Ch 11 - Reactions of Alcohols
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Ch 11 - Reactions of Alcohols

Course Number: CHEM 140, Spring 2010

College/University: Whitman

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R eacti ons of Alcohols 11 TABLE 1 1 -1 A lcohols are important organic compounds because the hydroxyl group is easily converted to almost any other functional group. In Chapter 10, we studied reac tions that form alcohols. In this chapter, we seek to understand how alcohols react and which reagents are best for converting them to other kinds of compounds. Table 11-1 summarizes the types of reactions alcohols...

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eacti R ons of Alcohols 11 TABLE 1 1 -1 A lcohols are important organic compounds because the hydroxyl group is easily converted to almost any other functional group. In Chapter 10, we studied reac tions that form alcohols. In this chapter, we seek to understand how alcohols react and which reagents are best for converting them to other kinds of compounds. Table 11-1 summarizes the types of reactions alcohols undergo and the products that result. Types o f Reactions o f Alcohols R - OH dehydration type of reaction Product 0 esterification R - OH alkenes R - OH RO-C-R' esters II R - OH oxidation ketones, aldehydes, acids R-X halides R - OH tosylation R - OTs tosylate esters (good leaving group) R - OH substitution R - OH reduction R-H alkanes R- OH ( I) form alkoxide (2) R'X R - O -R' ethers 11-1 Oxidation States of Alcohols a nd Related Fu nctional Grou ps O xidation of alcohols leads to ketones, aldehydes, and carboxylic acids. These func tional groups, in turn, undergo a wide variety of additional reactions. For these rea sons, alcohol oxidations are some of the most common organic reactions. In inorganic chemistry, we think of oxidation as a loss of electrons and reduction 6 as a gain of electrons. This picture works well for inorganic ions, as when Cr + is 3+ reduced to Cr . Most organic compounds are uncharged, however, and gain or loss of electrons is not obvious. Organic chemists tend to think of oxidation as the result of adding an oxidizing agent (2, Br2, etc.), and reduction as the result of adding a reducing agent (H2, NaBH4, etc.). Most organic chemists habitually use the following simple rules, based on the change in the formula of the substance: OXIDATION: a ddition of or 02; addition of X2 (halogens); loss of H 2. R EDUCTION: addition of H2 (or H -); loss of or 02; loss of X2. + N EITHER: addition or loss of H , H20, HX, etc . is neither an oxidation nor a reduction. 4 60 11-1 O xidation States of Alcohols and Related Functional Groups 4 61 H R- C - H I alkane no bonds to 0 H R - C - R' H H I ----i> [0] OH R-C- H I p rimary alcohol one bond to 0 OH R - C - R' H H I ----i> [ 0] o R-C-H II ----i> [0] o R - C - OH II + H 20 aldehyde two bonds to 0 carboxylic acid three bonds to 0 I I ----i> [0] I o I R- + alkane no bonds to 0 secondary alcohol one bond to 0 ketone two bonds to 0 .... Figure 11-1 Oxidation states of alcohols. An alcohol is more oxidized than an alkane, yet less oxidized than carbonyl compounds such as ketones, aldehydes, and acids. Oxidation of a primary alcohol leads to an aldehyde, and further oxidation leads to an acid. Secondary alcohols are oxidized to ketones. Tertiary alcohols cannot be oxidized without breaking carbon-carbon bonds. - R' H2 0 (no further oxidation) H R - C - R' I alkane no bonds to 0 R" I ----i> [0] OH R - C - R' I tertiary alcohol one bond to 0 R" I (usually no further oxidation) We can tell that an oxidation or a reduction of an alcohol has taken place by counting the number of C - 0 bonds to the carbon atom. For example, in a primary alcohol, the carbinol ( C - OH ) carbon atom has one bond to oxygen; in an aldehyde, the carbonyl carbon has two (more oxidized); and in an acid, it has three. Oxidation of an alcohol usu ally converts C - H bonds to C - 0 b onds. If we convert an alcohol to an alkane, the carbinol carbon loses its bond to oxygen and gains another bond to hydrogen. Figure 1 1- 1 c ompares the oxidation states of primary, secondary, and tertiary alcohols with those obtained by oxidation or reduction. The symbol [0] i ndicates an unspecified oxidizing agent. Notice that oxidation of a primary or secondary alcohol forms a carbonyl ( C = O ) group by the removal of two hydrogen atoms: one from the carbinol carbon and one from the hydroxyl group. A tertiary alcohol cannot easily oxidize because there is no hydrogen atom available on the carbinol carbon. P R O B L E M 1 1- 1 Classify each reaction as an oxidation, a reduction, or neither. o H-C-H II o ---7 H O-C-OH II 4 62 (c) (e) (g) UC"-H II Chapter H3C CH3 C CH3-C-C- H3 HO OH I I I I ------7 1 1 : Reactions of Alcohols --? H+ (k) 0 (i) 0 0 H Er ------? Os04 H 2 02 BR, THF 0H (XOH BH2 () aB:r (h) V rY0H O + H20 OH (X,".'OH 11-2 Oxidation of Alcohols Primary and secondary alcohols are easily oxidized b y a variety of reagents, including chromium reagents, permanganate, nitric acid, and even household bleach (NaOCl, sodium hypochlorite). The choice of reagent depends on the amount and value of the alcohol. We use cheap oxidants for large-scale oxidations of simple, inexpensive alco hols. We use the most effective and selective reagents, regardless of cost, for delicate and valuable alcohols. In this chapter, we study only the oxidants that have the widest range of uses and the best selectivity. An understanding of the most common oxidants can later be extended to include additional reagents. 1 1-2A Oxidation of Secondary Alcohols S econdary alcohols are easily oxidized to give excellent yields of ketones. The chromic acid reagent is often best for laboratory oxidations of secondary alcohols. OH R - CH- R ' secondary alcohol I Na2Cr20/ H2 S04 ) o R-C-R' ketone I Example cyclohexanol cyclohexanone (90%) The chromic acid reagent is prepared by dissolving sodium dichromate, ( Na2Cr207 ) in a mixture of sulfuric acid and water. The active species in the mixture is probably chromic acid, H2Cr04, or the acid chromate ion, HCrO;. Adding chromium trioxide ( Cr03 ) to dilute sulfuric acid achieves the same result. N Crz07 + H zO + 2H 2S04 2 H O - Cr - OH o o 1 1 -2 Oxidation of Alcohols + 2 Na+ + 2 HSO:;- 4 63 I s odium dichromate I Cr03 c hromium trioxide + o c hromic acid (H2Cr04) H zO HO - Cr - OH o I o II H+ + - O - Cr - OH o acid chromate ion I II chromic acid The mechanism of chromic acid oxidation probably involves the formation of a chromate ester. Elimination of the chromate ester gives the ketone. In the elimination, the carbinol carbon retains its oxygen atom but loses its hydrogen and gains the second bond to oxygen. Formation o the chromate ester f R' R-C-O-H H alcohol I I + o H - O - Cr - OH o I R' II R - C - O - Cr - OH H I 0 I I 0 I + chromic acid chromate ester Elimination o the chromate ester and oxidation o the carbinol carbon f f R' R - C - O - Cr - OH HzO ". I .. "0- II R' J 1--' " V II ..0.. R - C = O.: I . + "0. ' - ..0., '. /C r- OH C r ( VI) C r ( IV) T he chromium(IV) species formed reacts further to give the stable reduced form, chromium(III). Both sodium dichromate and chromic acid are orange, while chromic 3 ion (Cr + ) is a deep blue. One can follow the progress of a chromic acid oxidation by observing the color change from orange through various shades of green to a greenish blue. In fact, the color change observed with chromic acid can be used as a test for the presence of an oxidizable alcohol. 1 1 -2 8 Oxidation of Pri mary Alcohols Oxidation of a primary alcohol initially forms an aldehyde. Unlike a ketone, however, an aldehyde is easily oxidized further to give a carboxylic acid. OH R - CH - H primary alcohol I -7 [0] o R - C-H aldehyde II -7 [ 0] o R - C - OH carboxylic acid I Obtaining the aldehyde is often difficult, since most oxidizing agents strong enough to oxidize primary alcohols also oxidize aldehydes. Chromic acid generally oxidizes a primary alcohol all the way to the carboxylic acid. 4 64 C ha pte r 1 1: Reactions of Alcohols 0 Cr03Cl- CH,OH 0 II C - OH o Pyridinium chlorochromate ( PCC): <- } or - H Cr0p3 .yH+iCr03CHCr pyr dine' l- A better reagent for the limited oxidation of primary alcohols to aldehydes is p yridinium chlorochromate (PCC), a c omplex of chromium trioxide with pyridine and H CI. PCC oxidizes most primary alcohols to aldehydes in excellent yields. Unlike most other oxidants, PCC i s soluble in nonpolar solvents such as dichloromethane ( CH 2 Cl 2 ) , which is an excellent solvent for most organic compounds. PCC can also serve as a mild reagent for oxidizing secondary alcohols to ketones. cyclohexyl methanol cyclohexanecarboxylic acid ( 92%) I R-C-H I H OH o Cr03'pyridine'HCI (PCC) ) R-C-H I primary alcohol Example o aldehyde C H i CH 2)S - CH 20H C H iCH 2 ) S -C-H II I -heptanol 1 1 -2C heptanal ( 78%) Resistance of Tertiary Alcohols to Oxidation Oxidation of tertiary alcohols is not an important reaction in organic chemistry. Ter tiary alcohols have no hydrogen atoms on the carbinol carbon atom, so oxidation must take place by breaking carbon-carbon bonds. These oxidations require severe condi tions and result in mixtures of products. The chromic acid test for primary and secondary alcohols exploits the resist ance of tertiary alcohols to oxidation. When a primary or secondary alcohol is added to the chromic acid reagent, the orange color changes to green or blue. When a nonox idizable substance (such as a tertiary alcohol, a ketone, or an alkane) is added to the reagent, no immediate color change occurs. S ummary of Alcohol Oxidations To Oxidize 10 10 P roduct R eagent chr one 20 aallccohol ohol PCCoomicc acidd (or PCC) alkcardetehydeic acid chr mi aci alcohol boxyl Prediwitththe products of the reactions of the fol owing compounds with chromic acid and alsoccyclohexanol -mo hyicyclo 2cycletohexane ohexanol -m hylcycl cIaycletchexanonehexanol acetialdacid, cet ehyde, eth no PRO B L E M 1 1 -2 (c) P Cc. (a) (b) (d) (e) (g) a l (f) (h) C H3COOH C H3CHO 1 1 -3 A dditional M ethods for O xidizin g Alcohols 4 65 Many other reagents and procedures have been developed for oxidizing alcohols. Some are simply modifications of the procedures we have seen. For example, the C ollins reagent i s a complex of chromium trioxide and pyridine, the original version of PCc. The Jones reagent is a milder form of chromic acid: a solution of diluted chromic acid in acetone. Two other strong oxidants are potassium permanganate and nitric acid. Both of these reagents are less expensive than the chromium reagents, and both of them give byproducts that are less environmentally hazardous than spent chromium reagents. Both permanganate and nitric acid oxidize secondary alcohols to ketones and primary alcohols to carboxylic acids. If these strong oxidants are not carefully controlled, they will cleave carbon-carbon bonds. 