In the reaction of trans-cinnamaldehyde under the conditions of 2-propanol and gamma-Al2O3, the product yielded is cinnamyl alcohol.

Question

When the reaction was first performed the GC and IR showed signals of the formation of a carbocation-like intermediate on the phenyl ring. The 2-propanol used contained high amounts of HCl. How does the high amount of HCl in 2-propanol affect gamma-Al2O? Also, show the intermediate that was formed? 

The second reaction performed did not contain high amounts of HCl so the product formed was cinnamyl alcohol. Draw the proton NMR spectrum of cinnamyl alcohol.

Please can someone in detail explain these questions?

Answer 1

The high amount of HCl can react with Al₂O₃ to form AlCl₃ and water, thus preventing the reduction of the aldehyde by the alumina. The carbocation intermediate on the phenyl ring would be located at either the ortho- or para- position due to resonance after protonation of the aldehydic oxygen.

The H-NMR structure of cinnamyl alcohol would contain up to 7 distinct peaks, depending upon the NMR magnet strength and resolution. I did not include splitting values in the following estimations. The phenyl ring proton peaks tend to overlap on lower resolution NMR. You also may not need to ID complex splitting (doublet of triplets (dt), etc.) in your class; your professor may be fine with calling the peaks multiplets (m).

O-H: 1 ppm - 3 ppm (broad singlet, 1H)

CH₂: 4 ppm - 5 ppm (doublet, 2H)

CH: 5.5 ppm - 7 ppm (dt or td, 1H) - alkene CH nearest OH

CH: 5.5 - 7 ppm (d, 1H) - alkene CH nearest phenyl ring

CH (ortho): 7 ppm - 8 ppm (d or dd, 2H)

CH (meta): 7 ppm - 8 ppm (dd, 2H)

CH (para): 7 ppm - 8 ppm (d or dd, 1H)