So far we have mostly limited our good leaving groups to the very similar, stable halide anions, Cl -, Br - and I -.

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1 Extending the - pectrum: Making Alcohols () into eactive and E eactants (Two Ways) o far we have mostly limited our good leaving groups to the very similar, stable halide anions, -, Br - and I -. 1 Base (solvated) + Base strong acids (I, Br, ) have......very stable conjugate bases, which make very good leaving groups = I, Br,. A select number of acid K a /pk a values are provided below. The top five acids are all strong acids having excellent leaving group conjugate bases. The last example is the weak acid water, which means its conjugate base (hydroxide) is a relatively poor leaving group. onjugated Base onjugated Acid K a pk a (Leaving Group) I I very good Br Br very good very good very good very good poor ur general - pattern can be expanded with two new versions having an attached oxygen atom. This is accomplished by changing the poor leaving group () of an alcohol () into a good leaving group using two different strategies. There are many commercially available alcohols as starting structures and there are a large number of reactions known to produce alcohol structures from other functional groups, and alcohols can react in many different ways. This makes alcohols a central group in synthetic methodology, as the following examples demonstrate. 1. ulfonated Alcohols as Good Leaving Groups A common strategy to make the poor leaving group of an alcohol into a good leaving group is to convert it into a sulfonate ester group (- 2 ). The new leaving group is the conjugate base of a sulfonic acid (example four in the table, just above). The conjugate base and/or leaving group is referred to as a sulfonate anion. As an anion, its stability and appearance are very similar to the bisulfate conjugate base of strongly acidic sulfuric acid (see below), so it is not surprising that it is such a good leaving group and stable conjugate base. Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

2 ulfuric acid ionization to form its very stable conjugate base, bisulfate 2 sulfuric acid reference base bisulfate (very stable anion) K a pk a ulfonic acid ionization and its conjugate base, alkyl sulfonate (same as leaving group from carbon) sulfonic acids reference base ulfonates are stable anions, which make excellent leaving groups when bonded to carbon. K a pk a Alcohols are converted into sulfonates by a different type of substitution reaction than presented in this topic. A possible mechanism is presented below for their formation. There are many variations of the sulfonate leaving group, but the most common is the toluenesulfonyl ester (tosylate ester). All of the forms of sulfur, below, have close analogies with carbon functional groups that we will study in the coming chapters. The sulfur functional groups necessary to prepare tosylates are presented along with the analogous carbon functional groups. Formation of an inorganic sulfonate ester toluenesulfonyl chloride (tosyl chloride) alcohol with a poor "" leaving group pyridine (base) alkyl toluene sulfonate (tosylate ester) pyridinium hydrochloride (precipitates from solution) Formation of an analogous organic alkanoate ester carboxylic acid chloride ' alcohol tertiary amine (base) carboxylic ester ' trialkylammonium chloride Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

3 Possible Mechanism for Tosylate Formation from an Alcohol and Tosyl hloride (making a poor leaving group, -, into a good leaving group, - 3 ) 3 The sulfur atom is positively polarized in its highly oxidized state as an acid chloride. This is easily seen in the charged resonance structures below. ulfur has a +2 formal charge in the only structure showing an octet. We draw the other resonance structures because sulfur s valence electrons are in the n=3 subshell, where 3d orbitals are available. esonance structures of tosyl chloride before attack of alcohol +2 The sulfur still has an octet, but is highly positive and is strongly electrophilic. Also sulfur has available 3d orbitals and can exceed the octet rule. Because sulfur has these 3d orbitals available, it can expand its octet by taking on a fifth group. The negatively polarized alcohol oxygen atom s lone pairs are strongly attracted to the positively polarized sulfur forming a new bond. We already know that chloride, --, is a very good leaving group and it can leave allowing sulfur to go back to its normal tetravalent state. The alcohol oxygen atom now has three bonds and it has acquired a positive formal charge. The positive formal charge can be easily neutralized with any available base (pyridine is added for just this purpose). verall, the chlorine atom is replaced by the - of the alcohol group, forming an inorganic sulfur ester (analogous to an organic carbon ester). The that is also formed in the reaction is neutralized by the pyridine base. A Different Mechanism for ubstitution eaction at ulfur (addition of the nucleophile before the leaving group leaves) The second resonance structure emphasizes the electrophilic nature of the sulfur. nucleophilic oxygen attacks electrophilic sulfur chloride is eliminated by a negative push from oxygen Pyridine is added to neutralize any acidic protons. It is often called a "proton sponge". 1 E1 2 E2 Ts Tosylate leaving group, similar to halides = "Ts-". This becomes one of our "" leaving groups. = alkyl sulfonate, the alcohol oxygen is now a good leaving group salt (ppt) Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

