Question

The systematic nomenclature of the following spiro compound is:

A. 2- Methyl spiro [3.4] octane
B. 3-Methyl spiro [3.4] octane
C. 6-Methyl spiro [3.4] octane
D. 7-Methyl spiro [3.4] octane

Answer 1

Hint: The basic rules of IUPAC ( International Union of Pure and Applied Chemistry) will be applied here, there will only be a small difference here, since it is a spiro compound. Like, the spiro compound names start with the ‘spiro’ and counting will begin from the jointed carbons. Just remember these small details.

Complete Solution :
First, we will start by finding the longest chain in the compound and as stated earlier we will have to begin our counting from the joined carbon and first it will start from that side where there are a smaller number of carbons present.
- There are a maximum 8 carbons present. Now, after numbering we will find that substitute Methyl is lying on the 6th position and we will start the name with the prefix which is 6-Methyl and since it is a spiro compound, the spiro will be added. Now, in the square brackets it is compulsory to show the no. of carbons present in each cycle starting from the point where they are joined. Hence, on the left side, there are 3 carbons and, on the right, there are 4 carbons present, therefore, [3.4] we will right.
Now, there are 8 carbons in total and since there is no unsaturation ( double bond and triple bond) present, its suffix will be octane.
So, the correct answer is “Option C”.

Note: If you see clearly then you will notice that numbering is done from left hand side, if they started from right hand side then, the substituent methyl will get the 7th position which will break the law. Also, in the square brackets, the lowest number of carbons are mentioned first.