Q67E - Organic Chemistry (Mcmurry)

Question

Carboxylic acids $\mathbf{\left(}{\mathbf{RCO}}_{\mathbf{2}}\mathbf{H}\mathbf{\right)}$ react with alcohols (R'OH) in the presence of an acid catalyst. The reaction product of propanoic acid with methanol has the following spectroscopic properties. Propose a structure.    

                                                        

MS: ${\mathbf{M}}^{\mathbf{+}}$ = 88

IR: 1735 ${\mathbf{cm}}^{\mathbf{-}}$

${}^{\mathbf{1}}\mathbf{H}$NMR: 1.11$\mathit{\delta }$ (3 H, triplet, J 5 7 Hz); 2.32$\mathit{\delta }$ (2 H, quartet, J 5 7 Hz); 3.65 $\mathit{\delta }$(3 H, singlet) ${}^{\mathbf{13}}\mathbf{C}$  NMR: 9.3, 27.6, 51.4, 174.6 $\mathit{\delta }$

Step-by-Step Solution

Answer

Structure as follow:

Step-by-Step Solution Step 1: Mass spectra

It is a spectroscopy used to determine the molecular mass of the compound and the structure by ploting the ion signal as a function of mass to charge ratio.

Step 2: Identify compound from mass spectra

So as per the given data the molecular ion peak is at 88 that means the product formed from the reaction of an acid with alcohol must be an ether that the hydrogen is replaced by the methyl group.

Step 3: Identify compound from NMR spectra

As per the NMR data is given it is observed that the peak at 1.11 is due to the terminal methyl group and then the peak at the2.32 is due to the methylene group and the peak split into quartets due to the 3H of the methyl and the peak observed at 3.65 the singlet having 3H so it is very clear the H is replaced by methyl having three hydrogen next to the oxygen atom.

Step 4: Identify compound from IR

As the IR peak is observed at 1723${\text{cm}}^{\text{-1}}$ so the compound must be ester reaction as follow:

Formation of esters