Question

How many structural isomers of $C_9{H}_{18}$have a six member ring and trisubstituted?
(A) 3
(B) 4
(C) 5
(D) 6

Answer 1

Hint: As we know that isomerism is the phenomenon where more than one compound has the same chemical formula but completely different chemical structures. Also chemical compounds which have identical chemical formulas but differ in physical and chemical properties due to different arrangement of atoms in the molecule are called isomers.
Formula used: We will use the following formula:-
$(C+1)-dfracH2-dfracX2+dfracN2$

Complete answer:
Let us briefly discuss about isomerism and structural isomerism as follows:-
-Isomerism: It is the phenomenon where more than one compound has the same chemical formula but completely different chemical structures. It is generally of two types i.e., structural isomerism and stereoisomerism.
-Structural isomerism: It is also known as constitutional isomerism and it is the phenomenon where compounds have the same molecular formula but atoms or groups are linked in different manner.
-We can draw the all isomers of a chemical formula by using double bond equivalence as it informs us about the number of rings and pi-bond present in it. The formula is given below:-
Double bond equivalence = $(C+1)-dfracH2-dfracX2+dfracN2$
where,
C = total number of carbon atoms in the chemical formula
H = total number of hydrogen atoms in the chemical formula
X = total number of halogen atoms in the chemical formula
N = total number of nitrogen atoms in the chemical formula
-The given chemical formula is $C_9{H}_{18}$ where C = 9 and H = 18. So double bond equivalence of this chemical formula is:-
$\begin{array}{rl}& \Rightarrow (C+1)-dfracH2-dfracX2+dfracN2\\ & \Rightarrow (9+1)-dfrac182\\ & \Rightarrow 10-9=1\end{array}$
This means that compounds of this formula can either have 1 ring or 1 pi- bond at a time. Since we require a six member ring according to the requirements so there will be no pi-bond in the structural isomers. Let us draw all the possible structural isomers of $C_9{H}_{18}$ which have a ring along with 3 substituents of methyl (3 methyl substituents are preferred because only then the requirements can be fulfilled as 6 out of 9 carbons are already used in the six member ring).
  

Therefore, the total number of structural isomers of $C_9{H}_{18}$ which have a six member ring and trisubstituted is (D) 6.

Note:
-First draw all the compounds which are different and all methyl groups placed at different possible positions. Then combine two methyl groups together on a carbon and make another methyl group at ortho, para and meta with respect to it. 3 methyl groups can’t be combined as carbon can only make 4 bonds at a time.
-Also learn and understand about different types of isomerism shown by the compounds in organic chemistry to solve these types of questions.