Alkynes • Alkynes Are Molecules That Incorporate a C≡C Triple Bond

10.1 Alkynes • Alkynes are molecules that incorporate a C≡C triple bond.

• Given the presence of two pi bonds and their associated electron density, alkynes are similar to alkenes in their ability to act as a nucleophile.

• Converting pi bonds to sigma bonds generally makes a molecule more stable. WHY?

• Acetylene is the simplest alkyne. – It is used in blow torches and as a precursor for the synthesis of more complex alkynes. • More than 1000 different alkyne natural products have been isolated. – One example is histrionicotoxin, which can be isolated from South American frogs, and is used on poison-tipped arrows by South American tribes.

• An example of a synthetic alkyne is ethynylestradiol. • Ethynylestradiol is the active ingredient in many birth control pills.

• The presence of the triple bond increases the potency of the drug compared to the natural analog. • How do you think a C≡C triple bond affects the molecule’s geometry? Its rigidity? Its intermolecular attractions?

• Alkynes are named using the same procedure we used in Chapter 4 to name alkanes with minor modifications: 1. Identify the parent chain, which should include the C≡C triple bond. 2. Identify and name the substituents. 3. Assign a locant (and prefix if necessary) to each substituent, giving the C≡C triple bond the lowest number possible. 4. List the numbered substituents before the parent name in alphabetical order. Ignore prefixes (except iso) when ordering alphabetically. 5. The C≡C triple bond locant is placed either just before the parent name or just before the -yne suffix.

• Alkynes are named using the same procedure we used in Chapter 4 to name alkanes with minor modifications: 1. Identify the parent chain, which should include the C≡C triple bond.

• Alkynes are named using the same procedure we used in Chapter 4 to name alkanes with minor modifications: 3. Assign a locant (and prefix if necessary) to each substituent. giving the C≡C triple bond the lowest number possible.

– The locant is ONE number, NOT two. Although the triple bond bridges carbons 2 and 3, the locant is the lower of those two numbers. Copyright 2012 John Wiley & Sons, Inc. 10 -7 Klein, Organic Chemistry 1e 10.2 Alkyne Nomenclature

• Alkynes are named using the same procedure we used in Chapter 4 to name alkanes with minor modifications: 4. List the numbered substituents before the parent name in alphabetical order. Ignore prefixes (except iso) when ordering alphabetically. 5. The C≡C triple bond locant is placed either just before the parent name or just before the -yne suffix.

• In addition to the IUCAP naming system, chemists often use common names that are derived from the common parent name acetylene.

• You should also be aware of the terminology below.

• Name the molecule below.

• Recall that when triple bonds are drawn, their angles are 180°.

• Recall that terminal alkynes have a lower pKa than other hydrocarbons.

• Acetylene is 19 pKa units more acidic than ethylene, which is 1019 times stronger. – Does that mean that terminal alkynes are strong acids?

• Because acetylene (pKa=25) is still much weaker than water (pKa=15.7), a strong base is needed to make it react.

• Recall from Chapter 3 that we used the acronym ARIO to rationalize differences in acidity strengths. – Use ARIO to explain why acetylene is a stronger acid than ethylene which is stronger than ethane.

• Use ARIO to rationalize the equilibria below.

• A base’s conjugate acid pKa must be greater than 25 for it to be able to deprotonate a terminal alkyne. Copyright 2012 John Wiley & Sons, Inc. 10 -13 Klein, Organic Chemistry 1e 10.4 Preparation of Alkynes

• Like alkenes, alkynes can also be prepared by elimination.

• Such eliminations usually occur via an E2 mechanism: – GEMINAL dihalides can be used.

– VICINAL dihalides can also be used.

– E2 requires anti-periplanar geometry.

• Often, excess equivalents of NaNH2 are used to shift the equilibrium toward the elimination products.

1- • NH2 is quite strong, so if a terminal alkyne is produced, it will be deprotonated. • That equilibrium will greatly favor products.

• A proton source is needed to produce the alkyne.

• Predict the products in the example below.

• Like alkenes, alkynes can readily undergo hydrogenation.

• Two equivalents of H2 are consumed for each alkynealkane conversion.

• The cis alkene is produced as an intermediate. WHY cis?

Copyright 2012 John Wiley & Sons, Inc. 10 -18 Klein, Organic Chemistry 1e 10.5 Reduction of Alkynes – Poisoned Catalyst • A deactivated or poisoned catalyst can be used to selectively react with the alkyne.

• Lindlar’s catalyst and P-2 (Ni2B complex) are common examples of a poisoned catalysts.

• Reduction with H2 gives syn addition. • Dissolving metal conditions can give anti addition producing the trans alkene.

• Ammonia has a boiling point of –33°C, so the temperature for these reactions must remain very low. • Why can’t water be used as the solvent?

Copyright 2012 John Wiley & Sons, Inc. 10 -21 Klein, Organic Chemistry 1e 10.5 Reduction of Alkynes – Dissolving Metal Reductions • Mechanism—step 1: – Note the single-barbed and double-barbed (fishhook) arrows. – Why does Na metal so readily give up an electron?

• Why is the first intermediate called a RADICAL ANION? • The radical anion adopts a trans configuration to reduce repulsion.

• Draw the product for step 3 of the mechanism.

• Do the pKa values for NH3 and the alkene favor the proton transfer?

• Practice with CONCEPTUAL CHECKPOINT 10.10.

• Practice with CONCEPTUAL CHECKPOINT 10.11.

• Like alkenes, alkynes also undergo hydrohalogenation.