11-3 Add iti o n a l M ethods fo r Oxi d i z i n g Alcohols < )- t OH H - CH ; I -phenylethanol 0 \ 'I , // / '\ , II o C -"CH + "/ 3/" / /j acetopnenone (72%) 1 I / / C H3(CH2)4 - CH20H I -hexanol C H3(CH2)4 - C :- OH hexanoIc acid (80%) 0 1 / -I Perhaps the least expensive method for oxidizing alcohols is dehydrogenation: literally the removal of two hydrogen atoms. This industrial reaction takes place at high temperature using a copper or copper oxide catalyst. The hydrogen byproduct may be sold or used for reductions elsewhere in the plant. The primary limitation of dehydrogenation is the inability of many organic compounds to survive a reaction at 300C. Dehydrogenation is not well suitedfor laboratory syntheses. R- d] 0 -0 I R' o heat, CuO R - C - R1 I + r A s mall fuel cell i n this portable breath tester catalyzes the oxidation of ethanol by oxygen in the air. The oxidation generates an electric current that is proportional to the concentration of ethanol in the sample. Example CuO 300C cyclohexanol cyclohexanone (90%) C( 0 + H2 The Swern oxidation u ses dimethyl sulfoxide (DMSO) as the oxidizing agent to convert alcohols to ketones and aldehydes. DMSO and oxalyl chloride are added to the alcohol at low temperature, followed by a hindered base such as triethylamine. Sec ondary alcohols are oxidized to ketones, and primary alcohols are oxidized only as far as t he aldehyde. The byproducts of this reaction are all volatile and are easily separated from the organic products. OH - C-H I 0 + H3C-S-CH3 D MSO dimethyl sulfoxide I II 0 + C l -C-C-Cl (C O C lh oxalyl chloride II II (CH3CH2hN: , C H2Cl2 -C II \ 0 + H 3C-S-CH3 + CO2 + CO + 2 HCI dimethyl sulfide alcohol ketone or aldehyde 4 66 C hapter 1 1: R eactions of Alcohols Examples D MSO, (COClh E t3N, C H2C12, cyclopentanol - 60C ) cyclopentanone (90%) D MSO, (COCI)2 E t3N, C H2C12, -60C l -decanol ) II C H3(CH2)8-C \ H decanal (85%) o P R O B L E M 1 1- 3 The Swern oxidation provides a useful alternative to PCC that avoids using chromium reagents as oxidants. (a) Determine which species are oxidized and which are reduced in the Swern oxidation. (b) W e have seen dimethyl sulfide before (Section 8- I SB), added to an ozonide after ozonolysis. What was its function there? P R O B L E M 1 1-4 What is it about dehydrogenation that enables it to take place at 300C but not at 25C? ( a) Would you expect the kinetics, thermodynamics, or both to be unfavorable at 25C? (Hint: Is the reverse reaction favorable at 25C?) ( b) W hich of these factors (kinetics or thermodynamics) improves as the temperature is raised? (c) E xplain the changes in the kinetics and thermodynamics of this reaction as the tempera ture increases. The s u m m a ry ta ble on page 464 i s worth reviewing. Remember that permanga nate oxidizes a l kenes as we l l as a l cohols. PROBLEM-SOLVING HiltZ; P R O B L E M 1 1-5 G ive the structure of the principal product(s) when each of the following alcohols reacts with ( 1 ) Na2Cr207/H2S04, (2) PCC, (3) KMn04, -OH. (b) 3-octanol ( a) I -octanol (c) 2-cyclohexen- I -ol ( d) I -methylcyclohexanol S O LV E D P R O B L E M 1 1- 1 S uggest the most appropriate method for each of the following laboratory syntheses. (a) cyclopentanol -----? cyclopentanone SOLUTION M any reagents are available to oxidize a simple secondary alcohol to a ketone. For a laboratory synthesis, however, dehydrogenation is not practical, and cost is not as large a factor as it would be in industry. Most labs would have chromium trioxide or sodium dichromate available, and the chromic acid oxidation would be simple. PCC and the Swern oxidation would also work, although these reagents are more complicated to pre pare and use. cyclopentanol (or PCC, or Swern) cyclopentanone 1 1-4 B iological Oxidation of Alcohols 4 67 (b) 2-octen-I-ol SOLUTI O N ------'> 2-octenal ( structure below ) This synthesis requires more finesse. The aldehyde i s easily over-oxidized to a carboxylic acid, and the double bond reacts with oxidants such as KMn04' Our choices are limited to PCC or the Swern oxidation. CH20H 2-octen- l -ol P R O B L E M 1 1-6 P CC, CH?Cl? - (or Swern) - ); / H o C 2-octenal I Suggest the most appropriate method for each of the following laboratory syntheses. ------'> butanal, CH3 CH2CH2CHO (a) I -butanol butanoic acid, CH3CH2CH2COOH ( b) I. -butanol ------'> ( c) 2-butanol ------'> 2-butanone, CH3COCH2CH3 ------'> 2 -butenal, CH 3 CH CH -CHO (d) 2 -buten-I-ol ------'> (e) 2 -buten-I-ol 2 -butenoic acid, CH3CH -CH -COOH (1) l -methylcyclohexanol ------'> 2 -methylcyclohexanone (several steps) = Although it is the least toxic alcohol, ethanol is still a poisonous substance. When someone is suffering from a mild case of ethanol poisoning, we say that he or she is intoxicated. Animals often consume food that has fermented and contains alcohol. Their bodies must detoxify any alcohol in the food to keep it from building up in the blood and poisoning the brain. To detoxify ethanol, the liver produces an enzyme called alcohol dehydrogenase (ADH). Alcohol dehydrogenase catalyzes an oxidation: the removal of two hydrogen atoms from the alcohol molecule. The oxidizing agent is called nicotinamide adenine dinu + cleotide (NAD). NAD exists in two forms: the oxidized form, called NAD , and the reduced form, called NADH. The following equation shows that ethanol is oxidized to acetaldehyde, and NAD + is reduced to NADH. 11-4 Biolog ica l Oxidation of Alcohols OH C H3 - C - H H I I + AD H ethanol reduced form N AD + A DH = alcohol dehydrogenase oxidized form o C H3 - C - H II acetaldehyde oxidized form + + NADH reduced form 4 68 Chapter 1 1: Reactions of A c ohol s l A subsequent oxidation, catalyzed by aldehyde dehydrogenase (ALDH), converts acetaldehyde to acetic acid, a normal metabolite. ALDH ALDH NAD ' = aldehyde dehydrogenase o C H3 - C - OH acetic acid I + NADH T hese oxidations take place with most small primary alcohols. Unfortunately, the oxidation products of some other alcohols are more toxic than acetic acid. Methanol is oxidized first to formaldehyde and then to formic acid. Both of these compounds are more toxic than methanol itself. H H-C-O-H H methanol I I -----? [ 0) -2 H H - C =O H formaldehyde I --? [0) H-C=O O-H formic acid I Ethylene glycol is a toxic diol. Its oxidation product is oxalic acid, the toxic compound found in the leaves of rhubarb and many other plants. II H-C-C-H II HO OH ethylene glycol H H -----? 2[0) -4 H H - C - C - H + 2 H2 0 o II 0 II -----? 2 [0) H O - C - C - OH o 0 oxalic acid II II N AD+ i s derived from vitamin B3 known as niacin. There is a miscon ception that taking large supple ments of niacin before a night of drinking ethanol will lessen the severity of the "hangover" the next morning. Metabolism is more com plex than this simple idea suggests. COOH N J __ n icotinic acid (niacin) Many poisonings by methanol and ethylene glycol occur each year. Alco holics occasionally drink ethanol that has been denatured by the addition of methanol. Methanol is oxidized to formic acid, which may cause blindness and death. Dogs are often poisoned by sweet-tasting ethylene glycol when antifreeze is left in an open container. Once the glycol is metabolized to oxalic acid, the dog's kidneys fail, causing death. The treatment for methanol or ethylene glycol poi soning is the same. The patient is given intravenous infusions of diluted ethanol. The ADH enzyme is swamped by all the ethanol, allowing time for the kidneys to excrete most of the methanol (or ethylene glycol) before it can be oxidized to formic acid (or oxalic acid). This is an example of competitive inhib ition of an enzyme. The enzyme catalyzes oxidation of both ethanol and methanol, but a large quantity of ethanol ties up the enzyme, allowing time for excretion of most of the methanol before i t i s oxidized. 1 1 -5 A lcohols as Nucleophiles and Electrophiles; Formation of Tosylates P R O B L E M 1 1-7 4 69 A chronic alcoholic requires a much larger dose of ethanol as an antidote to methanol poi soning than does a nonalcoholic patient. Suggest a reason why a larger dose of the competi tive inhibitor is required for an alcoholic. P R O B L E M 1 1-8 U nlike ethylene glycol, propylene glycol (propane- l ,2-diol) is nontoxic because it oxidizes to a common metabolic intermediate. Give the structures of the biological oxidation products of propylene glycol. One reason alcohols are such versatile chemical intermediates is that they react as both nucleophiles and electrophiles. The following scheme shows an alcohol reacting as a weak nUcleophile, bonding to a strong electrophile (in this case, a carbocation). .. '" R - 0 : ------ C+ / 1 H weak nucleophile strong . .+ 1 R - O -C I) 1 H ..} R-O-H 1 .. R - O -C . . 1 11-5 Alcohols as N ucleoph i les and E l ectrophi les; Formation of Tosylates electrophile An alcohol is easily converted to a strong nucleophile by forming its alkoxide ion. The alkoxide ion can attack a weaker electrophile, such as an alkyl halide . R-O-H weak nucleophile ----7 Na ..- I R - 0: N a+ -C- X .. 1 \A strong nucleoph il e weak ------? .. 1 R - O-C .. 1 x- Na+ electrophile The 0 - H bond is broken when alcohols react as nucleophiles, both when an alcohol reacts as a weak nucleophile and when the alcohol is converted to its alkoxide that then reacts as a strong nucleophile. In contrast, when an alcohol reacts as an elec trophile, the C - O bond is broken. T his bond is broken when This bond is broken when - C - O ..2.... H A n alcohol is a weak electrophile because the hydroxyl group is a poor leaving group. The hydroxyl group becomes a good leaving group ( H20) when it is protonated. For example, HBr reacts with a primary alcohol by an SN2 attack of bromide on the proto nated alcohol. Note that the C - O bond is broken in this reaction. R 1 1 tH - O 2 poor .. R H 1+ .. I :Br: CH2- O - H .. l,,- . . good e lectrophile electrophile The disadvantage of using a protonated alcohol is that a strongly acidic solution is required to protonate the alcohol. Although halide ions are stable in acid, few other good nucleophiles are stable in strongly acidic solutions. Most strong nucleophiles are 4 70 Chapter 11 : Reactions of Alcohols also basic and will abstract a proton in acid. Once protonated, the reagent is no longer nucleophilic. For example, an acetylide ion would instantly become protonated if it were added to a protonated alcohol. PROBLEM-SOLVING rosyl group ROTs Ts 0 - -o-CH 0 Htnv ' II 0 IOsylalc ester TsOH R -O- 0 II IQsic acid TsCI I-I - O 0 0 -o_ CH' How can we convert an alcohol to an electrophile that is compatible with basic nucleophiles? We can convert it to an alkyl halide, or we can simply make its tosylate ester. A tosylate ester (symbolized ROTs) is the product of condensation of an alcohol with p-toluenesulfonic acid (symbolized TsOH). o o (Osyl chloride -OTs CI- 0 II losyl:uc ion - O- 0 -o_ 0 II -oCHl CH' R- O + H + HO + a lcohol -o- CH3 ( ' I _ o ) R -O- o -o- CH3 ' I _ + Hp T sOH p-toluenesulfonic acid alkyl tosylate, ROTs a p-toluenesulfonate ester 0 II -o- CHl The tosylate group is an excellent leaving group, and alkyl tosylates undergo substitu tion and elimination much like alkyl halides. In many cases, a tosylate is more reactive than the equivalent alkyl halide. II -C-CII OTs OH TsCl pyridine - C-C- I '--- I ,..