Complicated Organic Chemistry - Essay Example

ORGANIC CHEMISTRY Muhammad Shoaib Shakir Academia Research 29th May 2009 Compound A Percentage of Carbon as calculated is: 62 Dividing thesepercentages by their respective relative molecular masses we get the Ratio by mole as: 62.1/12 Percentage of Hydrogen is calculated as: 10.3% Ratio by mole is: 10.3/1 Percentage of Oxygen is: 27.6% Ratio by mole of oxygen is: 27.5/16 Solution obtained from these calculations show that the compound has a chemical formula of C3H6O Classes of organic compounds which give positive test when treated with silver nitrate are aldehydes but since this does not it can either be a ketone or an alcohol No reaction on the addition of sodium hydrogen carbonate clears the possibility that the compound is a carboxylic acid. This test is quite useful in distinguishing carbonyl compounds from alcohols. If the compound would have been an alcohol it would have been oxidised to its respective carbonyl compound. The inertness of the compound towards Tollen’s and Fehling’s reagent also shows that the compound can not be further oxidised. By the above tests it is confirmed that the compound C3H60 is a ketone known as propanone. The distinct large peak shown by the infra red analysis at 1716 per cm of the compound confirms the presence of the C=O, a double carbon-oxygen bond. If the relative molecular mass of the mass is calculated: (12*3)+(1*6)+16 is equal to 58 as shown by the mass spectrometer The singlet peak shown by the proton NMR at a chemical shift value of 2.1ppm confirms the presence of R-C=O with a methyl group attached Compound B Elemental analysis suggests that the presence of two oxygen atoms in the compound can either be a carboxylic acid or an ester. Presence of the carboxylic acid is confirmed with the addition of sodium hydrogen carbonate as all acids release the effervescence of carbon dioxide during the reaction. Neither alcohols nor carbonyl compounds react with sodium hydrogen carbonate. Hence it is confirmed that the compound is a carboxylic acid, propanoic acid. Peaks shown by the infra red spectrum analysis of the compound at 3600 per cm and 3100 per cm is due to the presence of the O-H hydrogen bond. Due to the strength of this bond the absorption peak is at a higher intensity. The absorption peak at 1716 per cm at the spectrum confirms the presence of the C=O bond. The M-1 peak of the compound is shown at 74 atomic mass unit which is the relative molecular mass of the compound. As far as peaks of other fragments are concerned 57 is due to the fragment of C3H50. Peak at 45 is due to the presence of the COOH functional group 12+32+1, peak at 29 confirms the presence of ethyl group of C2H5 (12*2) + (1*5), 28 is due to C2H4 (12*2) + (1*4), C2H3, (12*2) + (1*3) shows the peak at 27 whereas C2H, (12*2) + 1 shows at 26. A careful look at the peaks results displayed by the NMR of the substance is also required. A singlet peak at a chemical shift value of 11.1ppm confirms the presence of the COOH group. Peak at 2.4ppm is due to the presence of the group R-C=O attached to CH. Applying the n+1 rule we know that the methylene or the CH2 group will be able to interact with the three hydrogen of the methyl or the CH3 group, hence (3+1)=4 peaks or a quadruplet is obtained. Similarly the three hydrogen atoms of methyl are ale to interact with the 2 hydrogen of methylene, hence (2+1) =3 or a triplet peak is displayed. The ratio obtained is determined by the relative heights of the respective peaks. Compound C Percentage of Carbon as calculated is 60.0% Dividing the percentages of the elements present by their relative molecular masses we get Ratio by mole of Carbon as 60.0/12 Percentage of Hydrogen is 13.3% Ratio by mole of Hydrogen is 13.3/1 Percentage of Oxygen is 26.7% Therefore ratio by mole of Oxygen is 26.7/16 Carrying out the above calculations we get the chemical formula of the compound as C3H8O. Negative result of silver nitrate suggests that the compound can either be an alcohol or a ketone. No reaction on the addition of sodium hydrogen carbonate also maintains our assumptions. Reaction with acidified potassium dichromate and its subsequent reduction indicates the presence of an alcohol. Both alcohols and ketones are inert to Tollen’s and Fehling’s reagent. Reaction with ethanoic acid in the presence of sulphuric acid confirms our assumption about the presence of alcohol. The alcohol is propanol or propan-1-ol Peak at 3330 per cm at the infra red spectrum is due to the high absorption range of the O-H group in the compound, whereas peaks centered at 2900 per cm is due to the presence of C-H bonds in the compound. M-1 peak of 60 is the RMM of the compound (12*3) + (1*8) + 16= 60 atomic mass unit, peak at 31 confirms the presence of CH2OH: 12+2+16+1=31, C2H3 gives peak at 27: (12*2) + (1*3), C3H7O :( 12*3) + (1*7) + 16=59. 3.6ppm value at the NMR suggests the presence of R-OH functional group. The single hydrogen is able to interact with the neighboring hydrogen of the CH2 group hence a triplet is obtained. A singlet peak at a chemical shift of 2.2ppm shows the presence of the R-C-O fragment. 6 peaks at 1.6ppm are obtained due to the presence of R3CH where single hydrogen is able to interact with 5 neighboring hydrogen giving 6 peaks. Triplet of 0.9ppm is present due to R-CH3 as three hydrogen interact with the neighboring 2 of the CH2 giving a triplet peak (2+1). The ratio of 2:1:2:3 provides us the different environments of the protons. Compound D Percentages of carbon, hydrogen and an unknown element are 45.9%, 8.9% and 45.2% respectively. Dividing them by their respective relative molecular masses 45.9/12, 8.9/1 and 45.2/35.5, it is found out that the other element is chlorine as the percentage is closest to its RMM. Hence the chemical formula is C3H7Cl is alkyl halide or chloro propane. Preceding test with silver nitrate further confirms the presence of Chlorine as it gives a white precipitate when treated with it. Peaks at 2950 per cm and 2980 per cm as shown by the infra red spectrum show the presence of the C-H bonds present in the compound, whereas peak at 620 per cm confirm the C-Cl bond in the compound. The 3:1 ratio of the peaks shown by the mass spectrometer is due to the isotopic forms of chlorine which are Cl35 and Cl37 respectively. C3H7Cl 35(Isotope) gives a peak at 78: (12*3) + (1*7) + 35, whereas peak at 80 is at C3H7 Cl37. Similarly peaks at 63 and 65 are due to fragments C2H4Cl35: (12*2) + (1*4) + 35 and C2H4Cl 37 responsible for 65. C3H7: (12*3) + (1*7) = 43 and C3H5: (12*3) + (1*5) = 41. Chemical shift value at 4.2ppm is due to the presence of the R-CH2-Hal group where the methylene group interacts with the neighboring three hydrogen of the methyl group giving a (3+1) =4 peaks. Peak at 1.6ppm confirms the presence of R3CH group as it interacts with just one other proton it gives a (1+1) duplet peak. This might be due to a different in the functional group of the compound. The ratio of 1:6 provides the number of protons present in each environment. Necessary reference 1. http://www.scribd.com/doc/13629226/9746-Chem-Data-Booklet-A-level-H2-Chemistry-2008 Read More