396 CHAPTER 12 hexyne. Another round of hydrogenation (H2, Lindlar’s catalyst), anti-Markovnikov addition (HBr, peroxide) and substitution (sodium acetylide) lengthens the chain by two more carbons, giving 1-octyne. Deprotonation of this terminal alkyne, followed by alkylation with another equivalent of 1-bromohexane yields 7-tetradecyne. Hydrogenation with H2 and Pt produces the desired product, tetradecane. 12.33. The following synthesis is one suggested synthetic pathway. There are likely many other acceptable approaches that accomplish the same goal. Each of the target compounds has a nine carbon linear chain, and our reagents must be alkenes with fewer than six carbon atoms. Thus, it is clear that we will be forming new C-C bonds in the course of this synthesis. The figure below outlines a retrosynthetic analysis for our target compound. An explanation of each of the steps (a-h) follows. a. Either of the products can be prepared from a common synthetic intermediate, 1-nonyne, by hydration of the alkyne via (a) Markovnikov addition, or (a') anti-Markovnikov addition. b. The terminal alkyne can be prepared via reaction of 1-bromoheptane with an acetylide anion (formed by deprotonating acetylene). c. Acetylene is prepared via a double elimination from 1,2-dibromoethane. d. 1,2-Dibromoethane is prepared via bromination of ethylene. e. 1-Bromoheptane is prepared via anti-Markovnikov addition of HBr across 1-heptene. f. 1-Heptene is prepared via hydrogenation of 1-heptyne in the presence of a poisoned catalyst. g. 1-Heptyne is prepared via reaction of 1-bromopentane with an acetylide anion. h. 1-Bromopentane is prepared via anti-Markovnikov addition of HBr across 1-pentene. Now, let’s draw the forward scheme. In the presence of peroxides, the reaction of 1-pentene with HBr produces 1- bromopentane (via anti-Markovnikov addition). Subsequent reaction with acetylide [produced from ethylene as shown by bromination (Br2), double elimination and deprotonation (excess NaNH2)] provides 1-heptyne. Reduction to the alkene (H2 / Lindlar’s catalyst) followed by anti-Markovnikov addition (HBr / peroxide) yields 1-bromoheptane. This primary alkyl bromide can then undergo an SN2 reaction when treated with acetylide (prepared above), giving the
CHAPTER 12 397 common intermediate, 1-nonyne. A hydroboration / oxidation protocol (9-BBN then H2O2, NaOH) produces the target aldehyde. Acid and mercury catalyzed hydration gives the target ketone. 12.34. The synthesis developed below is only one suggested synthetic pathway. There are likely other acceptable approaches that accomplish the same goal. An analysis of the structure of the product suggests the following origins of each of the carbon atoms in the product. The following figure outlines a retrosynthetic analysis for our target molecule. An explanation of each of the steps (a-f) follows. O aO Br b O OH + c Br d + e a. The product 3-phenylpropyl acetate can be made via an SN2 reaction between the carboxylic acid (after deprotonation to make a competent nucleophile) and the primary alkyl bromide. b. The primary alkyl bromide can be made by anti-Markovnikov addition of HBr to the monosubstituted alkene. c. The alkene is made by reduction of the corresponding terminal alkyne. d. The terminal alkyne is made by alkylating acetylene (using sodium amide to deprotonate) with benzyl bromide. e. Benzyl bromide is made via radical bromination of toluene. (The carbon adjacent to the aromatic ring is activated toward bromination due to the resonance-stabilized radical intermediate that forms.)
