Empirical Formula Lab Report
Given the unknown colourless liquid it is essential that the structure is derived. Firstly it is important to determine the percentage composition of elements to work out the empirical formula. The empirical formula was found to be C10H12O and the mass of the unknown was 148.09 m/z which when calculating the molecular weight of the empirical formula it did equal 148.09 g mol -1. This means that the empirical formula is also the molecular formula. As 12.01x10 carbons +1.008x12 hydrogens + 16= 148.09 09 g mol -1.
From this knowledge the unknown molecule must include either an alcohol, aldehyde, ketone, ether or methoxy- functional group.
Secondly it is important to calculate the degrees of unsaturation in the molecule. The degrees of unsaturation …show more content…
will allow knowledge of the number of double bond equivalents in the molecule. These could be double bonds or a ring system. This was calculated via equation one and shows that there was 10 hydrogens short of having all single bonds in the structure. This means that there are five double bond equivalents in the molecule.
Using the IR the functional groups in the molecule were found. The most recognisable peak was at1698 cm-1 this is a carbonyl stretch which has a very strong intensity due to the change in electronics, because carbon and oxygen have different molecular weights and stretch differently. From the recognition of a carbonyl in the molecule this accounts for one double bond equivalent. From this the unknown compound must either be a ketone or aldehyde as an alcohol, ether and methoxy functional group only contain a C-O stretch which is not present in the IR that was obtained for the compound. There is also a C-H stretch present at 2981 cm-1 which is an alkyl carbon which is sp3 hybridised. In order for the molecule to have a sp3 hybridised carbon hydrogen bond, there must be either CH3 groups attached to CH2 or CH groups which do not involve a double bond. There is also a very distinct C-H stretch at 2850 cm-1 which is due to a C-H group of an aldehyde or a ketone. This is very distinct, it is two sharp peaks to the left of the sp3 hybridised peak at 2961 cm-1. Also present in the infrared spectrum are two peaks at 1605 and 1574 cm-1 which is characteristic of an aromatic skeletal structure. This is likely due to the aromatic benzene, which contains three double bonds and a ring which is four double bond equivalents. This means that with the ring system and the carbonyl allow for the five double bond equivalents calculated in equation 1. The final peak, which allows for initial characterisation of the unknown molecule is a peak at 826 cm-1. This peak is in the fingerprint region, which is notoriously difficult to characterise, but as the peak is strong it allows the characterisation of a 1,4 disubstituted benzene derivative. This peak is consistent with the aromatic skeletal structure which has already been identified in the infrared spectrum. Overall from the infrared spectrum the information that the unknown compound contains a benzene derivative which has 1,4 disubstitution, as well as having either a ketone or an aldehyde present. When comparing the IR spectrum to that of the molecule in figure 1 it is the same as the carbonyl stretch from the aldehyde has been identified as well as the C-C stretch from the aromatic ring and the disubstited benzene. (3)
When examining the 1H NMR there are 5 different peaks which correspond to five different proton environments in the unknown molecule.
The most distinct peak is that of a singlet at 9.95 ppm. This has a relative intensity of one. The more deshielded a molecule the higher the chemical shift. Because the oxygen of the carbonyl bond is inductively withdrawing the electron density will reside favourably on the oxygen, which means that the electron density associated with the hydrogen is reduced which results in a high chemical shift. A chemical shift of this is only typical for alcohols or aldehydes attached directly to an aromatic ring, or nitrogen containing molecules of which is not applicable for the unknown molecule are there is no nitrogen in the molecular formula. From this peak the only possible functional group is an aldehyde as a ketone does not have a hydrogen to give a peak at this chemical shift. The relative intensity does fit with the configuration of the carbon only being bound to one hydrogen. This means that the aldehyde group must be on one end of the molecule as an aldehyde cannot be attached to more than one carbon. This peak is a singlet which indicates that the aldehyde proton is not coupled to any other proton this is because the carbon adjacent does not contain any hydrogens. This would be consistent with the aldehyde group being directly attached to the benzene ring. If the peak was between 9.4 and 10 ppm this would be consistent with the aldehyde group being bound to a linear carbon system. As this peak is between 9.7 and 10.5 ppm2 this expected for an aldehyde proton attached to an aromatic
ring.
The peaks at 7.37 and 7.80 ppm are very typical of benzene hydrogens as in a saturated ring system the carbons are all sp2 hybridised which means that the there is an induced magnetic field which then creates a local magnetic field in the same direction as the external field outside of the aromatic ring which is where the protons are located; because of this, this leads to deshielding of the proton nuclei resulting in a much higher chemical shift compared to an C-H stretch of a CH3 group. Both of these peaks have an integration of two which is consistent with four hydrogens present in the aromatic ring. This is also consistent with the peak that was identified in the infrared spectrum, which indicated that the unknown compound was a disubstituted benzene. For a disubstituted benzene there is only four hydrogens present. Each of these peaks have an integration of two. As there is a difference in chemical shifts for the benzylic protons it indicates that the unknown compound is not symmetrical. Using equation 2 and 3 it is seen that the proposed molecule is highly likely as equation 2 shows that protons 1 and 5 is the higher chemical shift of the aromatic protons. This is expected due to the electron withdrawing group of the aldehyde which means that the protons 1 and 5 will have less electron density surrounding them, therefore, they are deshielded resulting in a higher chemical shift. For equation 3 it shows that the proposed molecule fits the chemical shift calculated where the isopropyl group is ortho to protons 2 and 4 and the aldehyde is meta to these protons. These protons are at a lower chemical shift as the isopropyl group is electron donating resulting in more electron density on the aromatic protons, so they are shielded resulting in a lower chemical shift. These peaks are both doubles which indicates that each hydrogen has one proton to which it is coupled to on an adjacent carbon. This also fits the molecule proposed in figure 1 as proton 1 couples with proton 2 and proton 4 couples to proton 5 and vice versa. The ring being disubstituted also fits this multiplicity as through the substitution it means that each proton only has one neighbouring proton.
From this knowledge the unknown molecule must include either an alcohol, aldehyde, ketone, ether or methoxy- functional group.
Secondly it is important to calculate the degrees of unsaturation in the molecule. The degrees of unsaturation …show more content…
will allow knowledge of the number of double bond equivalents in the molecule. These could be double bonds or a ring system. This was calculated via equation one and shows that there was 10 hydrogens short of having all single bonds in the structure. This means that there are five double bond equivalents in the molecule.
Using the IR the functional groups in the molecule were found. The most recognisable peak was at1698 cm-1 this is a carbonyl stretch which has a very strong intensity due to the change in electronics, because carbon and oxygen have different molecular weights and stretch differently. From the recognition of a carbonyl in the molecule this accounts for one double bond equivalent. From this the unknown compound must either be a ketone or aldehyde as an alcohol, ether and methoxy functional group only contain a C-O stretch which is not present in the IR that was obtained for the compound. There is also a C-H stretch present at 2981 cm-1 which is an alkyl carbon which is sp3 hybridised. In order for the molecule to have a sp3 hybridised carbon hydrogen bond, there must be either CH3 groups attached to CH2 or CH groups which do not involve a double bond. There is also a very distinct C-H stretch at 2850 cm-1 which is due to a C-H group of an aldehyde or a ketone. This is very distinct, it is two sharp peaks to the left of the sp3 hybridised peak at 2961 cm-1. Also present in the infrared spectrum are two peaks at 1605 and 1574 cm-1 which is characteristic of an aromatic skeletal structure. This is likely due to the aromatic benzene, which contains three double bonds and a ring which is four double bond equivalents. This means that with the ring system and the carbonyl allow for the five double bond equivalents calculated in equation 1. The final peak, which allows for initial characterisation of the unknown molecule is a peak at 826 cm-1. This peak is in the fingerprint region, which is notoriously difficult to characterise, but as the peak is strong it allows the characterisation of a 1,4 disubstituted benzene derivative. This peak is consistent with the aromatic skeletal structure which has already been identified in the infrared spectrum. Overall from the infrared spectrum the information that the unknown compound contains a benzene derivative which has 1,4 disubstitution, as well as having either a ketone or an aldehyde present. When comparing the IR spectrum to that of the molecule in figure 1 it is the same as the carbonyl stretch from the aldehyde has been identified as well as the C-C stretch from the aromatic ring and the disubstited benzene. (3)
When examining the 1H NMR there are 5 different peaks which correspond to five different proton environments in the unknown molecule.
The most distinct peak is that of a singlet at 9.95 ppm. This has a relative intensity of one. The more deshielded a molecule the higher the chemical shift. Because the oxygen of the carbonyl bond is inductively withdrawing the electron density will reside favourably on the oxygen, which means that the electron density associated with the hydrogen is reduced which results in a high chemical shift. A chemical shift of this is only typical for alcohols or aldehydes attached directly to an aromatic ring, or nitrogen containing molecules of which is not applicable for the unknown molecule are there is no nitrogen in the molecular formula. From this peak the only possible functional group is an aldehyde as a ketone does not have a hydrogen to give a peak at this chemical shift. The relative intensity does fit with the configuration of the carbon only being bound to one hydrogen. This means that the aldehyde group must be on one end of the molecule as an aldehyde cannot be attached to more than one carbon. This peak is a singlet which indicates that the aldehyde proton is not coupled to any other proton this is because the carbon adjacent does not contain any hydrogens. This would be consistent with the aldehyde group being directly attached to the benzene ring. If the peak was between 9.4 and 10 ppm this would be consistent with the aldehyde group being bound to a linear carbon system. As this peak is between 9.7 and 10.5 ppm2 this expected for an aldehyde proton attached to an aromatic
ring.
The peaks at 7.37 and 7.80 ppm are very typical of benzene hydrogens as in a saturated ring system the carbons are all sp2 hybridised which means that the there is an induced magnetic field which then creates a local magnetic field in the same direction as the external field outside of the aromatic ring which is where the protons are located; because of this, this leads to deshielding of the proton nuclei resulting in a much higher chemical shift compared to an C-H stretch of a CH3 group. Both of these peaks have an integration of two which is consistent with four hydrogens present in the aromatic ring. This is also consistent with the peak that was identified in the infrared spectrum, which indicated that the unknown compound was a disubstituted benzene. For a disubstituted benzene there is only four hydrogens present. Each of these peaks have an integration of two. As there is a difference in chemical shifts for the benzylic protons it indicates that the unknown compound is not symmetrical. Using equation 2 and 3 it is seen that the proposed molecule is highly likely as equation 2 shows that protons 1 and 5 is the higher chemical shift of the aromatic protons. This is expected due to the electron withdrawing group of the aldehyde which means that the protons 1 and 5 will have less electron density surrounding them, therefore, they are deshielded resulting in a higher chemical shift. For equation 3 it shows that the proposed molecule fits the chemical shift calculated where the isopropyl group is ortho to protons 2 and 4 and the aldehyde is meta to these protons. These protons are at a lower chemical shift as the isopropyl group is electron donating resulting in more electron density on the aromatic protons, so they are shielded resulting in a lower chemical shift. These peaks are both doubles which indicates that each hydrogen has one proton to which it is coupled to on an adjacent carbon. This also fits the molecule proposed in figure 1 as proton 1 couples with proton 2 and proton 4 couples to proton 5 and vice versa. The ring being disubstituted also fits this multiplicity as through the substitution it means that each proton only has one neighbouring proton.