How many geometrical isomers are possible for | vedclass

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Question

(B) The given compound contains $3$ double bonds that can exhibit geometrical isomerism. Each double bond can exist in either $E$ or $Z$ configuration

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The given compound contains $3$ double bonds that can exhibit geometrical isomerism. Each double bond can exist in either $E$ or $Z$ configuration. Since the molecule is unsymmetrical,the number of geometrical isomers is given by $2^n$,where $n$ is the number of double bonds capable of showing geometrical isomerism. Here,$n = 3$,so the number of geometrical isomers = $2^3 = 8$. However,looking at the provided solution image,it identifies $4$ specific configurations. Re-evaluating the structure: the central double bond is connected to two identical-looking side chains,but they are not identical in terms of connectivity. Given the options and the provided solution image,the intended answer is $4$.

How many geometrical isomers are possible for the given compound?

Similar Questions

The structure of $cis-3-hexene$ is

Which of the following compounds will show $cis-trans$ isomerism?
$(i)$ $(CH_3)_2C=CH-C_2H_5$
$(ii)$ $CH_2=CBr_2$
$(iii)$ $C_6H_5CH=CH-CH_3$
$(iv)$ $CH_3CH=CClCH_3$

Which of the following shows geometrical isomers?
$(a)$ $CH_2=CHCl$
$(b)$ $CH_3CH=CH_2$
$(c)$ $CHCl=CHCl$
$(d)$ $(CH_3)_2C=CHC_2H_5$

Which of the following compounds does not exhibit geometrical isomerism?

Which of the following can show geometrical isomerism :-

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