How many geometrical isomers are possible for the given compound?C{H_3} - CH = CH - CH = CH - {C_2}{H_5}A.FourB.ThreeC.TwoD.Five

Geometrical isomers arise due to the restricted rotation around a double bond, where different spatial arrangements of atoms can lead to distinct isomeric structures. In the given compound C{H_3} - CH = CH - CH = CH - {C_2}{H_5}, there are two double bonds, and each double bond can potentially have geometric isomers.

Let's label the carbons in the compound: A-B-C-D-E, where A is the leftmost carbon and E is the rightmost carbon.

For the first double bond (between carbons B and C):

The configuration can be cis (both substituents on the same side) or trans (substituents on opposite sides).
For the second double bond (between carbons D and E):

Similarly, the configuration can be cis or trans.
Combining the possibilities for both double bonds:

For each double bond, there are 2 possibilities (cis and trans).
Since there are 2 double bonds, the total number of geometrical isomers is 2 * 2 = 4.
So, the correct answer is A. Four geometrical isomers are possible for the given compound.