SYNTHESIS-KLEIN

12 Synthesis 12.1 One-Step Syntheses 12.2 Functional Group Transformations DID YOU EVER WONDER… 12.3 Reactions That Change the Carbon Skeleton what vitamins are and why we need them? 12.4 How to Approach a Synthesis Problem 12.5 Retrosynthetic Analysis Vitamins are essential nutrients that our bodies require in order 12.6 Practical Tips for Increasing Proficiency to function properly, and a deficiency of particular vitamins can lead to diseases, many of which can be fatal. Later in this chapter, we will learn more about the discovery of vitamins, and we will see that the laboratory synthesis of one particular vitamin represented a land- mark event in the history of synthetic organic chemistry. This chapter serves as a brief introduction to organic synthesis. Until this point in the text, we have only seen a limited number of reactions (a few dozen, at most). In this chapter, our modest repertoire of reactions will allow us to develop a methodical, step-by-step process for proposing syntheses. We will begin with one-step synthesis problems and then progress toward more chal- lenging multistep problems. The goal of this chapter is to develop the fundamental skills required for proposing a synthesis.

12.1 One-Step Syntheses 537 DO YOU REMEMBER? Before you go on, be sure you understand the following topics. If necessary, review the suggested sections to prepare for this chapter. 12.1 One-Step Syntheses The most straightforward synthesis problems are the ones that can be solved in just one step. For example, consider the following: Br Br This transformation can be accomplished by treating the alkene with Br2 in an inert solvent, such as CCl4. Other synthesis problems might require more than a single step, and those problems will be more challenging. Before approaching multistep synthesis problems, it is absolutely essential to become comfortable with one-step syntheses. In other words, it is critical to achieve mastery over all reagents described in the previous chapters. If you can’t identify the reagents necessary for a one- step synthesis problem, then certainly you will be unable to solve more complex problems. The following exercises represent a broad review of the reactions in previous chapters. These exercises are designed to help you identify which reagents are still not at the forefront of your consciousness. CONCEPTUAL CHECKPOINT H OH O ± En 12.1 O OH Br OH ± En Br OH Br OH OH ± En ± En Br OH Br ± En Br

538 CHAPTER 12 Synthesis 12.2 Br Br Br O CH3 O H Br Br Br Br Br Br OO Br ±C Br OH O 12.2 Functional Group Transformations In the previous few chapters, we developed several synthesis strategies that enable us to move the location of a functional group or change its identity. Let’s briefly review these techniques, as they will be extremely helpful when solving multistep synthesis problems. In Chapter 9, we developed a technique for changing the position of a halogen by perform- ing an elimination reaction followed by an addition reaction. For example: Br Br Elimination Addition In this two-step process, the halogen is removed and then reinstalled at a different location. The regiochemical outcome of each step must be carefully controlled. The choice of base in the elimination step determines whether the more substituted or the less substituted alkene is formed. In the addition step, the decision whether or not to use peroxides will determine whether a Markovnikov addition or an anti-Markovnikov addition occurs. NaOEt Br HBr HBr ROOR Br t-BuOK HBr ROOR Br

12.2 Functional Group Transformations 539 As we saw in Chapter 9, this technique must be slightly modified when the functional group is a hydroxyl group (OH). In such a case, the hydroxyl group must first be converted into a tosylate (a bet- ter leaving group), and only then can the technique be employed (elimination followed by addition): OH OTs Convert OH into a Elimination Addition HO better leaving group After converting the hydroxyl group into a tosylate, the regiochemical outcome for elimination and addition can be carefully controlled, as summarized below: OH 1) TsCl, pyridine H3O± 2) NaOEt 1) BH3 THF 2) H2O2, NaOH HO 1) Hg(OAc)2, H2O 2) NaBH4 1) TsCl, pyridine 1) BH3 THF HO 2) t-BuOK 2) H2O2, NaOH In Chapter 9, we also developed a two-step technique for moving the position of a double bond. For example: Br Addition Elimination Once again, the regiochemical outcome of each step can be controlled by choice of reagents, as summarized below: Br HBr t-BuOK NaOEt HBr NaOEt ROOR t-BuOK Br

540 CHAPTER 12 Synthesis BY THE WAY In Chapter 11, we developed one other important technique: installing functionality in a com- pound with no functional groups: Br Radical Elimination bromination This procedure, together with the other reactions covered in the previous chapters, enables the interconversion between single, double, and triple bonds: H2 , Lindlar’s Catalyst H2 , Pt or Na, NH3 1) Br2/CCl4 1) Br2, hn 2) xs NaNH2 3) H2O 2) NaOEt We will soon learn a new way of approaching synthesis problems (rather than relying on a few, precanned techniques). For now, let’s ensure mastery over the reactions and techniques that allow us to change the identity or position of a functional group. SKILLBUILDER 12.1 CHANGING THE IDENTITY OR POSITION OF A FUNCTIONAL GROUP LEARN the skill Br SOLUTION - 13 5 Br 1 2 3 4 5 24 C-1 C-2 and C-3 is functionalized are functionalized - -

12.2 Functional Group Transformations 541 anti Anti-Markovnikov Hofmann addition elimination X anti HBr ROOR Br (Racemic) 1) BH3 THF OH (Racemic) 2) H2O2, NaOH anti HBr t-BuOK ROOR Br 1) BH3 THF TsCl t-BuOK 2) H2O2, NaOH Pyridine OH OTs anti HBr Br ROOR

542 CHAPTER 12 Synthesis 1) HBr, ROOR 2) t-BuOK PRACTICE the skill 12.3 3) HBr, ROOR (a) Br (c) (e) OH 1) BH3 THF 2) H2O2, NaOH OH 3) TsCl, pyridine (g) 4) t-BuOK 5) HBr, ROOR APPLY the skill 12.4 - 12.5 12.6 Br Br (b) (d) Br OH OH (f) OH OH Br (h) need more PRACTICE? Try Problems 12.17, 12.21, 12.22

12.3 Reactions That Change the Carbon Skeleton 543 12.3 Reactions That Change the Carbon Skeleton In all of the problems in the previous section, the functional group changed its identity or location, but the carbon skeleton always remained the same. In this section, we will focus on examples in which the carbon skeleton changes. In some cases, the number of carbon atoms in the skeleton increases, and in other cases, the number of carbon atoms decreases. If the size of the carbon skeleton increases, then a C—C bond-forming reaction is required. Thus far, we have only learned one reaction that can be used to introduce an alkyl group onto an existing carbon skeleton. Alkylation of a terminal alkyne (Section 10.10) will increase the size of a carbon skeleton: HH –+ HHH HH HHH H C C C C Na ± X C C C H H C C C C C C C H ± NaX HH HHH HH HHH Four carbon atoms Three carbon atoms Seven carbon atoms Over time, we will see many other C—C bond-forming reactions, but for now, the knowledge that we have only seen one such reaction should greatly simplify the problems in this section, enabling a smooth transition into the world of synthetic organic chemistry. If the size of the carbon skeleton decreases, then a C—C bond-breaking reaction, called bond cleavage, is required. Once again, we have only seen one such reaction. Ozonolysis of an alkene (or alkyne) achieves bond cleavage at the location of the bond: H HHH CH HHH O H ± CH HCCCC 1) O3 HCCCC 2) DMS O HHH H HHH H One carbon atom Five carbon atoms Four carbon atoms Over time, we will see other reactions that involve C—C bond cleavage. For now, the knowledge that we have only seen one such reaction should greatly simplify the problems in this section. SKILLBUILDER Br 12.2 CHANGING THE CARBON SKELETON LEARN the skill SOLUTION R C C– Na± RX R C C R ± NaX Alkynide Alkyl halide