• Draw the final product for the reaction above. • Do the reactions above exhibit Markovnikov regioselectivity? Copyright 2012 John Wiley & Sons, Inc. 10 -28 Klein, Organic Chemistry 1e 10.6 Hydrohalogenation of Alkynes

• Modeled after the hydrohalogenation of alkenes, you might expect alkynes to react by the same mechanism.

• Yet, the mechanism above does not explain all observed phenomena: – A slow reaction rate – 3rd order overall rate law – Vinylic carbocations are especially unstable Copyright 2012 John Wiley & Sons, Inc. 10 -29 Klein, Organic Chemistry 1e 10.6 Hydrohalogenation of Alkynes

• Kinetic studies on the hydrohalogenation of an alkyne suggest that the rate law is 1st order with respect to the alkyne, and 2nd order with respect to HX.

• What type of collision would result in such a rate law? – Unimolecular, bimolecular, or termolecular?

• Reaction rate is generally slow for termolecular collisions. WHY? • Considering the polarizability of the alkyne, does the mechanism explain the regioselectivity? • May involve multiple competing mechanisms. Copyright 2012 John Wiley & Sons, Inc. 10 -31 Klein, Organic Chemistry 1e 10.6 Hydrohalogenation of Alkynes

• Peroxides can be used in the hydrohalogenation of alkynes to promote anti-Markovnikov addition just like with alkenes.

• Which product is E and which is Z? • The process proceeds through a free radical mechanism that we will discuss in detail in Chapter 11. • Practice with CONCEPTUAL CHECKPOINT 10.13.

• Like alkenes, alkynes can also undergo acid catalyzed Markovnikov hydration.

• The process is generally catalyzed with HgSO4 to compensate for the slow reaction rate that results from the formation of vinylic carbocation.

• HgSO4 catalyzed hydration involves the mercury (II) ion interacting with the alkyne. • Can you imagine what that interaction might look like and how it will increase the rate of reaction for the process?

• The enol/ketone TAUTOMERIZATION generally cannot be prevented and favors the ketone greatly.

• TAUTOMERS are constitutional isomers that rapidly interconvert. How is that different from resonance? • Practice with SKILLBUILDER 10.3. Copyright 2012 John Wiley & Sons, Inc. 10 -35 Klein, Organic Chemistry 1e 10.8 Hydroboration-Oxidation

• Hydroboration-oxidation for alkynes proceeds through the same mechanism, as for alkenes, giving the anti- Markovnikov product. • It also produces an enol that will quickly tautomerize.

• In this case, the tautomerization is catalyzed by the base (OH-) rather than by an acid.

• In general, we can conclude that a C=O double bond is more stable than a C=C double bond. WHY?

• After the –BH2 and –H groups have been added across the C=C double bond, in some cases, an undesired second addition can take place. H H H B H B H H R Undesired product

R H BH 2

• To block out the second unit of BH3 from reacting with the intermediate, bulky borane reagents are often used. Copyright 2012 John Wiley & Sons, Inc. 10 -38 Klein, Organic Chemistry 1e 10.8 Hydroboration-Oxidation

• Some bulky borane reagents are shown below.

• Practice with CONCEPTUAL CHECKPOINT 10.20.

• Predict products for the following reaction.

• Draw the alkyne reactant and reagents that could be used to synthesize the following molecule. O

• Markovnikov hydration leads to a ketone. • Anti-Markovnikov hydration leads to an aldehyde.

• Alkynes can also undergo halogenation. • Two equivalents of halogen can be added.

• You might expect the mechanism to be similar to the halogenation of alkenes, yet stereochemical evidence suggests otherwise.

• When one equivalent of halogen is added to an alkyne, both anti and syn addition is observed.

• The halogenation of an alkene undergoes anti addition ONLY. • The mechanism for alkyne halogenation is not fully elucidated.

• When alkynes react under ozonolysis conditions, the pi system is completely broken.

• The molecule is cleaved, and the alkyne carbons are fully oxidized. • Practice with CONCEPTUAL CHECKPOINT 10.25. Copyright 2012 John Wiley & Sons, Inc. 10 -44 Klein, Organic Chemistry 1e 10.10 Alkyne Ozonolysis

• Predict the product(s) for the following reaction.

H2O

• As acids, terminal alkynes are quite weak. • Yet, with a strong enough base, a terminal alkyne can be deprotonated and converted into a good nucleophile.

• Which has a higher pKa, NH3 or R-C≡C-H? WHY?

• The alkynide ion can attack a methyl or 1° alkyl halide electrophile.

• Such reactions can be used to develop molecular complexity.

• Alkynide ions usually act as bases with 2° or 3° alkyl halides to cause elimination rather than substitution.

• Acetylene can be used to perform a double alkylation.

– Why will the reaction be unsuccessful if the NaNH2 and Et-Br are added together? e . a 2 q N NH2 . 2 eq Et-Br

• Complex target molecules can be made by building a carbon skeleton and converting functional groups. • Practice with SKILLBUILDER 10.5.

• Recall the methods for increasing the saturation of alkenes and alkynes.

• Halogenation of an alkene followed by two dehydrohalogenation reactions can decrease saturation.

• We will have to wait until Chapter 11 to see how to convert an alkane into an alkene, but here is a preview.

Step A Step B

10.12 Synthetic Strategies

• In the alkene to alkyne conversion above, why is water needed in step 3) of that reaction?

• Practice with SKILLBUILDER 10.6.

• Give necessary reaction conditions for the multi-step conversions below.

+ Br HO

OH Br

+ En + En