The total number of geometrical isomers for t | vedclass

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(A) The structure of $hexa-1, 3, 5-triene$ is $CH_2=CH-CH=CH-CH=CH_2$. Geometrical isomerism occurs due to restricted rotation around double bonds. In

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$2$

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(A) The structure of $hexa-1, 3, 5-triene$ is $CH_2=CH-CH=CH-CH=CH_2$. Geometrical isomerism occurs due to restricted rotation around double bonds. In this molecule,there are $3$ double bonds. However,the terminal double bonds ($C_1=C_2$ and $C_5=C_6$) cannot exhibit geometrical isomerism because they are attached to two identical hydrogen atoms. Only the central double bond $(C_3=C_4)$ can exhibit geometrical isomerism ($cis$ and $trans$). Therefore,the total number of geometrical isomers is $2$ ($cis$ and $trans$).

The total number of geometrical isomers for the compound $hexa-1, 3, 5-triene$ is:

Similar Questions

Which compound shows $cis-trans$ isomerism?

Draw $cis$ and $trans$ isomers of the following compounds. Also write their $IUPAC$ names:
$(i)$ $CHCl=CHCl$
$(ii)$ $C_{2}H_{5}C(CH_{3})=C(CH_{3})C_{2}H_{5}$

In the given reaction: $C_6H_5-CH=O \xrightarrow{NH_2OH/H^{+}} [X]$. $[X]$ will be:

Geometrical isomerism is shown by

Identify the compound which can show geometrical isomerism:

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