Chemistry 421 Spring 2016 Applied Assignment 2: individual Spectroscopy problems

Question

Chemistry 421 Spring 2016 Applied Assignment 2: individual Spectroscopy problems (25 its) Due May 10 at 5 PM (not accepted late) The "Individual spectroscopy problems" are composed of a set.of FOUR spectroscopy unknown problem. Each student gets a different set of four problems (although some students will share one or tw problems with another). The problems provide the needed data to determine the structure of an "unknown". Your task is to use and interpret all of the data to (1) determine the structure of your unknown and draw it, and (2) interpret the appropriate peaks on the IR, 1H-NMR, 13C-NMR and MS spectra that support your assignment. You assigned problems are listed below. (Must pick up a hard copy in class in order to get assigned problems)-Turn in this sheet with your answers. This takes a while, so do NOT wait until the last minute to start. These problems are available at the following link: http://www3.nd.edu/-smithgrp/structure/workbook.html Click on "Do the Problems", and you can pull/print the data for your four problems. For full credit for a given problem, you must include all of the following in your answer 1. Problem number and molecular formula 2. Correct organic structure for unknown 3. Printouts of all spectra with peaks properly assigned 4. Brief (1 paragraph) description of how you arrived at your answer Grading: max score possible for each problem easy: 7 points medium: 8 points each (x 2 = 16 points) hard: up to 7 bonus points (total possible score: 32/25) This takes a while, sol do NOT wait until the last

Explanation / Answer

Molecular formulae is C9H10O3

It is obvious that the hydrocarbon part is not saturated as for saturated hydrocarbons general formulae is CnH2n+1.

So, the compound must be highly unsaturated hydrocarbon (with 4 double bonds or 2 double bond and 1 triple bond) or the hydrocarbon part must be aromatic.

But the 1H NMR spectra clearly shows highly symmetric signals in aromatic region. So this must be an aromatic compound. This is further supported by the IR spectra : 816 cm-1 ( out of plane C-H bending ); 3127 cm-1 (aromatic C-H stretch) and 1514 cm-1 (aromatic C-C stretch).

1H NMR clearly indicates the presence of a carboxyl group which gives signal in 12 ppm with deuterium exchange. It also gives signal in IR spectra at 1701 cm-1 due to carbonyl C-O stretch.

From the general formulae, we can easily figure out that we are left with one oxygen atom. This may come from carbonyl group, aldehyde group, alcohol group or ether group. There are no other signal in the range of 1690-1740 cm-1 in IR spectra, so aldehyde and ketone groups are ruled out. Alcohol group is not there otherwise there would have been a wide band in the range of 3100-3300 cm-1 inIR spectra. So, the left out oxygen must come from an ether group.

From the above discussion, we can surely say that the compound is 4-Methoxyphenylacetic acid. It has 7 non-equivalent carbon atoms as indicated by 13C NMR spectra and it fully justifies the 1HNMR spectra also.

(Sorry, I could not include pictorial representation due to some technical error. Just google the name and you will find the structure)

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