Question

Identify each compound from its molecular formula and itsNMR spectrum:

a) ${\mathbf{C}}_9{\mathbf{H}}_{12}$

b) ${\mathbf{C}}_5{\mathbf{H}}_{10}\mathbf{O}$

c)${\mathbf{C}}_9{\mathbf{H}}_{10}{\mathbf{O}}_2$

Short answer

a.

b.

c.

Step-by-step solution

Step-by-Step SolutionStep 1: Degree of unsaturation

Degree of unsaturation indicates the number of double bonds or number of rings present in the given structure.

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

The degree of unsaturation helps in determining the structure of the molecule.

Structure of the compound (a) C9H12

The degree of unsaturation for compound A is 4. It means the compound may contain a ring structure and a double bond.

The signal at near about 7.2 ppm indicates the presence of a benzene ring. So, maybe the ring is ${\mathrm{C}}_6{\mathrm{H}}_5$. The ring may contain a single substituent. So, we have a formula ${\text{C}}_{\text{9}}{\text{H}}_{\text{12}}$and we conclude the ring structure ${\mathrm{C}}_6{\mathrm{H}}_5$. We get,${\text{C}}_{\text{9}}{\text{H}}_{\text{12}}$- ${\mathrm{C}}_6{\mathrm{H}}_5$= ${\mathrm{C}}_3{\mathrm{H}}_7$. The NMR spectra show that it shows triplet at 0.9 ppm integrated to three protons which means it shows the presence of methyl group adjacent to methylene. The identical integration of signals at 1.6 ppm and 2.6 ppm indicates two methylene groups.

Thus, the given compound is propyl benzene.

Compound (a)

Structure of the compound (b) C5H10O

The degree of unsaturation for the given compound is 1, and it means the given structure contains one double bond.

The triplet that integrates with 3 protons and the quartet that integrates with 2 protons indicate the presence of ${\text{CH}}_{\text{3}}{\text{-CH}}_{\text{2}}$ group bonded to an atom that has no hydrogen atom. The molecular formula is ${\mathrm{C}}_5{\mathrm{H}}_{10}\mathrm{O}$which indicates the presence of diethyl ketone. The structure of the compound is as follows;

Compound (b)

Structure of the compound (c) C9H10O2

The degree of unsaturation for the compound is 5. The signals between 7 to 8 indicates the presence of benzene protons so, one mono substituted benzene ring is present in the structure. The substituent has a formula ${\mathrm{C}}_3{\mathrm{H}}_5{\mathrm{O}}_2$. The triplet integrates to 3 protons and quartet that integrates to 2 protons suggest that the presence of ${\text{CH}}_{\text{3}}{\text{-CH}}_{\text{2}}$group bonded to an atom that does not contain any hydrogen. So, the compound contains one of the following structure;

Compound (c)

The substituent which is attached to the benzene ring withdraws electron density while the substituents which is attached to ring B donates the electron density.

The peak shows that the signal is present at a higher frequency. It means the electron withdrawing group is present. Hence, compound A is the correct structure.