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Intermolecular forces essay
A term paper on intermolecular forces
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I believe my molecule could be a Ketone because of its very high boiling point and solubility. The boiling point is 145.5˚C. The molecule I have been given is very soluble in water. There are dispersion and Dipole-dipole forces between each Ketone molecule. Oxygen is more electronegative than carbon, so it tends to pull the electrons towards the oxygen. One of the two pairs of electrons from the carbon-oxygen bond is easily drawn towards the oxygen. This makes the bond highly polar. This is why the dipole-dipole forces in ketones are so strong.
Because the carbonyl group interacts with water by hydrogen bonding, ketones are typically more soluble in water than other molecules. I have predicted that my molecule is a ketone because of
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Immediately I know that this cannot be my molecule as the boiling point is almost a third of my given boiling point. This is because of the intermolecular forces in this molecule. There are no hydrogens directly attached to an oxygen so that cancels out the possibility of any hydrogen bonding. However, there is dispersion forces and dipole-dipole interactions. These are the two weakest intermolecular forces, causing the low boiling point. Esters are fairly soluble in water, but the solubility decreases with the chain length. For ethylmethanoate its solubility is 10.5g/100mL. The reason for the solubility is that esters cannot hydrogen bond with themselves, they can hydrogen bond with water however. The poor solubility of ethylmethanoate also clarifies that my molecule is not an …show more content…
The -diol end to the name means that there is two alcohol functional groups. An alcohol can hydrogen bond with other molecules. Being the strongest intermolecular forces, the energy needed to separate them will be a lot greater. Having two alcohol functional groups makes the energy needed greater again. The double hydrogen bonding makes the boiling point very high at 195˚C. There is no definite solubility for 1,1-cyclopropanediol so it cannot be commented on. According to the boiling point, 1,1-cyclopropanediol is not my molecule.
The final isomer that hasn’t been explored yet is 1,3-dioxolane. The cyclic shape of this isomer changes the intermolecular forces in the molecule. There are no functional groups that branch off of this isomer, so there cannot be any hydrogen bonds or dipole-dipole interactions. The only possible intermolecular force is dispersion forces. Being the weakest force, the molecules are don’t need much energy to separate each other. This is the reason for the low boiling point. 1,3-dioxolane has a boiling point of 73˚C. It is impossible for my molecule to be 1,3-dioxolane because of its boiling
The C-H (sp3) hydrogens from our product displayed at wavelength 2959 cm-1 correlates to the methyl groups located on the ends of isopentyl acetate4. A really prominent, strong peak located at 1742 cm-1 shows that a C=O ester stretch is located in the product, along with at 1244 cm-1 the spectrum shows a strong peak representing the C(=O)-O stretch that is crucial to the structure of isopentyl acetate. Shown in my IR spectrum is a weak O-H (H-bonded) peak at 3464 cm-1 which shows that I have an impurity of isopentyl alcohol in my product. Isopentyl alcohol has similar boiling points and density as my product so the impurity could have easily boiled out with the isopentyl acetate during distillation. The isopentyl alcohol was also present in my 1H-NMR spectrum backing up the impurity peak at 3464
The boiling point of the product was conducted with the silicone oil. Lastly, for each chemical test, three test tubes were prepared with 2-methylcyclohexanol, the product, and 1-decene in each test tube, and a drop of the reagent were added to test tubes. The percent yield was calculated to be 74.8% with 12.6g of the product obtained. This result showed that most of 2-methylcyclohexanol was successfully dehydrated and produced the product. The loss of the product could be due to the incomplete reaction or distillation and through washing and extraction of the product. The boiling point range resulted as 112oC to 118oC. This boiling point range revealed that it is acceptable because the literature boiling point range included possible products, which are 1-methylcyclohexene, 3-methylcyclohexene, and methylenecyclohexane, are 110 to 111oC, 104oC, and 102 to 103 oC. For the results of IR spectroscopy, 2-methylcyclocahnol showed peaks at 3300 cm-1 and 2930 cm-1, which indicated the presence of alcohol and alkane functional group. Then, the peak from the product showed the same peak at 2930 cm-1 but the absence of the other peak, which indicated the absence of the alcohol
As shown in figure 2, the percentage of each isomeric alcohol in the mixture had been determined. The hydrogen atom on the carbon atom with the hydroxyl group appear at around 4.0 ppm for borneol and 3.6 ppm for isoborneol. The product ratio has been determined by integrating the peaks. A ratio of 6:1 for the Isoborneol/borneol ratio was expected and is validated by the calculations shown above, with isoborneol percentage at 83.82% and 16.17% of borneol. A CHCl3 group noted at around 7ppm and a CH2Cl2 at around 3.5ppm.
