Comparing The Molecules In Two Interconvertible Conformation

Of the two alkane liquids studied n-Hexane was determined to have the stronger IMF of attraction by virtue of its experimentally determined lower ?T. Conversely n-Pentane has weaker IMF of attraction due to its larger ?T. As explained at length above, this is because it takes more energy to excite a molecule in liquid to leave (vaporize or become gaseous), when there are stronger IMF of attraction holding the molecules together in liquid form.

The 13C NMR Spectrum reveals the chemical environments of the C-H bonds. In this experiment, the carbons were assigned a number, from C1 to C4. C4 appeared the most downfield because it was the most deshielded because it is double bonded directly to an O and thus the O pulls the most on electron density. C3 is the most upfield because it is not bonded directly to any electron withdrawing groups or electronegative atom. C1 is the second most deshielded because this carbon is involved in a double bond which draws electron density. C2 appears second most upfield because this carbon is single bonded to an oxygen, but it is not as electron pulling as C4 or C1 that is double

In this experiment, the identity of an unknown compound will be deduced from analysis of physical properties, boiling points, IR spectra, chemical tests, and melting points of derivatives. This experiment allows for individual selection of chemical tests, order of steps, and additional analysis, as necessary, based on what is needed to determine identity of unknown.

A ramp rate was first used to obtain the approximate melting point of the unknown before obtaining a more accurate melting point range of the unknown substance. Based on the rapid heating rate of the test, it was determined that the unknown substance was either Trans-cinnamic acid or Benzoin due to the wider melting point range at 132℃. After the second, slower heating rate, it was determined that the unknown might be Benzoin due to the wider melting point range. However, when experimented with the unknown and the Benzoin, the 50/50 mixture melted before the other 2 compounds had begun to change states because it was acting as an impurity instead of melting at the same time. If the unknown and the possibly known substance were the same compound, those two compounds plus the mixture would have melted at the same time. The Trans-cinnamic acid was then compared to the unknown substance, which resulted in two compounds and the mixture melting at the same time, which had a melting point range of 133 to 136.2℃. According to the results of the melting point ranges given, the melting point range of Trans-cinnamic acid should be 132-135℃. The experimented results were 133-136.2℃, which are about a degree higher than what they should be, but the compounds and the 50/50 mixture all melted at the same time, making the unknown substance Trans-cinnamic

melting point of an unknown substance to that of a true compound. This method can be

After experiment A was completed it was exceptionally clear that there was an obvious difference between the melting points of all mixtures, yet Mixture 1 and 2’s melting point ranges are much narrower than that of Mixture 3’s, as seen on Table 1.A. The results from this experiment help to clearly identify the three unknown mixtures by comparing them to the established melting points of naphthalene and benzoic acid. Mixture 1’s melting point range of 123-125.5 degrees Celsius is in the vicinity of benzoic acid’s melting point of 122 degrees Celsius1; a reason for the discrepancy may be the Mel-Temp was set to rise in temperature too quickly. Mixture 2’s melting point range of 82.3-84.2 degrees Celsius matches closely to naphthalene’s melting point of about 80 degrees Celsius1; similarly this process would have been more accurate had the Mel-Temp been programmed to rise in temperature at a slower rate. And finally the composition of Mixture 3 is most likely a combination of the two, due to the low as well as very wide melting point range, both characteristics of a combination of different compounds2.

Also another factor is finding a solvent at which ones compound dissolves in when heat is introduced and insoluble at room temperature, but it should also crystalize when introduced to cold temperatures. My compound had the melting point of trans-stilbene at 121-123°C which, was just lower than the PubChem recorded melting temperature of 122-125°C. Concluded that my neutral organic compound was trans-stilbene because of the mixed melting points similarity.

The experimental value was found to be 120-123°C, while the literature value was found to be 123°C. This value is exact to the literature, showing that the unknown was extremely pure since it had a small range for its melting point and matched the literature value. There was a decent amount of solid formed in the final portion of the experiment. After reading the unknown, it was proven that the unknown was Isopropyl Isobutyrate. NMR as a system shows the number of signals of how many different kinds of protons are present, the location of the signals based on how shielded or deshielded the proton is, the intensity by showing the number

Understanding boiling point, density, mass, and volume and being able to classify them as either intensive or extensive properties will enhance our knowledge of these properties and ultimately innovate our daily lives. For instance, unknown material can be identified by knowing the boiling point, density, mass, and volume of the substance. Intensive properties aid the process of identifying unknown substances for the reason that they don’t depend on the quantity of the substance nor do they alter. Although, extensive properties do alter due to quantity that is present, they differentiate between matters. Therefore, both of the experiments will aid in the knowledge and

The known heat of vaporization value used for calculating the percent error was associated with pure n-heptane. The structure of n-heptane is a straight chain of carbons with their associated hydrogens attached. The straight structure of the n-heptane allows a large surface area for other molecules to layer upon. The greater surface area creates stronger attractive and binding forces. This requires more energy in order for the liquid molecules to become gaseous, thus a larger heat of vaporization when compared to non-straight heptane structures. The branched structures of heptane do not have as much surface area between the heptane molecules. Less surface area means less binding forces, thus a lower heat of vaporization. The lower experimental heat of vaporization of heptane may have been caused by a majority of the used heptane having a branched

A compound can be identified by its physical properties like melting point, boiling point, and density. The melting point can show how pure a compound is. The purer the compound, the higher the melting point and the narrower it’s melting point range. Adding continuous amounts of impurity to a pure compound will cause the melting point to decrease. In this experiment the melting point of crystalline substances are observed to see how pure the compound is. This was done by putting the compounds into a capillary tube. The capillary tube is then inserted into a melting point apparatus and heated. The temperature at which the compound begins to liquefy and when the compound becomes completely liquefied will show how pure the substance is. The melting

In 1,3-disubstitution pattern, if both groups are in cis formation, they will be equatorial. Therefore, in this case, cis isomer is more stable. Chair flipping provides a conformation of cis isomer with both

More recently, distinctions between states have been based on differences in molecular interrelationships. Solid is the state in which intermolecular attractions keep the molecules in fixed spatial relationships. Liquid is the state in which

The above principle may also be used to explain the thermochromism of iodine in solvents such as ethyl stearate. Iodine in ethyl stearate is brown at room temperature which on heating above 80[degrees]C becomes violet. At room temperature, the polar ends of the long chain molecules form CT complexes with iodine.

downfield in chemical shift and this was used to distinguish the ratio of product configuration