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Week 1

by: Hayley Lecker

Week 1 CHEM 2325 - 001

Hayley Lecker
GPA 3.42
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This is covering Week 1 (actually week 2). I did the homeworks early, so you can get help if you get it before its due. This covers L1,L2, and L3.
Organic Chemistry - 12551
James Salvador
Class Notes
Organic Chemistry




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This 17 page Class Notes was uploaded by Hayley Lecker on Saturday January 23, 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 467 views.


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Date Created: 01/23/16
OrganicChemistry 2Week 1 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! L1: 12.10 pages 621-622, 12.13 pages 635-636, 12.14 (Anchemeric Assistance) page 636 12.10 Ethers are unreactive compounds and require an acid to protonate the oxygen. They also need a good nucleophile to react. Commonly used is HI and HBr in hot concentrated solutions. If the ether is sterically open it will proceed with anNS 2 reaction, if it is hindered it will proceed wNth an S 1 reaction. The first step is the protonation of the ether oxygen. The second step in an N 2 reaction is the nucleophilic substitution by the halide ion. This reaction above formed an alkyl halide and an alcohol. Alcohols can react with more HBr to form a nd 2 alkyl halide (in this case an alkyl bromide). These kind of reactions are slow enough to isolate the alcohol. However, with HI (the better nucleophile) the alcohol cannot be isolated because of the rate of reaction. An exception to the unreactive ethers is oxirane rings. These are 3-membered rings that have a high strain (this means they want to be broken), because of the high strain they are very reactive to reactions that can open the ring. Oxirane can reaction with HCl directly to “cleave” the ring. This reaction proceeds like an S N reaction. However, the reaction reacts with the 2 Carbon because it has a larger partial positive charge than the primary carbon. 12.13 Cyanide (KCN) is good nucleophile and goes through a S 2 rNaction, this is useful with primary substrates and many secondary substrates. Cyanide ion substitution is difficult because Na and K cyanides are insoluble in most organic solvents. Phase-transfer catalyst(s): is the use of crown ethers or another member of these compounds to overcome the issue of cyanides being insoluble in organic solvents. Another phase-transfer catalyst used is quaternary ammonium salts (R N X) 4hese cary in solubility in different solvents with changes in the anion. Example: As you can see the Cl came off and the CN was picked up. Once the CN is in the organic solvent it reacts with the substrate releasing the halide ion, with makes Tetra butyl ammonium halide, then that migrates to the aqueous phase to carry another CN ion into the organic phase. When sodium amide (NaNH ) or a2Grignard reagent are used to react, terminal alkynes (RC=CH) (that was a triple bond) have enough acidity to form carbanions (these are also called acetylide ions). Due to the sp hybridized orbital on the triple bond makes acetylide easier to form and a more stable cabanion than sp or sp , however it is shorter and weaker. However, this makes them good nucleophiles and can react in S 2 Neactions. Sodium amide is prepared by adding Na to liquid ammonia in the presence of a catalytic quantity of iron (III) cations. This is a colorful reaction (deep blue) that changed to a colorless liquid with gray solids (this is when terminal alkyne and alkylating agent are added). Grignard reagents and organolithium reagents are not good with S 2 reNctions because strong bases promote elimination and not substitution. However, they do reaction with substitution on reaction substrates. Example of this is epoxides: As you can see the cyclohexane ring with MgCl is a Grignard reagent, with reacted with acid and ethane oxide, it MgCl is removed the epoxide ring is broken and replaces the MgCl. The acid is used to protonate the oxygen to break the ring (these are already very easy to break because of the high angle strain). 12.14 The formation of cyclic ethers can have a nucleophilic substitution and be an intramolecular process. In an intramolecular reaction (also known as neighboring group participation) the presence of the functional groups in a substrate effect the rate of reaction and the stereochemistry of the reaction (other than the leaving group). Homework examples (4 included): PLEASE BE CAREFUL IF IT SAYS PRODUCTS THERE WILL BE TWO MOLECULES DRAWN! Epoxides: Three membered ring with an oxygen in it. Also called oxirane or oxapropane. This is usually made from alkenes. Epoxides experience anchiomeric assistance also known as the neighboring group effect. The oxygen with block one side as if it was bonded to the one side so that a compound with attack from the other side. To name: Use the alkene it came from and add oxide as another word. Example: ethene oxide There can be chiral centers when you get to propene oxides and higher. Nucleophiles in SN2 reactions attack from the backside where the least sterics are found. Common nucleophiles that open epoxide rings include (via SN2): OH, OR, CN, SR, NH3 Alkoxides are deprotonated alcohols. Example 1: Draw the product(s) of cyclohexene oxide, sodium cyanide, and hydronium workup. In this reaction there will be TWO products. In the first one the oxygen (now alcohol) will be facing upward and in the second the mirror so downward. The carbon adjacent to it is the carbon that made the carbon cation, so the cyanide comes in and satisfies it. Example 2: Draw the product of 2-methylpropene oxide, water, and acid catalysis. The first step is the acid breaking the oxygen bond, this will go to the middle carbon. The water will then come in satisfy the carbon cation to the side of the middle carbon. The extra hydrogen on the water is then washed away like in every acid catalysis reaction. Example 3: Draw the product of (S)-propene oxide, CH3MgCl, and hydronium workup. First note that CH3MgCl is a Grignard reagent, so the only thing that matter is the CH3. The first step is the hydronium (H+) coming in and protonating the oxygen, thus breaking the bond to the middle (primary) carbon. The carbon cation then takes the CH3 of the Grignard reaction to become happy. Example 4: Draw of the product of 2-methylpropene oxide, hydrogen chloride in diethyl ether. DO NOT CONFUSE WITH just HCl reactions. In this reaction the oxygen doesn’t go to primary carbon, this is because it is sterically hindered. Instead it goes to the outside carbon. The middle carbon becomes a carbon cation and the chlorine satisfies it. L2 13.9 page 684-687 Dehydration of an alcohol makes an alkene and water. This reaction is reversible and has a small equilibrium constant meaning it favors the substrate over the product. Tertiary and secondary alcohol will reaction faster than primary alcohols. The rate of reaction is dependent on the stability on of carbocation. The example below is the simplest alcohol to alkene you can have. The rate for tertiary and secondary alcohols follows an E1 reactions. However, in primary alcohols the OH is a strong base and a poor leaving group. Alcohol dehydrations doesn’t occur in a basic reaction mixture. In primary alcohols the first step is to protonate the OH group. Then it will follow the typical elimation reaction. (Almost see homework examples for why this isn’t always true). An alt way is concerted arrangment. In this the molecule rearranges to make a more stable carbon cation since primary ones are poor. 1-butanol and 2-butanol form the same mixture, the rearrange the same way, so a mixture is made of alkenes. The best acid catalysts to use are sulfuric and phospheric because of the high affinity for water. Homework examples (3 included) I choose all concert arrangements since these are the hardest. : Rule to remember: Sayzeft Rule’s it means to take the hydrogen to make the double bond from the carbon with the least hydrogens. Example 1: Draw the rearrangement elimination product of 2,2,3,3-tetramethylbutan-1-ol, sulfuric acid, and heat. That is what the alcohol looks like, if you can see it’s a primary alcohol, and the place where it would make a double bond is already full. This is where concerted rearrangement comes in. First step is to draw the alcohol, protonate the water and remove it. Then you can look and see where you need to rearrange. In this case the t-butyl group was moved to the primary alcohols position by doing that a double bond could be made between carbon’s 2 and 3 (if you go from the right hand side). In concert arrangement the same molecule doesn’t remain the same this would no longer be the same as the above picture. Example 2: Draw the rearrangement elimination product of 2,2-dimethylpropan -1-ol, sulfuric acid, and heat. This one is much simpler with concert rearrangement, the only step after removing the alcohol is to move one methyl group so the first primary carbon, this creates a stable carbocation and can bond between positions 1 and 2. Example 3: Draw the rearrangement elimination product of 3,3-dimethyl-2,2-bis(1,1- dimethylethyl)butan-1-ol, sulfuric acid, and heat. Sorry for this being blurry, when I cropped it I didn’t know it would go blurry. So the double bond is now in the middle and 3 “branches” (that’s a term I use, not scientific at all) are coming off. Each branch is a t-butyl group. L3: All Homework problems since it is Chapters 12 and 13. In this I will do 5 problems and explain them. SN2 and E2 reactions need a basic solvent (also called aprotic solvent), because if it is too acidic it would stabilize the base/nucleophile. Strong nucleophiles suggest an SN2 reactions. Strong bases suggest E2. Look at the leaving group for clues to which reaction will occur: If a methyl is at the carbon with the leaving group: SN2 If it is a primary carbon: SN2 If it is a tertiary carbon: SN1 OR E1 it depends on the base. If it is a secondary carbon: Any of the reaction could occur. A good leaving group is necessary for SN1 OR E1 reactions. Tertiary carbons don’t like SN2 because there is too much steric hindrance that will repel any incoming nucleophile. However, tertiary carbons do favor SN1 reactions. Primary carbons favor SN2, it doesn’t favor SN1 or E1 because it would lead to a carbocation that was primary (those are not stable). Good nucleophiles are bas leaving groups, such as OH that will want to attach to something and not be on its own The more sterically hindered the base, the more likely it is to eliminate the substitute. Colder- temperatures favor SN product; hot favor elimination products. Example 1: Draw the major product(s) of 2-chloropropane and water. TAKE NOTE IT SAYS MAJOR PRODUCTS For this you are going to do hydrogen bonding, so the chlorine can be ionized off. So the oxygen will share a hydrogen just so the chlorine can come off the carbon, but then take it back. The next step is the oxygen takes the place of the chlorine but if see HCl is not produced that only happens in a workup. Example 2: Draw the major product(s) of 2-methylpropane and water. TAKE NOTE IT SAYS MAJOR PRODUCTS This is the same as the first, but just with a different structure. Example 3: Draw the major product(s) of iodomethane and sodium tert-butoxide. In this problem the tert-butoxide replaced the iodine which ionizes off. Example 4: Draw the major product(s) of 2-chloro-2-methylpropane and potassium hydroxide. In this problem it is an elimination while the other 3 examples were substitution. In this reaction the chlorine ionizes off, and the hydroxide takes a vicinal hydrogen to become water. The 1 and 2 carbons then create a double bond. Example 5: Draw the major product(s) of (1R,2R)-1-iodo-2-methylcyclohexane and potassium ethoxide. This is an elimination as well. The iodine is going to ionize off, and the ethoxide will take a vicinal hydrogen, which allows the double bond to form. PLEASE TAKE NOTE THE METHYL IS POINTING TO THE BACK YOU NEED TO MAKE SURE THE STEREOCHEMISTRY OF THE METHYL REMAINS R!


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