- OTs I I ( substitution) -C-C- I I I I + - OTs Nuc or elimination: -C-CH B:0 I IJ I U "" ( elimination) /C = C "" / + B -H + - OTs Tosylates are made from alcohols using tosyl chloride (TsCI) in pyridine, as shown next. This reaction gives much higher yields than the reaction with TsOH itself. The mechanism of tosylate formation shows that the C - O bond of the alcohol remains intact throughout the reaction, and the alcohol retains its stereochemical con figuration. Pyridine serves as an organic base to remove the HCl formed in the reac tion, preventing it from protonating the alcohol and causing side reactions. ll R-1 :s e H p -toluenesulfonyl chloride TsCI, " tosyl chloride" Q I 0 R - O- S = O + .. o II CH 3 ROTs, a tosylate ester o + H C I- 1 1 -5 Alcohols as Nucleophiles and Electrophiles; Formation of Tosylates 47 1 The following reaction shows the SN2 displacement of tosylate ion COTs) from (S)-2-butyl tosylate with inversion of configuration. The tosylate ion is a particularly stable anion, with its negative charge delocalized over three oxygen atoms. : .I. : - --""'-:",C i odide .. CH3C 9 - Ts (' H"' C H3 2 .. I - C" \' / CH2CH3 H -OTs lOsylate ion [-:o'-'o .. .,0" C I-I3 + t osylate i on - : OTs (S)-2-butyl tosylate ( R)-2-butyl iodide II Ij C H3 _ o=-o : 0 :- .. " 0,, II C H3 a .. _ Y-o' Ij : 0 :_ I C H3 _ ] resonance-stabilized anion Like halides, the tosylate leaving group is displaced by a wide variety of nucle ophiles. The SN2 mechanism (strong nucleophile) is more commonly used than the SN 1 in synthetic preparations. The following reactions show the generality of SN2 displacements of tosylates, In each case, R m ust be a primary or unhindered secondary alkyl group if substitution is to predominate over elimination, t S U M M ARY SN2 Reactions of Tosylate Esters R - OTs + hydroxide -OH ------'? R - OH R-C=N nitrile alkyl halide alcohol + + -OTs R - OTs + + + -C = N cyanide halide alkoxide ------'? -OTs R - OTs B r- ------'? R - Br + -OTs R - OTs R ' - O: NH ammonia ------'? R-O-R' R - NH! alkane elher + -OTs R - OTs R - OTs + ------'? amine salt -OTs + L iAlH4 LAH ------'? R-H + -OTs P R O B L E M 1 1-9 PROBLEM-SOLVING Predict the major products of the following reactions. ( a) e thyl tosylate + potassium t-butoxide (b) isobutyl tosylate + N aI (c) ( R)-2-hexyl tosylate + NaCN (d) the tosylate of cyclohexylmethanol + excess N H3 (e) n- bu tyl tosylate + s odium acetylide, H - C - e:- +Na P R O B L E M 1 1- 10 Tosylate esters are particularly usefu l : They a re great leaving g rou ps, often better than h a l ides. G rig nard reactions build a l cohols, which are easi Iy converted to tosylates for substitution o r e l i m i nation. HiltZ; Show how you would convert I -propanol (and whatever reagents are needed) to the following compounds using tosylate intermediates. (b) n -propylamine, C H3CH2CH 2NH2 (a) l -bromopropane (c) C H3CH2CH20CH2CH3 e thyl propyl ether (d) C H3CH2CH2CN butyronitrile 4 72 Chapter 1 1 : Reactions of Alcohols The reduction of alcohols to alkanes is not a common reaction because it removes a functional group, leaving fewer options for further reactions. R - OH ----?) 11-6 Red uction of Alcohols reduction R-H ( rare) We can reduce an alcohol in two steps, by dehydrating it to an alkene, then hydro genating the alkene. cyclopentanol cyclopentene cyclopentane Another method for reducing an alcohol involves converting the alcohol to the tosylate ester, then using a hydride reducing agent to displace the tosylate leaving group. This reaction works with most primary and secondary alcohols. o H H + C lcyclohexanol o II tosyl chloride, TsC I P R O B L E M 1 1- 1 1 -o - CH3 pyridine ) r'1-- 0 - T S V cyclohexyl tosylate (b) (d) LiAIH4 ) r'1-- H V cyclohexane (75%) Predict th e products of the following reactions. ( a) c yclohexylmethanol + TsCI/ pyridine (c) I -methylcyclohexanol + H2S04, heat product of ( a ) + L iAIH4 product of ( c ) + H 2, Pt 11-7 Reactions of Alcohols with Hyd ro h a l ic Acids Tosylation of an alcohol, followed by displacement of the tosylate by a halide ion, con verts an alcohol to an alkyl halide. This is not the most common method for converting alcohols to alkyl halides, however, because simple, one-step reactions are available. A common method is to treat the alcohol with a hydrohalic acid, usually HCI or HBr. In acidic solution, an alcohol is in equilibrium with its protonated form. Protonation converts the hydroxyl group from a poor leaving group COH) to a good leaving group (H20 ) . Once the alcohol is protonated, all the usual substitution and elimination reac tions are feasible, depending on the structure ( 1 , 2, 3) of the alcohol. R poor leaving group good leaving group M ost good nucleophiles are basic, becoming protonated and losing their nucle ophilicity in acidic solutions. Halide ions are exceptions, however. Halides are anions of strong acids, so they are weak bases. Solutions of HBr and HCI contain nucleophilic Br- and Cl- ions. These acids are commonly used to convert alcohols to the corre sponding alkyl halides. R eactions with Hydrobromic Acid {b I 'H R -X (jjj= R - OH + H Br/H20 R - Br Concentrated hydrobromic acid rapidly converts t-butyl alcohol to t-butyl bromide. The strong acid protonates the hydroxyl group, converting it to a good leaving group. The hindered tertiary carbon atom cannot undergo SN2 displacement, but it can ionize 11-7 Reactions of Alcohols with Hydrohalic Acids 473 to a tertiary carbocation. Attack by bromide gives the alkyl bromide. The mechanism is similar to other SN 1 m echanisms we have studied, except that water serves as the leaving group from the protonated alcohol. M ECHAN I SM 1 1- 1 R eaction of a Tertiary Alcohol with H B r (SN 1 ) A tertiary alcohol reacts with HBr by the SNI mechanism. E XAM PLE: C onversion of t-buty l a lcohol to t-buty l bromide. Step 1: Protonation converts the bydroxyl group to a good leaving group. H r-P;r: . H3 C- C -O-H .. . C H3 I Step 2: t-butyl alcohol CH3 I CH3 I + /H H3C - - R'-... + : 1?.r:H CH3 ? Water leaves, forming a carbocation. Step 3 : CH3 I '"+ /H "H 3c-c -'-b' I '-... H CH3 CH3 I H3 C - C + C H3 I + H20 Bromide ion attacks the carbocation. C H3 .. 1+ . Br .. H 3C - C I CH3 . CH3 _ H 3C - C Br : I .. CH3 - I .. t-butyl bromide Many other alcohols react with HBr, with the reaction mechanism depending on the structure of the alcohol. For example, I -butanol reacts with sodium bromide in concentrated sulfuric acid to give I -bromobutane by an SN2 displacement. The sodium bromide/sulfuric acid reagent generates HBr in the solution. I -butanol I -bromobutane ( 90%) Protonation converts the hydroxyl group to a good leaving group, but ionization to a primary carbocation is unfavorable. The protonated primary alcohol is well suited for the SN2 displacement, however. B ack-side attack by bromide ion gives I -bromobutane. MECHAN ISM 1 1 -2 R eaction of a Primary Alcohol with H Br (SN2) A primary alcohol reacts with HBr by the SN2 mechanism. E XAMPLE: C onversion of 1-buta nol to 1-bromobutane. Step 1: Protonation converts the hydroxyl group to a good leaving group. ( Continued) 4 74 Chapter 1 1 : Reactions of Alcohols Step 2 : B romide displaces water to give the alkylbromide. Secondary alcohols also react with HBr to form alkyl bromides, usually by the SN 1 mechanism. For example, cyclohexanol i s converted to bromocyclohexane using HBr as the reagent. ---? H Br PROBLEM-SOLVING M emorizing a l l these mechanisms is not the best way to study this materia l . Depend i n g on the substrate, these reactions can go by more than one mechanism. Gain experience working problems, then consider each i nd ividual case to propose a l i kely mechanism . Hi-ltv cyclohexanol bromocyclohexane (80%) P R O B L E M 1 1- 12 Propose a mechanism for the reaction of (a) cyclohexanol with HBr to form bromocyclohexane. (b) 2-cyclohexylethanol with HBr to form I -bromo-2-cyclohexylethane. R eactions with H yd roch loric Acid R - CI Hydrochloric acid (HCI) reacts with alcohols in much the same way that hydrobromic acid does. For example, concentrated aqueous HCl reacts with t-butyl alcohol to give t-butyl chloride. (CH3 h C - OH + H CI/H20 ( CH3hC - CI + H 20 I- butyl alcohol I-butyl chloride The reaction of t-butyl alcohol with concentrated Hel goes by the SN 1 mechanism. Write a mechanism for this reaction. P R O B L E M 1 1- 1 3 ( 98%) Chloride ion is a weaker nucleophile than bromide ion because it is smaller and less polarizable. An additional Lewis acid, such as zinc chloride (ZnCI2), i s some times necessary to promote the reaction of HCl with primary and secondary alcohols. Zinc chloride coordinates with the oxygen of the alcohol in the same way a proton does-except that zinc chloride coordinates more strongly. The reagent composed of HCI and ZnCl2 is called the Lucas reagent. Secondary and tertiary alcohols react with the Lucas reagent by the SN 1 mechanism. CH3 ' ZnCI 2 I / H -C-O-H I CH3 SNl reaction with the Lucas reagent (fast) / CH3 CH3 H -C+ '" carbocation .. - + HQ - ZnCI2 1 1 -7 R eactions of Alcohols with H ydrohalic Acids 4 75 SN2 reaction with the Lucas reagent (slow) When a primary alcohol reacts with the Lucas reagent, ionization is not possible the primary carbocation is too unstable. Primary substrates react by an SN2 mechanism, which is slower than the SN 1 reaction of secondary and teltiary substrates. For example, when I -butanol reacts with the Lucas reagent, the chloride ion attacks the complex from the back, displacing the leaving group. :j: / CI - C"," \H H transition state The Lucas Test CH2CH2CH3 + - :0: / " Z nCI? H - The Lucas reagent reacts with primary, secondary, and tertiary alcohols at fairly predictable rates, and these rates can be used to distinguish among the three types of alcohols. When the reagent is first added to the alcohol, the mix ture forms a single homogeneous phase: The concentrated HCI solution is very polar, and the polar alcohol-zinc chloride complex dissolves. Once the alcohol has reacted to form the alkyl halide, the relatively nonpolar halide separates into a second phase. The Lucas test involves adding the Lucas reagent to an unknown alcohol and watching for the second phase to separate (see Table 1 1 -2). Tertiary alcohols react almost instantaneously because they form relatively stable tertiary carbocations. Sec ondary alcohols react in about 1 to 5 m inutes because their secondary carbocations are less stable than tertiary ones. Primary alcohols react very slowly. Since the activated primary alcohol cannot form a carbocation, it simply remains in solution until it is attacked by the chloride ion. With a primary alcohol, the reaction may take from 1 0 minutes to several days. P R O B L E M 1 1- 14 TABLE 1 1 -2 Reactions of Alcohols with the Lucas Reagent A lcohol Type T ime to React (min) <I >6 1 -5 primary secondary tertiary Show how you would use a simple chemical test to distinguish between the following pairs of compounds. Tell what you would observe with each compound. (a) i sopropyl alcohol and t-butyl alcohol ( b) isopropyl alcohol and 2-butanone, CH3COCH2CH3 ( c) I -hexanol and cyclohexanol (d) a llyl alcohol and I -propanol (e) 2-butanone and t-butyl alcohol Limitations on the Use of Hydrohalic Acids with Alcohols The reactions of alcohols with hydrohalic acids do not always give good yields of the expected alkyl halides. Four principal limitations restrict the generality of this technique. 1. Poor yields of alkyl chlorides from primary and secondary alcohols. Primary and secondary alcohols react with HCI much more slowly than tertiary alcohols, even with zinc chloride added. Under these conditions, side reactions prevent good yields of the alkyl halides. 2. Eliminations. H eating an alcohol in a concentrated acid such as HCl or HBr often leads to elimination. Once the hydroxyl group of the alcohol has been pro tonated and converted to a good leaving group, it becomes a candidate for both substitution and elimination. 