544 CHAPTER 12 Synthesis from the perspective of the alkyne HCCH 1) NaNH2 HCC R Starting material 2) R X RX ±– RCCH Na C C H Starting material Br ±– C Na C C H C H PRACTICE the skill 12.7 (a) (b) Br Br APPLY the skill (c) O 12.8 (a) O H (b) Br Br Br (c) Br 12.9 need more PRACTICE? Try Problems 12.18, 12.19, 12.20, 12.23, 12.26

12.3 Reactions That Change the Carbon Skeleton 545 MEDICALLYSPEAKING Vitamins - OH - Retinol (vitamin A) Sources: Deficiency disease: NH2 ± Thiamine (vitamin B1) N OH N N S - Sources: - Deficiency disease: - O OH OH OH - HO - Sources: Vitamin C Deficiency disease: OH H HO H Sources: Ergocalciferol (vitamin D2) Deficiency disease: Phylloquinone (vitamin K1) O Amino group NH2 N ± N NS OH O Thiamine (vitamin B1) Sources: Deficiency disease: vital amine - -

546 CHAPTER 12 Synthesis 12.4 How to Approach a Synthesis Problem In the previous two sections, we covered two critical skills: (1) functional group transformations and (2) changing the carbon skeleton. In this section, we will explore synthesis problems that require both skills. From this point forward, every synthesis problem should be approached by asking the following two questions: 1. Is there a change in the carbon skeleton? Compare the starting material with the product to determine if the carbon skeleton is gaining or losing carbon atoms. 2. Is there a change in the identity or location of the functional group? Is one functional group converted into another, and does the position of functionality change? The following example demonstrates how these two questions should be applied. SKILLBUILDER 12.3 APPROACHING A SYNTHESIS PROBLEM LEARN the skill SOLUTION 1. Is there a change in the carbon skeleton? 5 24 246 13 1 3 57 2. Is there a change in the identity or location of the functional group? 5 24 246 13 1 3 57

12.4 How to Approach a Synthesis Problem 547 - trans 1) NaNH2 trans 2) Et Na , NH3 (l ) PRACTICE the skill 12.10 (b) O (a) Br OH Br (d) (c) (f) OH O H (e) APPLY the skill 12.11 OH HO 12.12 O need more PRACTICE? Try Problems 12.19–12.26 H

548 CHAPTER 12 Synthesis MEDICALLYSPEAKING The Total Synthesis of Vitamin B12 - Chlorophyll a Vitamin B12 - O N CN N NH2 O Co O O H2N NN O NN Porphyrin H2N H NH2 Mg ring system ±N O N Corrin NN H2N O ring system OH O O NH2 O O MeO2C NH HO O –O P O O - - - - A B Approach A D Approach Macrocyclization Macrocyclization occurs between ring A and ring D - occurs between ring A and ring B AB NN AB NN M M NN DC - N N D C - AB N M NN DC

12.5 Retrosynthetic Analysis 549 R - R ±C R - B NN Temporary tether R - R S R R B N C N -R CR N B N intra inter - - - - - - - 12.5 Retrosynthetic Analysis As we progress through the course and increase our repertoire of reactions, synthesis problems will become increasingly more challenging. To meet this challenge, a modified approach will be necessary. The same two fundamental questions (as described in the previous section) will continue to serve as a starting point for analyzing all synthesis problems, but instead of trying

550 CHAPTER 12 Synthesis to identify the first step of the synthesis, we will begin by trying to identify the last step of the synthesis. Analysis of the following synthesis problem will illustrate this process: ?OH An alcohol An alkyne Rather than focusing on what can be done with an alcohol that will ultimately lead to an alkyne, we instead focus on reactions that can generate an alkyne: OH Focus on this step In this way, we work backward until arriving at the starting material. Chemists have intuitively used this approach for many years, but E. J. Corey (Harvard University) was the first to develop a systematic set of principles for application of this approach, which he called retrosynthetic analysis. Let’s use a retrosynthetic analysis to solve the problem above. We must always begin by determining whether there is a change in the carbon skeleton or in the identity or location of the functional group. In this case, both the starting mate- rial and the product contain six carbon atoms, and the carbon skeleton is not changing in this instance. However, there is a change in the functional group. Specifically, an alcohol is converted into an alkyne but remains in the same location in the skeleton. We have not learned a way to do this in just one step. In fact, using reactions covered so far, this transfor- mation cannot be accomplished even in two steps. So we approach this problem backward and ask: “How are triple bonds made?” We have only covered one way to make a triple bond. Specifically, a dihalide undergoes two successive E2 eliminations in the presence of excess NaNH2 (Section 10.4). Any one of the following three dihalides could be used to form the desired alkyne: Geminal Br dibromide Br Vicinal Br dibromide Br Geminal Br dibromide Br

12.5 Retrosynthetic Analysis 551 The geminal dibromides can be ruled out, because we only saw one way to make a geminal dihalide—and that was starting from an alkyne. We certainly do not want to start with an alkyne in order to produce the very same alkyne: ?OH Br Br Therefore, the last step of our synthesis must be formation of the alkyne from a vicinal dihalide: ?OH 1) xs NaNH2 2) H2O Br Br A special retrosynthetic arrow is used by chemists to indicate this type of “backward” thinking: This alkyne this dibromide can be made from... Br Br Don’t be confused by this retrosynthetic arrow. It indicates a hypothetical synthetic pathway thinking backward from the product (alkyne). In other words, the previous figure should be read as: “In the last step of our synthesis, the alkyne can be made from a vicinal dibromide.” Now let’s try to go backward one more step. We have learned only one way to make a vicinal dihalide, starting with an alkene: Br Can be made from: Br alkene Notice again the retrosynthetic arrow. The figure indicates that the vicinal dibromide can be made from an alkene. In other words, the alkene can be used as a precursor to prepare the desired dibromide.

