Alkynes – Organic chemistry – solutions to problems

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Alkynes – solutions to problems

Solution 1:

Since this is a hydrogenation catalyzed by the Lindlar catalyst, an alkene will be produced whose stereochemistry will be Z as the two hydrogens enter from the same side of the molecule (syn addition).


Solution 2:

In this case it is a hydrogenation by an electron transfer process and the alkene with E stereochemistry is produced.


Solution 3:

Upon addition of a single mole of HCl and since the initial product (2-butene) is symmetrical a 2-chloropropene will be produced. Since this is a trans addition the stereochemistry of the same will be as indicated.


Solution 4:

If an excess of HCl is used in the addition of the second mole Markovnikov’s rule is satisfied and the two chlorines will be located on the same carbon atom.


Solution 5:

Markovnikov’s rule is satisfied and the two hydrogens will go on the more hydrogenated (less substituted) carbon producing 2,2-dibromo-3,3-dimethylbutane.


Solution 6:

This will be an anti-Markovnikov addition in this case producing 1,1-dibromo-3,3-dimethylbutane and the mechanism is radicalic as indicated in the scheme:


Solution 7:

The addition of one mole of chlorine to an alkyne leads to a dichloroalkene with trans stereochemistry. That of 2 moles leads to a tetrachloroalkane.


Solution 8:

Addition of water to an alkene yields an enol, which by tautomerism gives a ketone. When an alkene is involved, an alcohol is produced.


Solution 9:

Addition of water to an alkyne starts with the addition of a proton with Markovnikov regioselectivity, giving an alkenyl carbocation that traps a water molecule transforming into an enol, which by tautomerism gives the corresponding methylketone.


Solution 10:

The end result is the addition of water to the terminal triple bond but with an anti-Markovnikov regioselectivity, whereby the enol formed, by tautomerism would give an aldehyde.


Solution 11:

a) Addition of one mole of halogen on an alkyne leads to the formation of a vinyl dihalide. These compounds are very unstable, so they can undergo a second addition of halogen, even if the mole of reactant has not been consumed.


b) The hydroboration/oxidation reaction on but-1-ino results in the addition of a water molecule with an anti-Markovnikov-type regiochemistry, producing an enol, which rapidly evolves into a carbonyl compound, in this case an aldehyde, since the oxygen atom is on the terminal carbon of the chain.


c) Catalytic hydrogenation of the but-1-ino leads to the formation of the corresponding alkane of equal number of carbons, by addition of two moles of hydrogen on the triple bond.


d) The reduction with metals (Li or Na) in liquid ammonia of an alkyne leads to the partial reduction of the alkyne generating an alkene. If there is the possibility, the one with Z configuration is formed. In this case, being a terminal alkyne, the same product is obtained with this procedure or by hydrogenation with the Lindlar catalyst.


e) With potassium permanganate the oxidative cleavage of an alkyne is produced, to give the potassium salt of the corresponding carboxylic acid, which is released by acidifying. If the alkyne is terminal, CO2 is generated.


f) Alkyl halides are added to alkynes following Markovnikov’s rule sequentially. In a first step a vinyl halide is produced, which is then transformed into a geminal alkyl dihalide.


Solution 12:

a) The result of the addition of one mole of X2 on but-2-ino is the formation of (E)-2,3-dihalobut-2-ene. Note that the relative stereochemistry of the halogen atoms is trans, because the attack of the second halogen is on the opposite side where the intermediate cyclic halonium ion of the reaction is produced.


b) The hydroboration/oxidation reaction of disubstituted alkynes leads to the formation of ketones. If the triple bond is attached to two different substituents, a mixture of two ketones would be obtained, since there is no preference between one or the other carbon of the triple bond by the borane. As the alkyne is symmetrical, both ketones are equivalent, so that the reaction product is unique.


c) If no other particular is specified, the conventional catalytic hydrogenation of an alkyne produces the saturated hydrocarbon of equal number of carbon atoms. Only if a modified catalyst such as Lindlar’s catalyst is used, the partial triple bond hydrogenation occurs, generating, in addition, an alkene of Z-configuration.


d) The reduction of an alkyne by treatment with a metal (Na or Li) in liquid ammonia generates an alkene with E configuration.

The metal transfers an electron to the alkyne to give a radical anion, which captures a proton from the ammonia, according to an acid-base process. At this point a second metal atom intervenes and transfers another electron. In this way an anion is produced. Finally, the anion captures a proton from the solvent to give the alkene.