398 CHAPTER 12 Now, let’s draw the forward scheme. Toluene is brominated using NBS and heat. Reaction with sodium acetylide (made by deprotonating acetylene with sodium amide) produces the terminal alkyne. The alkyne is reduced to the alkene using molecular hydrogen and Lindlar’s catalyst. Anti-Markovnikov addition of HBr in the presence of peroxides produces the primary alkyl halide. SN2 substitution with the conjugate base of acetic acid (made by deprotonating acetic acid with sodium hydroxide) produces the desired product. 12.35. The synthesis developed below is only one suggested synthetic pathway. There are likely other acceptable approaches that accomplish the same goal. Acetic acid and ethylene each have two carbon atoms, and our product has eight carbon atoms. So our synthesis will need to involve a total of four equivalents of starting materials in order to produce a product with eight carbon atoms. A more detailed look at the product allows us to hypothesize where each of the two-carbon components will ultimately end up in the product (below). This is helpful in that it may allow us to determine which new bonds will be formed in the course of the reaction (i.e., those connecting each of the 2C components). By comparing the structures of the reactants and product, we can also make an initial guess on the origins of each of the 2C components (shown below). It seems reasonable to assume that the ester will be derived from acetic acid (as both of these have a carbonyl flanked by a methyl group and an oxygen), and the other three 2C components will be derived from ethylene (with appropriate functional group modification).
CHAPTER 12 399 When considering which types of reactions to use, we will connect these pieces using a number of substitution reactions. Also, the only way we have learned to produce a cis alkene is via hydrogenation of an alkyne using H2 and Lindlar’s catalyst, so this will clearly be one of our steps. The figure below outlines a retrosynthetic analysis for our target molecule. An explanation of each of the steps (a-j) follows. a. The indicated C-O bond (wavy line) can be made via an SN2 reaction between a carboxylate (the conjugate base of a carboxylic acid) and a substrate with an appropriate leaving group (e.g., tosylate). b. The carboxylate can be prepared from acetic acid (one of our given reactants) by treatment with a suitable base, such as NaOH. c. The tosylate can be prepared from the corresponding alcohol. d. The cis alkene can be produced from the corresponding alkyne (H2 / Lindlar’s catalyst). e. This retrosynthetic step is the key disconnection that utilizes the reaction described in the problem statement. We can make this internal alkyne/alcohol by the reaction of an alkynide ion (formed by deprotonating 1- butyne) and an epoxide. f. The epoxide is prepared from ethylene via epoxidation. g. 1-Butyne is prepared by alkylating the conjugate base of acetylene using bromoethane. h. Bromoethane is prepared via HBr addition to ethylene. i. Acetylene is prepared via a double elimination from 1,2-dibromoethane. j. 1,2-Dibromoethane is prepared via bromination of ethylene. Now, let’s draw out the forward scheme. This multi-step synthesis uses three equivalents of ethylene (labeled A, B, C in the scheme below) and one equivalent of acetic acid (labeled D). Ethylene (A) is converted to 1,2-dibromoethane upon treatment with bromine. Subsequent reaction with excess sodium amide produces an acetylide anion which is then treated with bromoethane [made from ethylene (B) and HBr] to produce 1-butyne. Deprotonation with sodium amide, followed by reaction with an epoxide [prepared by epoxidation of ethylene (C)] and water workup, produces a compound with an alkyne group and an alcohol group. Reduction of the alkyne to the cis alkene is accomplished with H2 and Lindlar’s catalyst, after which the alcohol is converted to a tosylate with tosyl chloride. Reaction with the conjugate base of acetic acid [produced by treating acetic acid (D) with NaOH] allows for an SN2 reaction, thus yielding the desired product, Z-hexenyl acetate.
