Chem 3010 Chapter 8: Alkynes Jan, 2001

Problems: 1, 3, 5a, 6-13, 19a-d, 21-25, 27-29, 30 a-c, 31-34, 36, 40, 41, 44.

8.1 Electronic structure of Alkynes.

Fig 8.1 - the orbital setup is a repeat of 1.19 for the -CºC-

The s framework is formed from two sp orbitals from each carbon -C-C-

these bonds are at 180^o to one another

Above and below is one p orbital formed from one p electron from each C

Inside and outside the page another p orbital formed from one p electron from each C

-CºC- is linear

The CºC bond length is smaller than the C=C which is in turn smaller than the C-C bond length

The bond dissociation energy of CºC is 835 kJ (don't memorize) but thought of as three components

the weaker p bond dissociation energy 224 kJ

the stronger p bond dissociation energy 236 kJ

the s bond dissociation energy 376 kJ

The weakest bond is the most vulnerable,

but not very different from the weakness of p bonds of alkenes

8.2 Naming Alkynes

Very much like naming alkenes. Use "yne", longest chain that includes the CºC, "di" as necessary, etc.

This time, no need to worry about cis and trans (Z or E)

Don't bother with "alkynyl"

8.3 Preparation of Alkynes: Elimination of Dihalides

We can apply the idea of CH-CX + KOH ® C=C + KX + H₂O from alkenes sect 7.1

The same sort of elimination reaction forms alkynes:

-HC=CX- + KOH ® CºC + KX + H₂O

The sequence leading to the alkyne actually begins with an alkene of the type -HC=CH- (p. 277)

an initial addition of two halogen atoms,

followed by sequential reactions eliminating two molecules of HX:

-HC=CH- + X₂ ® -XHC-CHX- then KOH ® -HC=CX- then more KOH ® -CºC-

The second KOH reaction normally requires heat

Stronger bases are often applied to the second elimination reaction: NaNH₂

Another important preparation route is via acetylide ions, Sect 8.9

8.4 Reactions of Alkynes:. Again the correspondence to C=C chemistry is helpful

Addition of HX: -CºC- + HX ® -HC=CX- (with anti stereochemistry, not explained)

More HX will further add: -HC=CX + HX ® -H₂C-CX₂-

A carbocation intermediate is a vinyl carbocation, -C⁺=CH- , with a double bond already in place

Regiospecific, Markovnikov thus R-CºC-H + HX ® RXC=CH₂ and not ® RHC=CHX

the Markovnikov regiospecificity is determined by R-C⁺=CH₂ being more stable than RHC=C⁺-H

Addition of and X₂

-CºC- + X₂ ® -HXC=CHX-

Analogous with the addition to C=C, the bromonium ion intermediate causes the anti stereochemistry

8.5 Hydration of Alkynes, mercuric ion catalyzed

Tautomerism. This term refers to the equilibrium

between an enol tautomer, and a carbonyl (book "keto") tautomer p. 280

normally the equilibrium favors the carbonyl form

(be careful that you do not confuse this equilibrium with a pair of resonance forms)

A normal addition of H-OH to a -CºC- would result in an enol

but this enol will tautomerize to a carbonyl compound

so the reaction is -CºC- + H₂O ® -C(=O)-CH₂- (Hg²⁺ catalyzed)

Fig 8.3 gives the mechanism, which is analogous to the mercury catalyzed hydration of alkenes

(some texts will include a mercurinium ion)

Unlike the oxymercuration of alkenes, NaBH₄ is not needed for -CºC- + H₂O ® -C(=O)-CH₂-

P. 282 a terminal alkyne will produce a methyl ketone and not an aldehyde:

R-CºC-H + H₂O ® -C(=O)-CH₃ and not -CH₂-CH(=O)

This is explained by comparing stabilities of the vinyl carbocation intermediates:

R-CºC-H + Hg²⁺ ®

Hydroboration - oxidation of alkynes

An alternative hydration method for conversion of alkynes to carbonyl compounds

The same reagents are used as in the alkene hydration: BH ₃ then H₂O₂, OH^-

For alkynes, the Markovnikov / non-Markovnikov issue arises when terminal alkenes are hydrated

Regiospecificity®	Markovnikov	non-Markovnikov
Hydration Reaction
alkyne to carbonyl	R-CºC-H ® R-C(=O)-CH3	R-CºC-H ® R-CH2-CH(=O)
reagents	HgSO4, H2O	BH 3 then H2O2, OH^-
alkene to alcohol	RHC=CH2 ® RCH(OH)-CH3 *	RHC=CH2 ® RH2C-CHOH *
reagents	Hg(OAc)2, H2O then NaBH4	BH 3 then H2O2, OH^-

* other alkene types are subject to Markovnikov regiospecificity: R₂C=CH₂ and R₂C=CHR

8.6 Reduction of Alkynes.

Analogous with alkenes, H₂ adds across the triple bond: -CºC- + H₂ ® -HC=CH- (metal catalyst)

But with the metal catalysts used for alkenes, this reaction will continue: -HC=CH- + H₂ ® -H₂C-CH₂-

If it is desired to stop at the alkene stage, a Lindlar catalyst prevents further reaction of the C=C

The Lindlar catalyst is palladium that has been deactivated; its catalytic power is reduced

Stereochemistry of hydrogenation with H₂: syn addition

This was also the case with H₂ + C=C (Fig 7.10)

Thus from R-CºC-R + H₂ (+ Lindlar catalyst) ® cis alkene only

Reduction of Alkynes with Li/NH₃

This is used to obtain trans alkenes from alkynes.
Thus the stereochemistry of this reduction is predominately anti.

Mechanism, Fig 8.4. Remember only the basic parts.

Electrons are transferred - not donated - from the lithium atoms to the -CºC-.

By receiving the electrons, -CºC- is reduced. (remember the oxidation-reduction terminology)

The extra electrons are used to form the new C-H bonds

The protons are transferred from NH₃ in acid-base reactions (curved arrows)

Since the H atoms are added sequentially to the -CºC-

and without a catalyst to block the direction of approach

steric effects will favor the production of the trans product

(In contrast, the catalyst in the H₂ hydrogenation blocks random approaches of H atoms

and enforces a syn stereochemistry)

8.7 Oxidative Cleavage of Alkynes with O₃ or KMnO₄

Formation of carboxylic acids from internal alkynes, R-CºC-R' ® R-COOH + HOOCR'

Terminal alkynes produce a carboxylic acid + CO₂: R-CºC-H ® R-COOH + CO₂

8.8 Alkyne Acidity: Formation of Acetylide Anions - Acid-Base Chemistry

Recall the relationship between pK_a and acidity, Sect 2.7

The pK_a of RCºC-H is about 25

Therefore it will be a strong acid in the presence of the conjugate base of ammonia (pK_a ~35)

RCºC-H + H₂N^- --D RCºC:^- + NH₃

s.a. pK_a ~25 s.b w.b. w.a. pK_a ~35

And a weak acid in the presence of the conjugate base of H₂O (pK_a ~15)

H₂O + RCºC:^- --D OH^- + RCºC-H

s.a. pK_a ~15 s.b. w.b. w.a. pK_a~25

Thus the acetylide ion, RCºC:^- , a useful intermediate in the synthesis of alkynes

must be generated from RCºC-H with conjugate bases of acids having pK_a's of > 25

Equilibrium will not favor the generation of RCºC:^- with conjugate bases of acids like water, etc.

Comparison of Acidity of CºC-H vs. C=C-H vs. Alkyl C-H

Like many comparisons of this sort, the focus is

not on how relatively strong these acids are: CºC-H vs. C=C-H vs. Alkyl C-H

but on how relatively stable their conjugate bases are: CºC:^- vs. C=C:^- vs. Alkyl C:^-

once their conjugate bases are ranked strongest, middle, weakest, then the original acids can be ranked

Fig 8.5 - in each case the non-bonded electron pairs of the bases are compared by their hybridizations

The more s character these electrons have, the more stable they are

(recall that unhybridized 2s is more stable than unhybridized 2p)

sp hybridized electrons have 50% s character. These are the non-bonded electrons in CºC:^-

sp² hybridized electrons have 33% s character. These are the non-bonded electrons in C=C:^-

sp³ hybridized electrons have 25% s character. These are the non-bonded electrons in Alkyl C:^-

Thus the sp hybridized electrons in CºC:^- are more stable than the sp² hybridized electrons in C=C:^-

and therefore CºC:^- is a weaker base than C=C:^-

if these two bases compete in an equilibrium,

the longer arrow always points to the weaker - more stable - species

and when their conjugate acids are included, their "strong" and "weak" assignments

must conform to the strong and weak bases already determined

this gives:

CºC-H + C=C:^- --D CºC:^- + C=C-H

s.a. s.b. w.b. w.a.

from this we conclude that CºC-H is a stronger acid than C=C-H

Similar logic yields Alkyl C:^- as a stronger base than C=C:^-

and therefore Alkyl C-H is a weaker acid than C=C-H

Combining all three acids: CºC-H is a stronger acid than C=C-H is stronger acid than Alkyl C-H

this is shown in Table 8.1:

the pK_a of HCºCH is ~25; the pK_a of H₂C=CH₂ is ~ 44; and the pK_a of CH₄ is ~60

In generating acetylide ions, CH₃Li is frequently used instead of NaNH₂. How can this be?

8.9 Alkylation of Acetylide Anions

Now that acetylide ions are formed, they can act as nucleophiles in a substitution reaction (p. 279):

CºC:^- + RCH₂-Br ® CºC-CH₂R + Br^-

In the mechanism, Fig 8.7, the non-bonded electrons in CºC:^- act as the Lewis base

and ^d+CH₃ of the CH₃-Br acts as the Lewis acid (curved arrows - the Na⁺ is a spectator ion)

Much more will be described about this kind of substitution reaction but three points are made now:

1. the CºC:^- can be H-CºC:^- or it can be any R-CºC:^- these are generated from the corresponding CºC-H

2. The RCH₂-Br can be H₃C-Br (as in Fig 8.7) or it can be any RCH₂-Br

but the alkyl halide cannot be 2^o (R₂CHBr) or 3^o (R₃CBr)

extra R groups prevent the CºC:^- from attacking the substituting C atom (Fig 8.7)

because of a steric effect

in the presence of CºC:^- 2^o and 3^o alkyl halides instead undergo elimination to form alkenes

compare p. 233. CºC:^- serves as the base instead of OH^-

3. This is one of only a few reactions in organic chemistry that joins one C to another C.

for this reason, the reaction is often incorporated in many synthetic sequences (see sect 8.10 below)

often one forms CºC:^- in two different steps of a sequence (example problem 8.11)

by starting with acetylene, H-CºC-H,

then forming acetylide ion then, treating it with RCH₂Br ® H-CºC-CH₂R

then repeating the same ion forming process ® ^-:CºC-CH₂R

then adding R'CH₂Br ® R'CH₂-CºC-CH₂R

8.10 Organic Synthesis

Reactions are sequenced to form complex products

you will need to use a reaction in one chapters and link its product to the a reaction in another chapter

(the reagents are often written 1…2…etc. above and below the reaction arrow to avoid excessive detail)

as more reactions are introduced in the book, synthetic problems will help you remember them

and a greater variety of products will be available

remember that there are often stereochemical and regiospecific implications

Also the problems may be turned around. Given a product how is it formed?

this way you need to remember the reagents for conversions

the problem set has some questions like these

if the product requires several reactions in sequence, the best way is by asking

"what is the immediate precursor to this product?"

then after identifying the reactant/reagents/conditions form the immediate precursor then

"what is the immediate precursor to the immediate precursor?"

Working backwards is a technique known as retro-synthesis

Organization: lists and/or flash cards - be able to know reactions backwards and forwards

for example, one side: "alkyl halide + ? ® alkene" other side: "alkyl halide + KOH/ethanol ® ?"

Practice problems in class

In the book problems, you may be asked to synthesize compounds like "hexanal" or "3-hexanone" or

"pentanedioic acid" these functional groups can be looked up in Table 3.1, but for the time being the "al"

"one" and "oic acid" need not be memorized. With Table 3.1 you can deduce that

Hexanal, an aldehyde, is CH₃CH₂CH₂CH₂CH₂CH(=O)

3-hecanone, a ketone, is CH₃CH₂C(=O)CH₂CH₂CH₃

and "pentanedioic acid, a di-carboxylic acid is HOOCCH₂CH₂CH₂COOH

For the present on exams and quizzes, structures will be given.

We will work with both names and structures in class.