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    oxidation with sodium hypochlorite Essay

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    final weightpercent yield 2,4-DNPTollen’s testpathway. 42g67%positivenegativeoxidation of secondary OHGood ThingsMy experiment went well. I began my experiment with . 64g of 2-ethyl-1,3-hexanediol. The molecular weight of this compound is 146.

    2g/mol. It is converted into 2-ethyl-1-hydroxyhexan-3-one. This compounds molecular weight is 144. 2g/mol. This gives a theoretical yield of . 63 grams.

    My actual yield was . 42 grams. Therefore, my percent yield was 67%. This was one of my highest yields yet. I felt that this was a good yield because part of this experiment is an equilibrium reaction. Hypochlorite must be used in excess to push the reaction to the right.

    Also, there were better ways to do this experiment where higher yields could have been produced. For example PCC could have been used. However, because of its toxic properties, its use is restricted. The purpose of this experiment was to determine which of the 3 compounds was formed from the starting material. The third compound was the oxidation of both alcohols.

    This could not have been my product because of the results of my IR. I had a broad large absorption is the range of 3200 to 3500 wavenumbers. This indicates the presence of an alcohol. If my compound had been fully oxidized then there would be no such alcohol present.

    Also, because of my IR, I know that my compound was one of the other 2 compounds because of the strong sharp absorption at 1705 wavenumbers. This indicates the presence of a carbonyl. Also, my 2,4-DNP test was positive. Therefore I had to prove which of the two compounds my final product was. The first was the oxidation of the primary alcohol, forming an aldehyde and a secondary alcohol.

    This could not have been my product because the Tollens test. My test was negative indicating no such aldehyde. Also, the textbook states that aldehydes show 2 characteristic absorptions in the range of 2720-2820 wavenumbers. No such absorptions were present in my sample.

    Therefore my final product was the oxidation of the secondary alcohol. My final product had a primary alcohol and a secondary ketoneBad ThingsI really dont have that many bad things to write about. I forgot to shake my sep funnel after my first addition of dichloromethane. I had to redo that step. Also, there is a very slight chance that a little bit of dichloromethane could have been left in my final product. When I was distilling, I thought I had boiled away everything in my flask.

    The dichloromethane distilled at 39. 5 degrees. To avoid distilling to dryness, I removed the heat. However, there was still some product left. I just couldnt see it very easily.

    I believe that most of the dichloromethane that was left after removing the heat was eventually removed while the flask was cooling. Also, I would like to have seen my ketone absorption a little higher in wavenumber. As stated in the book, an unknown at an absorption of 1750 wavenumber almost certainly is a ketone. The absorption of my ketone was at 1705.

    It could have been influence by something else. Perhaps it could have been dichloromethane. My yield could have been improved by being more careful with my washes. If I had been much more careful while separating layers, I could have had more product. This would have taken more time however. Main IdeaIn an oxidation reaction an alcohol is converted to carbonyl.

    This carbonyl can either be an aldehyde or a ketone. In organic chemistry it is important to be able to select certain groups over another. In this experiment, the secondary alcohol is selected over the primary alcohol. In many cases the primary alcohol can be oxidized all the way a carboxylic acid.

    In order to achieve selectivity, sodium hypochlorite is used. It is reacted with acetic acid to form HOCl. Then HOCl is reacted in excess with the alcohol compound to push the reaction to the right in which OCl replaces the OH. Then water is added. It abstracts a hydrogen and the hydrogens electrons form a double bond to the oxygen.

    The chlorine is expelled while this double bond is formed. Temperature is also important in this reaction. The temperature must stay below 30 degrees Celsius to retain the selectivity. PCC and PDC can also be used to do the same procedure but are cancer suspect agents. Using bleach is much safer. Three tests are used to identify the product.

    I briefly discussed them earlier in my first paragraph. An IR is first used. This tells me what functional groups are present. The two regions that I am looking at are 3200-3500 and 1600-1800. This lets me know if I have an alcohol and a carbonyl.

    I had both of these in my final product. Two classification tests were used to find out if I had certain functional groups. 2,4-DNP is the first. It tells me that I have an aldehyde or a ketone. In this case a precipitate forms after the nucleophilic nitrogen on the 2,4-DNP replaces the oxygen on the carbonyl.

    This derivative is an orange precipitate and is easily recognized. I had a precipitate and therefore I had either an aldehyde or a ketone. The last test is the Tollens test. The Tollens test is only a positive test for aldehydes.

    It forms silver when it further oxidizes the aldehyde. The silver forms a silver mirror on the bottom of the test tube. A positive test was easily recognized when I performed a test on benzaldehyde. My final product was definitely not positive for this test.

    By compiling the data from each of these tests, it is very easy to determine what functional group is selected over the other. The secondary alcohol is oxidized and the primary alcohol is not.

    This essay was written by a fellow student. You may use it as a guide or sample for writing your own paper, but remember to cite it correctly. Don’t submit it as your own as it will be considered plagiarism.

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