If rotation of light is anti-clockwise
laevo rotatory (l-form)
If rotation of light is clockwise
dextro rotatory substance(d-form)
If there is no rotation of light then substance is called optically inactive.
If all the four valencies of carbon are satisfied by four different atoms or four different group atom then carbon is known as chiral carbon.
► (2 Chiral carbon)
► Chiral carbon <w<¡us ds fy, igys point of doubt check djuk pkfg,
(no chiral carbon)
► For optically active isomer, the object and its mirror image must be non superimposable.
► If image and object are superimposable then they are not optically isomers to each other.
►To check, superimposable either of object or mirror image is rotated 180º along the mirror. After that mirror image is checked for superimposablility.
Object Mirror image
After 180º rotation, mirror image and object are identical.(mirror image of meso compound will be identical)
► If the compound have only one chiral carbon then it will be certainly optically active.
Object and its non-superimposible mirror image are called enantiomers to each other
32. A and B are enantiomers to each other. If specific rotation of A is 20º and rotation of mix of A and B is -10º then find the percentage of A and B in the mixture?
Sol. Let x mol A
(1-x) mol of B
total rotational of A = x × 20
total rotational of B = (1 - x)(-20)
Total rotation of the mixture = - 10
x × 20 + (1 - x) (-20) = - 10
20 x - 20+20x = - 10
40 x = 10 x = = 0.25
% A = 25%
% B = 75%
By fisher projection three dimensional structure is converted into 2D.
33. Write the Fisher projection of CH3CH(OH)COOH
► Maximum carbon must be in vertical line.
► Place higher priority carbon containing functional group on top of vertical line.
34. Write Fisher projection of
Sol. (i) Place higher priority carbon containing functional group on top of vertical line.
(ii) Arrange other group according to its clock wise or anti-clockwise position w.r.t. group on the top.
Wedge Dash Structure:
► dsoy Fourth priority order dks Cross dj ldrs gSa] during checking of clock or anticlockwise.
► To draw Fisher projection of such structure.
(1) Dotted group is placed below.
(2) The group attached by dark line is placed on the top.
(3) Remaining group are placed according to their clock wise or anti-clockwise position w.r.t. group in dark line.
35. Draw the fisher projection of:
If fourth valency is not given then we assume it to be hydrogen.
R and S form
R → Rectus → Right → Clock wise.
S → Sinister → Left → Anti-clockwise.
► Higher the atomic number of the element, greater will be its priority.
► In case of double bond ( = ) or triple ( ≡ ) bond
-CH = CH2 -C º CH
1 → 2 → 3 If clock wise ⇒ R
1 → 2 → 3 If Anti clock wise ⇒ S
⇒ Mirror image of R is S
► If lowest priority element is in horizontal then exact order is obtained by reversing the form (i.e. it comes R then exact form will be S and vice versa )
⇒ It comes as R but will be S form,
If lowest priority order element is in dotted then
1 → 2 → 3 If clock wise ⇒ R
and if 1 → 2 → 3 If Anti clock wise ⇒ S
► If lowest priority element is not in dotted then we bring it in dotted by doing even number of inter changes (2,4 or 6) as by even no. of inter change, the configuration does not change.
► After odd number of changes the configuration changes (If R then goes into S and vice versa).
After one change
After two interchanges
Ex.36 Draw the Fisher projection of
All the carbons are placed in vertical.
► Groups attached to the chiral carbons are placed on any side. After that they are adjusted whether they are R or S form w.r.t chiral carbons.
Ex.37 Draw the Fisher projection of the following.
Sol. Wedge Dash formula
It is an equimolar mixture of R and S or d and l. Racemic mixture is optically inactive.
► A compound is optically active due to:
(1) Absence of plane of symmetry (POS)
(2) Absence of centre of symmetry (COS)
(meso form )
► POS is an imaginary plane where if we place a mirror, mirror image will exactly overlap the other half.
► For meso form, there must be at least two identical chiral carbons.
Identical carbon ⇒ Chiral carbons having identical group attached.
► If compound has POS then it will be certainly optically inactive and will be called meso form.
► After finding two identical carbons. We assign them as R or S. If first part is R and other is S then they will rotate the light in opposite direction but to equal extent the compound will be optically inactive.
► If a compound have 'n' different chiral carbons then total no. of optically active isomers = 2n
No. of meso form = 0
no. of different chiral carbon = 4
total optical isomer = 2n = 24 = 16
► There will be no meso as the compound does not have identical chiral carbon.
► If a compound has n identical chiral centre (symmetrical) ⇒ There must be symmetry from some where.
(i) If n is even
optical isomer (a) = 2n-1
mesoform (m) = 2n/2-1
total optical isomer = a + m
(ii) If n is odd
Total optical isomer = = 2n-1
► When there is odd no. of identical carbon atom (i.e. symmetrical) then this compound will certainly contain pseudo chiral w.r.t. which compound be symmetrical (i.e. POS).
► Other meso compound of the above compound will form by changing the place Br and H around Pseudo Chiral carbon.
Total meso = 2
= 22 - 2
= 4 - 2 = 2
Total no. of different chiral carbon = 4
Total optical isomer = 24 = 16
Total no. of even chiral = 4
a = = = 23 = 8
m = = = 22-1 = 21 = 2
Total isomers = 8 + 2 = 10
For single chiral centre, there is no diastereoisomer. The stereoisomer which are not related as object and mirror image. They may be optically active or optically inactive.
► Fix one chiral carbon
After one inter-change
If (R, R) → (R, S)
For compound having 3 chiral carbon to get diastereoisomer, fix two chiral carbon and one interchange with left carbon or fix one chiral carbon and inter change with other two,
Total isomer = 23 = 8
(I) and (III), I and (IV), (II) and (III), (II) and (IV) are diastereoisomers.
What are 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
Ex.39 Find total isomers obtained by dichlorination of cyctopentane?
Total isomers = 3 + 3 + 1 = 7
Optically isomers = 6, Optically active isomers = 4
Ex.40 Find the total isomers obtained by trichlorination of propane.
Total isomers = 6
optically isomers = 2
Ex.41 Find total isomers obtained by dichlorination of n-butane
(3) (2 optically 1 meso )
Total isomers = 10 (6 optically active + 1 meso + 3 structural)
Ex.42 How many stereoisomers of 1,2,3-cyclohexantriol
No. of Chiral carbon = 3 (identical) symmetrical)
a = = = 4- 2
m = 2
total stereoisomers = 2 + 2 = 4
► Mesoform is optically inactive due to internal compensation and racemic mixture is optically inactive due to external compensation.
Ex.43 A and B are enantiomer of each other. Specific rotation of A is 20 º. Rotation of mixture of A and B = -5º what is the percentage of 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 = - =
What is rotation of mixture?
Sol. Rotation will be due to B only,
= 0.3 × (-20º)
= - 6º
► Chiral compound → optically active compound
Ex.45 Which of the following compound is Chiral (Optically active)?
(C) Both (D) None
They are non-superimposable mirror image
If no. of rings are even ⇒ optically active
If no. of rings are odd ⇒ Inactive
⇒ This is the even no. of double bonds case.
► For optical activity, the carbons at extreme position must have different group attached.
⇒ Planar compound
Always have POS, Optically inactive.
If biphenyl contain bulky group at its ortho position (only) then due to repulsion the planarity of compound disappears and its mirror image is non superimposable.
► In the biphenyls none of the two rings must have symmetry.
► (Optically active)
In 2º Amines.
Optically inactive due to formation of racemic mixture.
► Order of flipping in amines: 1º > 2º > 3º
D-Form :- For compound having one chiral carbon
(a) If OH is right side → D
(b) If OH is left side → L
(Note: → All the carbon must be in vertical having highest O. N. Carbon on the top)
Conversion of Fisher Projection in Sawhorse Structure
The above Fisher projection can also be written in Saw Horse form as follows
Conversion of Newmann Projection in to Sawhorse Projection
► If all the identical groups are same side or in exactly opposite direction to each other in Newmann projection then compound will be meso.