Respuesta :

The Mass Spectra of these two compounds is quite easy to be differentiated. I am discussing both spectra first, and then will highlight the main differences.

Vannilin:
             The main parent peak of this compound will appear at m/z = 152, and the characteristic base peak will appear at m/z = 151. And other peaks like m/z = 29 is well known for aldehydes.

Broovanillin:
                   The Parent peak will appear at m/z = 230 and 232 because Bromine has two stable isomers Br-79 and Br-81, and natural abundance of both isotopes is 50:50 means there will be two peaks at m/z = 230 and 232 of same relative abundance. The base peak will appear at m/z = 151. And other peaks are at m/z = 107, 109 and m/z = 123.

Difference:
                First difference is in the value of parent peak, Vanillin has parent peak at 152 while Bromovanillin has at m/z = 230 and 232 with same abudance. Next main difference is the presence and absence of peaks. Like the peak of m/z = 29 is present in Vanillin but absent in Bromovanillin, second m/z = 107 and 109 is present in BromoVanillin and absent in Vanillin.

The Fragmentation scheme for both compounds is shown below,
Ver imagen transitionstate
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