3. Rearrangements. Carbocation intermediates are always prone to rearrange ments. We have seen (Section 6 -15) that hydrogen atoms and alkyl groups can migrate from one carbon atom to another to form a more stable carbocation. This rearrangement may occur as the leaving group leaves, or it may occur once the cation has formed. 4 76 C hapter 1 1 : Reacti ons of Alcohols 4. Limited ability to make alkyl iodides. M any alcohols do not react with H I to give acceptable yields of alkyl iodides. Alkyl iodides are valuable intermediates, however, because iodides are the most reactive of the alkyl halides. We will dis cuss another technique for making alkyl iodides in the next section. S O LV E D P R O B L E M 1 1-2 When 3-methyl-2-butanol is treated with concentrated HBr, the major product is 2-bromo2-methylbutane. Propose a mechanism for the formation of this product. H C H3 -C-CH- CH3 C H3 SOLUTION 11 I OH ------7 Br HBr C H3 - C- CH,-CH3 C H3 2-bromo-2-methy Ibulane 1 I 3 -methyl-2-butanol The alcohol is protonated by the strong acid. This protonated secondary alcohol loses water to form a secondary carbocation. H C H 3 - C -CH-CH 3 CH 3 11 1 : O ( H+ ) H C H 3 - C - CH -CH 3 protonated alcohol CH 3 1 1 (I OH + 2 H C H 3 -C-CH-CH 3 secondary carbocation 1 1 + CH3 A hydride shift transforms the secondary carbocation into a more stable tertiary cation. Attack by bromide leads to the observed product. C H 3 - C- CH - CH 3 secondary carbocation + CH 3 1 C H - C -CH-CH 3 tertiary carbocation 3 : r : ! CH 3 1 C H 3 - C -CH-CH 3 observed product r CH 3 1 A lthough rearrangements are usually seen as annoying side reactions, a clever chemist can use a rearrangement to accomplish a synthetic goal. Problem 1 1 - 1 5 shows how an alcohol substitution with rearrangement might be used in a synthesis. P R O B L E M 1 1- 15 Neopentyl alcohol, ( CH3 hCCH20H, reacts with concentrated HBr to give 2-bromo2-methylbutane, a rearranged product. Propose a mechanism for the formation of this product. P R O B L E M 1 1- 16 Explain the products observed in the following reaction of an alcohol with the Lucas reagent. H 4CH3 OH P R O B L E M 1 1- 17 CH3 HCI/ZnCI, -) aa - - H . CH3 + H . C H3 ; C H3 CI ; CI CH3 When cis-2-methylcyc1ohexanol reacts with the Lucas reagent, the major product is l -chloro I -methylcyc1ohexane. Propose a mechanism to explain the formation of this product. 1 1-8 Reactions of Alcohols with Phosphorus Halides 4 77 Several phosphorus halides are useful for converting alcohols to alkyl halides. Phos phorus tribromide, phosphorus trichloride, and phosphorus pentachloride work well and are commercially available. 3 R - OH 3 R - OH + + PCl 3 PBr3 P CIs 3 R - Cl 3 R - Br + + + P (OH h P (OH)3 P OCI3 + 11-8 Reactions of Alcohols with Phosphorus H a l ides R - OH + R - CI H CI Phosphorus triiodide is not sufficiently stable to be stored, but it can be generated in situ (in the reaction mixture) by the reaction of phosphorus with iodine. 6 R - OH + 2P + 3 12 6 R-I + 2 P(OHh Phosphorus halides produce good yields of most primary and secondary alkyl halides, but none works well with tertiary alcohols. The two phosphorus halides used most often are PBr3 and the phosphorus/iodine combination. Phosphorus tribromide is often the best reagent for converting a primary or secondary alcohol to the alkyl bromide, especially if the alcohol might rearrange in strong acid. A phosphorus and iodine combi nation is one of the best reagents for converting a primary or secondary alcohol to the alkyl iodide. For the synthesis of alkyl chlorides, thionyl chloride (discussed in the next section) generally gives better yields than PCl 3 or PCIs, especially with tertiary alcohols. The following examples show the conversion of primary and secondary alcohols to bromides and iodides by treatment with PBr3 and P/I2. CH3 C H3 - C - CH 2 OH C H3 neopentyl alcohol + I I CH3 C H3 - C - CH 2Br CH3 neopentyl bromide (60%) I I P/I2 CH3(CH2)14 -CHzl (85%) Write balanced equations for the two precedi ng reactions. Mechanism o the Reaction with Phosphorus Trihalides f P R O B L E M 1 1- 18 T he mechanism of the reac tion of alcohols with phosphorus trihalides explains why rearrangements are uncom mon and why phosphorus halides work poorly with tertiary alcohols. The mechanism is shown here using PBr3 as the reagent; PCl3 and PI3 (generated from phosphorus and iodine) react in a similar manner. MECHANISM 1 1-3 Step 1: PBr3 is a strong electrophlle. An alcohol displaces bromide ion from PBr3 to R eaction of Alcohols with P Br3 give an excellent leaving group. .. I .. R -O: / : P -B r: " I l..I H : 1?r: : :8r: ::8i : / R - O -P: + .. + H I : Br: \ excellent leaving group ( Continued) 4 78 Chapter 1 1: Reactions of Alcohols Step 2: B romide displaces the leaving group to give the alkyl bromide. + : .8 r : .. / : O- p: H I : Br: \ Step 1 : D isplacement of bromide and formation of a leaving group. E XAMPLE: R eaction of {R)-2-pentanol with PBr3' leaving group CH3CH2CH2 : .8[ : " \ . . . I r.. H 3C' C- , l.,- H :.r : H :; Step 2: B romide displaces the leaving group to give (S)-2-bromopentane. (R)-2-pentanol ? l + + ( S)-2-brornopentane ' : :8 r: / .. : O-p: H I : Br: \ Rearrangements are uncommon because no carbocation is involved, so there is no opportunity for rearrangement. This mechanism also explains the poor yields with tertiary alcohols. The final step is an SN2 displacement where bromide attacks the back side of the alkyl group. This attack is hindered if the alkyl group is tertiary. In the case of a tertiary alcohol, an ionization to a carbocation is needed. This ionization is slow, and it invites side reactions. 1 1 -9 R eactions of Alcohols with Th ionyl Ch l oride Thionyl chloride ( SOCI2) is often the best reagent for converting an alcohol to an alkyl chloride. The byproducts (gaseous S02 and HCI) leave the reaction mixture and ensure there can be no reverse reaction. o R - OH + C I - S -Cl I heat -- R - CI + HCI R - '-----? s .: T H Cl .. Cl . t hionyl chloride "- . Under the proper conditions, thionyl chloride reacts by the interesting mecha nism summarized next. In the first step, the nonbonding electrons of the hydroxyI oxy gen atom attack the electrophilic sulfur atom of thionyl chloride. A chloride ion is expelled, and a proton is lost to give a chlorosulfite ester. In the next step, the chloro sulfite ester ionizes (when R 2 0 or 3 0), and the sulfur atom quickly delivers chloride to the carbocation. When R is primary, chloride probably bonds to carbon at the same time that the C - 0 bond is breaking. .. .. C 0: 0: .. .. .. I. .. -I' R -O- S-I' + HCI : R -O- S : R - O- S ':--- O : \ "+ I) "- CI CI H H Cl = +1 I "--- Cl - c hlorosulfite ester 1 1 -9 Reactions of Alcohols with Thionyl Chloride Y 4 79 R : C-: O ""Cl / S" = o.' R+ S = O.. .. Cl ion pair : 0 "-..[ : :') chlorosulfite ester Thi s mechanism resembles the SN 1 , except that the nucleophile is delivered to the carbocation by the leaving group, giving retention of configuration as shown i n the following example. (Under different conditions, retention of configuration may not be observed.) H , ./' OH / C ",,-CH3(CH2)4CH2 CH3 (R)-2-octanol SOC12 dioxane (solvent) (0) ) H / C ",,-CH3(CH2)4CH2 CH3 (R)-2-chlorooctane ( 84%) , .i/ CI S ummary of the Best Reagents for Converting Alcohols to Alkyl Halides C lass of Alcohol C hloride B romide I odide primary secondary tertiary "Works only in selected cases. S OCl2 SOCl2 HCl PBr3 or H Br* PBr3 HBr Pil2 PiJi HT* PROBLEM 1 1-19 Suggest how you would convert trans-4-methylcyclohexanol to (a) trans- l -chl oro-4-methylcy clohex ane . (b) cis- l -chloro-4-methylcyclohexane. P R O B L E M 1 1 - 20 Two products are observed in the following reaction. PROBLEM-SOLVING ( a) S uggest a mechanism to explain how these two products are formed. (b) Your mechanism for part (a) should be different from the usual mechanism of the reac T hionyl chloride reacts with a l cohols by various mechanisms that depend o n the su bstrate, the solvent, and the tem perature. Be cautious in predicting the structure and stereochemistry of a product u n l ess you know the actual mechanism. Htnt: tion of SOCl2 with alcohols. Explain why the reaction follows a different mechanism in this case. PRO B L E M 1 1-21 Give the structures of the products you would expect when each alcohol reacts with ( I ) H Cl, ZnCI2; (2) HBr; ( 3) P Br3; ( 4) P iT2; ( 5) S OCI2. ( a) I -butanol ( b) 2 -butanol ( c) 2-methyl-2-butanol ( d) 2 ,2-dimethy l- l -butan ol (e) cis-3-methylcyciopentanol 4 80 Chapter 1 1 : Reactions of Alcohols 1 1 - 1 0A Formation of Al kenes 1 1 -1 0 D ehyd ration Reactions of Alcohols We studied the mechanism for dehydration of alcohols to alkenes in Section 7 - 1 0 together with other syntheses of alkenes. Dehydration requires an acidic catalyst to proton ate the hydroxyl group of the alcohol and convert it to a good leaving group. Loss of water, followed by loss of a proton, gives the alkene. An equilibrium is estab lished between reactants and products. M ECHANISM 1 1 -4 (Review): Acid-Catalyzed Dehydration of an Alcohol Step 1: Protonation converts the hydroxyl group to a good leaving group. Dehydration results from E l elimination of the protonated alcohol Step 2: Water leaves, forming a carbocation. I I -C-CI I H H : H :OH I H+ -'" I I - C-CI I Step 3: Loss of a proton gives the alkene. I P -C-CI I H H :OH I - C -C I I H I + + H :Q -H I To drive this equilibrium to the right, we remove one or both of the products as they form, either by distilling the products out of the reaction mixture or by adding a dehydrating agent to remove water. In practice, we often use a combination of distilla tion and a dehydrating agent. The alcohol is mixed with a dehydrating acid, and the mixture is heated to boiling. The alkene boils at a lower temperature than the alcohol (because the alcohol is hydrogen-bonded), and the alkene distills out of the mixture. For example, cyclohexanol, bp 1 6 1 DC HH aH ex: + cyclohexene, bp 83DC (80%) (distilled from the mixture) 1 1 - 1 0 Dehydration Reactions of Alcohols A lcohol dehydrations generally take place through the E l mechanism. Protona tion of the hydroxyl group converts it to a good leaving group. Water leaves, forming a carbocation. Loss of a proton gives the alkene. 4 81 Figure 1 1 -2 shows the reaction-energy diagram for the E1 dehydration of an alcohol. The first step is a mildly exothermic protonation, followed by an endothermic, rate-limiting ionization. A fast, strongly exothermic deprotonation gives the alkene. Because the rate limiting step is formation of a carbocation, the ease of dehydration follows from the ease of formation of carbocations: 3 > 2 > 1 . As in other carbocation reactions, realTangements are common. With primary alcohols, reanangement and isomerization of the products are so common that acid-catalyzed dehydration is rarely a good method for converting them to alkenes. The following mechanism shows how I -butanol undergoes dehydration with reanangement to give a mixture of I -butene and 2-butene. The more highly sub stituted product, 2-butene, is the major product, in accordance with the Zaitsev rule (Section 6- 1 9). ionization of the protonated alcohol, with rearrangement I I I I I HHHH H-C-C-C-C-H H-O: H+ H H \.... H I I I I HI )=+J H H IHl H-C-C-C-C-H H-O+ H H H H I ( I. I I I I H20 leaves H : - migrates ) + H H H -C-C-C-C-H HHHH I I I I I I secondary carbocation >, 01) i3 I -C-CH I I OH H2S04 I I -C-C H I I OH2 HS04.... Figure 11-2 reaction coordinate --------------- --_ . - --- + I I Reaction-energy diagram for dehydration of an alcohol. 