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Week 2 Notes - L4,L5,L6

by: Hayley Lecker

Week 2 Notes - L4,L5,L6 CHEM 2325 - 001

Hayley Lecker
GPA 3.42

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This includes lecture notes, as well as the reading for L4-6. I include detailed examples from the homework as well.
Organic Chemistry - 12551
James Salvador
Class Notes
Chemistry, Organic Chemistry, Organic Chemistry 2
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This 16 page Class Notes was uploaded by Hayley Lecker on Friday January 29, 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 99 views.


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Date Created: 01/29/16
OrganicChemistry 2Week 2 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! L4: 13.12 pages 696-700 13.12 An oxidation reaction is the elimination of 2 hydrogens from a primary or secondary alcohol to form an aldehyde or ketone. A common oxidation technique is chromate oxidation which uses chromium trioxide or chromic acid to oxidize secondary alcohols to ketones. The mechanism for this type of reaction involves the initial formation of a chromate ester. The ester then undergoes E2 elimination to form a carbonyl group. A secondary alcohol goes to a ketone and primary to an aldehyde. After a primary alcohol turns into an aldehyde it can be changed further with aqueous acid to form a hydrate. Then Chromium trioxide can be used to turn the hydrate into a carboxylic acid like below. The problem with using chromium trioxide is it can oxidize other functional group, even double and triple bonds, because of this PCC can be used. PCC is soluble in low polarity solvents, the most common is methylene chloride. That is used specifically to turn primary and secondary alcohols to aldehydes and ketones. Homework examples (3 included) Quick Notes: Primary Alcohols: Aldehydes, if further Carboxylic Acids Secondary: Ketones PCC + water = Carboxylic Acid K2Cr2O7 + H2SO4 = Ketone PCC + DCM = aldehyde PCC + CH2CL2 = aldehyde H2CrO4=Remains ALCOHOL Chromic Acid= Carboxylic Acid or Ketone Example 1: Draw the major organic product of (R)-butan-2-ol and (K2Cr2O7 + H2SO4) In this example oxidizing a secondary alcohol, from the above information we know it will make a ketone. Example 2: Draw the major organic product of 2-methylpropan-2-ol and H2CrO4. I like this example because this is a secondary alcohol, but it can’t be oxidized. Why you may ask? Because the oxygen can’t many a double bond anywhere. This is an example of when the alcohol has no reaction. Example 3: Draw the major organic product of ®-2-methylbutan-1-ol and PCC in CH2Cl2. I choose this example to demonstrate both primary alcohols going to aldehydes and remembering to check for R and S. So we already know this goes to an aldehyde, but if you look closely it also remains R chirality. L5 13.3 pages 664-671 In an elimination reaction the substrate loses the leaving group and usually a hydrogen next to the leaving group. The leaving group can have more than one adjacent hydrogen and due to many factors either could leave. Elimination reactions can be either regiospecific or regioselective . Regiospecific means one isomer of an alkene is produced. Regioselective means many isomers are produced, but one isomer is greater over the other. Regiospecific or regioselective depends on whether one hydrogen is preferred or if a mix of hydrogens is preferred. If the alkene produced is high substituted it is considered a Saytzeff elimination. If the alkene produced is less substituted the alkene is said to be a Hoffman orientation. Most elimination reactions follow the Saytzeff elimination because it creates a more stable alkene, more stable = more substituted. Alkyl groups help stabilize the double bond, unlike hydrogens. Alkyl groups are slightly electron donating while hydrogens are slightly electron withdrawing. To measure stabilization, the amount of energy given off when hydrogen adds to the double bond to form an alkane is measured. This is convenient for comparing energy content of various alkenes. As energy needed to hydrogenate a double bond decreases the stability increases. Homework examples (3 included): Rule to remember: Sayzeft Rule’s it means to take the hydrogen to make the double bond from the carbon with the least hydrogens. (Yes this statement is from last week remember Saytzeff can be spelled many ways). Remember: Steric interactions are important. If the double bond is in the center it is more stable. Also reducing steric interactions increases stability. Example 1: Draw the least stable butane. First we need to know that to increase stability the double bond needs to be not in the middle so that leaves the first carbon-carbon area. Next we need to increase steric interactions, trans tends to be favored because of stability, cis increases interactions. If you look, the double bond is on the first carbon, and the fourth carbon is point downward so it can interact with the double bond instead of pointing up. Example 2: Draw the most stable hex-3-ene. In this example it is asking for the most stable. So we know the double bond must be in the middle, the only other choose is trans or cis? Well, trans is the best for stability, if you look at the image the hydrogens and alkyl groups are on opposite sides of each other. Example 3: Draw the most stable alkene with the methylyclohexane sigma framework. Don’t let the wording sike you out, this is just like the above problem except with a ring. So we want the most alkyl groups coming off the double bond (this means neighbors). So if you move around the ring, most areas have only two neighbors but if the double bond is by the methyl that creates a third neighbor, so it is more stable. L6: 2.9 pages 100-109 2.9 Alkenes and cycloalkenes are hydrocarbons that have 1 or more carbon-carbon double bonds. The equation for this is n H2n Alkenes and cycloalkanes are constitutional or structural isomers. Naming Alkenes: 1. Find the longest continuous chain containing the double bond, this will be the parent chain. 2. Replace –ane of the parent chain with –ene. 3. Indicate position of double bond by numbering parent chain to give the double bond the lowest number. Double bonds take priority over substituents. Example of Naming Alkenes: Naming cycloalkenes is similar to alkenes except you number the ring so the double bond is in between C1 and C2. Example below: Polyenes contain 2 or more double bonds. Dienes contain 2 double bonds and trienes, three. To Name Dienes or Triene- Follow alkene rules plus these addition rules: 1. Write numbers indication locations of double bond separated with common, then a hyphen. 2. After hyphen change parent ending from –ene to –diene if two, -triene if three and so forth. Example of Naming Dienes or Trienes: Special Types of Dienes Conjugated- double bonds separated by one single bond. CH =CH-CH=2H Chemical 2 properties differ greatly from simple alkenes. Cumulated- double bonds share a carbon CH =C=CH 2 2 A side chain containing a double bond is an alkenyl. Naming alkenyl: 1. Number carbons from the carbon that bonds the side chain to parent 2. When naming side chain use ending –enyl instead of –ene. Use parentheses so you know the number belongs to side chain. Alkynes are often called acetylenes because of the common name of the simplest compound of this type. Ethyne which common name is acetylene. IUPAC Naming of Alkynes: 1. Find parent chain, must contain the triple bond. 2. Add –yne to ending instead of –ane. 3. Number so triple bond has the lowest number. 4. Place number of triple bond first, then hyphen to parent chain. *If double bond and triple bond in competition, use the above procedure for naming the parent chain, then for determining bond position use the following: 1. Find longest continuous chain with both triple and double bond. 2. Number the chain so multiple bonds get lowest numbers. 3. Number position of the double bond to parent. 4. Use –en are the end of parent name. 5. For triple use –yne ending along with number identifying position. Number double bond so triple takes precedence over substituents. *If both groups could get the same number –en takes priority and gets the lowest number. Homework examples (5 included) Remember R and S, E and Z!: Example 1: Name the following structure: First step, find the longest chain with both the double and triple bond in it. That would be four so we have a butane structure. Now we have to decide on numbering, while the triple bond USUALLY has the priority, not in this structure since if we give the double bond the highest priority if gives the ethyl hanging off a lower number. E is trans, Z is cis! So the name of this structure is 3-ethylbut-en-3-yne! Example 2: Name the following structure: Since both groups are in the number one spot en is going to take the priority. So we know we have 3- methylpent-1-en-4-yne, but look closely there is a chiral center. The triple bond is priority one, followed by double, then the methyl group. This makes this structure an S. Example 3: Draw (Z)-pent-3-en-1-yne. First draw it out, then deal with the orientation. For these problems E and Z are used. Z is cis, E is trans. Example 4: Name the Structure: You will get easy problems like so: So we have 5 carbons. The en and yne are in the same location numbering wise, so the en takes priority. If you deduce things fall into place quickly. Example 5: Draw 4-methylpent-3-en-1-yne. In this problem we have a pent, so draw five carbon, then start adding in the triple and double bonds. st rd The triple is taking priority on the 1 carbon, so draw that first, then the double bond on the 3 carbon. Finally add in the methyl. Lecture Notes SN2 reactions usually lead to inversion means R to S. If you always come in from the back side you will get the right structure, since inversion does not always happen. This is when the priorities end up changing. SN1- needs acid SN2- needs base SN2 only occurs on primary and methyls. Strong acids have weaker Hydrogen-Halogen bond, so that means a weaker Carbon- Halogen bond. To determine the weaker halogen bond determine the halogens acid. So 1-bromobutane conjugate acid is HBr. Remember: Acidity increases as you go down the periodic table. So HBr is a stronger acid than HCl.


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