University of ManitobaCHEM 2210 - Introductory Organic Chemistry 1 Laboratory Report 3: Substitution Chemistry Student: Ygor Dantas Furtado Lab Section: B06, Room #: 290 October 8, 2015 For this reaction with have that the rate will depend on the [2-Naphtol] and [Iodobutane]. - The reaction to make the 2-butoxynaphthalene would undergo Sn2 mechanism. The NaOH will be responsible to deprotonate the 2-Napthol to react further with the Iodobutane. In addition. we also used a weak Nucleophile Cl-. So. tertiary Carbons are more likely to undergo Sn1 mechanism due to the high stability of the carbon cation and also its high reactivity.Moreover. Due to the fact that we’re using a protic Solvent (H2O) from the HCl concentrated solution. The first intermediate is and the second is b. Also. we also have a very good leaving group (OH). . Draw the reaction profile diagrams for both reactions. we have all the pre-requisites to confirm that the reaction to make the tert-butyl chloride undergo Sn1 mechanism. - The reaction to make the tert-butyl chloride would undergo Sn1 mechanism. For the experiment of Day 1 and 2: a. The explanation for that is a little bit more complex. This is not a general Sn2 reaction since we used a protic solvent (EtOH). Rationalize using theory whether the reactions are likely to undergo a Sn1 or Sn2 mechanism.1. What are four ways to speed up this reaction? Be sure to describe how these act to speed up the reaction. Why was the NaOH necessary? - The NaOH was necessary to deprotonate the 2-Naphtol to make it a better nucleophile. - Increasing the [2-Naphtol].Sn1 Mechanism Energy Diagram Sn2 Mechanism Energy Diagram 2. b. For the synthesis of 2-butoxynaphthalene: a. What other substitution product could there be? - Butanol c. rate = k[2-Naphtol][Iodobutane] . 4. Do you think tert-butyl chloride will react with water? If so.667 g . - Percent yield must be used since we’re talking about efficiency of a reaction. it would react with water in acid condition to form an alcohol.50 0. 3. high temperatures mean higher rates.- Increasing the [Iodobutane] [Iodobutane]. Increasing the temperature increases the frequency in which molecules collide with each other. rate = k[2-Naphtol] - Increase the temperature. - Yes. For 2-butoxynaphthalene yield= amount of product that we actually got amount of product expected (Theoretical) yield= 0. So.17 g x 100=25. Should a percent yield or percent recovery be reported for these reactions? Show your work on how that number is calculated for both experiments. show the mechanism of the reaction and expected product. 83 g x 100=64. We got 3 different signals on the product spot.70 5. melting point and IR techniques. The TLC plate showed that we still have some of the start materials. One of these signals represents .92 g Calculating the amount of product expected: 5. Using all possible evidence.Calculating the amount of product expected: For tert-butyl chloride yield= amount of product that we actually got amount of product expected (Theoretical) yield= 3. describe the purity of the products (2butoxynapthalene and tert-butyl chloride) For 2-butoxynaphthalene The purity was assigned by TLC. However. We can check that when we compare with the signal of just the start material (Rf= 0. The IR showed that we got our product.44). Then we know that there are more 2 compounds in our product.the 2-naphtol (Rf= 0. One of those 2 must be the iodobutane and the other one must be our desired product. the melting point analysis .44). This difference in results can be due to the fact that the TLC was analyzed using the reaction mixture while for the IR we just used the precipitated material. IR data Peak (cm-1) Functional group 3057/2956/2868 C-H 1627 C=C (aromatic) About 1200 C-O We did not see a broad peak about 3300 that would represent the (OH) group from the 2 naphtol so our product may not contain start material anymore. chemspider. which indicate that there was not water present.html (Retrieved Nov 11. So we can conclude that we made the tert-butyl chloride. We have a peak at 567 cm-1 that represents C-Cl. Attach your IR spectra for both products 1 Chem Spider Web site: http://www. No braod peak about 3300 that would represent the OH bond from the tert-butanol. 6. The solution was colorless and transparent.confirmed that the product was not pure. Another think that we can analyze is the aspect of our solution. The melting point observed was 32 0C while in the literature it must be between 35 to 36 0C1. For tert-butyl chloride The purity was assigned via IR spectroscopy.74597.com/Chemical-Structure.2014) .