4 82 C hapter 1 1 : Reactions of Alcohols Loss of either proton to give two products HH H + I I I H -C - C - C - C - H I Hb secondary carbocation B :- chJ H HH H HH I I I I I I H -C-C=C-C-H + H -C=C-C - C - H I I I H Hb H 2-butene (major, 70%) a disu bsti tuted alkene [ loss of H: [ l oss of H: I -butene (minor, 30%) a monosubstituted alkene chJ L et's review the utility of dehydration and give guidelines for predicting the products : 2 . D ehydration works best with tertiary alcohols and almost as well with secondary alcohols. Rearrangements and poor yields are common with primary alcohols. 1. D ehydration usually goes by the E l mechanism. Rearrangements may occur to form more stable carbocations. 3. (Zaitsev 's rule) I f t wo or more alkenes might be formed by deprotonation of the carbocation, the most highly substituted alkene usually predominates. Solved Problem 1 1 -3 shows how these rules are used to predict the products of dehydrations. The carbocations are drawn to show how rearrangements occur and how more than one product may result. PROBLEM-SOLVING Most alcohol dehydrations go by E I mechanisms i nvolving protonation of the OH group, fo l l owed by loss of water. HinZ; S O LV E D P R O B L E M 1 1- 3 Predict the products of sulfuric acid-catalyzed dehydration of the following alcohols. (b) neopentyl alcohol (a) I -methyIcyclohexanol SOLUTION ( a) I -Methylcyclohexanol reacts to form a tertiary carbocation. A proton may be abstract ed from any one of three carbon atoms. The two secondary atoms are equivalent, and abstraction of a proton from one of these carbons leads to the trisubstituted double bond of the major product. Abstraction of a methyl proton leads to the disubstituted double bond of the minor product. 1/ (XC",,+ H H H H H l -methylcyclohexanol protonated cation loss of Ha major product (trisubstituted) loss of Hb minor product (disubstituted) ( b) Neopentyl alcohol cannot simply ionize to form a primary cation. Rearrangement occurs as the leaving group leaves, giving a tertiary carbocation. Loss of a proton from the adjacent secondary carbon gives the trisubstituted double bond of the major 1 1 - 1 0 Dehydration Reactions of Alcohols product. Loss of a proton from the methyl group gives the disubstituted double bond of the minor product. C H -C-CH2OH C H3 4 83 n eopenlyl alcohol (2,2-dimcthyl- l -propanol ) 3 CH3 I I . :- w C H3 . CH 3 -C-CH,-OH, - .. CH3 i onization with rearrangement I I (" + C H3 CH3 -C -CH2 3 + C H3 c ation I I + H H loss of H; major product (trisubstituted) "" / / "" C =C l oss of C H,CH3 C H3 Hb m inor product (disubstitllted) P R O B L E M 1 1-22 PROBLEM-SOLVING Predict the products of the sulfuric acid-catalyzed dehydration of the following alcohols. When more than one product is expected, label the major and minor products. ( a) 2 -methyl-2-butanol ( b) I -pentanol (c) 2-pentanol (d) l -isopropylcyclohexanol (e) 2 -methylcyclohexanol P R O B L E M 1 1-23 D raw the carbocation, look for poss i b l e rearra ngements, then consider all the ways that the orig inal carbocation a n d any rearranged carbocation m i ght lose protons to g ive a l kenes. Za itsev's rule usually pred i cts the major product. HtftJ/ S ome alcohols undergo rearrangement or other unwanted side reactions when they dehydrate in acid. Alcohols may be dehydrated under mildly basic conditions lIsing phosphorus oxy chloride ( POCI3 ) in pyridine. The alcohol reacts with phosphorus oxychloride much like it reacts with tosyl chloride (Section 1 1 -5), displacing a chloride ion from phosphorus to give an alkyl dichlorophosphate ester. The dichlorophosphate group is an outstanding leaving group. Pyridine reacts as a base with the dichlorophosphate ester to give an E2 elimination. Propose a mechanism for the dehydration of cyclohexanol by POCI3 in pyridine. a CI /P" CI II I CI phosphorus oxychloride pyridine o N 1 1 -10B Bimolecu l a r Dehydration to Form Ethers (Ind ustrial) In some cases, a protonated primary alcohol may be attacked by another molecule of the alcohol and undergo an SN2 displacement. The net reaction is a bimolecular dehy dration to form an ether. For example, the attack by ethanol on a protonated molecule of ethanol gives diethyl ether. / , C H 3CH ? 0 - C . ' + C H3 nucleophilic electrophilic . - k protonated ether . 0! H "'H water d iethyl ether Bimolecular dehydration can be used to synthesize symmetrical dialkyl ethers from simple, unhindered primary alcohols. This method is used for the industrial synthesis of 484 Chapter 1 1 : Reactions of Alcohols diethyl ether ( CH3CH2 - O - CH2CH 3 ) and dimethyl ether (CH3 - O - CH 3 ) . Under the acidic dehydration conditions, two reactions compete: Elimination (to give an alkene) competes with substitution (to give an ether). Substitution to give the ethel; a bimolecular dehydration H2S04, 140C ) C H3CH2 - O - CH2CH3 ethanol diethyl ether + H 20 Elimination to give the alkene, a unimolecular dehydration CH2 = CH2 + H20 ethylene P R O B L E M 1 1-24 Contrast the mechanisms of the two preceding dehydrations of ethanol. H ow can we control these two competing dehydrations? The ether synthesis (substitution) shows two molecules of alcohol giving two product molecules: one of diethyl ether and one of water. The elimination shows one molecule of alcohol giv ing two molecules: one of ethylene and one of water. The elimination in results an increase in the number of molecules and therefore an increase in the randomness (entropy) of the system. The elimination has a more positive change in entropy ( ilS) than the substitution, and the - TilS term in the Gibbs free energy becomes more favorable for the elimination as the temperature increases. Substitution (to give the ether) is favored around 1 40aC and below, and elimination is favored around I 80aC and above. Diethyl ether is produced industrially by heating ethanol with an acidic catalyst at around 1 40aC. P R O B L E M 1 1-25 Explain why the acid-catalyzed dehydration is a poor method for the synthesis of an unsym metrical ether such as ethyl methyl ether, C H3CH2 - O - CH3. P R O B L E M 1 1 -26 Propose a detailed mechanism for the following reaction. H2S04, heat ) P ropos i n g Reaction Mech a n isms P R O B L E M - S O LV I N G S T R AT E G Y I n view of the large number of reactions we've covered, proposing mechanisms for reactions you have never seen may seem nearly impossible. As you gain experience in working mech anism problems, you will start to see similarities to known reactions. Let's consider how an organic chemist systematically approaches a mechanism problem. (A more complete ver sion of this method appears in Appendix 4.) Although this stepwise approach cannot solve all mechanism problems, it should provide a starting point to begin building your experience and confidence. D eterm i ning the Type of Mechanism First, determine what kinds of conditions and catalysts are involved. In general, reactions may be classified as involving (a) strong electrophiles (including acid-catalyzed reactions), 1 1 - 1 0 Dehydration Reactions of Alcohols ( b) strong nucleophiles (including base-catalyzed reactions), or (c) free radicals. These three types of mechanisms are quite distinct, and you should first try to determine which type is involved. (a) In the presence of a strong acid or a reactant that can dissociate to give a strong elec 4 85 trophile, the mechanism probably involves strong electrophiles as intermediates. Acid catalyzed reactions and reactions involving carbocations (such as the SNl , the E l , and most alcohol dehydrations) fall in this category. (b) In the presence of a strong base or a strong nucleophile, the mechanism probably involves strong nucleophiles as intermediates. Base-catalyzed reactions and those depending on base strength (such as the SN2 and the E2) generally fall in this category. (c) Free-radical reactions usually require a free-radical initiator such as chlorine, bromine, NBS, or a peroxide. In most free-radical reactions, there is no need for a strong acid or base. Once you have determined which type of mechanism you will write, there are gener al methods for approaching the problem. At this point, we consider mostly the electrophilic reactions covered in recent chapters. Suggestions for drawing the mechanisms of reactions involving strong nucleophiles and free-radical reactions are collected in Appendix 4. R eactions I nvolving Strong E lectroph iles When a strong acid or electrophile is present, expect to see intermediates that are strong acids and strong electrophiles; cationic intermediates are common. Bases and nucle ophiles in such a reaction are generally weak, however. Avoid drawing carbanions, alkox ide ions, and other strong bases. They are unlikely to coexist with strong acids and strong electrophiles. Functional groups are often converted to carbocations or other strong electrophiles by protonation or reaction with a strong electrophile; then the carbocation or other strong elec trophile reacts with a weak nucleophile such as an alkene or the solvent. 1 . Consider the carbon skeletons of the reactants and products, and decide which car bon atoms in the products are most likely derived from which carbon atoms in the reactants. 2. Consider whether any of the reactants is a strong enough electrophile to react with out being activated. If not, consider how one of the reactants might be converted to a strong electrophile by protonation of a Lewis basic site (or complexation with a Lewis acid). Protonation of an alcohol, for example, converts it to a strong electrophile, which can undergo attack or lose water to give a carbocation, an even stronger electrophile. Protona tion of an alkene converts it to a carbocation. 3. Consider how a nucleophilic site on another reactant (or, in a cyclization in another part of the same molecule) can attack the strong electrophile to form a bond needed in the product. Draw the product of this bond formation. If the intermediate is a carbocation, consider whether it is likely to rearrange to form a bond in the product. If there isn't any possible nucleophilic attack that leads in the direction of the product, consider other ways of converting one of the reactants to a strong electrophile. 4. Consider how the product of a nucleophilic attack might be converted to the final product (if it has the right carbon skeleton) or reactivated to form another bond needed in the product. To move a proton from one atom to another (as in an isomerization), try adding a proton to the new position, then removing it from the old position. 5 . Draw out all steps of the mechanism using curved arrows to show the movement of electrons. Be careful to show only one step at a time. C om mon M istakes to Avoid in Draw i n g Mechan isms 1. D o not use condensed or line-angle formulas for reaction sites. Draw all the bonds and all the substituents of each carbon atom affected throughout the mechanism. In ( Continued) 4 86 C hapter 1 1 : Reactions of Alcohols reactions involving strong electrophiles and acidic conditions, three-bonded carbon atoms are likely to be carbocations. If you draw condensed formulas or line-angle for mulas, you will likely misplace a hydrogen atom and show a reactive species on the wrong carbon. 