Alkynes C 17

Synthesis of Acetylene Heat coke with lime in an electric to form calcium carbide. Then drip water on the calcium carbide. coke lime *This reaction was used to produce light for miners’ lamps and for the stage. *

The Structure of Alkynes A triple bond is composed of a s bond and two p bonds

Question Arrange ethane, ethene, and ethyne in order of increasing C-C bond length. A) ethane < ethene < ethyne B) ethene < ethane < ethyne C) ethyne < ethene < ethane D) ethane < ethyne < ethene

Acidity of Acetylene and Terminal Alkynes H C 17

Acidity of Hydrocarbons In general, hydrocarbons are very weak acids Compound pKa HF 3.2 H2O 16 NH3 36 45 CH4 60 H2C CH2 20

Acetylene Acetylene is a weak acid, but not nearly as weak as alkanes or alkenes. Compound pKa HF 3.2 H2O 16 NH3 36 45 CH4 60 26 HC CH H2C CH2 20

Question Which one of the following is the strongest acid? A) water B) ammonia C) 1-butene D) 1-butyne

Figure Number: 06-02-041UN Title: Relative Acidities Caption: Electrostatic potential maps of various acids. Notes: Increasing amounts of blue color on hydrogen in the stronger acids indicate more positive charge on hydrogen due to increasing electronegativities of atoms attached to hydrogen.

Carbon: Hybridization and Electronegativity 10-60 C H H+ + : sp3 C H C : sp2 10-45 H+ + C C 10-26 C H C : sp H+ + Electrons in an orbital with more s character are closer to the nucleus and more strongly held. 21

Question Which one of the following statements best explains the greater acidity of terminal alkynes (RCºCH) compared with monosubstituted alkenes (RCH=CH2)? A) The sp-hybridized carbons of the alkyne are less electronegative than the sp2 carbons of the alkene. B) The two p bonds of the alkyne are better able to stabilize the negative charge of the anion by resonance. C) The sp-hybridized carbons of the alkyne are more electronegative than the sp2 carbons of the alkene. D) The question is incorrect - alkenes are more acidic than alkynes.

The stronger the acid, the weaker its conjugate base top 252

Sodium Acetylide Solution: Use a stronger base. Sodium amide is a stronger base than sodium hydroxide. NH3 NaNH2 + HC CH NaC .. .. – – H2N : + + H C CH H2N H : C CH stronger acid pKa = 26 weaker acid pKa = 36 Ammonia is a weaker acid than acetylene. The position of equilibrium lies to the right. 23

Question Which of the following bases is strong enough to completely deprotonate propyne? A) NH3 B) CH3OH C) NaNH2 D) NaOH

Preparation of Various Alkynes by alkylation reactions with Acetylide or Terminal Alkynes 26

Synthesis Using Acetylide Ions: Formation of C–C Bond

Alkylation of Acetylene and Terminal Alkynes H—C C—H R—C C—H R—C C—R 27

Alkylation of Acetylene and Terminal Alkynes – : H—C SN2 + X– : R X C—R H—C + The alkylating agent is an alkyl halide, and the reaction is nucleophilic substitution. The nucleophile is sodium acetylide or the sodium salt of a terminal (monosubstituted) alkyne. 28

Example: Alkylation of Acetylene NaNH2 HC CH HC CNa NH3 CH3CH2CH2CH2Br HC C CH2CH2CH2CH3 (70-77%) 29

Question Which alkyl halide will react faster with the acetylide ion (HCºCNa) in an SN2 reaction? A) bromopropane B) 2-bromopropane C) tert-butyl iodide D) 1-bromo-2-methylbutane

Example: Alkylation of a Terminal Alkyne CH (CH3)2CHCH2C NaNH2, NH3 CNa (CH3)2CHCH2C CH3Br (81%) C—CH3 (CH3)2CHCH2C 30

Example: Dialkylation of Acetylene H—C C—H 1. NaNH2, NH3 2. CH3CH2Br C—H CH3CH2—C 1. NaNH2, NH3 2. CH3Br C—CH3 CH3CH2—C (81%) 31

Effective only with primary alkyl halides Limitation Effective only with primary alkyl halides Secondary and tertiary alkyl halides undergo elimination 32

What is the product of the following reaction? Question What is the product of the following reaction?

