How many geometrical isomers are possible by H_(3)C-CH=CH-CH=CH-CH=C=

1.	How many geometrical isomers are possible by H_(3)C-CH=CH-CH=CH-CH=C=C=C-CH_(3) ?
Answer» Solution :Compound `H_(3)C-CH=CH-CH=CH-CH=C=C=C-CH_(3)` has three GEOMETRICAL centre and is UNSYMMETRICAL thus, totla G.I. = `2^(3)`=8.

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