Table of contents |
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Representation of Optical Isomers |
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Assigning R-S Configuration in Wedge Dash |
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Calculating the Total no. of Optically Active Isomers |
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Diastereoisomers through Fischer Projection |
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A Wedge Dash, the most popular three-dimensional depiction of a molecule on a two-dimensional surface is in projection (paper). This type of representation is typically used for molecules with chiral centres. This type of representation employs three different types of lines.
Fischer projections are best used to represent the straight-chain structures of monosaccharides and some amino acids. They represent structural forms that allow one to convey valuable stereochemical information by drawing 3D molecules as flat structures.
Fischer Projection
How to Draw Fischer Projection?
In a Fischer projection, the longest chain is drawn vertically. The horizontal lines indicate the bonds with hydrogen, hydroxyl, and amino groups. The four bonds to a chiral carbon make a cross, with the carbon atom at the intersection of the horizontal and vertical lines. The following steps can be employed for an aldohexose.
Aldohexose
Step 1: Arrange the molecule so that the chiral carbons and the longest continuous chain are in a vertical line. The aldehyde group representing carbon 1 goes at the top.
Step 2: Draw horizontal lines to make crosses at C-2, C-3, C-4, and C-5.
Step 3: Put the OH groups on the exact side of the cross.
Step 4: Remove C-2, C-3, C-4, and C-5, and the Fischer projection is obtained.
Fischer Projection Rules
The following rules should be kept in mind while working with Fischer projection.
Examples:
Q.1. Write the Fischer projection of CH3CH(OH)COOH
Sol.
Q.2 Write Fischer's projection of
Sol.
(i) Place a higher-priority carbon-containing functional group on top of the vertical line.
(ii) Arrange another group according to its clockwise or anti-clockwise position w.r.t. group on the top.
Q.3. Convert the following wedge-dash figure in the Fischer projection.Sol.
Q.4. Out of the two cross lines in the representation of Fischer projection, what does the horizontal line represent?
Ans: The Fischer Projection consists of both horizontal and vertical lines, where the horizontal lines represent the atoms that are pointed out of the plane while the vertical line represents atoms that are pointed away from the plane. The point of intersection between the horizontal and vertical lines represents the central carbon.
Q.5. Draw the fisher projection of:
Sol.
*
If fourth valency is not given then we assume it to be hydrogen.
R → Rectus → Right → Clockwise.
S → Sinister → Left → Anti-clockwise.
Note: Designations (R) and (S) bear no relationship to whether a molecule rotates plane-polarized light clockwise (+) or counterclockwise (-).
Let’s first consider the molecule below. The name of this molecule is (R)-1-fluoroethanol. It is listed below with priorities assigned based on atomic number. In this case F>O>C>H. So F is #1 and H is #4. The tricky part here is that the #4 priority is pointing out of the page (on a “wedge”).
You can “simply” rotate the molecule in your head so that the #4 priority is on a dash. Then you can traditionally assign R or S. This “simple” advice is not always an easy task for beginners.
Rotating the Molecule
Here’s another way around this. When the #4 priority is on a wedge you can just reverse the rules. So now we have two sets of rules:
If the #4 priority is on a dash:
If the #4 priority is on a wedge, reverse the typical rules:
If the Lowest Priority Group group is in the plane of the page, swapping any two groups will change the configuration from R to S or vice versa. To determine the configuration:
When the Lowest Priority Group is in the Plane of Page
1. If a compound has 'n' different chiral carbons then the total no. of optically active isomers = 2n
No. of meso form = 0
e.g.
no. of different chiral carbon = 4
total optical isomer = 2n = 24 = 16
Example: (Glucose)
Total no. of different chiral carbon = 4
Total Optical Isomers = 24 = 16
(i) If n is even
(ii) If n is odd
Total optical isomers = = 2n-1
Example 1:
Total meso forms = 2
Total optical isomers =
=
= 22 - 2
= 4 - 2 = 2
Here are the two Optical Isomers:
Example 2:
Total no. of even chiral = 4
Number of optical isomers = a = =
= 23 = 8
Meso forms = m = =
= 22-1 = 21 = 2
Total number of optical isomers = 8 + 2 = 10
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Representation of Optical Isomers
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For a single chiral center, there is no diastereoisomer. The stereoisomers which are not related as object and mirror images. They may be optically active or optically inactive.
Inactive
ActiveFor a compound having 2 chiral centers, Fix one chiral carbon
After one inter-change, if (R, R) → (R, S), then we get diastereomers.
For a compound having 3 chiral carbon to get diastereoisomer, fix two chiral carbon and one interchange with the left carbon or fix one chiral carbon and interchange with other two,
(I) (II)
Total isomer = 23 = 8
(III) (IV)
(I) and (III), I and (IV), (II) and (III), (II) and (IV) are diastereoisomers.
Q.1.
(I)
(II)
(III)
(IV) What is the relation among the above compounds?
Sol. I and II are identical
III and IV are identical
II and III are diastereo isomer
I and IV are diastereo isomer
Q.2. Find the total isomers obtained by dichlorination of cyclopentane.
Sol.
Total isomers = 3 + 3 + 1 = 7
Optically isomers = 6, Optically active isomers = 4
Q.3. Find the total isomers obtained by trichlorination of propane.
Sol.
Total isomers = 6
optically isomers = 2
Q.4. Find total isomers obtained by dichlorination of n-butane
Sol.
1
2
2
1
1
3 (2 optically 1 meso ) Total isomers = 10 (6 optically active + 1 meso + 3 structural)
Q.5. How many stereoisomers of 1,2,3-cyclohexantriol are there?
Sol.
No. of Chiral carbon = 3 (identical) symmetrical)
a = =
= 4- 2
m = 2
total stereoisomers = 2 + 2 = 4
Meso
Mesoform is optically inactive due to internal compensation and racemic mixture is optically inactive due to external compensation.
Q.6. A and B are enantiomers of each other. The specific rotation of A is 20 º. Rotation of mixture of A and B = -5º What is the percentage of the racemic part?
Sol. x mol A, 1-x mol B
x × 20 + ( 1- x ) (-20) = -5
20 x - 20 + 20x = - 5
40 x = 15 ⇒ x = 3/8 = 0.375
moles of A = 3/8
moles of B = =
moles of A and
moles of B will form racemic mixture.
Enantiomer excess or optical purity = -
=
Q.7.
What is the rotation of the mixture?
Sol. Rotation will be due to B only,
= 0.3 × (-20º)
= - 6º
► Chiral compound → optically active compound
Q.8. Which of the following compounds is Chiral (Optically active)?
(A)
(B)
(C) Both
(D) None
Ans. (D)
1. What is the significance of assigning R-S configuration in wedge dash representation of optical isomers? | ![]() |
2. How can we calculate the total number of optically active isomers? | ![]() |
3. What are diastereoisomers and how can they be represented through Fischer projection? | ![]() |
4. How can the representation of optical isomers be helpful in the JEE exam? | ![]() |
5. What are some common misconceptions or mistakes students make when dealing with optical isomers? | ![]() |