Week 11 Lecture Notes/ Readings
Week 11 Lecture Notes/ Readings CHEM 2321
Popular in Organic Chemistry I
verified elite notetaker
Popular in Chemistry
This 29 page Class Notes was uploaded by Hayley Lecker on Friday November 6, 2015. The Class Notes belongs to CHEM 2321 at University of Texas at El Paso taught by Dr. James Salvador in Fall 2015. Since its upload, it has received 34 views. For similar materials see Organic Chemistry I in Chemistry at University of Texas at El Paso.
Reviews for Week 11 Lecture Notes/ Readings
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 11/06/15
OrganicChemistry Week 11 Important Information: Professor’s Email: firstname.lastname@example.org Class Website: organic.utep.edu/courses/2324 Class Code (E-book): utep232Xfall2015 L22/M6 The Wittig Reaction reference pages 341-347 7.10 The Wittig Reaction In 1954 Georg Wittig discovered how to convert aldehyde/ketone to a carbon-carbon double bond. The Wittig reaction involves a nucleophilic addition of a phosphous-stabilizied carbanion to ketone or aldehyde, then elimination of phosophous gives an alkene. A phosphous stablized carbanion (ylide) has a negative charged carbon and a positive charged phosphous adjacent to each other, some ylides have sulfurs instead. The net charge is zero. The preparation of phosphous ylide has 2-step reaction of triphenylphosphine and alkyl halide. 1. A nucleophilic phosphous attacks the saturated carbon with halogen. The halogen-bearing carbon is usually a primary carbon. The product is an alky triphenylphosphonium salt. 2. The salt is treated with a strong base usually butyl lithium- this removes a proton from carbon. Ylide’s are reasonace hybrids with 2 reasonace structures. One has a charge of Phosphous and Carbon. The other has a double bond between phosphous and carbon. When converting the carbonyl into alkene, the ketone or aldehyde react with ylide in a typical nucleophlic addition reaction. The reaction produces an intermediate called betaine. Because of the close proximity of the postive charged Phosphouse and negatively charged Oxygen, phosphous and oxygen form a bond and create a 4 memebered ring. The ring rapidly collaspes to give alkene and triphenylphosphine oxide (a very stable and usually insoluble compounds in this reaction). Examples of Wittig reactions where cis and trans geometric isomers are found. Example of a solved wittig reactions: M7 Mechanism: The synthesis of caprolactone by the acid catalyzed Fischer esterification of 6- hydroxyhexanoic acid 8.1-8.4,8.7 reference pages 359-382, 399-402 8.1 The Aycl Tranfer Mechanism Nucleophilic reactions change direction based on the strength of the nucleophile. Stronger nucleophiles will favor the product. The example below is the reverse reaction, the loss of a nucleophile to reform the original carbonyl compound. The nucleophile substitution at a carbonyl group of a carboxylic acid or derivative combines 2 steps. In nucleophilic substitution the group that leaves is the electronegative group bonded to the carbonyl carbon, so the starting and ending compounds are different. A nucleophilic substitution involving a carbonyl group is often called acyl transfer reactions. The ease in which a leaving group, leaves is based on basicity, thus the more basic it is the more it won’t leave. A strong base is more willing to donate electrons to the carbonyl carbon, so the weaker the base the more readily it will leave. Acyl halides work well because they are weak bases. Acyl transfer does not occur with aldehydes or ketones because hydrides and carbanions are too strong of a base. The leaving group and nucleophile are both bases, weak bases are stable anions. The behavior of the leaving group affects the reaction. Acyl halides and anhydrides have most stable leaving group. Ester and carboxylic acids have intermediate stability, thus intermediate reactivity, Amides are the least stable. 