552 CHAPTER 12 Synthesis Therefore, our retrosynthetic analysis, so far, looks like this: Product Br Br Alkene This scheme indicates that the product (alkyne) can be prepared from the alkene. This sequence of events represents one of the strategies discussed earlier in this chapter—converting a double bond into a triple bond: H2 , Lindlar’s catalyst H2 , Pt or Na, NH3 1) Br2 /CCl4 1) Br2, hn 2) xs NaNH2 3) H2O 2) NaOMe In order to complete the synthesis, the starting material must be converted into the alkene. At this point, we can think forward, in an attempt to converge with the pathway revealed by the retrosynthetic analysis: OH ? 1) xs NaNH2 2)H2O Br Br2 Br CCl4 This step can be accomplished with an E2 elimination. Just remember that the hydroxyl group must first be converted into a tosylate (a better leaving group). Then, an E2 elimination will create an alkene, which bridges the gap between the starting material and the product: OH TsCl Br2 1) xs NaNH2 pyridine CCl4 2) H2O Br OTs t-BuOK Br The synthesis seems complete. However, before recording the answer, it is always helpful to review all of the proposed steps and make sure that the regiochemistry and stereochemistry of each step will lead to the desired product as a major product. It would be inefficient to involve

12.5 Retrosynthetic Analysis 553 any steps that would rely on the formation of a minor product. We should only use steps that produce the desired product as the major product. After reviewing every step of the proposed synthesis, the answer is recorded like this: 1) TsCl, py OH 2) t-BuOK 3) Br2 /CCl4 4) xs NaNH2 5) H2O SKILLBUILDER 12.4 RETROSYNTHETIC ANALYSIS LEARN the skill Propose a plausible synthesis for each of the following transformations: O (a) H (b) SOLUTION (a) First determine whether there is a change in the carbon skeleton. The starting com- pound has four carbon atoms while the product has six. This transformation therefore requires the installation of two carbon atoms: O 24 1 234 5 6H 13 As we have seen many times in this chapter, the only method we have learned for achiev- ing this transformation is the alkylation of a terminal alkyne. Our synthesis must therefore involve an alkylation step. This should be taken into account when performing a retrosyn- thetic analysis. Second, we need to determine whether there is a change in the identity or location of the functional group. In this case, the functional group has changed both its identity and its location. The functional group in the product is an aldehyde moiety. As seen in Section 10.7, an aldehyde can be made via hydroboration-oxidation of an alkyne. As shown with the retrosynthetic arrow, a terminal alkyne can be converted into an aldehyde. Can be made from: O 1 234 56H 2 45 6 13 Continuing with a retrosynthetic analysis, we must consider the step that might be used to produce the alkyne. Recall that our synthesis must contain a step involving the alkylation of an alkyne. We therefore propose the following retrosynthetic step: Can be made from: 2 45 6 1 2 4 X ± +– 13 3 Na C C H

554 CHAPTER 12 Synthesis ? - O anti +– (b) H X Na C C H 1) R2BH 2) H2O2 , NaOH 1) HBr, ROOR 2) H C CNa - 3) R2BH 4) H2O2 , NaOH O H 1 13 2 2 - -

12.5 Retrosynthetic Analysis 555 ? Na, NH3 (l ) 1) NaNH2 2) Me WATCH OUT Never draw a carbon atom with five bonds - Addition Br HBr, ROOR Elimination anti Br t-BuOK HBr ? Na, NH3 (l ) ROOR 1) NaNH2 Br 2) Me t-BuOK

556 CHAPTER 12 Synthesis 1) Br2/CCl4 - 2) xs NaNH2 O 3) H2O O PRACTICE the skill 12.13 Br OH 1) HBr, ROOR Br O Br Br 2) t-BuOK OH (a) 3) Br2 /CCl4 4) xs NaNH2 trans 5) H2O 6) NaNH2 7) Me 8) Na, NH3 (l ) (b) O (c) OH (d) (e) Br Br Br ± En (f) OH (g) (h) 12.14 APPLY the skill 12.15 cis 12.16 - need more PRACTICE? Try Problems 12.19–12.26 O H

12.6 Practical Tips for Increasing Proficiency 557 PRACTICALLYSPEAKING - - Retrosynthetic Analysis - - - - O O Me Me H Longifolene 12.6 Practical Tips for Increasing Proficiency Organizing a Synthetic “Toolbox” of Reactions All of the reactions in this course will collectively represent your “toolbox” for proposing syntheses. It will be very helpful to prepare two lists that parallel the two questions that must be considered in every synthesis problem (change in the carbon skeleton and change in the functional group). The first list should contain C—C bond-forming reactions and C—C bond-breaking reactions. At this point, this list is very small. We have only seen one of each: alkylation of alkynes for C—C bond forming and ozonolysis for C—C bond breaking. As the text progresses, more reactions will be added to the list, which will remain relatively small but very powerful. The second list should contain functional group transformations, and it will be longer. As we move through the text, both lists will grow. For solving synthesis problems, it will be helpful to have the reactions categorized in this way in your mind.

558 CHAPTER 12 Synthesis Creating Your Own Synthesis Problems A helpful method for practicing synthesis strategies is to construct your own problems. The process of designing problems often uncovers patterns and new ways of thinking about the reactions. Begin by choosing a starting compound. To illustrate this, let’s begin by choosing a simple starting compound, such as acetylene. Then choose a reaction expected for a triple bond, perhaps an alkylation: 1) NaNH2 2) Me Next, choose another reaction, perhaps another alkylation: 1) NaNH2 1) NaNH2 2) Me 2) Me Then, treat the alkyne with another reagent. Look at the list of reactions of alkynes and choose one; perhaps hydrogenation with a poisoned catalyst: 1) NaNH2 1) NaNH2 H2 2) Me 2) Me Lindlar’s catalyst Finally, simply erase everything except for the starting compound and the final product. The result is a synthesis problem: Once you have created your own synthesis problem, it might be a really good problem, but it won’t be helpful for you to solve it. You already know the answer! Nevertheless, the process of cre- ating your own synthesis problems will be very helpful to you in sharpening your synthesis skills. Once you have created several of your own problems, try to find a study partner who is also willing to create several problems. Each of you can swap problems, try to solve each other’s problems, and then get back together to discuss the answers. You are likely to find that exercise to be very rewarding. Multiple Correct Answers Remember that most synthesis problems will have numerous correct answers. As an example, anti-Markovnikov hydration of an alkene can be achieved through either of the two possible routes: 1) BH3 THF 2) H2O2, NaOH OH 1) HBr, ROOR 2) NaOH The first pathway represents hydroboration-oxidation of the alkene to achieve anti-Markovnikov addition of water. The second pathway represents a radical (anti-Markovnikov) addition of HBr followed by an SN2 reaction to replace the halogen with a hydroxyl group. Each of these path- ways represents a valid synthesis. As we learn more reactions, it will become more common to encounter synthesis problems with several perfectly correct answers. The goal should always be efficiency. A 3-step synthesis will generally be more efficient than a 10-step synthesis.

