups are closest to each other.
Example Question #7 : Help With Comformers
In regards to the Fischer projection shown, how are the hydroxide groups oriented?
The right one is going into the screen and the left one is coming out
Both are coming out of the screen
The left one is going into the screen and the right one is coming out
Both are going into the screen
Both are lying flat in the plane of the screen
Both are coming out of the screen
In a Fischer projection, the groups on the left and right are coming out of the screen and towards the viewer. On the other hand, the top and bottom groups are going into the screen away from the viewer. One way to help you remember this is to think of the Fischer projection as a skeleton wearing one or more bowties (the skeleton is vertical and its bonds are drawn as dashes, which are going into the plane of the page/screen, and the horizontal bonds are drawn as wedges, which are coming out of the plane of the page/screen).
Example Question #1 : Organic Oxidizing Agents
What is the product when 1-propanol is treated with potassium permanganate?
Propanoic acid
None of these
No reaction occurs
1-propanone
Propanal
Propanoic acid
Potassium permanganate is a strong oxidizing agent. It can convert secondary alcohols to ketones. It can also convert primary alcohols to carboxylic acids. 1-propanol has a hydroxy group on carbon 1, so it is primary; thus it will be converted to propanoic acid.
Example Question #1 : Using Pcc
What is the product when 2-butanol is treated with PCC?
None of these
2-butanol
1-butanone
2-butanone
1-butanol
2-butanone
PCC is an oxidizing agent. It converts alcohols to ketones, but is not strong enough to convert primary alcohols to carboxylic acids. 2-butanol has a hydroxy group on its carbon 2. The addition of PCC will convert this hydroxy group into a carbonyl, producing 2-butanone.
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