400 CHAPTER 12 12.36. The following synthesis is one suggested synthetic pathway. There are likely other acceptable approaches that accomplish the same goal. Take note that the reactant and product each have six carbon atoms. This suggests that our synthetic plan will not necessarily involve any C-C bond-forming reactions. However, there is a change in the carbon skeleton, and we will need a C-C bond-breaking reaction to convert the cyclic starting material into an acyclic product. The product contains two ketone groups, which is suggestive of an ozonolysis. The figure below outlines a retrosynthetic analysis for our target molecule. An explanation of each of the steps (a-e) follows. a. The two ketone groups can be prepared via ozonolysis of 1,2-dimethylcyclobutene. This is a key disconnection as it brings us to a synthetic intermediate with the same basic connectivity as the provided reactant (3,4-dimethylcyclobutene). The remaining steps involve reactions to move the position of the double bond. b. 1,2-dimethylcyclobutene is prepared via elimination from a suitable alkyl halide (e.g., 1-bromo-1,2- dimethylcyclobutane). c. The tertiary alkyl bromide is prepared via Markovnikov addition of HBr to 1,4-dimethylcyclobutene. d. This alkene can be prepared via Zaitsev elimination of 1-bromo-2,3-dimethylcyclobutane. e. This secondary alkyl bromide can be made via addition of HBr to 3,4-dimethylcyclobutene (our provided reactant). Now, let’s draw out the forward scheme. HBr converts 3,4-dimethylcyclobutene to 1-bromo-2,3-methylcyclobutane. Zaitsev elimination using sodium ethoxide affords 1,4-dimethylcyclobutene, which is subsequently converted to 1- bromo-1,2-dimethylcyclobutane using HBr (Markovnikov addition). Elimination, followed by azonolysis of the resulting alkene gives the product, 2,5-hexanedione.
CHAPTER 12 401 12.37. The following synthesis is one suggested synthetic pathway. There are likely other acceptable approaches that accomplish the same goal. Thus far, we have learned two ways to make aldehydes: (i) anti-Markovnikov hydration of a terminal alkyne or (ii) ozonolysis of an alkene, either of which is potentially reasonable here. However, in order to produce both of these compounds from a single synthetic protocol, a key recognition is that they can be produced from ozonolysis of the following disubstituted alkene. The following figure outlines a retrosynthetic analysis for our target molecule. An explanation of each of the steps (a-k) follows. H a H b c Br H d + O O j H OH f e Br + 3o i h Br g k OH Br OH 2o 1o a. The mixture of aldehydes can be made by ozonolysis of this disubstituted alkene (trans-2-methyl-4-decene), as described above. (Note that the E alkene is shown here, but ozonolysis of the Z alkene would also produce the same two aldehydes.) b. Thus far, we have learned two methods to make an alkene: 1) reduction of an alkyne or 2) elimination. In this case, the better choice to make this alkene is by reduction of the corresponding internal alkyne. The reason for this is explained in the next step. c. We have to start our synthesis with compounds with fewer than six carbons, so this alkyne is a useful intermediate because we know how to make bonds between sp and sp3 hybridized carbon atoms. This internal alkyne can thus be made from 4-methyl-1-pentyne (which must be deprotonated to produce a nucleophile) and 1-bromopentane. d. The terminal alkyne can be made from acetylene (which must be deprotonated to produce a nucleophile) and 1-bromo-2-methylpropane. e. Recall that we need to start with one 1°, one 2° and one 3° alcohol. The synthetic intermediate 1-bromo-2- methylpropane is the only one with a 3° carbon, so it follows that this compound is the one produced from a 3° alcohol. With this in mind, the alkyl halide can be produced from anti-Markovnikov addition of HBr to an alkene. f. The alkene can be made from acid-catalyzed dehydration of the 3° alcohol. g. Acetylene has only two carbons, so the only type of alcohol that can be used to make it is a 1° alcohol. Acetylene is thus made from double elimination of 1,2-dibromoethane. h. 1,2-Dibromoethane is produced by bromination of ethylene. i. Ethylene is produced by acid-catalyzed elimination from ethanol, a 1° alcohol. j. 1-Bromopentane is made by anti-Markovnikov addition of HBr to 1-pentene. k. 1-Pentene is made from the 2° alcohol by tosylation followed by reaction with a bulky base to give the less substituted product.
