1.

Calculate the number of geometrical isomers in the following polyenes. (i) H_(3)C-CH=CH-CH=CH-CH=CH-CH=CH-Br (ii) H_(3)C-CH=CH-CH=CH-CH=CH-CH=CH-CH_(3) (iii) C_(6)H_(5)-CH=CH-CH=CH-CH=CH-C_(6)H_(5) (iv) C_(6)H_(5)-CH=CH-CH=CH-Cl

Answer»

Solution :(i) The MOLECULE has FOUR double bonds and cannot be divided into two EQUAL HALVES unsymmetrical the number of GEOMETRICAL isomers `=2^((n-1))+2^((n//2)-1)`
`=2^(3)+2^(1)=8+2=10`
(iii) The molecule has three double bonds (odd number).
The number of geometrical isomers `=2^((n-1))+2^((n+1)/(2)-1)`
`=2^(2)+2^(2-1)=4+2=6`
(iv) The molecule has two double bonds and is unsymeetrical The number of geometrical isomers `=2^(n)=2^(3)=4`


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