Q84E - Organic Chemistry (Mcmurry)

Question

Propose structures for ketones or aldehydes that have the following ${}^{\mathbf{1}}\mathbf{H}$ NMR spectra:

 (a)C₁₀H₁₂O 

$\mathit{I}\mathit{R}\mathbf{:}\mathbf{ }\mathbf{1710}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$

(b) ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{3}}$

$\mathbf{IR}\mathbf{:}\mathbf{ }\mathbf{1715}\mathbf{ }{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$

Step-by-Step Solution

Answer

The structure of compound is:

a.

b

Step 1: Identify compound a from IR

As the IR peak is observed at 1710 CM^- so the compound must be contain the ketonic functional group or a derivate of the carbonyl compound.

Step 2: Identify the compound A from NMR and degree of unsaturation

As per the NMR graph the peak at 7.28 is due to the presence of the aromatic ring and the peak at 1.01 is due to the 3H and shows triplet due to the adjacent methylene group and that group show peak at 2.47 that is quartet due to adjacent 3H of methyl group and the peak at 3.66 is due to the adjacent carbonyl and the degree of unsaturation is 1 as per molecular formula so that only 1 ring is present so, the structure must be:

Step3: Identify compound b from IR

As the IR peak is observed at 1715 cm^-so the compound must contain the ketonic functional group or a derivate of the carbonyl compound and the presence of ether also possible.

Step 4: Identify the compound B from NMR

As per the NMR graph, the peak at 3.37 is the longest peak due to the presence of the ether oxygen and the peak at 4.79 is showing triplet which means adjacent methylene group is present and the peak at 2.74 is doublet that is methylene group is present with CH group and sharp peak at the 2.18 is due to the methyl group adjacent to the methyl group so, the compound can be:

               NMR peak