2. Do not show more than one step occurring at once. Do not show two or three bonds chang ing position in one step unless the changes really are concerted (take place simultaneous ly). For example, protonation of an alcohol and loss of water to give a carbocation are two steps. You must not show the hydroxyl group "jumping" off the alcohol to join up with an anxiously waiting proton. Remember that curved arrows show movement of electrons, always from the nucleophile (electron donor) to the electrophile (electron acceptor). For example, protonation of a double bond must show the arrow going from the electrons of the double bond to the pro ton- never from the proton to the double bond. Resist the urge to use an arrow to "point out" where the proton (or other reagent) goes. 3. S AMPLE PROBLEM To illustrate the stepwise method for reactions involving strong electrophiles, we will devel op a mechanism to account for the following cyclization: OR The cyclized product is a minor product in this reaction. Note that a mechanism problem is different from a synthesis problem: In a mechanism problem, we are limited to the reagents given and are asked to explain how these reactants form these products under the conditions shown. Also, a mechanism problem may deal with how an unusual or unexpected minor product is formed. In the presence of sulfuric acid, this is clearly an acid-catalyzed mechanism. We ex pect strong electrophiles, cationic intermediates (possibly carbocations), and strong acids. Carbanions, alkoxide ions, and other strong bases and strong nucleophiles are unlikely. 1 . Consider the carbon skeletons of the reactants and products, and decide which carbon atoms in the products are most likely derived from which carbon atoms ' in the reactants. Drawing the statting material and the product with all the substituents of the affected carbon atoms, we see the major changes shown here. A vinyl hydrogen must be lost, a C - C bond must be formed, a methyl group must move over one carbon atom, and the hydroxyl group must be lost. = 2. Consider whether any of the reactants is a strong enough electrophile to react with out being activated. If not, consider how one of the reactants might be converted to a strong electrophile by protonation of a Lewis basic site (or complexation with a Lewis acid). The starting material is not a strong electrophile, so it must be activated. Sulfuric acid could generate a strong electrophile either by protonating the double bond or by proto nating the hydroxyl group. Protonating the double bond would form the tertiary carboca tion, activating the wrong end of the double bond. Also, there is no good nucleophilic site 1 1 - 1 0 Dehydration Reactions of Alcohols on the side chain to attack this carbocation to form the correct ring. Protonating the dou ble bond is a dead end. 4 87 d oes not lead toward product The other basic site is the hydroxyl group. An alcohol can protonate on the hydroxyl group and lose water to form a carbocation. 3. Consider how a nucleophilic site on another reactant (or, in a cyclization, in anoth er part of the same molecule) can attack the strong electrophile to form a bond needed in the product. Draw the product of this bond forrnation. The carbocation can be attacked by the electrons in the double bond to form a ring; but the positive charge is on the wrong carbon atom to give a six-membered ring. A favorable rearrangement of the secondary carbocation to a tertiary one shifts the positive charge to the correct carbon atorn and accomplishes the methyl shift we identified in Step 1 . Attack by the (weakly) nucleophilic electrons in the double bond gives the correct six-membered ring. 4 . Consider how the product of nucleophilic attack might be converted to the final product (if it has the right carbon skeleton) or reactivated to forrn another bond needed in the product. Loss of a proton (to HS04 or H20, but not to -OH, which is not compatible!) gives the observed product. 5. Draw out all steps of the mechanism using curved arrows to show the movement of electrons. Combining the equations written immediately above gives the complete mechanism for this reaction. The following problems require proposing mechanisms for reactions involving strong electrophiles. Work each one by completing the five steps just described. ( Continued) 4 88 C ha p er Reactions of Alcohols Propose a mechanism for each re c io 0 OH P R O B L E M 1 1-27 (a) ( b) t 11: (c) 00CH3 U I H2S04, heat Hp W a t n. ) ero CHpH o Q--CH, I I + + + (f 1 1 -1 1 U n iq u e Reactions of Diols 1 1-1 1 A The Pinacol Rea rrangement U sing our knowledge of alcohol reactions, we can explain results that seem strange at first glance. The following dehydration is an example of the pinacol rearrangement: 11 H3 C - C - C - CH3 11 pinacol (2,3-dimethyl-2,3-butanediol) H3C CH3 H 3C - - - CH3 + H20 o pinacolone (3,3-dimethyl-2-butanone) HO OH CH3 CH3 T 1 T he pinacol rearrangement is formally a dehydration. The reaction is acid catalyzed, and the first step is protonation of one of the hydroxyl oxygens. Loss of water gives a tertiary carbocation, as expected for any tertiary alcohol. Mi gration of a methyl group places the positive charge on the carbon atom bearing the second - OH group, where oxygen' s nonbonding electrons help to stabilize the charge through resonance. This extra stability i s the driving force for the rearrangement. Deprotonation of the resonance-stabilized cation gives the product, pinacolone. M ECHANISM 1 1-5 Step 1: T he PililaGol Rearrangement Step 2: Loss of water gives a carbocation. Protonation of a hydroxyl group. H3C CH3 I I HQ : : QTT 11 H3 C - C - C - CH3 / + H+ H3C CH3 I I H3C-C-C-CH3 HO : +OH2 .. 1 I) H 3C CH3 1 / H3C -C-C+ "1 CH3 HO : + H20 Step 3: M ethyl migration fOnTIS a resonance-stabilized carbocation. (methyl migration) - C H3 ) Step 4: Deprotonation gives the product. [ 1 1- 1 1 Unique Reactions of Diols I H 3C-C-C-CH 3 II H -Q: CH 3 + CH 3 H 3C-C- -CH 3 II I H-O+ CH3 CH 3 l 489 resonance-stabilized carbocation H 3C-C-C-CH 3 H + CH 3 I )0: I CH 3 I Hp:-...H ...- .. I H 3C-C-C-CH 3 II CH 3 CH 3 <3-0: H 3C-C-C-CH 3 II I CH 3 pinacolone CH 3 ..0.. I resonance-stabilized carbocation Pinacol-like rearrangements are common in acid-catalyzed reactions of diols . One of the hydroxyl groups protonates and leaves as water, forming a carbocation. Rearrangement gives a resonance-stabilized cation with the remaining hydroxyl group helping to stabilize the positive charge. Problem 1 1 -28 shows some additional exam ples of pinacol rearrangements. P ROBLE M 1 1 - 2 8 Propose a mechanism for each reaction. {\ ./OH (b) P h Ph H2S04 /\ C-OH -9+<;+ (yO Ph Ph P ROBLEM 1 1 - 2 9 The following reaction involves a starting material with a double bond and a hydroxyl group, yet its mechanism resembles a pinacol rearrangement. Propose a mechanism, and point out the part of your mechanism that resembles a pinacol rearrangement. c><0 H By analogy with the pinacol rearrangement, watch for carbocation rearrangements that place the + c harge on a carbinol carbon atom. PROBLEM-SOLVING Hinz; CH=CH2 11-11 B Period ic Acid Cleava g e of Gl ycol s (ij= """"' C=C....... "'/ ,../ a l kene -----;> H z02 os04 -C-CO H OH I I I I -----;> HI04 """"' C = O ,../ + "'/ O=C..... glycol ketones and aldehydes 490 C h apter 1 1 : Reactions of Alcohols 1 ,2-Diols (glycols), such as those formed b y hydroxylation o f alkenes, are cleaved by periodic aci d ( HI04 ) . The products are the same ketones and aldehydes that would be formed by ozonolysis-reduction of the alkene. Hydroxylation followed by periodic acid cleavage serves as a useful alternative to ozonolysis, and the periodate cleavage by i tself is useful for determining the structures of sugars (Chapter 23). Periodic acid cleavage of a glycol probably involves a cyclic periodate interme diate like that shown here. H OH alkene Q c is - CH, g l yco l OH FOO =< cyclic periodate intermediate + HI03 H C H3 keto-aldehyde P eriodic acid cleaves a diol to give the same products as ozonolysis-reduction (03 followed by Me2S) of the alkene. PROBLEM-SOLVING Htltt: P ROBLEM 11-3 0 Predict the products formed by periodic acid cleavage of the following dials. CHpH (a) C H3CH(OH)CH(OH)CH3 (c) Ph - C - CH( O H)CH? CH3 ( b) OH I CH3 I / (;>H (d) H H ___ U OH 11-12 E ste r ificat i o n o f Alco h o l s To an organic chemist, the term ester normally means an ester of a carboxylic acid, unless some other kind of ester is specified. Replacing the -OH group of a car boxylic acid with the - OR group of an alcohol gives a carboxylic ester. The follow ing reaction, called the Fischer esterification, s hows the relationship between the alcohol and the acid on the left and the ester and water on the right. R -O alcohol -EJ + I H -O+ - R' acid o o R -O-C-R' ester II + I H -O-H I For example, if we mix isopropyl alcohol with acetic acid and add a drop of sul furic acid as a catalyst, the following equilibrium results. C H3 H- - o CH3 isopropyl al cohol i -EJ acetic acid H3 H - C -O - C -CH I C H3 isopropyl acetate 3 + H -O-H water 1 1 - 1 3 Esters of Inorganic Acids 491 Because the Fischer esterification is an equilibrium (often with an unfavorable equilibri um constant), clever techniques are often required to achieve good yields of esters. For example, we can use a large excess of the alcohol or the acid. Adding a dehydrating agent removes water (one of the products), dJiving the reaction to the right. There is a more powerful way to form an ester, however, without having to deal with an unfavor able equilibrium. An alcohol reacts with an acid chloride in an exothermic reaction to give an ester. The alcohol groups of unpleasant tasting drugs are often converted to esters in order to mask the taste. In most cases, the ester has a less unpleasant taste than the free alcohol. R -O The mechanisms of these reactions that form acid derivatives are covered with similar mechanisms in Chapter 21. P ROBLEM 1 1 - 3 1 alcohol -EJ + - R' acid chloride o o R-O- -R' ester + Show the alcohol and the acid chloride you would use to make the following esters. 0 o (a) C H3CH2CH2C -OCH2CH2CH, n-propyl butyrate II (c) H3C --< p-tolyl isobutyrate }- o (d) O- - CH(CH3)2 ! ( b) C HiCH2)3-0- C - CH2CH3 II-butyl propionate II C>-O-!--< > o cyclopropyl benzoate In addition to forming esters with carboxylic acids, alcohols form inorganic esters with inorganic acids such as nitric acid, sulfuric acid, and phosphoric acid. In each type of ester, the alkoxy ( -OR) g roup of the alcohol replaces a hydroxyl group of the acid, with loss of water. We have already studied tosylate esters, composed of para-toluenesulfonic acid and alcohols (but made using tosyl chloride, Section 1 1-5). Tosylate esters are analogous to sulfate esters (Section 1 1 - 1 3A), which are composed of su lfuric acid and alcohols. 