Nitration of a halobenzene was performed in lab. The halogen group attached to t

Question

Nitration of a halobenzene was performed in lab. The halogen group attached to the benzene I think is bromine. Due to having a melting point of 127-130 degrees celsius for a nitrated halobenzene.

Given the HNMR spectra determine the percentage of ortho, meta, para substitution, and determine the percentage of mono- and dinitration of your halobenzene?

Once at least one proton from each compound is identified( and where more can, they should be identified), the integrals can be used to calculate the percent of each compound in the mixture.

Please fully interpret the spectrum to the best of your ability. Show which proton from a structure is represented by which signal. clearly show the calculations of percent composition.

THANK YOU!!!

Here is a link to the HNMR in case the picture does not show up below.

http://i65.tinypic.com/wak6ex.jpg

Explanation / Answer

On nitration of bromobenzene we can get substitution at ortho, meta and para positions, As Br being an ortho-para director, so we get substitution of nitro group at ortho and para positions preferably in comparison to meta substitution. Generally we get para substituted product with higher % (around 56%) and ortho (around 43%), and least stable meta subsituted product with 1% yield. On again nitration of para subtituted product, we get dinitrated product with nitro group at both ortho and para positions. From the given spectra, the most intense peaks arises around 8.101-8.0073 ppm and 7.721--7.684 ppm is a doublet of triplet due to para substituted nitro products as there are 2 types of H, first doublet of triplet is due to proton near nitro group and second doublet of triplet is due to proton near Br. Other peaks of subsequently less intensity are due to ortho substituted product in which there are four different types of H, in which we get peaks around 7.836 due to proton at C-6 position, 7.744 due to proton around C-3 position, 7.47 due to H at C-4 position and around 7.45 due to H at C-5 position. similarly in meta subsituted product there are also 4 different protons, peak around 8.378 due to H at C-2 postion, 8.176 due to H at C-6 position, 7.834 due to H at C-4 position, and 7.458 due to H at C5 position. last peak at 7.240 is due to proton of chloroform used in analysis.

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