4 This mechanism is not possible at tetravalent carbon (sp 3 ) because carbon cannot bond to five groups or have more than eight electrons in its valence. owever, there is a similar mechanism at carbon where trivalent carbon (sp 2 ) of an acid chloride becomes tetravalent (sp 3 ) in an intermediate and then goes back to being trivalent (sp 2 ). That mechanism is presented in just a bit as a problem for you to solve. The important result of this transformation is that a poor leaving group, the - of an alcohol, is made into a very good leaving group, a sulfonate group - 3. The reaction presented above is a general reaction of acid chlorides, which can produce esters (of carbon = organic, nitrogen = inorganic, sulfur = inorganic, phosphorous = inorganic and others). The oxygen atoms and the chlorine atom inductively pull electron density from the central atom (,, or P) and make it very partial positive. This inductive effect is reinforced by a resonance effect where we explicitly write a positive charge on the central atom as electrons are shifted to oxygen. In the case of carbon, theoretical calculations estimate this contribution at percent. It is definitely reasonable to write it and is very helpful in understanding the electrophilic nature of carbonyl carbons (=). nce the tosylate is made, and the alcohol oxygen is attached to the sulfonyl portion, the oxygen has become a very good leaving group and can leave as a very stable anion. (Think again of a stable base vs. its strong acid K a /pk a.) 4 The stable anion leaving group makes ionization process much easier in 1/E1 reactions and it is also a very good leaving group in 2/E2 reactions. Its conjugate acid has a pk a = -3. ompare this to the leaving group of an alcohol. conjugate acid ( 2 ) pka = 16 "" is a poor leaving group. This is a difficult reaction because both products are high energy intermediates. Problem 1 - how analogous resonance structures for the carbon, nitrogen and phosphorous acid chlorides to those given above for tosyl chloride, which emphasizes their electrophilic character. a b c 3 P Problem 2 - how a similar mechanism of substitution (to that of tosyl chloride, above) for each of the above acid chlorides with methanol as the attacking alcohol (add the nucleophile and then eliminate the leaving group). Use a generic base, B:, to remove the acidic proton. What is the nucleophile? What is the electrophilic atom in each example? What is the leaving group in each example? Problem 3 - how how each of the following compounds can react in an 2 reaction (mechanism). Identify the nucleophile, substrate, leaving group and product in each equation ( = -Ts) Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

5 5 a b 3 2 Ts c 3 Ts Problem 4 - What alcohol and reactions would generate each of the above tosylate esters? Write out the reaction equation, including the necessary reagents to produce the desired transformation. (Don t forget the proton sponge.) Problem 5 - Alcohols can be converted into inorganic esters and organic esters. An example of each possibility is shown below. an inorganic ester an organic ester alcohol tosyl chloride py alkyl tosylate alcohol ethanoyl chloride (acetyl chloride) alkyl ethanoate (alkyl acetate) Use these two reactions and the fact that -Ts is a good leaving group and is a good nucleophile to propose a synthesis of both enantiomers shown below from the chiral alcohol shown. assify all chiral centers as or Problem 6 - ()-2-butanol is subjected to the following reaction sequence. The reaction with hydroxide in the second step occurs at carbon. What is the absolute configuration of the final 2-butanol product? Draw out the reaction in 3D ()-2-butanol Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

