Identity Of 2-Pentanol
Based on the spectra generated by C5H12O, the identity of the compound was 2-pentanol. The first step in identifying the compound was finding the degree of unsaturation. For C5H12O, the degree of unsaturation was zero, which means that the compound only has single bonds. Knowing that there were only single bonds led to the assumption that the O in the compound was either an alcohol or an ether. The IR spectrum helped in determining which functional group was present. The peak that determined the identity of the functional group was peak A at 3352.71 cm-1. The peak corresponds to an alcohol functional group because alcohol spectra have an O-H stretch around 3200-3700 cm-1. Additionally, peak A was smooth and broad, which is another characteristic
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In the13C NMR, there were 5 peaks, which meant that there were five different non-equivalent carbons. With this information, it can be deduced that the compound does not have a methyl group attached to a four-carbon chain because the methyl carbon and the first carbon on the chain would be equivalent. This would result in an 13C NMR with 4 peaks, so the compound would need to have a five-carbon chain. With the assumption that the structure was five-carbon chain, it could be deduced that the structure was either a primary or secondary alcohol because straight chained alkanes cannot make tertiary …show more content…
The chemical shift for the C-O ranges between 50-80 ppm. Carbon 5 has a ppm of 67, meaning it represents the carbon attached to oxygen. If the alcohol was primary, then the carbon bonding would be RH2C-O. If the alcohol was secondary, then the carbon bonding would be R2HC-O. RH2C-O has a chemical shift of 45-65 ppm while R2HC-O has a chemical shift of 60-80 ppm. Carbon 5’s ppm value of 67 was within the R2HC-O chemical shift meaning that the alcohol was secondary. This meant that the OH group could only be bonded to the second or third carbon, which makes the compound either 2-pentanol or 3-pentanol. The 1H NMR was used to determine which carbon was bonded to the OH
In the13C NMR, there were 5 peaks, which meant that there were five different non-equivalent carbons. With this information, it can be deduced that the compound does not have a methyl group attached to a four-carbon chain because the methyl carbon and the first carbon on the chain would be equivalent. This would result in an 13C NMR with 4 peaks, so the compound would need to have a five-carbon chain. With the assumption that the structure was five-carbon chain, it could be deduced that the structure was either a primary or secondary alcohol because straight chained alkanes cannot make tertiary …show more content…
The chemical shift for the C-O ranges between 50-80 ppm. Carbon 5 has a ppm of 67, meaning it represents the carbon attached to oxygen. If the alcohol was primary, then the carbon bonding would be RH2C-O. If the alcohol was secondary, then the carbon bonding would be R2HC-O. RH2C-O has a chemical shift of 45-65 ppm while R2HC-O has a chemical shift of 60-80 ppm. Carbon 5’s ppm value of 67 was within the R2HC-O chemical shift meaning that the alcohol was secondary. This meant that the OH group could only be bonded to the second or third carbon, which makes the compound either 2-pentanol or 3-pentanol. The 1H NMR was used to determine which carbon was bonded to the OH