In this lab 4-tert-butylcyclohexanone is reduced by sodium borohydride (NaBH4) to produce the cis and trans isomers of 4-tert-butylcyclohexanol. Since the starting material is a ketone, NaBH4 is strong enough to perform a reduction and lithium aluminum hydride is not needed. NaBH4 can attack the carbonyl group at an equatorial (cis) or axial (trans) position, making this reaction stereoselective. After the ketone is reduced by the metal-hydride, hydrochloric acid adds a proton to the negatively charged oxygen to make a hydroxyl group. The trans isomer is more abundant than the cis based on the results found in the experiment and the fact that the trans isomer is more stable; due to having the largest functional groups in equatorial positions.
The IR spectrum that was obtained of the white crystals showed several functional groups present in the molecule. The spectrum shows weak sharp peak at 2865 to 2964 cm-1, which is often associated with C-H, sp3 hybridised, stretching in the molecule, peaks in this region often represent a methyl group or CH2 groups. There are also peaks at 1369 cm-1, which is associated with CH3 stretching. There is also C=O stretching at 1767 cm-1, which is a strong peak due to the large dipole created via the large difference in electronegativity of the carbon and the oxygen atom. An anhydride C-O resonates between 1000 and 1300 cm-1 it is a at least two bands. The peak is present in the 13C NMR at 1269 and 1299 cm-1 it is of medium intensity.
By adding the antifreeze, it interfered with the vaporization of the water therefore lowering the amount of evaporation that occurred which then lowered the vapour pressure. Boiling point elevation is the difference in temperature between the boiling point of a solution and the boiling point of the pure solvent
Methanol and Ethanol have differences as Methanol melts at a higher temperature and boils at a lower temperature than Ethanol. Higher alcohols, which include Butanol and Propanol, have a higher molecular weight and this is why Butanol is used in perfumes. Ethanol, which is sugar based, with its low freezing point, has a specific use as an antifreeze for cars and other vehicles. GRAPH Tripod Matches Goggles Method: To begin with, I choose one of the four different alcohols. I weigh beforehand in the spirit burner.
activation energy than what is required to turn camphor into borneol. Though borneol is the more stable product, the energy requirements to form isoborneol are lower because the borohydride is adding to the less sterically hindered point on the carbonyl carbon. The product made is then mostly (85%) isoborneol. More borneol would be expected if more energy was available during the reaction.
The solvent should be easily removed from the purified product, not react with the target substances, and should only dissolve the target substance near it’s boiling point, but none at freezing. A successful recrystallization uses minimum amount of solvent, and cools the solution slowly, if done to fast, many impurities will be left in the crystals. Using the correct solvent, in this case ice water and ethyl acetate, the impurities in the compound can be dissolved to obtain just the pure compound. A mixed solvent was used to control the solubility of the product. The product is soluble in ethanol an insoluble in water. Adding water reduced solubility and saturates the solution and then the crystals
Therefore because the hydrogen and oxygen atoms are different in size and electronegativity the water molecule is non-linear and dipolar. When two water molecules get close to one another, the oppositely attracted parts of the molecules attract each other. This type of attraction is called hydrogen bonding. This polarity means that individual water molecules can from hydrogen bonds with up to four other water molecules. Although these individual hydrogen bonds are weak, together they make water a much more stable substance.
Methanol was the most polar among 3 alcohols used in this part, hence was soluble in water as both water and methanol were polar. However, methanol was partially soluble in hexane because the Van der Waals interaction between methanol
molecules its size it would have a boiling point of -75øC and a freezing point of -125øC4.
The difference in boiling point can be attributed to the differences in functional groups, intermolecular forces, as well as a molecules size and shape. The first separation we can create between the allocated molecules, to identify which has the highest boiling point, is done by identifying and using the functional groups of the molecules. Pentanal, is an aldehyde, whilst 1-pentanol, 1-hexanol, and 3-methyl-1-butanol, are all alcohols, which shows that pentanal, has the weakest intermolecular force (dipole-dipole interactions) present in the allocated molecules. Thus shows that pentanal would have the lowest boiling point, which is shown by the literature as well.
The obtained NMR spectra was very similar to that of the predicted results. The methyl hydrogen group had a single peak at 2.153 ppm. The results were a little off from the predicted shift of 2.2 – 2.9 ppm but not by much. Next in the spectra, were the hydrogens in the aromatic ring with a range of 7.065-7.436 ppm consisting of multiple peaks, these also matched up to the predicted shift of 6.5-8.0 ppm. The last signal belonged to that of the hydrogen in the amide that had a peak at 7.571 ppm which is well in the range of the predicted shift it 5.0 – 9.0 ppm. Overall the NMR proved that the purified acetanilide had very little impurities shown in the