The reaction is stereospecific, since only one of the possible isomers is obtained. The key to the process is the greater stability of the trans-configuration radical anion than the cis-configuration one.

e) The oxidative cleavage with potassium permanganate of but-2-ino generates two equal fragments. Once acidified, 2 moles of acetic acid per mole of substrate are obtained.


f) The addition of hydrogen halides to alkynes leads to the formation of geminal dihalides following a two-step ionic mechanism. In a first step a vinyl halide is produced via an intermediate vinyl carbocation, formed by proton attack on the starting alkyne. Next, a second HX molecule is added, forming a carbocation on which the attack of the second X ion takes place. Thus, both halogens are attached to the same carbon. Since this molecule is symmetrical, a single product is obtained.


Solution 13:

Compound A) The addition of water to a triple bond, catalyzed by acids and in the presence of mercuric salts, leads to the formation of ketones, via an enol. In this case a mixture of isomers will be obtained.


Compound B) Ozonolysis of 4-methylpent-2-ino produces two carboxylic acids, since the starting alkyne is not a symmetrical compound: acetic acid and 2-methylpropanoic acid (isobutyric acid).


Compound C) Alkali metals (Li or Na) in liquid ammonia produce alkenes of trans configuration. This reaction applied to 4-methylpent-2-ino leads to the formation of (E)-4-methylpent-2-ene.


Compound D) The addition of HX to an alkyne in the presence of radicals follows an anti-Markovnikov type mechanism whereby the halogen is added on the least substituted carbon. Addition of the nucleophile to disubstituted alkenes does not show such a preference, so a mixture of products will be obtained.


Compound E) Alkynes treated with hydrogen in the presence of a metal catalyst under conventional conditions add two moles of hydrogen to give an alkane.


Compound F) Permanganate oxidation of 4-methylpent-2-ino under controlled conditions (pH and temperature) leads to the formation of a diketone, in which carbonyl groups are generated on the carbons forming part of the triple bond.

826828Compound G) The addition of bromine on a substituted alkyne produces a tetrabrominated derivative according to the following reaction scheme:

Compound H) Addition of 1 mol of HBr on alkynes in the absence of peroxides follows Markovnikov’s rule. In a disubstituted alkyne there is no preference for proton attack on any of the carbons supporting the triple bond, so a mixture of products will be produced.


Solution 14:

Product a) can be obtained either by ozonolysis or by oxidation with potassium permanganate in acidic medium of any of the proposed alkynes.


Product b) can be prepared from III by two procedures: acid catalyzed addition of water in the presence of mercuric salts or by borane reaction followed by oxidation with hydrogen peroxide in a basic medium. Both reactions show an inverse regiochemistry, appreciable in terminal alkynes, but since the alkyne is symmetrical, both reactions lead to the formation of the same product.


Product c) can be obtained by a sequence of reactions starting from alkyne I. The terminal alkynes treated with a base such as sodium amidide give the corresponding acetylide. These acetylides treated with a primary or secondary alkyl halide give substituted alkynes. For the present case the sequence of reactions would be as follows:


The preparation of product d) requires, again, a sequence of reactions. Starting from I, an alkyne is constructed with the number of carbons necessary to obtain D through an acetylide, followed by a substitution reaction on ethyl bromide, and then the triple bond is transformed into a double bond with trans stereochemistry. For this purpose the intermediate alkyne is treated with Nao in liquid NH3.


Product e) comes from the hydration reaction of I by a hydroboration/oxidation process. In this reaction an enol is produced with the hydroxyl in the primary position (anti-Markovnikov), which rapidly evolves to the carbonyl compound which in this case is an aldehyde.


Product f) is prepared from I by acid-catalyzed electrophilic addition of water in the presence of Hg(II) salts. The reaction shows an orientation (Markovnikov) opposite to that of the previous case. Likewise, an enol is obtained which evolves into a ketone by a process of equilibrium displaced towards the formation of the carbonyl compound, since the OH group is on a secondary carbon.


Product g) is obtained from the partial hydrogenation reaction of III, carried out with hydrogen, in the presence of the Lindlar catalyst which makes the stereochemistry of the double bond cis-type.


Product h) is prepared from terminal alkyne I by addition of HBr in the presence of peroxides. The reaction is an anti-Markovnikov-type addition.


