Q19E - Organic Chemistry (Mcmurry)

Question

Nicotine is a diamino compound isolated from dried tobacco leaves. Nicotine has two rings and =162.1157 by high-resolution mass spectrometry. Give a molecular formula for nicotine, and calculate the number of double bonds.

Step-by-Step Solution

Answer

The molecular formula for Nicotine is ${\mathrm{C}}_{10}{\mathrm{H}}_{14}{\mathrm{N}}_2$. Nicotine contains three double bonds and two rings.

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 formula for Nicotine

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

Consider mass of Nicotine = 162

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 162 by 13

$\frac{162}{13}=13\begin{array}{c}12\\ \overline{)\begin{array}{l} 162\\ \underset{—}{-156}\\ 6\\ \end{array}}\end{array}$

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

Here n=12 and r=6

${\mathrm{C}}_{12}{\mathrm{H}}_{(12+6)}={\mathrm{C}}_{12}{\mathrm{H}}_{18}$

(n= quotient, r=remainder)

 

Replacing 2 CH₄ molecule with two nitrogen atom 

So, the molecular formula becomes ${\mathrm{C}}_{12}{\mathrm{H}}_{18}-2{\mathrm{CH}}_4+2\mathrm{N}={\mathrm{C}}_{10}{\mathrm{H}}_{14}{\mathrm{N}}_2$

Step 3: Determining rings of Nicotine 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,${\mathrm{C}}_{10}{\mathrm{H}}_{16}\mathrm{O}({\mathrm{M}}^{+}=152.1201)$

                                         $\begin{array}{l}\mathrm{DBE}=\frac{2\left(10\right)+2+2-\mathrm{X}-14}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=\frac{10}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=5\end{array}$

The actual formula, ${\mathrm{C}}_{10}{\mathrm{H}}_{14}{\mathrm{N}}_2({\mathrm{M}}^{+}=162.1157)$ corresponds to five degree of unsaturation - two of them due to two rings and other three due to the double bonds.

The structure of nicotine is shown below:

Nicotine