SkillBuilder Review 559 REVIEW OF CONCEPTS AND VOCABULARY - - SECTION 12.1 bond cleavage SECTION 12.2 SECTION 12.4 - - SECTION 12.3 SECTION 12.5 retrosynthetic analysis KEY TERMINOLOGY bond cleavage retrosynthetic analysis SKILLBUILDER REVIEW 12.1 CHANGING THE IDENTITY OR POSITION OF A FUNCTIONAL GROUP Change the position of a halogen. Br Br Elimination Addition Change the position of a bond. Br Addition Elimination Install a functional group. Br Elimination Radical bromination Interconvert between single, double, and triple bonds. H2, Pt H2, Lindlar’s Catalyst or 1) Br2, hn Na, NH3 2) NaOEt 1) Br2/CCl4 2) xs NaNH2 3) H2O Try Problems 12.3–12.6, 12.17, 12.21, 12.22

560 CHAPTER 12 Synthesis 12.2 CHANGING THE CARBON SKELETON C!C bond formation. HH – HHH HH HHH HCCCC ± XCCCH HCCCC CCCH HH HHH HH HHH Four carbon atoms Three carbon atoms Seven carbon atoms C!C bond cleavage. H HHH CH HHH O H ± CH HCCCC 1) O3 HCCCC 2) DMS O HHH H HHH H One carbon atoms Five carbon atoms Four carbon atoms Try Problems 12.7–12.9, 12.18, 12.19, 12.20, 12.23, 12.26 12.3 APPROACHING A SYNTHESIS PROBLEM BY ASKING TWO QUESTIONS Try Problems 12.10–12.12, 12.19–12.26 12.4 RETROSYNTHETIC ANALYSIS Can be made Can be made from Br from Can be made from Br Starting Product material Try Problems 12.13–12.16, 12.19–12.26

Practice Problems 561 Note: PRACTICE PROBLEMS WileyPLUS Br 12.17 Br OH OH OH OH Br Br OH Br 12.18 Br O O CH3 H Br Br Br Br Br Br Br Br Br Br OO OH ± C O

562 CHAPTER 12 Synthesis 12.19 O H (c) Br SS 12.20 (d) O 12.21 12.24 O OH O RR RS (a) OH 12.25 OH O SR 12.26 H OO (b) O (a) (b) H 12.22 OH (c) 12.23 O (d) (a) O (e) (b) Br OH INTEGRATED PROBLEMS 12.27 N- O H C C– O OH – OH C OH C C O+ C H H HH

Challenge Problems 563 12.28 - O OH – O RR HCC – H OH O± O HH RR C RR C C H C H CHALLENGE PROBLEMS 12.31 12.29 O Br O HO ± En 1,4-Dioxane O 12.30

Chapter 12 Synthesis Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 12. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary.  The position of a halogen can be moved by performing __________ followed by ______________.  The position of a π bond can be moved by performing ___________ followed by ______________.  An alkane can be functionalized via radical ________________.  Every synthesis problem should be approached by asking the following two questions: 1. Is there any change in the ____________________? 2. Is there any change in the identity or location of the _________________?  In a _________________ analysis, the last step of the synthetic route is first established, and the remaining steps are determined, working backwards from the product. Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 12. The answers appear in the section entitled SkillBuilder Review. 12.1 Changing the Identity or Position of a Functional Group

CHAPTER 12 375 12.2 Changing the Carbon Skeleton 12.3 Approaching a Synthesis Problem by Asking Two Questions 12.4 Retrosynthetic Analysis Useful Reagents: This chapter does not cover any new reactions. As such, there are no new reagents in this chapter. The problems in this chapter require the use of the reagents covered in previous chapters (specifically, Chapter 3 and Chapters 7-11). For each of those chapters, a summary of reagents can be found in the corresponding chapters of this solutions manual. Common Mistakes to Avoid When proposing a synthesis, avoid drawing curved arrows (unless the problem statement asks you to draw a mechanism). So often, students will begin drawing mechanism, rather than a synthesis, when asked to propose a synthesis. If you look through all of the solutions to the problems in this chapter, you will find that none of the solutions exhibit curved arrows. Also, avoid using steps for which you have no control over the regiochemical outcome or the stereochemical outcome. For example, acid-catalyzed hydration of the following alkyne will produce two different ketones. This process should not be used if only one of these ketones is desired.

376 CHAPTER 12 Solutions 12.1. The reagents for these reactions can be found in the summary material at the end of Chapter 9. They are shown again here: 12.2. The reagents for these reactions can be found in the summary material at the end of Chapter 10. They are shown again here: Br O O HBr CH3 H Br HBr H2SO4, H2O H2 Br HgSO4 Lindlar's 1) xs NaNH2 1) 9-BBN Catalyst 2) H2O 2) H2O2, NaOH 1) xs NaNH2 1) O3 1) NaNH2 H2 2) H2O 2) H2O 2) CH3I Pt Br2 Br Na Br xs Br2 NH3(l) Br Br OO Br Br Br +C OH O Br

CHAPTER 12 377 12.3. base cannot be sterically hindered. Appropriate choices (a) We begin by analyzing the identity and location of include hydroxide, methoxide, and ethoxide. the functional group in both the starting material and the product: The identity of the functional group has changed (double (d) We begin by analyzing the identity and location of bond  triple bond), but its location has not changed (it the functional group in both the starting material and the remains between C1 and C2). The conversion of a product. In doing so, we immediately realize that the double bond to a triple bond can be accomplished via the starting material has no functional group (it is an alkane), following two-step process (bromination, followed by so the first step of our synthesis must be the installation elimination): of a functional group. This can be accomplished via radical bromination, which selectively installs a bromine (b) We begin by analyzing the identity and location of atom at the tertiary position. The resulting tertiary alkyl the functional group in both the starting material and the halide can then be converted into the product via an product. The identity of the functional group (Br) has elimination process. The more substituted alkene is not changed, but its location has changed. We have seen desired (trisubstituted rather than disubstituted), so the that this transformation can be accomplished via a two- base cannot be sterically hindered. Appropriate choices step process (elimination, followed by addition). For the include hydroxide, methoxide, and ethoxide. elimination process, there is only one possible regiochemical outcome; nevertheless, tert-butoxide is (e) The identity of the functional group (OH) has not used as the base (rather than hydroxide, methoxide, or changed, but its location has changed. We have seen that ethoxide) in order to suppress the SN2 reaction, which this transformation can be accomplished via a two-step will otherwise dominate for a primary bormide. For the process (elimination, followed by addition). Elimination next step of our synthesis, Markovnikov addition is can be achieved either by heating the starting alcohol required, so we use HBr without peroxides. with concentrated sulfuric acid (E1), or by converting the alcohol to a tosylate followed by treatment with a strong base (E2). The resulting alkene can then be converted to the desired product via an addition process. Specifically, a hydration reaction must be performed (addition of H and OH) in a Markovnikov fashion. This can be achieved by treating the alkene with dilute acid (acid- catalyzed hydration): (c) We begin by analyzing the identity and location of the functional group in both the starting material and the product. The identity of the functional group (a  bond) has not changed, but its location has changed. We have seen that this transformation can be accomplished via a two-step process (addition, followed by elimination). For the first step of our synthesis, Markovnikov addition is required, so we use HBr without peroxides. For the next step of our synthesis, the more substituted alkene is desired (trisubstituted rather than disubstituted), so the