6 Problem 7 - uggest a mechanism for each of the following transformations. 6 / () Ts () achiral ( / ) racemic Problem 8 - Predict reasonable products (major and/or minor) for each of the following reaction conditions. If certain choices are excluded, state why. a b Ts / Ts / c Ts d Ts 2. Protonated Alcohols - Water as a Good Leaving Group A second method for making an alcohol,, into a good leaving group involves a very different set of conditions. trong protic acids are used to extensively protonate the alcohol. When the alcohol is protonated, the leaving group is water, not hydroxide. Water s conjugate acid is 3 +, (pk a = -2). If substitution is the desired goal, then the strong halide acids are normally used,, Br or I. If elimination is the desired goal, then concentrated sulfuric acid ( 2 4 ) and/or concentrated phosphoric acid ( 3 P 4 ) are used at an elevated temperature ( ). Using the hydrohalic acids (, Br or I), very polar, strongly acidic conditions encourage 1 reactions, and these are assumed to be operating at all tertiary and secondary alcohol () centers. earrangements are frequently observed under these conditions. When methyl or primary alcohols are used, the large energy expense of carbocation formation prevents the escape of water on its own. The 2 at a primary 2 + is assumed to be pushed off by the conjugate halide base partner of the strong acid used to form a methyl or primary alkyl halide without rearrangement ( 2). Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

7 a. 1 o, 2 o and 3 o reacted with acids (, Br, I) - usually chemistry 7 i. primary alcohols ( 2 emphasized, no rearrangement, similar at methanol) conc. Br Br 2 Br 3 ii. secondary alcohols ( 1 emphasized, rearrangements possible) I I top and bottom attack I I 2 3 iii. tertiary alcohols ( 1 emphasized, rearrangements possible) top and bottom attack but can't tell in this example b. 1 o, 2 o and 3 o reacted with 2 4 and/or 3 P 4 and high temperature - E1 chemistry Using strongly acidic sulfuric acid, 2 4, at elevated temperatures favors E1 reactions and these are assumed to be operating in all of the reactions below. earrangements are possible and observed. i. primary alcohols (with high temperature, E1 is proposed, rearrangements possible) conc. 2 4 very high Temp (difficult) distills out ii. secondary alcohols (E1 emphasized, rearrangements possible) 2 4 distills out Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

8 iii. tertiary alcohols (E1 emphasized, rearrangements possible) distills out major (>90%) We have some, limited control to direct the alcohol functionality toward or E choices. The conditions to effect these different pathways are important, so you must be aware of the details mentioned above (halide acids = reactions and 2 4 / = E1 reactions). eat is a crucial aspect of the E1 reaction, since it allows the lower boiling alkene to escape from the reaction mixture by distillation, while the higher boiling alcohol or inorganic ester remains, in the reaction pot to reestablish equilibrium by forming more alkene, which distills...etc. The alkene boils at a much lower temperature than the alcohol it came from because it does not have an to form hydrogen bonds with. Examples of Boiling Point Differences Between Alcohols and Possible Alkene Products boiling points of alcohols ( o ) boiling point of alkenes ( o ) T bp (79 o ) 2 2 (-104 o ) (183 o ) (82 o ) (-47 o ) (129 o ) (97 o ) (-47 o ) (144 o ) (100 o ) (-6 o ) (106 o ) (1 o ) (96 o ) (4 o ) (99 o ) (161 o ) (83 o ) (77 o ) Problem 9 ()-2-butanol retains its optical activity indefinitely in aqueous base ( - ), but is rapidly converted to optically inactive 2-butanol (racemic) when in contact with dilute sulfuric acid ( ). Explain with detailed mechanisms. Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

9 9 Problem 10 - ummary problem, uggested trategy 1. Look at the reaction conditions first. A strong base/nucleophile favors the bimolecular processes ( 2,E2). A weak base/nucleophile favors the unimolecular processes ( 1,E1) oxygen examples of strong base/nucleophiles in our course: oxygen examples of weak base/nucleophiles in our course: 2. ext, look to the reactant structures ( 3, 1 o, 2 o, 3 o, hindered 1 o, allylic, benzylic, etc.) a. Methyl i. bimolecular conditions - only 2 ii. unimolecular conditions no reaction, we don t propose methyl carbocations b. 1 o i. bimolecular conditions mainly 2, some E2, The amount of E2 increases with substitution at β or increasing bulk in the base/nucleophile (i.e. much more E2 with t-butoxide than with less sterically bulky alkoxides, such as ethoxide). Allylic and benzylic are especially reactive by 2. If β is fully substituted then neither 2 or E2 is possible at 1 o. ii. Unimolecular conditions - no reaction due to the high energy of 1 o +. c. 2 o = i. bimolecular conditions with hydroxide and alkoxides there is generally more E2 than 2, but significant amounts of both observed. If the nucleophile is less basic, like carboxylates, more 2 is observed (i.e. more 2 with acetate than ethoxide). In our course we will assume mainly 2 with acetate and mainly E2 with hydroxide and alkoxides. Allylic and benzylic are especially reactive by 2. If β is fully substituted then 2 is not possible. E2 may be possible at the other β position. ii. unimolecular conditions generally more 1 than E1. If a highly substituted alkene can form, it can become a significant product. Greater amounts of more substituted alkenes form than less substituted alkenes. earrangements are common. In our course we will assume that an alcohol reacts by E1 in strong acid/high temperature conditions ( 2 4,/ 3 P 4 / ) and the alkene is distilled out (E1). d. 3 o i. bimolecular condition - only E2 products are obtained (no 2). The most stable alkene is usually the major product, but other alkene products obtained too. ii. unimolecular conditions - generally 1 > E1. ee 2 o comments above. Because + s are flat (sp 2 ), β -s can be lost from either side, leading to a greater variety of elimination products. earrangements are common. Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