8.2 Water and Alcohol Nucleophiles Esterifcation is the nucleophilic substitution reaction that converts a carboxylic acid or derviative to an ester. The reaction requires a substitution of hydroxy group in the carboxylic acid with an alkoxy group from an alcohol. Reverse reaction called hydrolysis. Equilibrium constants for esterfication reactivity are small. Below is an example, the constant is 4. The pKa of the nucleophilic and leaving groups helps understand the equailibrium constant. Ethanol has a pKa of 16.3 and water 15.7, there is little difference, so little thermodynamic preference between substrate and product. st In the 1 step, nucleophilic oxygen of the alcohol attacks the carbonyl carbon. In the 2 step of the reaction the proton bonded to the oxygen from alcohol is transferred to the oxygen that was the carbonyl carbon’s oxygen. Proton transfer requires a base, so water is used to remove the proton from one oxygen to another. In the 3 intermediate the compound will lose a hydroxide ion, thus an ester is formed. In the 4 step the hydroxide ion removes a proton from carboxyl group, to remove water from esterfication, distillation is used. Example of this below, you should have done distillation by now in lab. Hydroxide ion is a poor leaving group so to increase the rate of esterfication, catalyic quantities of acid are added. Acid pronates the leaving group, allowing water to leave, known as Fischer esterfication. The acid increases reactivity of carbonyl group, the acid pronates oxygen, and enhances carbon’s reactivity to nucleophile. The pronoted carboxylic acid is resonance stablized, look below. Resonance allows serveral atoms to bear partial chage, so ion has more stability and less reactivity than st rd pronated aldehydes and ketones. The 1 2 reasonace contributors are equal in energy, 3 is minor. The pronated carboxylic acid then adds the alcohol to form hydrate of ester. After proton transfer from alkyoxy group to hydroxy group, the intermediate loses water to form ester. Most popular acid to use: *Very strong. It is very strong because conjuated base is reonance stablized with negative charge distrubed on 3 oxygen bonds, bonded to sulfur. Fischer works will for most primary alochols because not sterically hindered. However, secondary and teritary alochols are sterically hindered and usually have lower equilibrium constants and low st concentrations of the ester at equlibrium. So cehmist 1 convery carboxylic acid to acid chloride, Cl is a good leaving group, the acid chloride rapidly forms ester. With ester hydrolysis reactions there are 2 mechanisms. 1 step is the formation of tetrahedral intermediate, that occurs in an esterfication, in this the nucleophile and leaving group both bond to the same carbon, however this has never been isolated, so truly been identified. Please see page 367-368 for rest of mechanism. Base assited ester hydrolysis follows same path as acid catalyzed hydrolysis st 1 in the reaction is the reaction of ester with nucleophilic reaction of base at carbonyl carbon. 2 is the loss of an alkoxide ion. rd 3 is a proton exchange it leaves the alcohol and carboxylate anion. Little reversal because alcohol is too weak to be a nucleophile. Lact ones are cyclic esters. 5/6 membered lactone rings occur from compounds containing hydroxy group and carboxylic acid group. Water and alcohols react very rapidly with acyl halides and fast with anhydrides. Leaving group is a stable anion, it is either halide or carboxylate (RCOO-) anion. Amides are less reactive than any other carboxylic acid derviatives. Hydrolysis of amide with either acid or base equires heat and longer reaction time than others when producing carboxylic acid. Below are examples of reactions: 8.3 Halide and Carboxylic Acid Nucleophiles Acyl halides and anhydrides are the most reactive members of carboxylic acid derviatives because they create the most stable leaving groups. The leaving group for acyl halide is a halide anion (example is Cl-) and anhydride is carboxylate anion (RCOO-). Leaving group of acyl halide is a conjugate base of a strong acid (pKa less than 1). Anhydride is a weaker acid (pKa roughly 5). The equlibrium constant is more favorable of acl halides than anhydrides. They are synthesized as intermediates than end products. Usualy only acyl chlorides are used, bromides and iodines are expensive to make and unstable. To synthesize acyl chlorides from carboxylic acid, thionyl chloride (SOCl2) or phosphous chlorides (PCl2, PCl3) are used. All reagents are acid halides or inorganic acid. Mechanism: Reaxtion of Carboxylic Acid with SOCl2. st 1 step: Produces a mixed anhydride (consists of organic aicds and inorganic Cl). nd 2 step: Follows typical nucleophilic substitution on carbonyl. As the reaction progressed SO2 leaves and 2 Cl- leaves as HCl. Both are gases so they bubble out. Oxalyl chloride (ClCOCOCl) is a becoming reagent of chloride for synthesis of acyl halides because easy to handle. The by products are CO2,CO, and HCl. Anydrides are made limitly because acyl halides are readily avialable and more reacrive. To make an anhydride, it involves an acyl halide with a carboxylic acid in presence of nonnucleophilic base (usually pyridine). Mechanism: Anhydride A weak base such as pyridine reacts with acidic proton of carboxylic acid forming carboxylate anion. The negatively charged oxygen reacts with the carbonyl carbon of the acyl halide. Heating carboxylic acids does not form anhydride unless 5-6 membered ring can form. This will be seen a lot in next week’s Monday homework. Many anhydrides form via anhydride exchange, it involves heating the acetic anhydride with carboxylic acid. Example of reaction: 8.4 Reactions with Nitrogen Nucleophiles Nitrogen nucleophiles studied most in carbonyl chemistry are ammonia (NH3) and primary amines RNH2 and seconary amines R2NH. All react with carboxylic acids to form amides. To prepare amides most common is to react carboxylic acid or acyl halide with NH3 or amine. Acid or acyl, anhydrides react similar to acyl halides but less reaction. The mechanism for formation of amide by reaction of amine with acid anhydride similar to above. The amine Nitrogen reacts with carbonyl carbon to form tetrahedral intermediate. Tetrahedral intermediate losses a caboxylate ion to form amide. Direct reaction of amine with carboxylic acid, can form amide. 8.7 Nitriles Nitriles considered carboxylic acid derviative because reactivity resembles carbonyl group. Nitriles hydrolyze form amides or carboxylic acids/derviatives. To form carboxylic acid, the nitrile is hydrolyze with water with either a base or acid catalyst, if mild conditions hydrolysis stops at amide, if vigous conditions hydrolyzes to amide then carboxylic acid (vigous means higher concentrations of acid or base and heat). st To hydrolyse with base, 1 A nucleophilic attack of hydroxide ion on the electrophilic carbon of nitrile group. It picks up a proton from water and produces unstable isoamide and regenerates hydroxide catalyst. Isoamides are unstable so the rapidly isomerize to form amide. The isomerize by transferring proton from oxygen to nitrogen. There is a shift in the pi bond. The equlibrium is to the right, the rapid interconversion of functional groups is called tautomerism. The reaction is unfavorable equilibruim because nitrogen not very basic, so does not pronate easily. After Nitrogen nitrile is protaned, water is added via a nucleophilic addition to carbon. Intermediate losses proton and tautomerizes to form amide. It needs heat to proceed, heat increases the rate of hydrolysis of amide. Hydride addition to nitriles used to snythesize primary amines. Proceeds by 2 successive additions of hydrides LiAl hydride to carbon. Addition of water to dianion hydrolyzes to primary amine. Nitriles with Grignard reagent and organolithium to produce ketones. M8: Mechanism: The reaction of propanoic acid and methanamine with DCC 23.6 refereance pages 1209-1216 23.6 Peptide Synthesis Amides are readily synthesize from amines and acid chlorides, does not work well for peptide chains because amino acids have additional functions groups that can react with acid chloride. So, protecting groups are added to protect active functional groups with protected groups the reaction forms an amide bond. Two groups used to protect carboxylic acid functional groups are methyl esters and benzyl ester. Benzyl and methyl esters are usually removed by hydrolyzing in an aqeous base. An alternative way to remove benzyl ester is through hydrogenation, to cleave the weak benzylic C-O bond. To protect the amine end of the amino acid a carbamate ester group is used (benzyl and tert-butyl esters most commonly used). st In 1 reaction the product is an N-benzyloxycarbonyl derviative of the amino acid (abbr. N-Cbz, or Cbz). In the 2 , the product is a N-tert-butoxycarbonyl dervative (abbr. N-Boc, or Boc). Benzy and tert-butyl esters are both good for protecting because stable carbocations or radicals. Cbz is removed by treating solution by HBr in acetic acid. Boc is similar but best acid to use is trifluoroacetic acid in dry methylene chloride or HCl in anhydrous ether. Formation of peptide bond is accomplished by reaction involving amino acid with protected carboxylic acid, another amino acid with a protect amine and DCC. DCC converts unprotected carboxylic acid to intermediate with chemical proerties similar to acid anhydride. On next page. Lecture Notes Moodle Answers: methyl 3-methylbutanoate + diisobutylaluminum hydride – 3-methylbutanal 3-methylbutanoyl chloride + lithium aluminum hydride – 3-methylbutan-1-ol 3-methylbutanenitrile + propyllithium – 2-methylheptan-4-one, N,N,3-trimethylbutanamide + diisobutylaluminum hydride – 3-methylbutanal, 3-methylbutanamide + diisobutylaluminum hydride – 3-methylbutanal, 3-methylbutanoic anhydride + lithium aluminum hydride – 3-methylbutan-1-ol, 3-methylbutanenitrile + lithium aluminum hydride – 3-methylbutanamine, N,3-dimethylbutanamide + lithium aluminum hydride – N,3-dimethylbutanamine, 3-methylbutanenitrile + diisobutylaluminum hydride – 3-methylbutanal, methyl 3-methylbutanoate + excess propyllithium – 2-methyl-4-propylheptan-4-ol, sodium 3-methylbutanoic acid + propyllithium – 2-methylheptan-4-one, 3-methylbutanoic acid + excess propyllithium – 2-methylheptan-4-one, 3-methylbutanenitrile + hydrogen gas + palladium catalysis – 3-methylbutanamine, 3-methylbutanamide + lithium aluminum hydride – 3-methylbutanamine, 3-methylbutanoic acid + lithium aluminum hydride – 3-methylbutan-1-ol, 3-methylbutanoyl chloride + lithium dipropylcuprate – 2-methylheptan-4-one, N,N,3-trimethylbutanamide + lithium aluminum hydride – N,N,3-trimethylbutanamine, sodium 3-methylbutanoate + lithium aluminum hydride – 3-methylbutan-1-ol, 3-methylbutanoyl chloride + lithium tri-tert-butoxyaluminum hydride – 3-methylbutanal, N,3-dimethylbutanamide + diisobutylaluminum hydride – 3-methylbutanal, methyl 3-methylbutanoate + lithium aluminum hydride – 3-methylbutan-1-ol Rules for Lithium Aluminum Hydride: LiAlH4 reduces all carboxylic acid derviatives. Esters and carboxylic acids give primary alcohols. When a propyl is used, you take a original compound + 3, so if you look above a butane turns into a heptane. Carboxylic acid and carboxylic salts are turned into ketones. Nitriles are turned into ketones. The x axis of the infrared spectra in the text are units of? Frequency Please note the correction to twisting 4, not 3. Hook’s Law ???? = ????????????????( ) ???? Where K = number of bonds and M= reduced ????????????????ℎ????∗???????????????? ????????????????ℎ???? ???????????????? ????????????????ℎ????+???????????????? ????????????????ℎ???? Example C-C 12∗12 ???? = ???????????????? 1/( ) = 0.408 12+12 Bending needs 3 atoms, so three points of reference. Stretching needs 2 atoms. Twisting needs 4. If frquency goes up bond tighter, more eletronegative stiff bond. Stiff bond does not mean stronger. Hydrogen bonded – nondiluted Nonhydrogen bonded- diluted Carboxylic acids have a broad peak because they exist in diamers, meaning 2 interacting with each other. Primary amines – RNH2 has two peaks because two hydrogens. Secondary Amines- R2NH has only hydrogen so one peak Primary Amide- RCONH2 2 hydrogens, so two peaks. Secondary Amide- RCONHR’ has 1 hydrogen, so one peak. When looking at the peaks only looking at the N-H bond. Tertiary amines will not peak because there are no N-H bonds.
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'