Question

What is the IUPAC name of the given compound:${\mathrm{C}}_{10}{\mathrm{H}}_{12}\mathrm{O}$?

Answer 1

Step 1: Determination of the structure of ${\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{12}}\mathbf{O}$

• In order to determine the structure from the formula, we need to calculate the double bond equivalent (DBE).
• DBE is the degree of unsaturation or we can say that it is the number of hydrogen atoms added in order to convert all the pi bonds in the compound to a single bond and all the rings to an acyclic structure.
• It can be calculated using the formula given below,

$DBE=\frac{No.ofhydrogenrequiredforasaturatedchain-No.ofhydrogenpresent}{2}$

• We know that the general formula of an alkane is $C_n{H}_{2n+2}$. So for a saturated chain with $10$ carbon atoms, we need $(10\times 2)+2=22$ hydrogen atoms. Thus the DBE is,

$$\begin{gathered} DBE=\frac{22-12}{2} \\ =\frac{10}{2} \\ =5 \end{gathered}$$

• Since the DBE is $5$ there is a possibility for a benzene ring and one extra double bond. Because DBE is the total number of double bonds and rings present in the compound. Benzene contains a ring and $3$ double bond. So the extra double bond will be out of the ring and since the oxygen is present we can assume it is the ketone group.
• Six carbon atoms and five hydrogen atoms will be utilized in the benzene ring the rest four carbon and seven hydrogens are out of the ring. Thus the structure will be as follows,

Step 2: IUPAC name of the given compound: ${\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{12}}\mathbf{O}$

• The first step in naming is to find the longest chain (parent chain). Here the longest chain contains four carbon atoms so the root word will be butan-. (Note that we took butan instead of but- because the functional group is a ketone, so while naming we have to just replace ‘e’ of butane by -one by mentioning its position.)
• Determination of name and position of locants (functional group) - Here the functional group attached are phenyl ring and a ketone. The position of the phenyl ring is $1$ and that of the ketone is $2$ . ( Numbering is done in such a way that the sum of the position of functional groups gets the lowest possible number.)
• Determination of suffix- Here the main functional group determines the suffix of the name. Since ketone is the more preferred functional group present here, the name will end with -one after the position number.

Therefore the IUPAC name of ${\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{12}}\mathbf{O}$ is 1-phenylbutan-2-one.