402 CHAPTER 12 Now let’s draw the forward scheme. The 3° alcohol is converted to 2-methylpropene using strong acid. Anti- Markovnikov addition of HBr (with peroxides) produces 1-bromo-2-methylpropane. Subsequent reaction with sodium acetylide (produced from the 1° alcohol by dehydration, bromination and double elimation/deprotonation as shown) produces 4-methyl-1-pentyne. Deprotonation with sodium amide followed by reaction with 1-bromopentane (made from the 2° alcohol by tosylation, elimination and anti-Markovnikov addition) yields 2-methyl-4-decyne. Reduction using sodium in liquid ammonia produces the E alkene. Ozonolysis followed by treatment with dimethylsulfide produces an equimolar ratio of the two products, 3-methylbutanal and hexanal. 12.38. The following synthesis is one suggested synthetic pathway. There are likely other acceptable approaches that accomplish the same goal. An analysis of the structure of the product reveals that the five-carbon alkyl group (highlighted below) matches the skeletal structure of 2-methylbutane, the given starting material. This indicates which C-C bond (arrow) must be made in the course of the synthesis. The figure below outlines a retrosynthetic analysis for our target compound. An explanation of each of the steps (a-h) follows.
CHAPTER 12 403 a. The only way we have learned to make an ester (so far) is via a reaction of a carboxylate nucleophile (the conjugate base of a carboxylic acid) and an alkyl halide (in this case, bromomethane). b. The only way we have learned to make a carboxylic acid (so far) is by ozonolysis of an alkyne. An alternative intermediate to the terminal alkyne shown would be the symmetric internal alkyne, 2,9-dimethyl- 5-decyne (not shown), which would produce two equivalents of the target carboxylic acid upon ozonolysis. c. The alkyne is prepared from the reaction of 1-bromo-3-methylbutane with an acetylide anion (formed by deprotonating acetylene). This alkyl halide has the same carbon skeleton as our given starting material (2- methylbutane), so our remaining steps involve primarily functional group manipulation. d. Knowing that the first synthetic step must be radical halogenation of 2-methylbutane to produce the tertiary alkyl halide (the only useful reaction of alkanes), we need to migrate the functionality back toward the tertiary carbon in this retrosynthetic analysis. Thus, 1-bromo-3-methylbutane can be prepared via anti- Markovnikov addition of HBr to 3-methyl-1-butene. e. 3-Methyl-1-butene is prepared via elimination (with a sterically hindered base) from 2-bromo-3- methylbutane. f. 2-Bromo-3-methylbutane is prepared via anti-Markovnikov addition of HBr to 2-methyl-2-butene. g. 2-Methyl-2-butene is prepared via Zaitsev elimination from 2-bromo-2-methylbutane. h. 2-bromo-2-methylbutane is made from our given starting material, 2-methylbutane, via radical bromination. Now, let’s draw the forward scheme. Radical bromination of 2-methylbutane produces the tertiary alkyl halide, selectively. Then, elimination with NaOEt, followed by anti-Markovnikov addition (HBr / peroxides), and then elimination with tert-butoxide, followed by another anti-Markovnikov addition (HBr / peroxides) produces 1-bromo-3- methylbutane. This alkyl halide will then undergo an SN2 reaction when treated with an acetylide ion to give 5-methyl- 1-hexyne. Ozonolysis of this terminal alkyne cleaves the CC triple bond, producing the carboxylic acid. Deprotonation (with NaOH) produces a carboxylate nucleophile that subsequently reacts with bromomethane in an SN2 reaction to give the desired ester. Br2 Br NaOEt HBr Br t-BuOK h EtOH ROOR HBr ROOR O 1) NaOH O 1) O3 HC CNa Br O 2) CH3Br OH 2) H2O 12.39. The synthesis developed below is only one suggested synthetic pathway. There are likely other acceptable approaches that accomplish the same goal. The following figure outlines a retrosynthetic analysis for our target molecule. An explanation of each of the steps (a-h) follows. OH a OH b O c + d h + Br e Br f g
