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chimanshu_007

here is a method to calculate the no. of optically active isomers , meso forms and racemic mixture ina a given compound.....

here

a - optically active isomer

m - meso form

r - racemic mixture...

okay in case 1

when there are n chiral carbon atoms bt there is no plane of symmetry

then ,

a = 2ⁿ

m = zero

r = "a"/2

take n = 1,2,3,4,5,6,7............................

case 2

when compounds have n chiral carbon atoms , and plane of symmetry is possible.....

then it further have 2 cases

1) IF N IS EVEN

a = 2^n-1

r = a/2

m = 2^{(n/2) - 1}..........(whole is in power okay?>???)

2) IF N IS ODD

a = 2^n-1 - 2^(n-1)/2

r = a/2 (always....)

m = 2^(n-1)/2........(whole in power again)

so by these ways , u can calculate no. of optically active isomers , meso forms n racemic mixture...

and 1 more thing....

how to calculate optical isomers.....

in simple compounds.....

1st check the chiral carbon.....

if its nt present , den its optically inactive

if its present

then

a) If 1 chiral carbon atom , the compound is 100% optically active.....

b)If more dan 1 chiral carbon atom ,

den check plane of symmetry

i)If plane of symmetry present , compound is optically inactive

ii) If plane of symmetry absent , compound is optically active.....

OPTICALLY ACTIVE COMPOUNDS , WICH ARE MIRROR IMAGES OF EACH OTHER ARE ENANTIOMERS....(AND ENANTIOMERS ALWAYS EXIST IN PAIRS)

and the configurational isomers wich arent the mirror images of each other are called diastereomers

racemic mixture - equimolar mixture of 2 enantiomers (as i have already said dat dey exist in pairs) is called racemic mixture...

seperation of racemic mixture into individual components is called RESOLUTION.....

NOW THIS IS IMPORTANT TOO......

1) If in a ficher projection we do odd no. of exchanges around all Chiral carbons , we get an enantiomer

2)If in a ficher projection , we do even no. of changes around all chiral carbon atoms , we get the same molecule....

3)If in a ficher projection , we do odd no. of exchanges on some chiral carbons n even on other , we get diastereomers.....

these can be well explained by diagrams......will try to show u.....

but u can do this method wen we get a qs like a compound is given , and its written....wich of the foll. are enantiomers of given comp.

so here are the diagrams.....

now look compound 1 is given to us.......okay

compound 2,3,4,5 are the options n in qs it is asked....

WICH OF DA FOLL. ARE NOOOOT THE ENANTIOMERS OF THE COMPOUND 1??????

so in case 2) (compound 2)

in 1st carbon we gt 1 change (odd)

in 2nd carbon we gt 1 change (H n Br interchanged da places) (odd)

in 3rd carbon again we got 1 change (OH n CH3) (odd)

okay na....

so odd in all da places , compound is enantiomer only...

compound 3rd

1st carbon - zero change on 1st carbon (we consider it as even)

2nd carbon - 1 change

3rd carbon - 0 again

so odd on 1 even on 2 , its a diastereomer

compound 4)

1st carbon - 1 change

2nd carbon - 0 change

3rd carbon - 1 change

again some odd some even

compound 5th

carbon 1 - 1 change

carbon 2 - 1 change

carbon 3 - 2 changes (firstly OH n CH3 interchanged there places , then Et n OH so 2 changes)

again some odd some even
so answer WICH OF THE FOLL. ARE NT ENANTIOMERS OF THE COMPOUND 1 IS

B , C , D i.e compound 3 , 4 , 5

hope it help u

ab i dun have to explain wat is carbon 1 or 2 or 3 , FIR BI I HAVE EXPLAINED IT IN LAST DIAG.

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