Q15-15E - Organic Chemistry (Mcmurry)

Question

1,6-Methanonaphthalene has an interesting 1H NMR spectrum in which the eight hydrogens around the perimeter absorb at 6.9 to 7.3 d, while the two protons absorb at 20.5$\mathit{\delta }$. Tell whether it is aromatic, and explain its NMR spectrum.

Step-by-Step Solution

Answer

Yes, it’s aromatic due to 10 π-electron and absorbs at 6.5 -8$\delta$ but ${\text{CH}}_2$ is absorbed at -0.5$\delta$.

Step 1: Huckel rule

It is the rule to check the aromaticity of the compound by knowing the number of delocalized electrons. The rule is called 4n+2 where the n value must be a whole number like when 6$\mathit{\pi }$ electrons are present n=1.

$\begin{array}{c}\mathbf{4}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{6}\\ \mathbf{4}\mathbf{n}\mathbf{=}\mathbf{6}\mathbf{-}\mathbf{2}\\ \mathbf{4}\mathbf{n}\mathbf{=}\mathbf{4}\\ \mathbf{n}\mathbf{=}\mathbf{1}\end{array}$

Step 2: Aromatic nature of 1,6-methanonaphthalene

As seen from the structure below 1,6-methanonapthalene has 5 double bonds which means it has 10 pi electrons which follow Huckel’s rule that is 4n+2 and it has n=2 so it is aromatic.

Step 3: H NMR 1

Nuclear magnetic resonance is the technique that is used to identify the hydrogen in the compound and produce the corresponding peaks for its detection.

Step 4: H NMR 1 for aromatic compounds

Aromatic compounds which are planer in shape and follow Huckel’s rule and the aromatic compounded due to its plane nature show peaks at 6.5 -8$\delta$ but to the applied magnetic field of the π-electrons, a ring current is generated which strongly yields the ${\text{CH}}_2$ proton that’s which it shows the peak far up the field at -0.5$\delta$.