Essay Example on The next signal is the methyl group closest to the Aldehyde

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The degrees of unsaturation calculated was one which again eliminates the possibility of a ring with the peak at 1700 cm 1 C O There is also C H stretches which indicates an aldehyde On the H NMR graph the singlet at around 10 ppm confirms that The peak at 1ppm is a triplet which illustrates the outer methyl group The second signal indicates protons on the carbon next the outer methyl group The next signal is the methyl group closest to the aldehyde and is a sextet The calculated degrees of unsaturation for this formula is 1 which most likely eliminates the possibility of there being a ring due to the oxygen and the lack of a 6 8ppm peak on the H NMR spectrum The first peak on the IR graph shows sp3 hybridization which also confirms no aromatic ring There is also a peak at 1700 cm 1 which indicates a C O The H NMR graph illustrated two peaks The first peak is a triplet which illustrates the outer methyl groups The symmetrical signals also show the structure is symmetrical The second signal around 2 5 ppm illustrates the inner protons nearest to the double bond 



The degrees of unsaturation calculated was 1 which eliminated the possibility of there being a ring because of the two oxygens most likely formed a carboxylic acid The H NMR graph proved it correct as no peaks were shown from the 6 8 ppm region The first signal in the spectrum was a triplet portraying the outer methyl group while the next signal was the CH2 next to the methyl group and the next signal was the carbon closest the carboxylic acid The peak at 11 ppm was a singlet which proved a carboxylic acid exists The IR shows a broad peak which is evident an alcohol group was present while showing a sp3 hybridization It also showed a C O since it shows a sharp peak around 1700 cm 1 Above is the structure concluded from both the graphs The degrees of unsaturation are 4 which indicates a very high chance that the structure has a ring The peak shows both a sp2 and sp3 hybridization which is evident there is a ring Also the H NMR graph also shows two peaks between the 6 8 ppm region which is where the aromatic rings show up Furthermore the peaks in the region indicate two difference attachments because of the two peaks The very first signal is a methyl group that is the least shielded because it is a triplet 



The next signal is a quartet and from concluding from the formula given it would be an ethyl group and the halogen being attached on another part of the aromatic ring Above is the deduced structure of C4H9Br The degrees of unsaturation calculated was 0 which eliminates the choice of any rings or double bonds The IR graph illustrated that as sp3 hybridization was show by the peak being behind the 3000 cm 1 line The H NMR graph helped further propose a structure The first signal was a triplet which was the outer methyl group The second signal at about 1 5 ppm illustrated the methyl nearest to the carbon holding the bromine The third signal at 1 7 ppm illustrated the carbon between the carbon holding halogen and the outer methyl group The last signal at 3 7 ppm is a sextet and illustrates the carbon holding the halogen The structure above was the deduced structure of C4H8O2 after comparing both the IR graphs and 1H NMR graphs The degree of unsaturation was first calculated which was 1 The IR graph should a sp3 hybridization which eliminated any choice of a ring also with the fact that a C O was a present at the sharp peak at 1700 cm 1

The H NMR graph portrayed 3 difference signals The outer first signal was at 1 3 ppm was a triplet which was the less shielded proton in the structure The methyl group closest to the C O Carbonyl group was the second peak at about 2 1 ppm The last signal was quartet around 4 1 ppm which portrayed the protons between the carbonyl group and the first signal methyl The obtained Infrared spectrum from the unknown sample is displayed above Figure 1 At about 3500 cm 1 there is a broad peak which indicates an alcohol group Due to the fact the degree of unsaturation was calculated to 5 it was anticipated that there would be a carbonyl group to fulfill the 5 degrees of unsaturation This is because a benzene ring was predicted due to the formula and the C H stretching on the NMR graph The 1H NMR further validated the structure The peaks 6 8 ppm was evidence that a benzene ring was present The ring at about 7 3 ppm was a triplet which portrayed the protons furthest from the carbonyl group The second peak was doublet which are the protons closest to the carbonyl group The peak at about 12 ppm was a singlet which was evident that the carboxylic acid was present as they occur from 10 13 ppm


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