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 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. Conjugate base of 2. Carbonyl with a an amine C-N bond O

N C R1 H H R3 N R2

University of Illinois Slide 26 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 26 Other Commonly Encountered Bases

+ Li H H C C H + N H C C Li+ + N

lithium diisopropyl amide pKa = 36 (LDA)

+ + H C C H + CH3CH2CH2CH2 Li H C C Li + CH3CH2CH2CH3

n-butyl lithium pKa = 62 (n-BuLi)

+ + H C C H + Na H H C C Na + H2

sodium hydride pKa = very large (NaH)

University of Illinois Slide 27 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 27 Acetylides are Nucleophiles!

• Nucleophiles Add to Electrophiles • First C-C bond forming reaction we’ve learned • Acetylides are small - good nucleophiles

H C C + E LG H C C E + LG

University of Illinois Slide 28 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 28 Self-Test Question

OCH3

CH A C Predict the product. H3CO CH3 OCH3

B H CO C 3 C CH3 H OCH3

OCH3 CH3 C H3CO 1. HBr, peroxides C CH3 CH CH + OCH 2 2. NH3, H C C Na 3 H3CO CH3 H CO Br D 3 C CH3 CH

University of Slide CHEM 232, Spring 2010 29 Illinois at Chicago UIC Lecture 19: March 16 29 CHEM 232 University of Illinois Organic Chemistry I at Chicago UIC

Alkylation of Terminal Alkynes and Acetylene

Chapter 9: Sections 9.6

30 Preparation of Alkynes

There are two main methods for the preparation of alkynes: 1. Alkylation of acetylide anions (C-C bond formation)

OCH3 OCH3 H C CNa

H CO Br H CO C 3 3 C CH3 CH3 H

2. Functional group transformations (Thursday’s topic)

University of Illinois Slide 31 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 31 Alkylation of Acetylene and Terminal Alkynes alkyl group: any group derived from an alkane alkylation: process of adding an alkyl group alkylating reagent: electrophilic reagent causing alkylation

SN2 + H C C Na + H3C Br H C C CH3 alkylation acetylide alkylating reagent (nucleophile) (electrophile)

• the alkylating reagent is a primary alkyl or methyl halide • the reaction is nucleophilic substitution • the nucleophile is an acetylide (carbanion)

University of Illinois Slide 32 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 32 Conditions/Example

H C C C C C C NaNH2, NH3 Br

• must start with acetylene or terminal alkyne • most common base = sodium amide (NaNH2) • solvent is typically ammonia (NH3) • alkylating reagent must be a 1º alkyl halide or methyl halide

University of Illinois Slide 33 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 33 Conditions/Example

H C CNa C C C C NaNH2, NH3 Br

positive metal counterion = negative carbanion

• must start with acetylene or terminal alkyne • most common base = sodium amide (NaNH2) • solvent is typically ammonia (NH3) • alkylating reagent must be a 1º alkyl halide or methyl halide

University of Illinois Slide 34 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 34 Dialkylation of Acetylene

HC CH

1. NaNH2, NH3 2. CH3CH2Br

CH3CH2 C CH

1. NaNH2, NH3 2. CH3Br

CH3CH2 C C CH3 internal alkyne

University of Illinois Slide 35 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 35 Limitation: Acetylide is a Strong Base

• acetylide is a stronger base than hydroxide = • E2 predominates over SN2 when alkyl halide is 2º or 3º • acetylides can only be alkylated with 1º alkyl or methyl halides

X E2 H C C + C C C C + H C C H H 2º or 3º

University of Illinois Slide 36 CHEM 232, Spring 2010 at Chicago UIC Lecture 19: March 16 36 Self-Test Question

Predict the major Cl organic product. A

B

Cl

tosylate (TsO–) C excellent LG O O Cl S CH3 + C CNa O S CH3 D O O (1.0 equivalent) (1.0 equivalent)

E O O S CH3 O

University of Slide CHEM 232, Spring 2010 37 Illinois at Chicago UIC Lecture 19: March 16 37 Self-Test Question

A Predict the major O O organic product. B

C CH O O 1. NaNH2, NH3 2. 1-bromobutane (1.0 equivalent) CH HC CH D O O 3. NaNH2, NH3 4. E Cl CH + O O O O

University of Slide CHEM 232, Spring 2010 38 Illinois at Chicago UIC Lecture 19: March 16 38 CHEM 232 University of Illinois Organic Chemistry I at Chicago UIC

Next Lecture. . .

Chapter 9: Sections 9.7-9.13

Quiz This Week. . .

Chapter 8

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