378 CHAPTER 12 (f) We begin by analyzing the identity and location of (h) We begin by analyzing the identity and location of the functional group in both the starting material and the the functional group in both the starting material and the product. The identity of the functional group has product. The identity of the functional group (OH) has changed (Br  OH), and its location has changed as not changed, but its location has changed. We have seen well. We have seen that this type of transformation can that this type of transformation can be accomplished via be accomplished via a two-step process (elimination of H a two-step process (elimination, followed by addition). and Br, followed by addition H and OH). For the first step of our synthesis, the more substituted alkene is For the first part of our synthesis (elimination), we must desired (trisubstituted rather than monosubstituted), so first convert the alcohol to a tosylate before treating with the base cannot be sterically hindered. Appropriate a strong base. The use of a sterically hindered base is not choices include hydroxide, methoxide, and ethoxide. A required (E2 = major product; while SN2 = minor hydration reaction must be performed (addition of H and product). Nevertheless, a sterically hindered base will be OH) in a Markovnikov fashion. This can be achieved by helpful here as it will suppress the competing SN2 treating the alkene with dilute acid (acid-catalyzed reaction. So tert-butoxide is used in the synthesis below. hydration): Note that an E1 process should be avoided, because heating the alcohol with concentrated sulfuric acid will likely involve a carbocation rearrangement (methyl shift). For the second part of our synthesis, we must perform an anti-Markovnikov addition of H and OH. This can be achieved via hydroboration-oxidation, or alternatively, it can be achieved by anti-Markovnikov addition of HBr followed by an SN2 process to give the desired alcohol: (g) We begin by analyzing the identity and location of 12.4. Let’s begin by drawing the starting material and the functional group in both the starting material and the the product, so that we can see the desired transformation product. The identity of the functional group has more clearly: changed (OH  Br), and its location has changed as well. We have seen that this type of transformation can In this case, the identity and the location of the be accomplished via a two-step process (elimination of H functional group have changed. During our synthesis, the and OH, followed by addition H and Br). For the first functional group (Br) must be relocated, AND it must be part of our synthesis (elimination), we must first convert converted into a triple bond. Moving the functional the alcohol to a tosylate before treating with a strong group can be achieved via elimination (to give the base. The use of a sterically hindered base is not required Zaitsev product) followed by addition (in an anti- (E2 = major product; while SN2 = minor product). Markovnikov fashion). Nevertheless, a sterically hindered base will be helpful here as it will suppress the competing SN2 reaction. So tert-butoxide is used in the synthesis below. Note that an E1 process should be avoided, because heating the alcohol with concentrated sulfuric acid will likely involve a carbocation rearrangement (methyl shift). For the second part of our synthesis, we must perform an anti-Markovnikov addition of H and Br, so we use HBr in the presence of peroxides.

CHAPTER 12 379 Then, changing the identity of the functional group can (b) Alkenes can be formed via elimination reactions, so be achieved via the following three-step process: we will need a compound with a leaving group, such as an alkyl halide, in order to make the desired product. If we use the tertiary alkyl halide from the previous exercise (12.6a), we can make the desired cycloalkene via an E2 process. A strong, non-hindered base should be used, such as hydroxide, methoxide, or ethoxide. The necessary reagents are shown here: (c) In exercise 12.6b, we saw that methylcyclohexane can be converted to 1-methyl-cyclohexene via a two step process: This cycloalkene can be converted into the product by moving the position of the  bond. We have seen that this can be accomplished via a two-step process (addition, followed by elimination): 12.5. We have not seen a way to convert an alkene During the addition step, we must use a reaction that directly into a geminal dihalide. But we have seen a way gives an anti-Markovnikov addition (such as to convert an alkyne into a geminal dihalide. So, we hydrobromination in the presence of peroxides). During must convert the starting alkene into an alkyne. This can the elimination step, we need the Hofmann product (less be accomplished via bromination, followed by substituted alkene), so we use tert-butoxide as the base. elimination with excess sodium amide. The resulting The entire synthesis is shown here: alkyne then undergoes two successive addition reactions upon treatment with excess HBr to give a geminal dihalide. 12.6. (a) Radical bromination of 1-methylcyclohexane will result in the following tertiary alkyl bromide: Alternatively, the end of our synthesis (moving the position of the  bond) can be accomplished via hydroboration-oxidation, followed by conversion of the

380 CHAPTER 12 resulting alcohol to a tosylate, followed by an (c) The starting material has eight carbon atoms, and the elimination reaction: product has only seven carbon atoms. One carbon atom must be removed, which can be achieved via ozonolysis: 12.8. (a) The starting material has four carbon atoms, and the product has six carbon atoms. We can install two carbon atoms by treating the starting material with sodium acetylide. The resulting alkylation product (a terminal alkyne) can then be reduced to an alkene via hydrogenation with a poisoned catalyst, such as Lindlar’s catalyst: This alternate answer illustrates a very important point. (b) The starting material has five carbon atoms, and the With synthesis problems, there are often multiple product has only four carbon atoms. One carbon atom acceptable answers. must be removed, which can be achieved via ozonolysis of an alkene: 12.7. (a) The starting material (acetylene) has two carbon The alkene can be prepared from the starting material via atoms, and the product has five carbon atoms. an elimination reaction with a sterically hindered base Therefore, we must change the carbon skeleton by (giving the less substituted alkene): forming carbon-carbon bonds. Specifically, we must install a methyl group and an ethyl group. Each of these alkylation reactions can be achieved upon treatment of the alkyne with sodium amide followed by the appropriate alkyl halide. Notice that each alkyl group must be installed separately, and it does not matter which alkyl group is installed first and which is installed second. (b) The starting material (benzyl bromide) has seven (c) The starting material has three carbon atoms, and carbon atoms, and the product has nine carbon atoms. the product has five carbon atoms. We can install two Therefore, we must install two carbon atoms. This carbon atoms by treating the starting material with alkylation process can be achieved in one step, by sodium acetylide. The resulting alkylation product (a treating the starting material with sodium acetylide: terminal alkyne) can then be converted to the desired product (a geminal dihalide) upon treatment with excess HBr:

CHAPTER 12 381 12.9. The starting material (3-bromo-3-ethylpentane) is a (c) The starting material has one more carbon atom than the product. Therefore, our synthesis must employ an tertiary substrate and will not readily undergo an SN2 ozonolysis process. Since the product is a carboxylic reaction. Under these conditions, the acetylide ion acid (rather than an aldehyde or ketone), we can conclude that the last step of our process must involve functions as a base, rather than a nucleophile, giving an ozonolysis of an alkyne, rather than ozonolysis of an alkene: E2 reaction, instead of SN2. The product is an alkene, while the base (acetylide) is protonated to give acetylene as a byproduct: 12.10. This alkyne can be prepared directly from the starting (a) The starting material has five carbon atoms, and the material upon treatment with excess sodium amide product has eight carbon atoms. So our synthesis must (followed by water workup). The complete synthesis is involve the installation of three carbon atoms. In shown here: addition, the identity of the functional group must be changed (from a triple bond to a double bond). Reduction of the alkyne (to give an alkene) must be the last step of our synthesis, because if we first reduce the triple bond, then we would not be able to perform the alkylation step. The alkylation step must be performed first, followed by reduction of the alkyne (with a poisoned catalyst), as shown here: (d) The starting material has six carbon atoms, and the product has nine carbon atoms. So our synthesis must involve the installation of three carbon atoms. Also, the location of the functional group has been changed. The product is a trans alkene, which can be made if the last step of our synthesis is a dissolving metal reduction: (b) The starting material has five carbon atoms, and the This alkyne can be made from the starting alkene via an product has seven carbon atoms. So our synthesis must anti-Markovnikov addition of HBr, followed by involve the installation of two carbon atoms. If we treatment with the appropriate alkynide ion, as shown simply alkylate the starting alkyne (by treating with here: sodium amide followed by ethyl iodide), then we will need to move the location of the triple bond. Instead, we 1) HBr, ROOR can convert the starting alkyne into an alkyl halide (via hydrogenation with a poisoned catalyst followed by an 2) H3C C CNa anti-Markovnikov addition of HBr). If this alkyl halide 3) Na, NH3(l) is treated with acetylide, the desired product is formed: Na HBr, Br Na, ROOR NH3 (l) H3C C C

