Q16E - Organic Chemistry (Mcmurry)

Question

Camphor, a saturated monoketone from the Asian camphor tree, is used among other things as a moth repellent and as a constituent of embalming fluid. If camphor has M⁺=152.1201 by high-resolution mass spectrometry, what is its molecular formula? How many rings does camphor have?

Step-by-Step Solution

Answer

The actual formula for camphor is ${\mathrm{C}}_{10}{\mathrm{H}}_{16}\mathrm{O}({\mathrm{M}}^{+}=152.1201)$corresponding to three degrees of unsaturation. Camphor have two rings.

Step-by-step Solution Step 1: Nitrogen rule

According to nitrogen rule an organic compound having odd molar mass contain odd number of nitrogen atoms and compound having even molar mass contain even number of nitrogen atoms.

Step 2: Calculating Molecular mass of camphor

High resolution mass spectrometry calculate exact mass of the fragments upto 5 ppm level.

We have given M⁺= 152.1201

Consider mass of camphor is 152.

Thus given mass is even it means camphor molecule contain zero or even number of nitrogen atoms, according to nitrogen rule.

According to rule of 13, divide 152 by 13

$\frac{152}{13}=13\begin{array}{c}11\\ \overline{)\begin{array}{l}152\\ \underset{\_}{-143}\\ 9\\ \end{array}}\end{array}$

The remainder is 9. The basic C-H formula is: ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{(\mathrm{n}+\mathrm{r})}$

Here n=11 and r=9

${\mathrm{C}}_{11}{\mathrm{H}}_{(11+9)}={\mathrm{C}}_{11}{\mathrm{H}}_{20}$

(n= quotient,  r=remainder)

As we know camphor is monoketone, it contain one carbonyl oxygen and so make addition of that oxygen to above CH formula by replacing with CH₄.

So, the molecular formula becomes: ${\mathrm{C}}_{11}{\mathrm{H}}_{20}-{\mathrm{CH}}_4+\mathrm{O}={\mathrm{C}}_{10}{\mathrm{H}}_{16}\mathrm{O}$

Step 3: Determining rings of camphor by Double Bond equivalent (DBE)

Degree of unsaturation is also known as double bond equivalent (DBE). If molecular formula is given plug in the number into these formula.

     $\mathrm{DBE}=\frac{2\mathrm{C}+2+\mathrm{N}-\mathrm{X}-\mathrm{H}}{2}$

Where, C is the number of Carbons

N is the number of nitrogen’s

H is the number of hydrogens

X is the number of halogens

For  ${\mathrm{C}}_{10}{\mathrm{H}}_{16}\mathrm{O}({\mathrm{M}}^{+}=152.1201)$ 

                                $\begin{array}{l}\mathrm{DBE}=\frac{2\left(10\right)+2+0-\mathrm{X}-16}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=\frac{6}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=3\end{array}$

The actual formula, ${\mathrm{C}}_{10}{\mathrm{H}}_{16}\mathrm{O}({\mathrm{M}}^{+}=152.1201)$corresponds to three degree of unsaturation. The ketone functional group accounts for one of these. Since camphor is a saturated compound, the other two degree of unsaturation are due to two rings. Hence, the camphor containing two rings.

Camphor