What is the product of the following reaction? Answer What is the product of the following reaction? B) SEE: Skillbuilder 10.5.

E2 predominates over SN2 when alkyl halide is secondary or tertiary. Acetylide Ion as a Base E2 predominates over SN2 when alkyl halide is secondary or tertiary. H C X C – : H—C E2 + C H—C —H X– : 33

Question Consider the reaction of each of the following with cyclohexyl bromide. For which one is the ratio of substitution to elimination highest? A) NaOCH2CH3, ethanol, 60°C B) NaSCH2CH3, ethanol-water, 25°C C) NaNH2, NH3, -33°C D) NaCºCH, NH3, -33°C

Preparation of Alkynes by Elimination Reactions 34

Preparation of Alkynes by "Double Dehydrohalogenation" X C H X C H Geminal dihalide Vicinal dihalide The most frequent applications are in preparation of terminal alkynes. 35

Geminal dihalide  Alkyne (CH3)3CCH2—CHCl2 1. 3NaNH2, NH3 2. H2O (56-60%) (CH3)3CC CH 36

Geminal dihalide  Alkyne (CH3)3CCH2—CHCl2 NaNH2, NH3 (slow) (CH3)3CCH CHCl NaNH2, NH3 (CH3)3CC CH (slow) H2O NaNH2, NH3 (CH3)3CC CNa (fast) 37

Question In addition to NaNH2, what other base can be used to convert 1,1-dichlorobutane into 1-butyne? A) NaOCH3 B) NaOH C) NaOCH2CH3 D) KOC(CH3)3

Vicinal dihalide  Alkyne CH3(CH2)7CH—CH2Br Br 1. 3NaNH2, NH3 2. H2O (54%) CH3(CH2)7C CH 38

Question Which of the following compounds yield 1-heptyne on being treated with three moles of sodium amide (in liquid ammonia as the solvent) followed by adding water to the reaction mixture? A) 1,1,2,2-tetrachloroheptane B) 1-bromo-2-chloroheptane C) 1,1,2-trichloropentane D) all of the above

Reactions of Alkynes 1

Reactions of Alkynes Acidity Hydrogenation Metal-Ammonia Reduction Addition of Hydrogen Halides Hydration Addition of Halogens Ozonolysis 2

Hydrogenation of Alkynes 3

Atomic Force Microscopy of Acetylene Lawrence Berkeley Laboratory (LBL) H

Imaging: acetylene on Pd(111) at 28 K H Molecular Image Tip cruising altitude ~700 pm Δz = 20 pm Why don’t we see the Pd atoms? Because the tip needs to be very close to image the Pd atoms and would knock the molecule away p orbital pz TIP H O + Surface atomic profile Tip cruising altitude ~500 pm Δz = 2 pm Calculated image (Philippe Sautet) If the tip was made as big as an airplane, it would be flying at 1 cm from the surface and waving up an down by 1 micrometer The STM image is a map of the pi-orbital of distorted acetylene 1 cm (± 1 μm) M. Salmeron (LBL)

Excitation of frustrated rotational modes in acetylene molecules on Pd(111) at T = 30 K Tip e- ((( ) ( ))) M. Salmeron (LBL)

Hydrogenation of Alkynes cat RC CR' + 2H2 RCH2CH2R' catalyst = Pt, Pd, Ni, or Rh alkene is an intermediate 4

Partial Hydrogenation RC CR' cat H2 RCH CHR' cat H2 RCH2CH2R' Alkenes could be used to prepare alkenes if a catalyst were available that is active enough to catalyze the hydrogenation of alkynes, but not active enough for the hydrogenation of alkenes. 8

Lindlar Palladium RC CR' cat H2 RCH CHR' cat H2 RCH2CH2R' There is a catalyst that will catalyze the hydrogenation of alkynes to alkenes, but not that of alkenes to alkanes. It is called the Lindlar catalyst and consists of palladium supported on CaCO3, which has been poisoned with lead acetate and quinoline. syn-Hydrogenation occurs; cis alkenes are formed. 9