404 CHAPTER 12 a. 1-Penten-3-ol can be made by reduction of the corresponding terminal alkyne. b. The terminal alkyne can be made from acetylene (after deprotonation to form a nucleophile) and the aldehyde shown. c. The aldehyde can be made by ozonolysis of (E)-3-hexene. d. (E)-3-Hexene is prepared via reduction of 3-hexyne. e. 3-Hexyne is made by alkylating 1-butyne with bromoethane. f. Bromoethane is made by HBr addition to ethylene. g. Ethylene is made by reduction of acetylene. h. 1-Butyne is made by alkylation of acetylene using bromoethane (made as described above in steps f-g). Now, let’s draw the forward scheme. Acetylene is reduced to ethylene using molecular hydrogen and Lindlar’s catalyst. Addition of HBr affords bromoethane. Reaction with sodium acetylide (prepared from acetylene and sodium amide) gives 1-butyne. Deprotonation with sodium amide followed by reaction with bromoethane produces 3-hexyne, which is subsequently reduced to (E)-3-hexene using a dissolving metal reduction. Alternatively, hydrogenation in the presence of Lindlar’s catalyst will provide the Z alkene, which will also lead to the product via the same steps. Ozonolysis produces two equivalents of the desired aldehyde. Reaction with sodium acetylide gives the alkyne/alcohol which is reduced to the product using molecular hydrogen and Lindlar’s catalyst. 12.40. The following synthesis is one suggested synthetic pathway. There are likely other acceptable approaches that accomplish the same goal. The target molecule, (E)-2-hexenal, is bifunctional - containing both an alkene group and an aldehyde group. We have learned two ways to make aldehydes: (i) ozonolysis of alkenes and (ii) anti-Markovnikov hydration of terminal alkynes. In this case, it is not immediately apparent which of these methods we should use. For example, ozonolysis of the compound below would not be a good approach, as both C=C bonds are susceptible to cleavage, yielding the following three products. Anti-Markovnikov hydration of a terminal alkyne also does not appear to be a viable approach, as there is no obvious precursor that would allow installation of the C=C double bond adjacent to the aldehyde group. Likewise, we know two ways to make an alkene: (i) reduction of an alkyne and (ii) elimination. We could potentially make the target molecule from the corresponding alkyne, but it is unclear what the next retrosynthetic step should be. (Note that in Chapter 13, we will learn reactions to make this approach possible.)
CHAPTER 12 405 Installing the alkene group by elimination turns out to be a viable approach in this case, as described below. The figure below outlines a retrosynthetic analysis for our target molecule. An explanation of each of the steps (a-f) follows. a. The target compound can be made via elimination of the alcohol under acidic conditions. b. The aldehyde can be made by ozonolysis of the alkene shown here. Note that this compound only has one C=C double bond, so we avoid the problem described above with ozonolysis of a diene. c. The alkene can be made by partial reduction of the corresponding alkyne. d. This compound can be made from the reaction between an acetylide ion and the aldehyde shown (see problem 12.28). e. The aldehyde is made from anti-Markovnikov addition of water to 1-pentyne. f. 1-Pentyne is made from 1,1-dibromopentane by double elimination. Now let’s draw the forward scheme. 1,1-Dibromopentane is converted to 1-pentyne by reaction with excess sodium amide (to afford double elimination followed by deprotonation of the resulting alkyne), followed by aqueous workup to protonate the terminal alkynide. 1-Pentyne is converted to the aldehyde via hydroboration/oxidation. Subsequent reaction with sodium acetylide, followed by aqueous workup, produces an alcohol. Reduction with H2 and Lindlar’s catalyst converts the alkyne group to an alkene group. Ozonolysis converts the alkene to an aldehyde. Reaction with concentrated acid allows for elimination of the alcohol, producing the target compound. Br 1) NaNH2 1) 9-BBN O 1) HC CNa OH Br (excess) 2) H2O2 2) H2O H2 2) H2O NaOH Lindlar's catalyst O H2SO4 O 1) O3 H H H (conc) 2) DMS H H OH heat OH
406 CHAPTER 12 12.41. Treatment of compound 1 with BH3 affords a chiral organoborane (2). Compound 3 is an alkyl halide, which is expected to react with an acetylide ion in an SN2 reaction, to give compound 4. Upon treatment with the strong base BuLi, compound 4 is deprotonated to give an alkynide ion, which then serves as a nucleophile in an SN2 reaction to give compound 5. As described in the problem statement, treatment of compound 5 with TsOH in methanol gives an alcohol (6). In the final step of the sequence, the triple bond in compound 6 is reduced to give a cis alkene (compound 7).
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