Organic Chemistry II Course Assignment - Chapter 9 PDF
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This document is a course assignment matrix for Organic Chemistry II, Chapter 9. It includes details about quizzes, exams, homework assignments, and participation grades, along with guidelines for homework completion.
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Organic Chemistry II Course Assignment Matrix Assignment Week Due Percent Topic 1 Quiz (Chapter 9) 3 4 Exam I (Chapters 10, 11, 14) 6...
Organic Chemistry II Course Assignment Matrix Assignment Week Due Percent Topic 1 Quiz (Chapter 9) 3 4 Exam I (Chapters 10, 11, 14) 6 15 Exam II (Chapter 15, 16, 17) 9 20 Exam III (Chapters 18 and 19) 12 10 Exam IV (Chapters 20, 21, 22) 15 15 ACS Exam 15 10 HW 11 Participation 15 Total Points 100 (1000) Note: Exams are closed-book and closed-note. A Reaction Sheet will be provided. HW Guidelines Homework is 50-100 questions per chapter. (see calendar) Questions can come from Participation, Review Packet, Old Exams. Indicate where the problem is taken from. Include a one-two sentence rationale for the correct answer. Need to average approximately 10 questions per day. “A” Student HW HW Grading 5 points for completion (100 problems) 5 points for accuracy and format. Participation Policy Complete Participation WS DUE Monday. Participation Questions are included in the Explore Review. If you are struggling to complete Participation, go to ExploreMore. Testmoz link will be posted. DO NOT POST ON CHEGG Chapter 9 Structure of Alkynes Alkyne Nomenclature Properties and Spectroscopy of Alkynes Alkyne Synthesis and Alkyne Reactions Synthesis Practice Alkynes contain a triple bond. General formula is CnH2n-2. Two elements of unsaturation for each triple bond. Reactions similar to alkenes, such as addition and oxidation. Alkyne specific reactions such as substitution. Structure 1. Triple Bond 2. Linear 3. sp hybridization 4. no cis-trans possibilities 5. 2 bonds are perpendicular to each other. More reactive than alkenes and are useful intermediates in organic synthesis. 1. Find the longest chain containing the triple bond. 2. Change -ane ending to -yne. 3. Number the chain, starting at the end closest to the triple bond. 4. Give branches a number to locate their position. Example If more than one triple bond use di, tri, tetra, etc. Cycloalkynes: At Least 9 Carbons Required Multiple Functional Groups 1. If other functional group is alkene enyne. 2. Number such that first unsaturated bond gets the lowest number. If numbering is equal start with alkene (alphabetical). What is the IUPAC name for the molecule in the box? If molecule has OH (alcohol), double and triple bonds are considered branches. (Alcohol has higher priority) OH OH Common Names: alkylacetylene or dialkylacetylene CH C CH 3 Methylacetylene (propyne) (terminal alkyne) (terminal alkyne) internal alkyne Nonpolar, insoluble in water. Soluble in most organic solvents. Boiling points are similar to alkane of same size. Less dense than water. Up to four carbons, gas at room temperature. Synthesis of Alkene: Single Elimination Synthesis of Alkyne: Double Elimination Triple-bonded carbons have sp hybrid orbitals. A sigma bond is formed between the carbons by overlap of the sp orbitals. Sigma bonds to the hydrogen atoms are formed by using the second sp orbital. Since the sp orbitals are linear, acetylene will be a linear molecule. Each carbon in acetylene has two unhybridized p orbitals with one non-bonded electron. It is the overlap of the parallel p orbitals that form the triple bond (2 pi orbitals). Triple bonds are shorter than double or single bonds because of the two pi overlapping orbitals. 1. Terminal alkynes, are more acidic than other hydrocarbons due to the higher s character of the sp hybridized carbon. (Conjugate base is stabilized by positive nucleus). 2. Terminal alkynes can be deprotonated quantitatively with strong bases such as sodium amide (NaNH2). 3. Hydroxide and alkoxide bases are not strong enough to deprotonate the alkyne quantitatively. Synthesis of Terminal Alkynes H+ can be removed from a terminal alkyne by sodium amide: NaNH2. The acetylide ion is a strong nucleophile that can easily do addition and substitution reactions. Synthesis of substituted alkynes: Nucleophilic attack by the acetylide ion on an unhindered alkyl halide. SN2 reaction with 1 alkyl halides lengthens the alkyne chain. 1 alkyl halides only. Complete the Following Synthesis. Acetylide ions are also strong bases. 2° and 3° elimination (E2) will occur. Zipper Reaction Internal alkynes (A) can be isomerized to terminal alkynes (G) using a strong base such as potassium 1,3-diaminopropanide (KAPA). Complete the mechanism by adding curved arrows. Nucleophiles can attack the carbonyl carbon forming an alkoxide ion which on protonation will form an alcohol. Determine the product for the following synthesis. An alkyne is an electron-rich molecule and therefore reacts as a nucleophile. Similar to addition to alkenes. bonds becomes two bonds. First bond is easier to reduce. One or two molecules may add. Two molecules of hydrogen can add across the triple bond to form the corresponding alkane. A catalyst such as Pd, Pt, or Ni needs to be used for the reaction to occur. Under these conditions the alkyne will be completely reduced; the alkene intermediate cannot be isolated. Alkynes reduce more readily than alkenes. Second bond is weaker. H2 PtO2, acetic acid The catalyst used for the hydrogenation reaction is partially deactivated (poisoned), the reaction can be stopped after the addition of only one mole of hydrogen. The catalyst used is commonly known as Lindlar's catalyst and it is composed of powdered barium sulfate, coated with palladium poisoned with quinoline. The reaction produces alkenes with cis stereochemistry. Both substrates, the hydrogen and the alkyne, have to be adsorbed on the catalyst for the reaction to occur. Once adsorbed, the hydrogen atoms add to the same side of the double bond (syn addition) giving the product a cis stereochemistry. Trans alkene, two hydrogen atoms must be added to the alkyne anti stereochemistry, so this reduction is used to convert alkynes to trans alkenes. Hydrogen atoms come from ammonia. Step 1: An electron adds to the alkyne, forming a radical anion. Step 2: The radical anion is protonated to give a radical. Step 3: An electron adds to the alkyne, forming an anion. Step 4: Protonation of the anion gives an alkene. Chapter 9 Na, NH3 Br2 2-butyne H2, Lindlar catalyst Br2 2-butyne + Na, NH3 OsO4 2-butyne + H2O2 H2, Lindlar catalyst OsO4 2-butyne H2O2 Addition of H2O Remember: 1. Hg(OAc)2, H2O 2. NaBH4 OH 1. BH3. THF OH 2. NaOH, H2O2 1a. Mercuric sulfate in aqueous sulfuric acid adds H—OH to one bond 1b. Markovnikov orientation forming a vinyl alcohol (enol) 1c. Rearranges to a ketone. 2a. Hydroboration–oxidation adds H—OH 2b. Anti-Markovnikov orientation enol. 2c. Rearranges to an aldehyde. (Note: Alkynes react slower than alkenes because intermediate is vinyl cation.) Thermodynamic Data for Tautomerization (The driving force for the following reaction is making C=O). H O O H C C H = 611 kJ / mol H = 735 kJ / mol C O O H = 380 kJ / mol H = 368 kJ / mol C C C O H H = 426 kJ /mol H = 420 kJ /mol C H Fill in the Boxes Internal Alkynes Not Useful Reaction Mixture of Products Equivalent Carbocations HgSO4 O H2SO4 O H2O + HgSO4 O O H2SO4 + H2O Fill in the Boxes Alkynes can be hydrated anti-Markovnikov by using the hydroboration–oxidation reaction. A hindered alkyl borane needs to be used. This prevents double addition of borane to the triple bond. Disiamylborane has two bulky alkyl groups. (Steric effect). Electronic effect. Borane adds to the less cationic carbon. If a terminal alkyne is used, the borane will add to the least substituted carbon. Mechanism for Enol Formation OH 1. Sia2BH 2. NaOH enol Mechanism for Enol-Keto Tautomerzation OH -OH H O Tautomers: Different Location of Protons (Structural Isomers) Example: Enol and Ketone/Aldehyde Resonance Structures: Different Location of Electrons Example: Enolates Oxidative Cleavage of Alkenes Remember: O O 1. O3 H + 2. (CH3)2S O O OH KMnO4 + Similar to oxidation of alkenes. Dilute, neutral solution of KMnO4 oxidizes internal alkynes to a diketone. Terminal = keto-acid. Warm, basic KMnO4 cleaves the triple bond. Ozonolysis, followed by hydrolysis, cleaves the triple bond. Under neutral conditions, a dilute potassium permanganate solution can oxidize a triple bond into an diketone. The reaction uses aqueous KMnO4 to form a tetrahydroxy intermediate, which loses two water molecules to produce the diketone. KMnO4 O Neutral H2O CO2H O KMnO4 Neutral H2O O If potassium permanganate is used under basic conditions or if the solution is heated too much, an oxidative cleavage will take place and two molecules of carboxylic acids will be produced. Ozonolysis of alkynes produces carboxylic acids (alkenes gave aldehydes and ketones). O O (1) O 3 CH 3 C C CH 2 CH3 CH3 C OH + HO C CH 2 CH3 (2) H 2O Used to find location of triple bond in an unknown compound. Oxidative Cleavage of Terminal Alkynes Br Br Br B r2 B r2 CH3C CH CH3C CH CH3C CCH3 CH2Cl2 CH2Cl2 Br Br Br Cl2 and Br2 add to alkynes to form vinyl dihalides. One mole of HCl, HBr, and HI add to alkynes to form vinyl halides. If two moles of HX is added, product is a geminal (1,1) dihalide. The addition of HX is Markovnikov. Alkenes are reactive than alkynes. The triple bonds abstract a proton from the hydrogen halide forming a vinyl cation. The proton adds to the least substituted carbon. The second step of the mechanism is the attack by the halide. By using peroxides, hydrogen bromide can be added to a terminal alkyne anti-Markovnikov. The bromide will attach to the least substituted carbon. Complete the following reactions.