Which of the following compounds exists as ge | vedclass

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(B) For a compound to exhibit geometric isomerism,each carbon atom of the double bond must be attached to two different groups. $2-$pentene has the st

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$2-$pentene

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(B) For a compound to exhibit geometric isomerism,each carbon atom of the double bond must be attached to two different groups. $2-$pentene has the structure $CH_3-CH=CH-CH_2-CH_3$. In this molecule,the first carbon of the double bond is attached to a $-H$ and a $-CH_3$ group,and the second carbon is attached to a $-H$ and a $-CH_2CH_3$ group. Since both carbons have different groups attached,it can exist as $cis$ and $trans$ isomers. $2-$pentyne is an alkyne,$2-$methylpropene has two identical $-CH_3$ groups on one carbon,and $2-$methyl$-2-$butene also has two identical $-CH_3$ groups on one carbon,so they do not show geometric isomerism.

Which of the following compounds exists as geometric isomers?

Similar Questions

For which of the following compounds are geometrical isomers possible?

The alkene that exhibits geometrical isomerism is:

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

$CH_2 = CH - CH = CH - CH = CH_2$
How many geometrical isomers are possible for this compound?

How many forms of benzaldoxime exist?

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