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Week 3 Notes

by: Hayley Lecker

Week 3 Notes CHEM 2325 - 001

Hayley Lecker
GPA 3.42

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About this Document

This includes examples for L7 AND L8 and the steps for S1 as it is the foundations for the other synthesis, so I took my time to try to explain what is going on.
Organic Chemistry - 12551
James Salvador
Class Notes
Organic Chemistry, Chemistry, Ochem
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This 10 page Class Notes was uploaded by Hayley Lecker on Friday February 5, 2016. The Class Notes belongs to CHEM 2325 - 001 at University of Texas at El Paso taught by James Salvador in Fall 2015. Since its upload, it has received 110 views.


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Date Created: 02/05/16
OrganicChemistry 2Week 3 Important Information: Professor’s Email: Class Website: Class Code (E-book): utep2325spring2016 Includes examples of the Homework! Posted the week before the homework’s are due to help everyone get 100’s! L7: Alkene Addition Reactions Homework examples (4 included) Quick Notes: While doing this you want to create the most stable carbocation that means a carbon with 3 bonds, is more stable then one with two and a primary carbocation is very unstable in relation. The products will either be syn or anti, syn means same side and anti means they are trans to each other. If you see peroxy ethanoic acid know this is a type of acid used to make expoxides. Any “per” acids are actually. When you have to use remove the double bond, take those electrons and attach it to the electrophile. Example 1: Draw the first product of 1-methylcyclohexene and ozone. First draw out both starting materials. Ozone is an oxygen double bonded to another oxygen, then another single bond to another oxygen. Next the ozone will attach to the “richer gets richer spot (meaning carbon with more hydrogens). As you can see in the bottom picture the ozone creates a second ring. Example 2: Draw the product of 1-methylcyclohexene and Hg(OAc)2/H20 followed by NaBH4/NaOH If you remember the reaction examples from class about B2H6 and NaO2H this is similar as the ending product is an alcohol, however the OH ends up on the methyl carbon and NOT the carbon with more hydrogens. As you can see you can’t determine if it will be syn or anti because the carbon adjacent to the methyl carbon has two hydrogens. Example 3:Draw the product of 1-methylcyclohexene, BH3, NaOOH. This is exactly like the problem we did in class, just with a cyclic ring. The ring will attack the BH3 to create a stable carbocation at the methyl carbon, then the BH3 gives a hydrogen almost at the same time to satisfy the carbocation. Then a “replacement” happens where the oxygen on the NaOOH comes in and inserts itself between C and B, a secondary NaOOH reacts to take off the BH2 and leave behind an alcohol. Because the hydrogen is facing up, the alcohol will as well. Example 4: Draw the product of 1-methylcyclohexene and dichlorobene. This example was shown in class. I screenshotted the picture so you can see that the carbon with the two chlorines points up while the methyl and hydrogen are downward (Cis). L8: Alkyne Addition Reactions Homework examples (5 included): If a sodium metal is present a trans alkene will be made. The metal adds one electron to the antibond orbital. Tunneling is the process of moving electrons from one side to another, this is what makes trans alkenes possible. Hg is important for reaction as it will stabilize the carbocation. Example 1: Draw the product of but-2-yne and excess HI When HI is in excess it removes the triple bond completely, and you get a carbon with two iodines attached to it. If it is not in excess there will be a double bond formed, the next example will show this. Example 2: Draw the first product of but-2-yne and hydrogen iodide. Because the HI is not in excess, it can only react once so the triple bond will be broken into a double bond. Now this reaction will be trans (the methyls will be opposite each other). As in previous lessons we know that trans is more stable that cis bonds. Example 3: Draw the first product of but-1-yne and HI. This like before but instead of but-2-yne it is but-1-yne. As you can see the triple bond becomes a double bond. The hydrogen attacks the first carbon, the second carbon becomes a carbcation that the iodide can attach to. Example 4: Draw the product of but-1-yne and dicyclohexyborane. Now this may look complicated, but it really isn’t. First draw out your dicyclohexyborane, the triple bond will become a double bond. The borane in this case attacks the first carbon (the second one makes the most stable carbocation). The hydrogens are also trans because as we know trans=more stable. Example 5: Draw the product of but-1-yne, HBr, benzyl peroxide and heat. In a normal HBr reaction, the bromine reactions twice, and in excess FOUR times. So with the addition of benzyl peroxide and heat it makes sure the HBr reacts only once. The bromine will attach to the first carbon breaking the triple bond, then the hydrogen will satisfy the carbocation on the second carbon. S1: Synthesis: Leaf Alcohol All steps included : For this I am going to walk through the steps as they are repeated in many synthesizes. Step 1: You are given NaNH2 and ethyne. The product of this is a negatively charged carbon tripled bonded to another carbon and a positively charged sodium atom. Step 2: I recommend drawing your first product and adding iodoethane in. Iodine likes making ionic bonds, so it will go with the Na (you can delete it out), then the ethane will come in and satisfy the negative carbon. Step 3: Now step one is repeated but with the other side of the triple bond. Step 4: This though isn’t like step 2 with an ethane coming in, instead it’s an epoxide. The epoxide ring will break and attach to the negative carbon. The sodium atom in this instance does not get deleted instead it makes an ionic relationship the negatively charged oxygen. Step 5: In this the hydronium workup gets rid of the sodium atom and a hydrogen attaches to the oxygen. Step 6: Now Lindlar Catalyst is hydrogen, transition metal, barium sulfate, and quinolone, this breaks triple bonds into double bonds. The hydrogen gas then will come in and attach to the carbons on the double bond. If you can see they attach as cis (same side). Now you have leaf alcohol. In the other synthesis this won’t be your end product but instead the reaction will continue.


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