Example CH3(CH2)3C C(CH2)3CH3 + H2 Lindlar Pd CH3(CH2)3 (CH2)3CH3 C H (87%) 10

Figure: 09-02-10UN Caption:

Metal-Ammonia Reduction of Alkynes Alkynes  trans-Alkenes 11

Partial Reduction RC CR' RCH CHR' RCH2CH2R' Another way to convert alkynes to alkenes is by reduction with sodium (or lithium or potassium) in ammonia. trans-Alkenes are formed. 8

Figure: 09-03-02UN Caption: To form a trans alkene, two hydrogens must be added to the alkyne with anti stereochemistry, so this reduction is used to convert alkynes to trans alkenes.

Example CH3CH2C CCH2CH3 Na, NH3 CH3CH2 H C CH2CH3 H (82%) 10

Figure: 09-03-02UN Caption: To form a trans alkene, two hydrogens must be added to the alkyne with anti stereochemistry, so this reduction is used to convert alkynes to trans alkenes.

Question How would you accomplish the following conversion? A) NaNH2 B) H2, Lindlar Pd C) Na, NH3 D) either B or C

Mechanism Metal (Li, Na, K) is reducing agent; H2 is not involved; proton comes from NH3 four steps (1) electron transfer (2) proton transfer (3) electron transfer (4) proton transfer 13

Question Select the most effective way to synthesize cis-2-pentene from 1-propyne. A) 1) NaNH2 2) CH3CH2Br 3) H2, Pd B) 1) NaNH2 2) CH3Br 3) H2, Lindlar Pd C) 1) NaNH2 2) CH3CH2I 3) H2, Lindlar Pd D) 1) NaNH2 2) CH3CH2Br 3) Na,NH3

Question Which reagent would accomplish the transformation of 3-hexyne into trans-3-hexene? A) H2/Ni B) H2, Lindlar Pd C) Na, NH3 D) NaNH2, NH3

Problem Suggest an efficient syntheses of (E)- and (Z)-2- heptene from propyne and any necessary organic or inorganic reagents. 18

Problem Strategy 19

Problem Strategy 19

Problem Synthesis 1. NaNH2 2. CH3CH2CH2CH2Br H2, Lindlar Pd Na, NH3 19

Question Which would be the best sequence of reactions to use in order to prepare cis-3-nonene from 1-butyne? A) 1. NaNH2 in NH3; 2. 1-bromopentane; 3. H2, Lindlar Pd B) 1. NaNH2 in NH3; 2. 1-bromopentane; 3. Na, NH3 C) 1. H2, Lindlar Pd; 2. NaNH2 in NH3; 3. 1- bromopentane D) 1. Na, NH3; 2. NaNH2 in NH3; 3. 1- bromopentane

Addition of Hydrogen Halides to Alkynes 21

Follows Markovnikov's Rule HBr CH3(CH2)3C CH CH3(CH2)3C CH2 Br (60%) Alkynes are slightly less reactive than alkenes 22

Two Molar Equivalents of Hydrogen Halide CH3CH2C CCH2CH3 2 HF F C H CH3CH2 CH2CH3 (76%) 24

Free-radical Addition of HBr CH3(CH2)3C CH CH3(CH2)3CH CHBr peroxides (79%) regioselectivity opposite to Markovnikov's rule 25

Hydration of Alkynes expected reaction: H+ RC CR' H2O + OH RCH CR' enol observed reaction: H+ + RC CR' H2O RCH2CR' O ketone 27

enols are regioisomers of ketones, and exist in equilibrium with them OH RCH CR' RCH2CR' O enol ketone enols are regioisomers of ketones, and exist in equilibrium with them keto-enol equilibration is rapid in acidic media ketones are more stable than enols and predominate at equilibrium 28

Figure: 09-03-02UN Caption: To form a trans alkene, two hydrogens must be added to the alkyne with anti stereochemistry, so this reduction is used to convert alkynes to trans alkenes.

