Alkynes, Part I Reading Study Problems Key Concepts and ...

[Pages:6]Alkynes, Part I

Reading: Wade chapter 9, sections 9-1- 9-8 Study Problems: 9-29, 9-32, 9-36

Key Concepts and Skills:

?

Explain why alkynes are more acidic than alkanes or alkenes; show how to

generate nucleophilic acetylide anions and heavy-metal derivatives.

?

Propose effective single-step and multi-step syntheses of alkynes.

?

Show how the reduction of an alkyne leads to an alkene with the desired

stereochemistry

Lecture Topics: I. Structure of Alkynes

Alkynes contain the carbon-carbon triple bond; each triple bond introduces two degrees of unsaturation, and thus a compound containing only one triple bond has the molecular formula CnH2n-2.

Nomenclature Again, we find the longest chain containing the alkyne and use that as a base name. Number the substituents as usual. Compounds also containing double bonds are named as enynes, and containing alcohols (which have a higher priority than triple bonds) are called ynols. Examples:

H3C

2-pentyne

CH3

H3C

H3C

CH3

Br

6-bromo-2-methyl-3-heptyne

H2C H3C

CH3

2-methyl-1-penten-3-yne 2-methylpent-1-en-3-yne

H3C

CH3 OCH3

4-methoxy-2-hexyne

H3C

H HO

3-butyn-2-ol 3-hydroxy-1-butyne

Alkynes are linear with sp hydbridization at carbon; each sp carbon has two unhybridized p-orbitals which overlap to form two pi bonds perpendicular to each other. The electron density of the two pi bonds blend to form a cylinder of electron density around the sigma bond axis. Note also that C-C triple bonds are shorter than C-C single or C-C double bonds because of the high degree of s character (50%) for the overlapping sp orbitals

which form the central C-C sigma bond. Remember that s orbitals are closer to the nucleus, and thus electrons in s orbitals are more tightly held.

HC

CH

sp sp sp sp

HC

CH

bond

H

HH

CC

H H

H

1.54?

blending of pi orbitals

HC

CH

H

H

CC

H

H

1.33?

HC C H

1.20?

cylinder of electron density

II. Acidity of Terminal Alkynes

HA

A-

s character pKa

H

HH

CC

H

HH

CC

25%

50

H H

H

H H

sp3

H

H

CC

sp2

H CC

33%

44

H

H

H

H

HC C H

HC C

sp

50%

25

Terminal alkynes are more acidic than alkenes or alkanes; the negative charge on the conjugate base is in a carbon sp orbital. The negative charge in an sp hybrid orbital is more stable than in an sp2 or sp3 orbital because it is closer to the nucleus. Electrons in orbitals with high s character have less charge separation. Note: NH3 (HA) NH2- (A-) (pKa=35) CH3OH (HA) CH3O- (A-) (pKa=16)

Thus, terminal alkynes may be deprotonated by H2N-, but not by CH3OOnly terminal alkynes can be deprotonated, internal alkynes are unaffected by base treatment:

H3C C CH

NaNH2 H3C

C C Na+

H3C

NaNH2

C C CH3

No Reaction

Testing for terminal alkynes:

H3C C CH

CuNO3 H3C C C Cu non-nucleophilic

R3N precipitate

H3C C CH

AgNO3 H3C R3N

C C Ag non-nucleophilic precipitate

H3C

dilute

C C Cu

HCl

H3C C CH

Terminal alkynes also react with Cu(I) and Ag(I) salts to form copper acetylides and silver acetylides, respectively. These insoluble compounds form characteristic precipitates that indicate the presence of terminal (and not internal) alkynes. A mild acidification with dilute HCl re-forms the terminal alkyne.

III. Alkyne Synthesis

a. Alkylation of acetylide anions

Alkali-metal acetylides (Li, Na) are good nucleophiles that can react with unhindered

(methyl, primary) alkyl halides to form internal alkynes in good yields. With secondary

alkyl halides, E2 elimination often takes place.

H

H3C

C CH

NaNH2 H3C

H

I

H

C C Na+

H3C

C C CH3 + NaI

HC

NaNH2

CH

HC C

CH3CH2Br

H3C C

C

CH2CH3

CH3I

HC

C CH2CH3

NaNH2

C

C CH2CH3

Br

H3C

H

C C Na+

H3C

C CH

H

+ H2C

CH3

(2? halide)

Alkynes can also attack the carbonyl groups of aldehydes and ketones to form primary, secondary, and tertiary alcohols. Note that the dipole of the carbonyl group places a partial positive charge on the carbonyl carbon, thus making that carbon a good electrophile for attack by the nucleophilic alkynyl anion:

+

-

Nu:

C

O

H3C C

CH

NaNH2

H3C C

H

O

C Na+ H

H3C C

formaldehyde

H C C O-

H

H3C C

H3C

O

C Na+ H

H3C C

acetaldehyde

H3C C

H

H

O

H H

C C OH

H

a primary alcohol

H

H

O

CH3 C C O-

H H3C C

H

CH3 C C OH

H

a secondary alcohol

H3C

O

C

C

Na+ H3C

acetaldehyde

H

H

O

CH3

H

C

C C O-

CH3

CH3 C C C OH

CH3

a tertiary alcohol

b. Synthesis of alkenes by elimination reactions (E2)

Elimination of two molecules of H-X from vicinal or geminal dihalides occurs under basic conditions. Low yields are sometimes encountered because of base-catalyzed rearrangements, leading to multiple products. With KOH at 200?C, the most stable (internal) alkyne is produced in excess. With NaNH2, which can deprotonate terminal alkynes irreversibly, the terminal alkyne is produced in excess:

B:

H

H

R

R'

X

X

a vicinal dihalide

B:

H

X

R

R'

H

X

a geminal dihalide

Br

R X

R H

B:

Br

Br Br Br

Br Br

Br

B:

H R

R'

X R

R'

H3C

KOH 200?C

NaNH2 150?C

H

R' R'

CH3 CH3

Base-promoted isomerization: only NH2- irreversibly deprotonates terminal alkyne, removing it from the equilibria:

CC

+NH3(g)

H

CH3

H2N-

H

H

H?O?H

H

HH

CH3

-OH

HO-

H

H?O?H

H

CH3 H

H3C

CH3

H

CH3

CH3

H

an allene

H

-OH

Additional Problems for practice:

1. Provide proper IUPAC names for the following:

a. CH3CH2C CC(CH3)3

b. CH3C CCH2C

c. CH3CH==C(CH3)C CCH(CH3)2

d. H2C==CHCH==CHC CH

e. CH3CH2CH(CH2CH3)C CCH(CH2CH3)CH(CH3)2

CH2CH3

2. How might you prepare the following compounds from acetylene and any needed

alkyl halide?

b. a.

H2C

CH3

H3CH2C

CH2CH2CH(CH3)2

c.

cyclodecyne

3. How would you accomplish the following conversion (2 steps):

H H

4. Assuming that H2/Pd can reduce alkynes to alkanes, how would you synthesize the following compound starting with styrene, and using only the reagents Br2, HBr/peroxides, NaNH2, and H2/Pd? More than one step is needed.

H CH2

styrene

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