The number of acyclic structural isomers (inc | vedclass

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(D) The molecular formula for pentene is $C_5H_{10}$. The acyclic structural isomers are: $1$. $Pent-1-ene$ $(CH_2=CH-CH_2-CH_2-CH_3)$ $2$. $Pent-2-en

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$6$

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(D) The molecular formula for pentene is $C_5H_{10}$. The acyclic structural isomers are: $1$. $Pent-1-ene$ $(CH_2=CH-CH_2-CH_2-CH_3)$ $2$. $Pent-2-ene$ $(CH_3-CH=CH-CH_2-CH_3)$ $3$. $2-Methylbut-1-ene$ $(CH_2=C(CH_3)-CH_2-CH_3)$ $4$. $3-Methylbut-1-ene$ $(CH_2=CH-CH(CH_3)_2)$ $5$. $2-Methylbut-2-ene$ $(CH_3-C(CH_3)=CH-CH_3)$ Now,considering geometrical isomers: $Pent-2-ene$ exists as $cis$ and $trans$ isomers. Therefore,the total number of acyclic isomers is $5$ (structural) $+ 1$ (additional geometrical isomer for $pent-2-ene$) $= 6$ isomers. These are: $1$. $Pent-1-ene$ $2$. $cis-Pent-2-ene$ $3$. $trans-Pent-2-ene$ $4$. $2-Methylbut-1-ene$ $5$. $3-Methylbut-1-ene$ $6$. $2-Methylbut-2-ene$ Thus,the total count is $6$.

The number of acyclic structural isomers (including geometrical isomers) for pentene $(C_5H_{10})$ are $......$

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