11-13 Este rs of I n orga n ic Acids R -O -EJ -EJ + L II 0 alcohol 0 para-tollleneslllfonic acid (TsOH) -o _ _ --CH3 ( 0 ) R -O- 0 para-tollleneslllfonate ester (ROTs) _ II -o0 CH3 + I H20 I I H CI I Made using tosyl chloride R -O + @3-- II 0 alcohol 0 para-toillenesllifonyl chloride (TsCI) -o- CH3 pyridine R -O- 0 tosylate ester (ROTs) I -o_ CH3 + 492 Chap ter 11: Reactions of Alcohols 11-13A Su lfate Esters II S-OH II s ulfuric acid o o A sulfate ester i s like a sulfonate ester, except there is no alkyl group directly bonded to the sulfur atom. In an alkyl sulfate ester, alkoxy groups are bonded to sulfur through oxygen atoms. Using methanol as the alcohol, I CH3 -O-S-OH I m ethyl sulfate o o + The body converts the hydroxyl groups of some drugs to their sul fate derivatives in order to produce water-soluble compounds that are readily excreted. The reaction is not as common as it might be because of limited availability of inorganic sulfate in the body. S ulfate ions are excellent leaving groups. Like sulfonate esters, sulfate esters are good electrophiles. Nucleophiles react with sulfate esters to give alkylated products. For example, the reaction of dimethyl sulfate with ammonia gives a sulfate salt of methylamine, CH3NHt CH30S03". H"" I CH3 -O-S-O-CH3 H -N: H/ 10 I ..0.. .. 1 H20 1 CH3 -O- S-O-CH3 Hp d imethyl sulfate o o II II + 0 H H . H-N-CH3 m ethylammonium ion 1+ 1 :O-S-O-CH3 m ethylsulfate ion .. .. "0 . a mmonia dimethyl sulfate ..0. . I II P ROB LEM 1 1 - 3 2 Use resonance structures to show that the negative charge in the methylsulfate anion is shared equally by three oxygen atoms. 11-13 B Nitrate Esters Nitrate esters are formed from alcohols and nitric acid. R -O- Nt" ""0- + a lcohol nitric acid alkyl nitrate ester I H- O - H I . The best known nitrate ester is "nitroglycerine," whose systematic name is glyceryl trinitrate. G lyceryl trinitrate results from the reaction of glycerol ( 1 ,2,3-propanetriol) with three molecules of nitric acid. CH -O-NO2 Z CHz- 0- N02 I llustration of Alfred Nobel operating the apparatus he used to make nitroglycerine. The temperature must be monitored and controlled carefully during this process; therefore, the operator's stool has only one leg to ensure that he stays awake. g lycerol (glycerine) nitric acid g lyceryl trinitrate (nitroglycerine) iI H-0-N02 + 3 H20 1 1 N itroglycerine was first made in 1847 and was found to be a much more pow erful explosive than black powder, which is a physical mixture of potassium nitrate, sulfur, and charcoal. In black powder, potassium n itrate i s the oxidizer, and sulfur and charcoal provide the fuel to be oxidized. The rate of a black powder explosion is 1 1 - 1 3 Esters of I norganic Acids l imited by how fast oxygen from the grains of heated potassium nitrate can diffuse to the grains of sulfur and charcoal. A black powder explosion does its work by the rapid increase in pressure resulting from the reaction. The explosion must be con fined, as in a cannon or a firecracker, to be effective. In nitroglycerine, the nitro groups are the oxidizer and the CH and CH 2 groups are the fuel to be oxidized. This intimate association of fuel and oxidizer allows the explosion to proceed at a much faster rate, forming a shock wave that propagates through the explosive and initiates the reaction. The explosive shock wave can shatter rock or other substances without the need for confinement. Because of its unprecedent ed explosive power, nitroglycerine was called a high explosive. Many other high explo sives have been developed, including picric acid, TNT (trinitrotoluene), PETN (pentaerythritol tetranitrate), and RDX (research department explosive). Nitroglycerine and PETN are nitrate esters. 493 02N OH N02 Nitroglycerin i s used t o relieve angina, a condition where the heart is not receiving enough oxygen. Angina is characterized by severe pain in the chest, often triggered by stress or exercise. Workers with angina at the Nobel dynamite plant discovered this effect. They noticed that their symptoms improved when they went to work. In the mitochondria, nitroglycerin is me tabolized to NO (nitric oxide), which regulates and controls many meta bolic processes. N02 PETN picric acid RDX Pure nitroglycerine is hazardous to make, use, and transport. Alfred Nobel's family were experts at making and using nitroglycerine, yet his brother and several workers were killed by an explosion. In 1 866, Nobel found that nitroglycerine soaks into diatomaceous earth to give a pasty mixture that can be molded into sticks that do not detonate so easily. He called the sticks dynamite and founded the firm Dynamit Nobel, which is still one of the world's leading ammunition and explosives manufac turers. The Nobel prizes are funded from an endowment that originated with Nobel's profits from the dynamite business. 1 1 -1 3C Pho s p h ate Ester s Alkyl phosphates are composed of 1 m ole of phosphoric acid combined with moles of an alcohol. For example, methanol forms three phosphate esters. II P - OH I OH o phosphoric acid + o + CH 3-O -P - OH HzO OH CH} - O-P - OH HzO O - CH} " I mono methyl phosphate + I I o 1 , 2, o r 3 " I o CH3 - O -P-O-CH3 HzO O - CH3 " By controlling the formation of phosphate esters on key proteins, the body is able to regulate many cellular processes. Any disruption of these processes can result in numerous health problems, includ ing cancer, diabetes, and obesity. I dimethyl phosphate trimethyl phosphate Phosphate esters play a central role in biochemistry. Figure 1 1 -3 s hows how phosphate ester linkages compose the backbone of the nucleic acids RNA (ribonucleic acid) and DNA (deoxyribonucleic acid). These nucleic acids, which carry the genetic informa tion in the cell, are discussed in Chapter 2 3. 494 Chapter 11: Reactions of A lcohols P hosphate ester groups bond the individual nucleotides together in DNA. The "base" on each of the nucleotides corresponds to one of the four heteroxyclic bases of DNA (see Section 23-20). Figure 11-3 11-14 R eactions of Alkoxides I n Section 1O-6B, we learned to remove the hydroxyl proton from an alcohol by reduc tion with an "active" metal such as sodium or potassium. This reaction generates a sodium or potassium salt of an alkoxide ion and hydrogen gas. R-O-H R-O-H .. .. + + Na K The reactivity of alcohols toward sodium and potassium decreases in the order: methyl > 1 > 2 > 3. S odium reacts quickly with primary alcohols and some secondary alcohols. Potassium is more reactive than sodium and is commonly used with tertiary alcohols and some secondary alcohols. Some alcohols react sluggishly with both sodium and potassium. In these cases, a useful alternative is sodium hydride, usually in tetrahydrofuran solution. Sodium hydride reacts quickly to form the alkoxide, even with difficult compounds . R-O-H alcohol .. + NaH sodium hydride THF sodium alkoxide hydrogen S odium metal reacts vigorously with simple primary alcohols such as ethanol. "N-::"''''''' KEY MECHANISM 11-6 Step 1: The alkoxide ion is a strong nucleophile as well as a powerful base. Unlike the alcohol itself, the alkoxide ion reacts with primary alkyl halides and tosylates to form ethers. T his general reaction, called the Williamson ether synthesis, is an SN2 dis placement. The alkyl halide (or tosylate) must be primary so that a back-side attack is not hindered. When the alkyl halide is not primary, elimination usually results. This is the most important method for making ethers. + N a (or NaH or K) Form the alkoxide of t he alcohol having the more hindered g roup . R - O: - Na+ +H2 R-O-H + alkoxide ion ! Step 2: T he alkoxide displaces the leaving group of a good SN2 substrate. .. .. (' R - Q -CH2 -R' R -O:,-:Na R' --: _____ CH2 - X 1 1 - 14 Reactions of Alkoxides 495 + NaX alkoxide ion 1: primary halide or tosylate ether EXAMPLE: Synthesis of cyclopentyl ethyl ether Step Form the alkoxide of the alcohol with the more h i n dered gro up . Step 2: The alkoxide displaces the leaving group of a good SN2 substrate. PROBLEM: Why is the cyclohexyl group chosen for the alkoxide and the ethyl group chosen for the halide? Why not use cyclohexyl bromide and sodium ethoxide to make cyclopentyl ethyl ether? O -CH2 -CH3 ------? + Na+ Br- In using the Williamson ether synthesis, one must remember that the alkyl halide (or tosylate) must be a good SN2 substrate, usually primary. In proposing a Williamson synthesis, we usually choose the less hindered alkyl group to be the halide (or tosylate) and the more hindered group to be the a l ko xi de i on . PROBLEM 1 1 - 3 3 A good Williamson synthesis of ethyl methyl ether would be sodium ethoxide melhyl iodide ethyl melhyl ether What is wrong with the following proposed synthesis of ethyl methyl ether? First, ethanol i s treated with acid to protonate the hydroxyl group (making it a good leaving group), and then methoxide is added to displace water. (incorrect synthesis of ethyl methyl ether) PROBLEM 1 1 - 3 4 Show how ethanol and cyclohexanol may b e used to synthesize cyclohexyl ethyl ether (tosylation followed by the Williamson ether synthesis). ( b) Why can't we synthesize this product simply by mixing the two alcohols, adding some sulfuric acid, and heating? PROBLEM 1 1 - 3 5 (a) A student wanted to make (R)-2-ethoxybutane, using the Williamson ether synthesis. He remembered that the Williamson synthesis involves an SN2 displacement, which takes place with inversion of configuration. He ordered a bottle of (S)-2-butanol for his chiral starting material. He also remembered that the SN2 goes best on primary halides and tosylates, so he I n using the Williamson ether synthesis to make R- 0 -R', choose the less hindered al kyl group to serve as the alkyl halide (R' -X), because it will make a better SN2 substrate. Choose the more hindered alkyl group to form the alkoxide (R-O-), because it is less sensitive to steric hindrance in the reaction. R-O+ PROBLEM-SOLVING Hi-ItZ; R'-X ---'> R-O-R' 496 C hapter 1 1 : Reactions of Alcohols made ethyl tosylate and sodium (S)-2-butoxide. After warming these reagents together, he obtained an excellent yield of 2-ethoxybutane. (a) W hat enantiomer of 2-ethoxybutane did he obtain? Explain how this enantiomer results from the SN2 reaction of ethyl tosylate with sodium (S)-2-butoxide. (b) What would have been the best synthesis of (R)-2-ethoxybutane? (c) H ow can this student convert the rest of his bottle of (S)-2-butanol to (R)-2-ethoxybutane? P R O B L E M 11-36 The anions of phenols (phenoxide ions) may be used in the Williamson ether synthesis, espe cially with very reactive alkylating reagents such as dimethyl sulfate. Using phenol, dimethyl sulfate, and other necessary reagents, show how you would synthesize methyl phenyl ether. PRO B LEM-SOLVING STRATEGY Multistep Synthesis Chemists use organic syntheses both to make larger amounts of useful natural compounds and to invent totally new compounds in search of improved properties and biological effects. Synthesis also serves as one of the best methods for developing a firm command of organic chemistry. Planning a practical multistep synthesis requires a working knowledge of the applications and the limitations of a variety of organic reactions. We will often use synthesis problems for reviewing and reinforcing the reactions we have covered. We use a systematic approach to solving multistep synthesis problems, working back ward, in the "retrosynthetic" direction. We begin by studying the target molecule and con sidering what final reactions might be used to create it from simpler intermediate compounds. Most syntheses require comparison of two or more pathways and the interme diates involved. Eventually, this retrosynthetic analysis should lead back to starting materi als that are readily available or meet the requirements defined in the problem. We can now extend our systematic analysis to problems involving alcohols and the Grignard reaction. As examples, we consider the syntheses of an acyclic diol and a disubsti tuted cyclohexane, concentrating on the crucial steps that assemble the carbon skeletons and generate the final functional groups. SAMPLE PROBLEM O ur first problem is to synthesize 3-ethyl-2,3-pentanediol from compounds containing no more than three carbon atoms. CH3 -CH-C-CH 2 -CH OH OH 3-ethyl-2,3-pentanediol I CH,CH3 II 3 1. Review the functional groups and carbon skeleton of the target compound. The compound is a vicinal diol (glycol) containing seven carbon atoms. Glycols are com monly made by hydroxylation of alkenes, and this glycol would be made by hydroxyla tion of 3-ethyl-2-pentene, which effectively becomes the target compound. CH2-CH3 I CH3- CH=C-CH2 -CH3 3 -ethyl-2-pentene KMn04 cold, dilute (or other methods) CH2-CH3 I CH3-CH-C-CH2-CH3 I I OH OH 3 -ethyl-2,3-pentanediol 2. Review the functional groups and carbon skeletons of the starting materials (if spec ified), and see how their skeletons might fit together into the target compound. The limitation is that the starting materials must contain no more than three carbon atoms. To form a 7-carbon product requires at least three fragments, probably a 3-carbon fragment and two 2-carbon fragments. A functional group that can be converted to an 1 1 - 1 4 Reactions of Alkoxides alkene will be needed on either C2 or C3 of the chain, since 3-ethyl-2-pentene has a dou ble bond between C2 and C3. CH3-CH=C-CH2-CH3 CH?