10 In each of the following substrates consider two sets of conditions: (1) -- / (strong) and (2) (weak). Also remember that cyclohexanes have two possible chair conformations and each conformation should be considered for the E2 reactions (anti elimination required). In all unimolecular reactions only consider up to one rearrangement and only if a more stable carbocation forms. 10 a b c d e f g h hallenging arbocation earrangements and eactions Problem 11 - Propose an arrow pushing mechanism for the following transformations. What is the purpose of the strong acid ( 3 + or BF 3 )? What is the good leaving group in a, b, c and d? Part e is less obvious, but the purpose of the BF 3 is similar to the purpose of 3 +. Both Lewis acids make oxygen a better leaving group when bonded to an oxygen lone pair. You need to make a good leaving group, have it leave to form the best possible carbocation. In parts d and e, there are multiple rearrangements and ring strain is part of the problem. In part d form the most stable carbocation first. In part e you don t have a choice. a b Why does methyl migrate instead of phenyl? c. d d. 3 BF 3 (strong Lewis acid) no Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

11 Additional carbocation rearrangements to consider. (ome of these are tough.) 11 - (acid) 3 + Ts BF 3 Et 2 BF 3 is a Lewis acid similar role to Problem 12 In the body cholesterol is constructed from eighteen 2 carbon acetate units (mevalonate is an intermediate) that are converted into a 30 carbon hexaene (squalene, with 6 double bonds), which is then oxidized to an epoxide (squalene oxide) and protonated in an acid catalyzed reaction (enzyme) involving four carbocation rearrangements to form the cholesterol precursor, lanosterol. Lanosterol becomes chloesterol after 19 additional steps! holesterol is the lead steroid for all of the body s other steroids. 3 oa 8 x acetyl oa via mevalonate, isoprenoids, geranyl farnesyl structures oxidation squalene squalene oxide -Enzyme 1. protonated epoxide ring opening, 2. pi bond nucleophilic attack, 3. rearrangement and 4. E1 reaction other steroids estrogen testosterone cortisol bile acids etc. additional steps cholesterol many additional steps lanosterol Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

12 i. The cyclization step to form lanosterol involves enzyme acid catalyzed protonation to make a good leaving group, which undergoes subsequent reactions you have studied. The complex system has been simplified below. how how this key cyclization process begins by writing an arrow pushing mechanism. int In acid what is your first expectation? Look for the bonds that have changed (broken and formed). onsider acid/base, carbocations, pi bond nucleophiles etc. 12 -Enzyme squalene oxide analog ii. how how the tetracyclic carbocation rearranges to lanosterol. (Four rearrangements and an elimination reaction.) 3 3 B lanosterol precursor lanosterol Problem 13 A spectacular example of rearrangement in an acid catalyzed alcohol dehydration is shown below. When 3ß-friedelanol is treated with acid, a carbocation forms, and then a total of seven methyl and hydrogen migrations occur before the loss of a proton produces 13(18)-oleanene. Each migration is a 1,2-shift of a methanide ion ( 3 : -- ) or hydride ion (: -- ). how how each of these migrations occur. A 3ß-friedelanol 13(18)-oleanene. Z:\classes\ow to ure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp & E #4 (,summary) for web.doc

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