382 CHAPTER 12 (e) The product has two more carbon atoms than the bromide, performing an alkylation process and then starting material. The installation of two carbon atoms converting the triple bond into the desired alcohol: can be achieved via an alkylation process. In fact, the desired product is a terminal alkyne, which can be made from the following alkyl halide in just one step: This alkyl halide can be made from the starting material by first moving the location of the  bond, followed by anti-Markovnikov addition: The following reagents can be used to achieve the desired transformations: (f) The starting material has one more carbon atom than the product. Therefore, our synthesis must employ an ozonolysis process. Since the product is an aldehyde, we can conclude that the last step of our process must involve ozonolysis of an alkene: This alkene can be prepared from the starting material by converting the alcohol to a tosylate, and then performing an E2 reaction with a sterically hindered base (giving the less substituted alkene): It should be noted that there are other acceptable answers. As one example, the last part of our synthesis (hydroboration-oxidation) could be replaced with anti- Markovnikov addition of HBr to give a primary alkyl bromide, followed by an SN2 process with hydroxide as the nucleophile: 12.11. The desired transformation involves the 12.12. We have not learned a direct way of installing installation of two carbon atoms, as well as a change in only one carbon atom. That is, two carbon atoms are the location of the functional group. This can be installed (not one) if we convert the starting alkene into achieved by converting the alcohol into a primary alkyl an alkyl halide (via an anti-Markovnikov addition), and then treat the alkyl halide with sodium acetylide.

CHAPTER 12 383 However, after installing two carbon atoms, we can (b) The product is an epoxide, which can be made from remove one of them with an ozonolysis procedure. In an alkene: order to obtain an aldehyde, the last step must be ozonolysis of an alkene, so the second-to-last step (called So the last step of our synthesis will likely be conversion the penultimate step) must be reduction of the alkyne to of the alkene into the epoxide. an alkene in the presence of a poisoned catalyst: Now let’s work forward from the starting material. The starting material has only four carbon atoms, while the product has six carbon atoms, so two carbon atoms must be installed. This can be achieved via an alkylation process: 12.13. Now we must bridge the gap (between the terminal (a) The product is a halohydrin, which can be made alkyne and the alkene). This can be achieved in one from an alkene: step, via hydrogenation with a poisoned catalyst. In summary, the desired transformation can be achieved So the last step of our synthesis will likely be conversion with the following synthesis: of the alkene into the halohydrin. Now let’s work forward from the starting material. The 1) HC CNa O starting material has only two carbon atoms, while the Br 2) H2, MCPBA product has four carbon atoms, so two carbon atoms must be installed. This can be achieved via an alkylation Lindlar's cat. process: 3) MCPBA HC C Na H2, Lindlar's cat. (c) The product is a methyl ketone, and we have seen that a methyl ketone can be prepared from an alkyne (via acid-catalyzed hydration): Now we must bridge the gap (between the terminal This alkyne can be prepared directly from the starting alkyne and the alkene). This can be achieved in one material upon treatment with excess sodium amide step, via hydrogenation with a poisoned catalyst. In (followed by water workup). In summary, the desired summary, the desired transformation can be achieved transformation can be achieved with the following with the following synthesis: synthesis:

384 CHAPTER 12 (d) The starting material has six carbon atoms and the in an anti-Markovnikov fashion (via hydroboration- product has three carbon atoms, indicating that an oxidation): ozonolysis reaction will be necessary. The product can be made from the following alkene: So the last step of our synthesis will likely be ozonolysis It should be noted that there are other acceptable of this alkene. answers. As one example, the last part of our synthesis Now let’s work forward from the starting material. The (hydroboration-oxidation) could be replaced with anti- starting material is an alkane (no functional group), so Markovnikov addition of HBr to give a primary alkyl we must first install a functional group. Radical bromide, followed by an SN2 process with hydroxide as bromination will selectively install a bromine atom at a the nucleophile: tertiary position: Now we must bridge the gap (between the alkyl halide (f) The starting material has one more carbon atom than and the alkene). This can be achieved in one step, via an the product. Therefore, our synthesis must employ an E2 reaction with a strong base (such as hydroxide, ozonolysis process. Since the product is a carboxylic methoxide, or ethoxide). In summary, the desired acid (rather than an aldehyde or ketone), we can transformation can be achieved with the following conclude that the last step of our process must involve synthesis: ozonolysis of an alkyne, rather than ozonolysis of an alkene: This alkyne can be prepared directly from the starting material upon treatment with excess sodium amide (followed by water workup). The complete synthesis is shown here: (e) The starting material is an alkane (no functional group), so we must first install a functional group. Radical bromination will selectively install a bromine atom at a tertiary position: Converting this compound into the product requires (g) The product is a halohydrin, which can be made changing both the identity and the location of the from the following trans alkene: functional group. This can be achieved via a two-step process, involving elimination followed by addition. The elimination process must be performed with a sterically hindered base so that the less substituted alkene is produced, and the addition process must be performed

CHAPTER 12 385 So the last step of our synthesis will likely be conversion of this alkene into the halohydrin. Now let’s work forward from the starting material. The starting material has only two carbon atoms, while the product has six carbon atoms. Two alkylation processes are required: And then finally, the double bond can be converted to the triple bond via the following two-step process: Now we must bridge the gap (between the internal In summary, the desired transformation can be achieved alkyne and the trans alkene). This can be achieved in with the following synthesis: one step, via a dissolving metal reduction. In summary, the desired transformation can be achieved with the following synthesis: (h) In this case, the carbon skeleton remains the same. 12.14. The product is a trans alkene, which can be made The starting material is an alkane (no functional group) from an alkyne. So the last step of our synthesis might so we must begin our synthesis by installing a functional be a dissolving metal reduction to convert the alkyne group. Radical bromination will selectively install a below into the product. This alkyne can be made from bromine atom at the tertiary position, giving the acetylene and 1-bromobutane via alkylation processes: following tertiary alkyl halide: Now we must change both the location and the identity of the functional group. We can move the functional group into the right location through the following series of reactions:

386 CHAPTER 12 1-Bromobutane can be made from 1-butyne, which can NaNH2 be made from acetylene and ethyl bromide via an alkylation process: 1) H2, HC C Na Lindlar's cat. Br And ethyl bromide can be made from acetylene: 2) HBr In summary, the desired transformation can be achieved Acetylene with the following synthesis: NaNH2 HBr, H2, Br ROOR Lindlar's HC C H2, cat. NaNH2 Lindlar's HBr, cat. ROOR HC C Na Br 1) NaNH2 2) Br H2, Lindlar's cat. 12.16. The starting material has two carbon atoms, and the product has five carbon atoms. So, we must join three fragments together (each of which has two carbon atoms), and then we must remove one of the carbon atoms. The latter process can be achieved via ozonolysis. Since the product is an aldehyde, it is reasonable to explore using ozonolysis as the last step of our synthesis: 12.15. We use the same approach taken in the previous This alkene can be prepared from an alkyne, which can problem (12.14). All carbon-carbon bonds are prepared be prepared from acetylene and 1-bromobutane: via alkylation of an alkynide ion with the appropriate alkyl halide. Each alkyl halide must be prepared from acetylene. The last step of the synthesis is the reduction of an alkyne to a cis alkene via hydrogenation with a poisoned catalyst:

CHAPTER 12 387 1-Bromobutane can be made from 1-butyne, which can be made from acetylene and ethyl bromide via an alkylation process: And ethyl bromide can be made from acetylene: In summary, the desired transformation can be achieved with the following synthesis: This synthesis represents just one correct answer to the problem. There are certainly other acceptable answers to this problem. 12.17. Most of the following reactions are addition reactions, which can be found in Chapter 10. The following reagents can be used to achieve each of the transformations shown:

388 CHAPTER 12 12.18. Most of the following reactions involve alkynes, which can be found in Chapter 11. The following reagents can be used to achieve each of the transformations shown: 12.19. The product can be made from 1-butene, which can be made from 1-butyne: 1-Butyne can be made from acetylene and ethyl bromide via an alkylation process. And ethyl bromide can be made from acetylene: 12.20. 1-Bromobutane can be made from 1-butyne, which can be made from acetylene and ethyl bromide via an alkylation process: In summary, the desired transformation can be achieved with the following synthesis:

CHAPTER 12 389 And ethyl bromide can be made from acetylene: In summary, the desired transformation can be achieved with the following synthesis: 12.21. 12.22. We must move the location of the  bond, and (a) The identity of the functional group (OH) has not we have seen that this can be achieved via a two-step changed, but its location has changed. We have seen that process (addition, followed by elimination). The this type of transformation can be accomplished via a addition step must occur in an anti-Markovnikov two-step process (elimination, followed by addition). fashion, which can be achieved by treating the starting For the first part of our synthesis (elimination), we must material with HBr in the presence of peroxides. The first convert the alcohol to a tosylate before performing elimination process must give the less substituted alkene, the elimination process. Then, for the elimination, we so a sterically hindered base is required: must use a sterically hindered base, tert-butoxide, in Alternatively, the addition of HBr can be replaced with order to obtain the less-substituted alkene. For the hydroboration-oxidation (addition of H and OH), which second part of our synthesis, we must perform an anti- is also an anti-Markovnikov addition. In that scenario, Markovnikov addition of H and OH. This can be the resulting alcohol must first be converted to a tosylate achieved via hydroboration-oxidation, or alternatively, it before the elimination step can be performed. can be achieved by anti-Markovnikov addition of HBr followed by an SN2 process (with hydroxide as a 12.23. nucleophile) to give the desired alcohol: (a) The starting material has one more carbon atom than the product. Therefore, our synthesis must employ an ozonolysis process. Since the product is a carboxylic acid (rather than an aldehyde or ketone), we can conclude that the last step of our process must involve ozonolysis of an alkyne, rather than ozonolysis of an alkene: This alkyne can be made from the starting alkene via a two-step process (bromination, followed by elimination with excess sodium amide): (b) The product is a methyl ketone, which can be made from an alkyne (via acid-catalyzed hydration): This alkyne can be made from the starting alkene via a two-step process (bromination, followed by elimination with excess sodium amide):

390 CHAPTER 12 1) Br2, h 2) NaOMe (b) We have not learned a direct way of installing only one carbon atom. That is, two carbon atoms are installed 3) HBr, ROOR (not one) if we use the alkyl halide in an alkylation process (with sodium acetylide). However, after Br2 4) HC CNa H2, installing two carbon atoms, we can remove one of them h 5) H2, Lindlar's cat. Lindlar's cat. with an ozonolysis procedure, giving the desired product: Br 1) NaOMe HC C Na Br 2) HBr, ROOR 12.24. We begin by drawing the desired products: (c) We have not learned a direct way of installing only These compounds have five carbon atoms, but our one carbon atom. That is, two carbon atoms are installed starting materials can contain no more than two carbon (not one) if we use the alkyl halide in an alkylation atoms. So our synthesis must involve the formation of process (with sodium acetylide). However, after carbon-carbon bonds. This can be accomplished via the installing two carbon atoms, we can remove one of them alkylation of acetylene (a compound with two carbon with an ozonolysis procedure. In order to obtain an atoms). The location of the functional groups (C2 and aldehyde, the last step must be ozonolysis of an alkene, C3) indicates that we need two alkylation processes (one so the penultimate step must be reduction of the alkyne to install a methyl group and the other to install an ethyl to an alkene in the presence of a poisoned catalyst: group). This places the triple bond between C2 and C3, which enables the installation of the functional groups at (d) The starting material has four carbon atoms, and those locations. Conversion of the internal alkyne into the product has six carbon atoms. Therefore, our the desired product requires the addition of H and H to synthesis must employ an alkylation process. The give an alkene, followed by the addition of OH and OH. starting material (2-methylpropane) cannot be converted In order to achieve the correct stereochemical outcome, into an alkyne without giving five bonds to the central one of these addition processes must be performed in a carbon atom (which is impossible). Therefore, the syn fashion, while the other must be performed in an anti starting material must be converted into an alkyl halide fashion. That is, we can perform an anti addition of H (so that it can be treated with sodium acetylide to give an and H, followed by a syn addition of OH and OH, or we alkylation process). Radical bromination provides a can perform a syn addition of H and H, followed by an tertiary alkyl halide, which must be converted to a anti addition of OH and OH, as shown: primary alkyl halide. That is, the position of the functional group must be moved, which can be accomplished via a two-step process (elimination followed by anti-Markovnikov addition). The primary alkyl halide is then treated with sodium acetylide to give an alkylation process. The resulting terminal alkyne can be reduced in the presence of a poisoned catalyst to give the desired alkene:

CHAPTER 12 391 12.25. We begin by drawing the desired products: These compounds have five carbon atoms, but our (b) The starting material has four carbon atoms, and the starting materials can contain no more than two carbon product has six carbon atoms. So our synthesis must atoms. So our synthesis must involve the formation of involve the installation of two carbon atoms. Also, the carbon-carbon bonds. This can be accomplished via the location of the functional group has been changed. The alkylation of acetylene (a compound with two carbon product is an aldehyde, which can be made from a atoms). The location of the functional groups (C2 and terminal alkyne (via hydroboration-oxidation): C3) indicates that we need two alkylation processes (one to install a methyl group and the other to install an ethyl As seen in the previous problem (12.26a), this alkyne group). This places the triple bond between C2 and C3, can be made from the starting alkene via an anti- which enables the installation of the functional groups at Markovnikov addition of HBr, followed by treatment those locations. Conversion of the internal alkyne into with sodium acetylide, as shown here: the desired product requires the addition of H and H to give an alkene, followed by the addition of OH and OH. In order to achieve the correct stereochemical outcome, both of these addition processes must be performed in an anti fashion, or both must be performed in a syn fashion. That is, we can perform an anti addition of H and H, followed by an anti addition of OH and OH, or we can perform a syn addition of H and H, followed by a syn addition of OH and OH, as shown: HBr, 1) HBr, ROOR H ROOR 2) HC CNa O 3) 9-BBN 1) 9-BBN 4) H2O2, NaOH 2) H2O2, NaOH Na Br H C C 12.26. (c) The starting material has four carbon atoms, and the (a) The starting material has four carbon atoms, and the product has five carbon atoms. We have not learned a product has six carbon atoms. So our synthesis must direct way of installing only one carbon atom. That is, involve the installation of two carbon atoms. Also, the two carbon atoms are installed (not one) if we convert location of the functional group has been changed. The the starting alkene into an alkyl halide (via an anti- product is a methyl ketone, which can be made from a Markovnikov addition), and then treat the alkyl halide terminal alkyne (via acid catalyzed hydration): with sodium acetylide. However, after installing two carbon atoms, we can remove one of them with ozonolysis, giving the product: This alkyne can be made from the starting alkene via an anti-Markovnikov addition of HBr, followed by treatment with sodium acetylide, as shown here:

392 CHAPTER 12 (d) See the solution to Problem 12.12. That problem is OH OH extremely similar to this one. The solution is also Br + O extremely similar: + HBr, 1) HBr, ROOR O ROOR 2) HC CNa H 3) H2, 1) O3 Lindlar's cat. 2) DMS 4) O3 5) DMS Br H2, HC C Na Lindlar's cat. This retrosynthetic analysis gives the following synthesis: (e) The starting material is cyclic (it contains a ring) and the product lacks a ring. So, we must break one of the carbon-carbon bonds of the ring. We have only learned one way (ozonolysis) to break a carbon-carbon bond. So, the last step of our synthesis is likely the following reaction: This cycloalkene can be prepared from the starting material in just two steps. First, radical bromination can be used to selectively install a bromine atom at the tertiary position. And then, the resulting alkyl halide can be treated with a strong base (such as hydroxide, methoxide, or ethoxide) to give an E2 reaction: 12.28. There are certainly many acceptable answers to this problem. The following retrosynthetic analysis employs the technique described in the problem statement: 12.27. There are certainly many acceptable answers to this problem. The following retrosynthetic analysis employs the technique described in the problem statement:

CHAPTER 12 393 This retrosynthetic analysis gives the following 12.30. synthesis: (a) The desired compound can be prepared from acetylene in just one step (via acid-catalyzed hydration): Acetylene NaNH2 OH Acetylene Alternatively, this transformation can also be achieved H2, H2, via hydroboration-oxidation of acetylene. Lindlar's Lindlar's (b) The following synthesis represents just one correct cat. cat. answer to the problem. There are certainly other acceptable answers to this problem. OH We have seen in previous problems that 1-butyne can be prepared from two equivalents of acetylene: 1) O3 O 1) HC CNa 2) DMS HH 2) H2O 12.29. The key to solving this problem is recognizing that the cyclic product can be made from the following acyclic compound (which can be prepared from the starting material in just one step): This reaction is similar to halohydrin formation: The product can be made from 1-butyne in just two steps (hydrogenation, followed by ozonolysis): O H H2, 1) O3 2) DMS Lindlar's cat. The  bond reacts with molecular bromine to give a In summary, the following synthesis can be used to make bromonium ion, which is then attacked by the OH group the desired compound from acetylene: in an intramolecular process. You may find it helpful to build molecular models to help visualize the stereochemistry of the ring-closing step. According to the retrosynthetic analysis above, the desired transformation can be achieved in just two steps, shown here:

394 CHAPTER 12 (c) The synthesis developed below is only one suggested This alkene can be prepared from an alkyne, which can synthetic pathway. There are likely other acceptable be prepared from acetylene and 1-bromobutane: approaches that accomplish the same goal. We have seen in previous problems that 1-butyne can be prepared from two equivalents of acetylene: And the product can be made from 1-butyne via 1-Bromobutane can be made from 1-butyne, which can hydroboration-oxidation: be made from acetylene and ethyl bromide via an alkylation process: In summary, the following synthesis can be used to make the desired compound from acetylene: And ethyl bromide can be made from acetylene: In summary, the desired transformation can be achieved with the following synthesis: (d) The synthesis developed below is only one suggested synthetic pathway. There are likely other acceptable approaches that accomplish the same goal. The starting material has two carbon atoms, and the product has five carbon atoms. So, we must join three fragments together (each of which has two carbon atoms), and then we must remove one of the carbon atoms. The latter process can be achieved via ozonolysis. Since the product is an aldehyde, it is reasonable to explore using ozonolysis as the last step of our synthesis:

CHAPTER 12 395 12.31. The key to solving this problem is recognizing that the cyclic product can be made from the following acyclic compounds via two SN2 reactions: Each of these starting materials can be made from acetylene, as seen in the following synthesis: 12.32. The synthesis developed below is only one suggested synthetic pathway. There are likely other acceptable approaches that accomplish the same goal. As seen in Table 4.1, tetradecane is a saturated hydrocarbon with 14 carbon atoms (with no branching), and our source of carbon (acetylene) has two carbon atoms, so we will likely use 7 equivalents of acetylene in this synthesis. There are a number of different approaches to complete this synthesis, including connecting two carbon atoms at a time sequentially from one end, or disconnecting it symmetrically from the center. The retrosynthesis below takes the latter of these two tactics. The figure below outlines a retrosynthetic analysis for our target molecule, employing squiggly lines to indicate C-C bonds that are disconnected in the retrosynthetic direction. An explanation of each of the steps (a-j) follows. a. Tetradecane can be made via hydrogenation of 7-tetradecyne. b. 7-Tetradecyne can be made by sequentially alkylating both sides of acetylene with 1-bromohexane. c. 1-Bromohexane is made via an anti-Markovnikov addition of HBr across 1-hexene. d. 1-Hexene is made by reduction of 1-hexyne using H2 and Lindlar’s catalyst. e. 1-Hexyne can be produced from acetylene (after deprotonation to make a nucleophile) and 1-bromobutane. f. 1-Bromobutane is made via an anti-Markovnikov addition of HBr across 1-butene. g. 1-Butene is made by reduction of 1-butyne using H2 and Lindlar’s catalyst. h. 1-Butyne is made from acetylene (after deprotonation) and 1-bromoethane. i. 1-Bromoethane is made by addition of HBr across ethylene. j. Ethylene is made by reduction of acetylene using H2 and Lindlar’s catalyst. Now, let’s draw out the forward scheme. Acetylene is reduced to ethylene using H2 and Lindlar’s catalyst. HBr addition, followed by SN2 substitution with an acetylide nucleophile (made by deprotonation of acetylene with sodium amide) gives 1-butyne. Reduction to 1-butene with H2 and Lindlar’s catalyst followed by anti-Markovnikov addition of HBr in the presence of peroxide produces 1-bromobutane. A substitution reaction with sodium acetylide gives 1-