Solution 15:


a) In the reaction described, catalytic hydrogenation of the triple bond occurs, but it has not been taken into account that under these conditions the addition of hydrogen over the double bond occurs simultaneously, so that the product obtained is the corresponding saturated hydrocarbon.


b) The addition of 1 mol of bromine on a triple bond produces a dibrominated neighboring derivative and not a geminal one (on the same carbon). This is because the reaction intermediate is a cyclic bromonium ion, which is attacked by Br, on the opposite side, with the consequent opening of the cyclic bromonium ion, so that the bromine atoms are attached to different carbons. The postulated mechanism makes it impossible for both bromine atoms to be attached to the same carbon.


c) The formation of a disubstituted alkyne from a monosubstituted one is represented. The conventional procedure consists of the formation of an acetylide by treatment with a base such as sodium amidide, followed by reaction with an alkyl halide. For this reaction to be feasible, the alkyl halide used in the second part must be primary or secondary. In this case there will be no substitution reaction and possibly an elimination reaction on the halide.


d) The addition of HBr to alkynes in the absence of peroxides is a process that follows a Markovnikov-type mechanism, and in which sequentially two moles of HBr are added. In a first step a vinyl halide is formed and in a second step the addition of another mole of HBr leads to the formation of an alkyl dihalide. According to the mechanism postulated for the reaction, the reaction intermediate for both processes is a carbocation. In both cases, the more stable carbocation will occur on the secondary carbon, whereby the nucleophile (Br) will attack the secondary position. The product obtained is a geminal dibrominated derivative (on the same carbon).


Solution 16:

Addition of bromine to an alkyne will generate the cyclic bromonium cation which in the presence of water will give the corresponding bromo enol which tautomerizes the corresponding bromoketone:


Solution 17:

In the following scheme, the main reactions of terminal alkynes are illustrated: (a) chain elongation prior formation of the acetylide; (b) partial hydrogenation to alkene with E (trans) stereochemistry; (c) Markovnikov-type addition of one mole of HBr; (d) bromination; (e) oxidative cleavage with permanganate; (f) formation of acetylides with copper salts; (g) Diels-Alder reaction to give cyclohexadienes; (h) hydrogenation to alkane; and (i) hydration catalyzed by mercuric salts yielding methylketones.


Solution 18:

  • A) Conversion from alkene to E-alkene is carried out with Lio / NH3(l).
  • B) Conversion from alkyne to aldehyde is achieved by hydration with anti-Markovnikov regioselectivity and will be carried out with BH3 / HO- followed by H2O2.
  • C) Oxidation to diketone (equivalent to alkene dihydroxylation) can be performed with KMnO4 / pH=7.
  • D) Conversion from alkyne to ketone will be performed by hydration, i.e. BH3 / HO followed by H2O2 or H2O / H2SO4 / HgSO4.
  • E) The alkyne to terminal dibromo conversion (anti-Markovnikov regioselectivity) will be performed with 2 moles HBr in the presence of peroxides.
  • F) The alkyne to dibromo conversion will be performed with the addition of 2 moles HBr (Markovnikov regioselectivity).


Solution 19:

The order of acidity reflects the stabilization of the carbanion formed. The acidity of the protons in the alkane-alkene-alkyne series increases in this order as the hydrogen-bearing carbon becomes more electronegative.


Solution 20:

Like alkenes, alkynes undergo oxidative cleavage when treated with ozone.


Solution 21:

A) will be the hexabrominated derivative as a result of the addition of bromine to the double and triple bond; B) the addition of HBr follows Markovnikov’s rule; C) hydrogenation of an alkene with alkali metal in liquid ammonia produces the alkene with E (trans) stereochemistry; D) sodium amidide produces the acetylide which reacts with iodoalkane giving the disubstituted alkyne; E) hydration of the disubstituted alkynes will produce two different enols which tautomerize to the two possible ketones; F) dihydration of the dialkyne will produce the corresponding diketone.


Solution 22:

The pKa of the terminal alkynes is 26, therefore only H and NH2 could be used, as they come from weaker acids than alkynes (higher pKa values).

Acid formBasepKa

Solution 23:

A) To perform the conversion from alkyne to aldehyde, we must perform an anti-Markovnikov hydration with borane and subsequent oxidation with H2O2 in basic medium; B) trans alkene will be obtained by hydrogenation using lithium in liquid ammonia; C) degradation to carboxylic acid will be by ozonolysis or with permanganate; D) hydration of terminal alkynes yields enols that tautomerize to methylketones; E) obtaining the geminal dibromo is achieved by Markovnikov addition of HBr; F) obtaining the derived tetrabromo involves the addition of two moles of bromine.


Solution 24:

The reaction conditions for starting from acetylene (ethene) to obtain the desired products are shown in the scheme:


Solution 25:

The key to the synthesis are two reactions: the elongation of the alkyne chain and the nucleophilic addition of an acetylide to a carbonyl compound (propalan).


Solution 26:

Hydration of an alkyne produces an enol which tautomerizes the corresponding carbonyl compound (aldehyde or ketone):


Solution 27: