Question

Provide a stable structure for the following compound: C9H10O3; IR: 2300-3200, 1710, 1600 cm-1; 1H NMR spectrum:

Answer 1

Concepts and reason

The concept used to solve this problem is identification of compound by IR and NMR technique.

NMR is a major analytical technique which is used for the determination of structure of organic molecule. IR spectroscopy is the absorption spectroscopy and is used to determine the functional groups.

Fundamentals

Nuclei has spin and when magnetic field is applied, it aligns itself either in the direction or against the direction of magnetic field and shows signals in NMR spectra. Nuclei in the same environment produce same signal while non-equivalent nuclei shows different signals. The $^{\rm{1}}{\rm{H}}$ is the NMR active nuclei. The IR spectroscopy measures the vibrational frequency in the bond. Therefore, it gives the finger print region for the functional group. The double bond equivalent $\left( {{\rm{DBE}}} \right)$ are calculated by using the relation as follows:

${\rm{DBE}} = {\rm{C}} - \frac{{\rm{H}}}{2} - \frac{{\rm{X}}}{2} + \frac{{\rm{N}}}{2} + 1$

Here, ${\rm{C}}$ is the number of carbon atoms, ${\rm{H}}$ is the number of hydrogen atoms, ${\rm{X}}$ is the number of halogen atoms, and ${\rm{N}}$ is the number of nitrogen atoms in the compound.

Calculate the number of double bonds equivalent $\left( {{\rm{DBE}}} \right)$ in ${{\rm{C}}_9}{{\rm{H}}_{10}}{{\rm{O}}_3}$ by using the relation as follows:

${\rm{DBE}} = {\rm{C}} - \frac{{\rm{H}}}{2} - \frac{{\rm{X}}}{2} + \frac{{\rm{N}}}{2} + 1$

Substitute 9 for ${\rm{C}}$ , 10 for ${\rm{H}}$ , 0 for ${\rm{X}}$ and 0 for ${\rm{N}}$ in the above equation for the calculation of ${\rm{DBE}}$ as follows:

$\begin{array}{c}\\{\rm{DBE}} = 9 - \frac{{10}}{2} - \frac{0}{2} + \frac{0}{2} + 1\\\\ = 5\\\end{array}$

The compound, ${{\rm{C}}_9}{{\rm{H}}_{10}}{{\rm{O}}_3}$ has 5 double bonds.

The IR frequency between $2300 - 3200{\rm{ c}}{{\rm{m}}^{ - 1}}$ shows the presence of ${\rm{OH}}$ functional group. The frequency ${\rm{1710 c}}{{\rm{m}}^{ - 1}}$ shows the presence of ${\rm{C = O}}$ functional group and the frequency ${\rm{1600 c}}{{\rm{m}}^{ - 1}}$ shows the presence of ${\rm{C = C}}$ bonds.

Consider the NMR data as follows:

The ${\rm{1H}}$ singlet at $11.7{\rm{ ppm}}$ shows the presence of ${\rm{OH}}$ functional group.

The multiplet ${\rm{H}}$ NMR spectra at $7.4{\rm{ ppm}}$ shows the presence of aromatic ring.

The ${\rm{2H}}$ singlet at $2.8{\rm{ ppm}}$ shows the presence of ${\rm{C}}{{\rm{H}}_2}$ functional group.

The ${\rm{2H}}$ singlet at $4.2{\rm{ ppm}}$ shows the presence of ${\rm{C}}{{\rm{H}}_2}$ functional group attached to electronegative atom.

The structure of the compound is as follows:

Ans:

The structure of the compound is as follows: