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Chem 232 Lecture 19 CHEM 232 University of Illinois Organic Chemistry I at Chicago UIC Lecture 19 Organic Chemistry 1 Professor Duncan Wardrop March 16, 2010 1 CHEM 232 University of Illinois Organic Chemistry I at Chicago UIC Chapter 9 Alkynes Sources and Nomenclature Sections 9.1 - 9.6 2 CHEM 232 University of Illinois Organic Chemistry I at Chicago UIC Exam Two • Monday, April 5 • 6:00-7:15 p.m. • 250 SES • Chapters 6-10, 13 • Makeup Exam: Monday, April 12, time t.b.a. Makeup policy: There are no makeup exams without prior approval. Only students showing proof of a class con!ict will have the option to take a makeup exam. To be added to the makeup list, you must email me no later than Friday, April. 2. 3 Self-Test Question Which column lists the correct mechanistic symbol for each description? A B C D E Methyl halides react with sodium ethoxide in ethanol only by this mechanism. SN2 SN1 E2 SN2 SN1 Unhindered primary halides react with sodium ethoxide in ethanol mainly by this mechanism. SN2 SN2 SN2 SN2 SN1 Cyclohexyl bromide reacts with sodium ethoxide in ethanol, mainly by this mechanism. E2 SN1 E2 SN2 SN1 The product obtained by solvolysis of tert-butyl bromide in ethanol arises by this mechanism. SN1 SN2 SN2 SN2 E2 In ethanol that contains sodium ethoxide, tert-butyl bromide in ethanol reacts mainly by this mechanism. E2 E2 SN1 SN1 E2 University of Slide CHEM 232, Spring 2010 4 Illinois at Chicago UIC Lecture 19: March 16 4 Substitution/Elimination Review SN2 H3C Cl + OCH2CH3 H3C OCH2CH3 Methyl halides react with sodium ethoxide in • methyl halide = ethanol only by this mechanism. no carbocation intermediate = Unhindered primary halides react with sodium • ethoxide in ethanol mainly by this mechanism. • 2nd order mechanism Cyclohexyl bromide reacts with sodium ethoxide in ethanol, mainly by this mechanism. The product obtained by solvolysis of tert-butyl bromide in ethanol arises by this mechanism. • only one carbon = In ethanol that contains sodium ethoxide, tert-butyl • must be substitution bromide in ethanol reacts mainly by this mechanism. University of Illinois Slide 5 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 5 Substitution/Elimination Review S 2 OCH CH + OCH CH N 2 3 Cl 2 3 Methyl halides react with sodium ethoxide in • primary halide = ethanol only by this mechanism. • no carbocation intermediate = Unhindered primary halides react with sodium ethoxide in ethanol manily by this mechanism. • 2nd order mechanism Cyclohexyl bromide reacts with sodium ethoxide in ethanol, mainly by this mechanism. • primary halide and small The product obtained by solvolysis of tert-butyl nucleophile = low steric bromide in ethanol arises by this mechanism. hinderance = In ethanol that contains sodium ethoxide, tert-butyl substitution is favored bromide in ethanol reacts mainly by this mechanism. • University of Illinois Slide 6 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 6 Substitution/Elimination Review Br E2 + OCH2CH3 H Methyl halides react with sodium ethoxide in • secondary halide = ethanol only by this mechanism. • carbocations usually favored Unhindered primary halides react with sodium only when tertiary = ethoxide in ethanol manily by this mechanism. • mostly a 2nd order mechanism Cyclohexyl bromide reacts with sodium ethoxide in ethanol, mainly by this mechanism. The product obtained by solvolysis of tert-butyl • secondary halide = significant bromide in ethanol arises by this mechanism. steric hinderance In ethanol that contains sodium ethoxide, tert-butyl elimination is favored bromide in ethanol reacts mainly by this mechanism. • University of Illinois Slide 7 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 7 Substitution/Elimination Review Br SN1 CH3CH2OH OCH2CH3 (–H+; not shown) Methyl halides react with sodium ethoxide in • tertiary halide = ethanol only by this mechanism. • carbocation intermediate = Unhindered primary halides react with sodium • 1st order mechanism ethoxide in ethanol manily by this mechanism. • tertiary halide = significant steric hinderance Cyclohexyl bromide reacts with sodium ethoxide in + ethanol, mainly by this mechanism. • pKa (CH3CH2OH2 ) = -1.7 (solvolysis = very weak base) The product obtained by solvolysis of tert-butyl bromide in ethanol arises by this mechanism. • substitution favored; only bases whose conjugate acids have pKa In ethanol that contains sodium ethoxide, tert-butyl ≥ 15.7 do elimination bromide in ethanol reacts mainly by this mechanism. University of Illinois Slide 8 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 8 Substitution/Elimination Review Br E2 H + OCH2CH3 Methyl halides react with sodium ethoxide in • tertiary halide = larger steric ethanol only by this mechanism. hinderance= Unhindered primary halides react with sodium • elimination favored when base ethoxide in ethanol manily by this mechanism. is strong (conjugate acid pKa ≥ 15.7) Cyclohexyl bromide reacts with sodium ethoxide in ethanol, mainly by this mechanism. • Although carbocation The product obtained by solvolysis of tert-butyl bromide in ethanol arises by this mechanism. intermediate is possible, E2 is generally faster than E1, even for In ethanol that contains sodium ethoxide, tert-butyl 3º alkyl halides bromide in ethanol reacts mainly by this mechanism. University of Illinois Slide 9 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 9 CHEM 232 University of Illinois Organic Chemistry I at Chicago UIC Next Quiz (after Spring Break). Three synthesis questions from the synthesis handout. pdf available at http://www.chem.uic.edu/chem232/ 10 Nomenclature acetylene and ethyne are both acceptable IUPAC names for the molecule below H C C H University of Illinois Slide 11 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 11 Nomenclature All nomenclature rules remain the same as for alkenes except -yne is used as the parent suffix; Alkenes have higher priority than alkynes. 1 H HO 4 3 H2C 2 CH 2 1 5 3 4 H 1-buten-3-yne 1-pentyne 2-methyl-3-butyn-2-ol 4 CH3 11 3 1 OH 3 HO 2 H H 2 1 2-methyl-1,4-pentadiyne (R)-cycloundec-3-ynol (R)-2-propyl-3-butyn-1-ol University of Illinois Slide 12 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 12 Self-Test Question Which structural formula is (2S,3S)-(E)-3-propylhept-4-en-6-yn-2-ol? OH OH CH3 A C OH E OH OH B D University of Slide CHEM 232, Spring 2010 13 Illinois at Chicago UIC Lecture 19: March 16 13 CHEM 232 University of Illinois Organic Chemistry I at Chicago UIC Chapter 9 Properties of Alkynes Sections: 9.3-9.5 14 Structure of Alkynes University of Illinois Slide 15 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 15 Orbitals on sp-Hybridized Carbons University of Illinois Slide 16 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 16 Triple Bond: 2 pi-bonds & 1 sigma-bond A triple bond is formed by three orbital overlaps 2 pi (π) bonds: side-to- side overlap of two sets of p-orbitals; 4 π-electrons 1 sigma (σ) bond: head-to-head overlap of two sp2-orbitals University of Illinois Slide 17 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 17 Hybridization Two p-orbitals are Reserved–Not Hybridized University of Illinois Slide 18 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 18 Valence Model of Bonding in Acetylene (with Hybridization) H H C C H H 1s 1s 2p 2p 2p 2p 2sp 2sp 2sp 2sp C C University of Illinois Slide 19 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 19 Bond Strength triple bonds = stronger and shorter University of Illinois Slide 20 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 20 Comparison of Ethane, Ethylene and Ethyne Ethane Ethylene Ethyne C-C distance 153 pm 134 pm 120 pm C-H distance 111 pm 110 pm 106 pm H-C-C angles 111.0º 121.4º 180º C-C BDE 368 kJ/mol 611 kJ/mol 820 kJ/mol C-H BDE 410 kJ/mol 452 kJ/mol 536 kJ/mol hybridization of C sp3 sp2 sp % s-character 25% 33% 50% pKa 62 45 26 University of Illinois Slide 21 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 21 Acidity increased s-character = increased electronegativity of carbon = electrons closer to the nucleus = stronger acid pKa = 26 sp H C C H H C C + H carbanion, acetylide H H H 2 pKa = 45 sp C C C C + H H H H H H H H Increasing Acid Strength Increasing 3 pKa = 62 sp H C C H H C C H + H H H H H University of Illinois Slide 22 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 22 Aqueous Acid-Base Equilibria Water is not a strong enough base to significantly deprotonate terminal alkynes. (pKa = 15.7) -[pKa (acid left) - pKa (acid right)] Keq = 10 -(26 - –1.7) Keq = 10 -27.7 Keq = 10 University of Illinois Slide 23 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 23 Aqueous Acid-Base Equilibria Even hydroxide (the strongest base that can exist in water) is not a strong enough base to significantly deprotonate terminal alkynes. -[pKa (acid left) - pKa (acid right)] Keq = 10 -(26 - 15.7) Keq = 10 -10.3 Keq = 10 University of Illinois Slide 24 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 24 Amide Deprotonates Terminal Alkynes Sodium amide (NaNH2) is the most common base for completely deprotonating terminal alkynes. -[pKa (acid left) - pKa (acid right)] Keq = 10 -(26 - 36) Keq = 10 10 Keq = 10 University of Illinois Slide 25 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 25 Diambiguation of Amide 1.
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