Mechanism of conversion of enol to ketone .. : O H C H + : O H H 30

Mechanism of conversion of enol to ketone + : .. 30

Mechanism of conversion of enol to ketone .. : O H H H C C + : O : H 30

Mechanism of conversion of enol to ketone .. O : : O H H H C C + 30

Key Carbocation Intermediate Carbocation is stabilized by electron delocalization (resonance). .. O C H .. + : O H H C C + 30

Mechanism of conversion of enol to ketone : + .. 30

Useful for symmetrical starting alkynes to produce a single product. Unsymmetrical starting alkynes that are not terminal produce a mixture of ketones…non-regioselectively.

Markovnikov's rule followed in formation of enol, Regioselectivity Markovnikov's rule followed in formation of enol, Useful with terminal alkynes. O H2O, H2SO4 CH3(CH2)5C CH CH3(CH2)5CCH3 HgSO4 (91%) via CH3(CH2)5C CH2 OH 32

Aldehyde vs. Ketone

Question What is the product of the acid catalyzed hydration of 1-hexyne? A) B) C) D)

I) Which reactions give ketones? II) Which reactions give aldehydes? Question I) Which reactions give ketones? II) Which reactions give aldehydes? A. I = C, D, E, F; II = A, B B. I = B, D, E, F; II = A, C C. I = B, C, D, F; II = A, E D. I = A, D, E, F; II = B, C E. I = A, C, D, F; II = B, E

Addition of Halogens to Alkynes 26

Example Cl (63%) C Cl2CH CH3 HC CCH3 + 2 Cl2 36

Addition is anti CH3CH2 Br Br2 C CH3CH2C CCH2CH3 Br CH2CH3 (90%) 37

NBS = N-bromosuccinimide NBS Example AgNO3 HC CCH3 + NBS Br-C CCH3 O NBr NBS = N-bromosuccinimide 36

gives two carboxylic acids by cleavage of triple bond Ozonolysis of Alkynes gives two carboxylic acids by cleavage of triple bond 38

Example CH3(CH2)3C CH 1. O3 2. H2O + CH3(CH2)3COH (51%) O HOCOH 39

Question What product is formed when 2-butyne is subjected to ozonolysis? A) B) C) D)

Answer What product is formed when 2-butyne is subjected to ozonolysis? A) B) C) D)

Alkynes Synthesis & Functions 26

Can you identify and name the function? Figure: 09-03-02UN Caption: To form a trans alkene, two hydrogens must be added to the alkyne with anti stereochemistry, so this reduction is used to convert alkynes to trans alkenes.

Example

Question What is the structure of Compound Y in the following synthetic sequence? A) pentane B) cis-2-pentene C) trans-2-pentene D) 2-pentyne

Answer What is the structure of Compound Y in the following synthetic sequence? A) pentane B) cis-2-pentene C) trans-2-pentene D) 2-pentyne

Natural Products 26

Polyynes

Compound 1 is isolated from the root and bark of mistletoe, Paramacrolobium caeruleum (Loranthaceae). The stems and leaves of members of this family have been used for the treatment of cancer in Indonesia.

Compound 2, thiarubrine B, has been isolated from Giant Ragweed, Ambrosia trifida. Native cultures in Canada and Africa use plants with similar natural products to treat skin infections and intestinal parasites.

Compound 3, oplopandiolacetate, is found in the bark and roots of Devil's club, Oplopanax horridus. It is used medicinally by native Americans to treat a variety of ailments. Other polyynes found in plants include oenanthotoxin, cicutoxin, and falcarinol .

Compound 4, dihydromatricaria acid, is found in the soldier beetle, Cantharidae, who are related to the Lampyridae or firefly family, but unable to produce light. They provide biological control of a number of insect pests including grasshoppers, aphids, caterpillars and other soft bodied insects.

Histrionicotoxin is isolated from the skin of frogs in the Dendrobatidae family. It comes from insects in their diet and is used by indigenous South American tribes as a poison on arrows.

Polyyne Cu (I) Coupling Reactions AgNO3 HC CCH3 + NBS Br-C CCH3 36