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Essay on Qualitative Synthesis Lab Report

The purpose of this experiment was to find the identity of the unknown compound #12. This was done both qualitatively and quantitatively. The qualitative observations were used to determine the functional group that was in the unknown, the quantitative results were used to determine the molar mass through Duma’s method. The types of compounds that were used in testing were all functional groups, Aldehyde, Ketone, Alcohol and, Ester. Functional groups are a portion of a molecule that give the molecules its properties, they are the centers of chemical reactivity.

The structures of the functional roups used in testing are displayed in Figure 1. . Figure 1. 0 Structure of Functional Groups to Distinguish from Within the Lab [1] To identify each of the functional groups 4 different tests were used. All of these tests involved comparison between the unknown #12 and the known functional group. For all of the tests besides the ester test, it involved comparisons between how certain reagents reacted with the known compound and the unknown. If the same reaction occurs in both it’s assumed that they both have the same functional groups, however if they are different reactions then they do not have the same unctional groups.

For the ester functional group, a simple fragrance test is suitable. If the fragrance is the same then it can be assumed esters are present in the unknown. Experimental Method First the qualitative analysis was started. For each of these tests, except the ester test, one sample of the unknown and another for the known was made, and then the 2 test tubes were compared. For aldehydes Tollen’s reagent was used, 1 mL was placed into both test tube along with one have 2 drops of the unknown #12 and another with 2 drops of the aldehyde.

For Ketones a reagent called phenylhydrazine (C6H5NHNH2) was sed, 1 mL of it was added to both test tubes with one test tube having 2 drops of the unknown and the other the 2 drops of ketone. For the alcohols 3 test tubes were used, 2 mL of Lucas reagent was added to each test tube with one having 5 drops of a primary alcohol, another with secondary alcohol and the last one with unknown #12. Next they were put into boiling water and observed until the secondary alcohol turned cloudy. The final qualitative test was for the Esters, both the unknown and the known’s fragrance were compared. Next, was the quantitative test.

An Erlenmeyer flask with aluminum foil and a ubber band mass was measured and then filled with 4 mL of unknown #12. The flask was then covered and had a hole poked through the foil. Next, it was placed into boiling water. After the liquid was fully evaporated it was taken out, wiped dry and, allowed to cool. Then the mass was measured. Then the lab station was cleaned and the waste properly disposed of. [1] Discussion The aldehyde test involved Tollen’s reagent with an aldehyde. Tollen’s agent contains Ag(NH3)2+ (aq), this within the solution would oxidize and aldehyde into a carboxylic acid [2].

What also ccurs within the reaction is that the silver ions would become reduced into a metallic silver [2]. This is what causes the test tube to turn a silver colour as the metallic silver attaches onto the sides of the test tube. For ketones the reagent that was used was phenylhydrazine (C6H5NHNH2). The reaction between the two is called a nucleophilic addition- elimination reaction [3]. When the two are added together the phenylhydrazine breaks the double bond of the oxygen to the central carbon of the ketone by losing 2 hydrogens of its own resulting in a solution of solid phenylhydrazone and water [3].

This explains why solids are visible for the qualitative tests. For alcohols the reagent is Lucas reagent, this is a solution of zinc chloride and concentrated HCI [1]. The reaction that occurs between Lucas reagent and secondary alcohols instead of primary alcohols is a nucleophilic substitution [4]. Only alcohols that generate stable carbocation intermediates are able to undergo the reaction [4] thus explaining why primary alcohols are unable to react with it. The acid protonates the OH group of the alcohol creating C- OH2+, this then reacts with the Cl- creating an alkyl halide [4].

The reason the solution becomes cloudy is due to the fact that alkyl chloride builds up due to it being not water soluble [4]. The ester test was a simple smell comparison between the two. As explained above that was the chemical break down of what occurs within those reactions, now to compare to the actual observations to determine the unknown. The determined compound was composed of a ester, this was determined throughout the qualitative analysis. First, the aldehyde test was performed. As shown in the data sheet the known solution of aldehyde began turning silver and the silver began sticking to he sides of the test tube.

For the unknown it was perfectly clear as it was before. Therefore, aldehydes are ruled out. Next the test for ketones, the reagent phenylhydrazine has an orange colour and when reacted with a ketone forms solid crystalline phenylhydrazone. In the unknown however, the solution just turned orange with no crystal formations. Thus, ketones are also ruled out. For the alcohol tests the primary alcohol did not become cloudy, the secondary did and the unknown did not become cloudy as well. This means that a secondary alcohol can be ruled out however, the possibility of a primary alcohol cannot e ruled out immediately.

The last test was for esters and involved smelling and comparing the fragrance, the odours matched almost exactly the same, the unknown and the ester smelled similar to nail polish remover and both were a very strong smell. This ruled out the primary alcohol and confirmed the functional group was an ester. Dumas’ method was a very simple method using PV=nRT in order to calculate to molar mass of the unknown. This method assumes that the liquid neither dissociates or associates in going from liquid to vapour, thus the molecular mass of the liquid would be the same of the apour [1].

With Dumas’ method the resulting molar mass was 73. 44 g/mol. This does make sense because in the results it was determined that the functional group was an ester and there are two esters that are very close in molar mass, methyl acetate and ethyl formate. The methyl acetate and the ethyl formate both have a molar mass of 74. 08 g/mol [5][6]. The last step is to distinguish between the two. Both methyl acetate and ethyl formate are described as having a “fruity odour”, relatively close boiling points and the same colourless appearance [5][6].

However, there is one source that says that methyl acetate has a “characteristically pleasant smell reminiscent of some glues and nail polish removers” [7]. For the ester test in the lab, the unknown and the ester did smell a lot like nail polish remover. One method of discovering the specific ester one had is to taste it. Each ether has a different taste, for methyl acetate it’s described as a “fleeting fruity taste” [5], and for ethyl formate it’s described as having a slightly bitter taste [6]. The main potential source of error is when using the Dumas’ method one must poke a hole into the tin foil.

This is effective in the case that it allows one to use the atmospheric pressure within the calculations however its potentially ineffective for the reason that some particles may be able to escape from the hole. Overall, this would result in a lower mass and cause the molar mass to be a lower number than it should have been. Conclusion The purpose of this was to determine the identity of the unknown compound #12. This was completed successfully through qualitative and quantitative analyses. The identity was be methyl acetate; the determined ucture is displayed in figure 2. 0.

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