Correct Answer - C
First calculate the degree of unsaturation or double bond equivalents `DBEs`:
`DBE =(sumn(v-2))/(2) +1`
`= (4(4-2)+7(1-2)+1(1-2))/(2)+1`
`= (8-7-1)/(2) +1 =1`
This implies the presence of a `pi` bond or a cyclie structure.
Acyclie isomers
Two skeletons are possible `C-underset(C )underset(|)C-C`and`C-C-C-C`
Branched skeleton
`{:(CH_(3)-underset(CH_(3))underset(|)C=CHCI,,CICH_(2)underset(CH_(3))underset(|)C=CH_(2),,):}`
unbranched skeleton
`{:(underset((cis,trans))(CH_(3)CH_(2)CH)=CHCI,,CH_(3)CH_(2)overset(CI)overset(|)(C)=CH_(2),,):}`
`{:(underset(Chiral(R,S))(CH_(3)overset(CI)overset(|)CHCH)=CH_(2),,CICH_(2)CH_(2)CH=CH_(2),,):}`
`{:(underset((cis-trans))(CH_(3)CH=CHCH_(2)CI),,underset(cis,trans)(CH_(3)CH=overset(CI)overset(|)C CH_(2)),,):}`
Cyclic isomers
Thus there are `12` acyclic isomers and `7` cyclie isomers.
