Organic Chemistry II -- Lab This nitration of methyl Benzoate is a typical elect
ID: 955515 • Letter: O
Question
Organic Chemistry II -- LabThis nitration of methyl Benzoate is a typical electrophilic aromatic substitution or EAS reaction. In this experiment, we used an EAS reaction to nitrate methyl benzoate (oil of wintergreen). The electrophile in this reaction is the nitronium ion, NO2+, which is formed in situ by the reaction of concentrated nitric acid with concentrated sulfuric acid.
The nitration of methyl benzoate could occur at the ortho, meta, or para position of the aromatic ring. The position of the ring that has the greatest electron density will be the most likely spot for the reaction with the electrophile to occur. We w ill use the melting point of the product to determine which isomer was formed in the reaction.
Data Table of Nitration Mass of methyl benzoate: 0.21 g Mass of nitro methyl benzoate: 0.05g Melting point of nitro methyl benzoate: 60-75 degrees Celsius
Questions: 1) Theoretical Yield of nitro methyl benzoate (please, show calculation) 2) % yield of nitro methyl benzoate (please, show calculation) 3) Which isomer of nitro methyl benzoate was formed from the reaction?
Please, see the picture of the reaction below. Thank You in Advance! duction: The nitration of methyl benzoate is a typical electrophilic aromatic subst eaction. Refer to Chapter 17 of your textbook for more information on EA ns. In this experiment, you will use an EAS reaction to nitrate methyl benzo green). The electrophile in this reaction is the nitronium ion, NO2 , which by the reaction of concentrated nitric acid with concentrated sulfuric ac HNO3 OCH3 H2SO4 OCH3 NO2 The nitration of methyl benzoate could occur at the ortho, meta, or p romatic ring. The position of the ring that has the greatest electron most likely spot for the reaction with the electrophile to occur. The s resent on the aromatic ring initially will determine where the electro is highest and thus where substitution will occur. You will use the
Explanation / Answer
ph-(C=O)-OCH3 ....HNO3/H2SO4.....> NO2-Ph-(C=O)-OCH3
no.of moles of methyl benzoate = (wt/mol.wt) = (0.21/136.15)= 1.54*10^-3 moles
no.of moles of nitro methyl benzoate = (0.05/181.15) = 0.276*10^-3 moles
From the above balanced equation 1 mole of methyl benzoate can give 1 mole of nitro methyl benzoate
so 1.54*10^-3 moles of methyl benzoate can give ...........?
= 1.54*10^-3*1/1 = 1.54*10^-3 moles of nitro methyl benzoate
theoritical yield = no.of moles formed * mol.wt = 1.54*10^-3*181.15 = 0.279 gm.
2)
% yield = (practical mass of nitromethylbenzoate/theoretical mass of nitromethylbenzoate)*100
= (0.05/0.279)*100 = 17.92%
% yield = 17.92 % or 18 %
3)
on methyl benzoate meta position is having more electron density so the incomming electrophile(NO2^+) can attacks on meta position to give meta nitro methylbenzoate
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