-CH3 497 3. Compare methods for assembling the carbon skeleton of the target compound. Which I- ones provide a key intermediate with the correct carbon skeleton and functional groups correctly positioned for conversion to the functionality in the target molecule? At this point, the Grignard reaction is our most powerful method for assembling a carbon skeleton, and Grignards can be used to make primary, secondary, and tertiary alcohols (Section 1 0-9). The secondary alcohoI3-ethyl-2-pentanol has its functional group on C2, and the tertiary alcohol 3-ethyl-3-pentanol has it on C3. Either of these alcohols can be synthesized by an appropriate Grignard reaction, but 3-ethyl-2-pentanol may dehydrate to give a mixture of products. Because of its symmetry, 3-ethyl-3-pentanol dehydrates to give only the desired alkene, 3-ethyl-2-pentene. It also dehydrates more easily because it is a tertiary alcohol. CH2-CH3 CH3-CH-CH-CH2-CH3 OH I I 3-ethyl-2-pentanol ! CH3-CH= -CH2-CH3 CH2-CH3 (major) 3-ethyl-2-pentene 2: + CH3=CH-C-CH2-CH3 CH2-CH3 (minor) 3-ethyl-l-pentene I PrefelTed synthesis: CH3-CH2- -CH2-CH3 CH2-CH3 r 3-ethyl-3-pentanol 4. Working backward through as many steps as necessary, compare methods for syn OH I H2S04 CH3-CH2=C-CH2-CH3 CH2-CH3 (only product) 3-ethy 1-2-pentene I thesizing the reactants needed for assembly of the key intermediate. (This process may require writing several possible reaction sequences and evaluating them, keep ing in mind the specified starting materials.) The key intermediate, 3-ethyl-3-pentanol, is simply methanol substituted by three ethyl groups. The last step in its synthesis must add an ethyl group. Addition of ethyl magne sium bromide to 3-pentanone gives 3-ethyl-3-pentanol. CH3-CH2-C-CH2 -CH3 o 3-pentanone II ( 1 ) CH3CH2-MgBr (2) H30+ ) CH3-CH2-C-CH2-CH3 OH CH2-CH3 I I 3-ethyl-3-pentanol The synthesis of 3-pentanone from a three-carbon fragment and a two-carbon fragment requires several steps (see Problem 11-37). Perhaps there is a better alternative, consider ing that the key intermediate has three ethyl groups on a carbinol carbon atom. Two sim ilar alkyl groups can be added in one Grignard reaction with an acid chloride or an ester (Section 1 O-9D). Addition of 2 moles of ethyl magnesium bromide to a three-carbon acid chloride gives 3-ethyl-3-pentanol. CH3-CH2-C-CI o II ( 1) 2 CH3CH2-MgBr (2) H30+ CH3-CH2- -CH2-CH3 OH propionyl chloride 3-ethyl-3-pentanoI ( Continued) T CH2-CH3 I 4 98 C hapter 1 1 : Reactions of Alcohols 5. Summarize the complete synthesis in the forward direction, including all steps and all reagents, and check it for errors and omissions. propionyl chloride CH3 -CT-f2-C-CH2-CH3 " CH?-CH3 I - 3-ethyl-3-pentanol KMn04 (cold, dilute) OH I ) C H3-CH -C-CH2 -CH3 CH?-CH3 I - 3 -ethyl - 2- pentene P ROBLEM 1 1 - 3 7 3-ethyl - 2,3-pentanediol OH I OH I To practice working through the early parts of a multistep synthesis, devise syntheses of (a) 3 -ethyl-2-pentanol from compounds containing no more than three carbon atoms. (b) 3-pentanone from alcohols containing no more than three carbon atoms. SAMPLE PROBLEM As another example of the systematic approach to multistep synthesis, let's consider the synthesis of l-bromo-2-methylcyclohexane from cyclohexanol. 2. Review the functional groups and carbon skeletons of the starting materials (if specified), and see how their skeletons might fit together into the target compound. 1. Review the functional groups and carbon skeleton of the target compound. The skeleton has seven carbon atoms: a cyclohexyl ring with a methyl group. It is an alkyl bromide, with the bromine atom on a ring carbon one atom removed from the methyl group. G H OH Ct H H3 Br 3. Compare methods for assembling the carbon skeleton of the target compound to determine which methods provide a key intermediate with the correct carbon skele ton and functional groups at the correct positions for being converted to the func tionality in the target molecule. The stru.ting compound has only six carbon atoms; clearly, the methyl group must be added, presumably at the functional group. There are no restrictions on the methylating reagent, but it must provide a product with a functional group that can be converted to an adjacent halide. 6 CH3MgBr O nce again, the best choice is a Grignard reaction, but there are two possible reactions that give the methylcyclohexane skeleton. A cycJohexyl Grignard reagent can add to formaldehyde, or a methyl Grignard reagent can add to cyclohexanone. (There ru.-e other possibilities, but none that are more direct.) + H H / '" ( 1 ) ether C=O (2) H3O+ ) + (10 (1) et her ) (2) Hp+ Neither product has its alcohol functional group on the carbon atom that is functionalized in the target compound. Alcohol C needs its functional group moved two carbon atoms, cf Ct CH?OH H to '" "m'tiom""'" alcohol C OH to be function ali zed alcohol D 1 1 -14 Reactions of Alkoxides but alcohol D needs it moved only one carbon atom. Converting alcohol D to an alkene functionalizes the correct carbon atom. Anti-Markovnikov addition of HBr converts the alkene to an alkyl halide with the bromine atom on the correct carbon atom. ROOR 499 HEr ) rf 4. Working backward through as many steps as necessary, compare methods for syn alcohol D target compound Br H CH3 thesizing the reactants needed for assembly of the key intermediate. All that remains is to make cyc1ohexanone by oxidation of cyciohexanol. OH 5. Summarize the complete synthesis in the forward direction, including all steps and all reagents, and check it for errors and omissions. (t H ROOR HEr ) riH Problem 11-38 provides practice intermediates. PROBLEM 11-38: 111 m ultistep syntheses and using alcohols as VBr CH3 Develop syntheses for the following compounds. As starting materials, you may use cyc1opentanol, alcohols containing no more than four carbon atoms, and any common reagents and solvents. (a) trans-cyciopentane-l,2-diol (b) l -chloro-l-ethylcyciopentane OCH2CH3 (d) (e) C 1. (,) i:3 SUMMARY =="\1 o CH'CH3 Reactions of Alcohols O xidation-reduction reactions a. Oxidation of secondary alcohols to ketones (Section 11-2A) OH R -CH-R' I NazCrz07, H2S04 ) II R - C - R' o o Example OH CH3 -CH -CH2CH3 2-butanol Na2Cr207, H2S04 I ) CH3-C -CH2CH3 2-butanone II ( Continued) 5 00 b. Chapter 11: Reactions of Alcohols Oxidation of primary alcohols to carboxylic acids (Section 11-2B) R-CH2-OH Na2Cr207, H2S04 l o R-C-OH II Example CH3(CH2)4-CH2-0H I-hexanol o CH3(CH2)4-C-OH II c. Oxidation of primary alcohols to aldehydes (Section 11-2B) R-CH2-OH Example I-hexanol PCC II R-C-H o hexanoic acid hexanal d. Reduction of alcohols to alkanes (Section 1 1 -6) R-OH (1) T sCl / pyridine ) R-H (2) LiAIH4 Example (1) cycIohexanol (2) LiAl H 4 TsCI/pyridine ) 2. cycIohexane a. Cleavage of the alcohol hydroxyl group -C+O-H Conversion of alcohols to alkyl halides (Sections 11-7 through 11-9) --=-R-OH ---------' ---->l R-Cl R-OH -----'---l--> R-Br HCl or SOCI R-OH - ------->l R - I - --=- -'HI or HBror PBr3 2ipyridine piI 2 Examples (CH3hC-OH t-butyl alcohol (CH3hCH - CH2OH isobutyl alcohol PBr3 HCl (CH3hC-CI t-butyl chloride ) ) (CH3hCH-CH2Br isobutyl bromide CH3(CH2)4-CH20H I-hexanol b. piI2 CH3(CH2)4 -CH21 l-iodohexane Dehydration of alcohols to form alkenes (Section I I - I DA) H OH H2 S04 or H3P04 II ) -C-C- I I / " C=C " / + HzO Example cycIohexanol cycIohexene o c. Industrial dehydration of alcohols to form ethers (Section 11-lOB) 2 CH3CH20H ethanol 2 R-OH __ 1 1-14 Reactions of Alkoxides 501 + Example CH3CH 2-O-CH 2CH 3 + H 20 diethyl ether R -O-R + H 20 3. I Cleavage of the hydroxyl proton -C -O+ H I a. Tosylation (Section 11-5) + ' R-OH CI--o-CH3 alcohol tosyl chloride (TsCl) _ o o II pyridine) Example b. Acylation to form esters (Section 11-12) ( CH 3hCH -OH isopropyl alcohol -----';) TsCl / pyridine R-O--o-CH3 HCI ' o alkyl tosylate _ o II + (CH 3hCH -OTs isopropyl tosylate R-OH Example cyclohexanol R-OH + N a (orK) (acyl chloride) II R'-C-Cl o ) (acetyl chloride) c. Deprotonation to form an alkoxide (Section 11-14) R-OH + N aH ('I<O-C-CH3 HCI H V o R-O-C-R' HCI ester o II + II cyclohexyl acetate + R-O- +Na R -O- a + + 4 H 2! H 2! Example d. Williamson ether synthesis (Sections 11-14 and 14-5) CH 3 -CH 2 -OH + Na ethanol Na+-O -CH 2 -CH3 sodium ethoxide Example Na+-O - CH2CH 3 + C H31 sodium ethoxide methyl iodide R-O- + R' X R-O -R' + X (R' must be primary ) CH3CH2-O -CH 3 ethyl methyl ether + NaI 502 C hapter 1 1: Reactions of Alcohols alcohol dehydrogenase (ADH) Chapter 11 Glossary An enzyme used by living cells to oxidize ethyl alcohol to acetaldehyde. (p. 467) alkoxide ion The anion formed by deprotonating an alcohol. (p. 494) R-O-H + Na The solution formed by adding sodium or potassium dichromate (and a small amount of water) to concentrated sulfuric acid. (p. 462) chromic acid test: When a primary or secondary alcohol is warmed with the chromic acid reagent, the orange color changes to green or blue. A nonoxidizable compound (such as a tertiary alcohol, a ketone, or an alkane) produces no color change. (p. 464) Collins reagent (Cr03' 2 pyridine) A complex of chromium trioxide with pyridine, used to oxidize primary alcohols selectively to aldehydes. (p. 465) ester A n acid derivative formed by the reaction of an acid with an alcohol with loss of water. The most common esters are carboxylic esters, composed of carboxylic acids and alcohols. (p. 490) II R -O - C - R' carboxylic ester o chromic acid reagent Fischer esterification: The acid-catalyzed reaction of an alcohol with a carboxylic acid to give an ester. (p. 490) o R -C a cid ether glycol II + O-R' alcohol o R-C-O-R' ester II + I H 20 I A compound containing an oxygen atom bonded to two alkyl or aryl groups. (p. 494) S ynonymous with dio!. The term "glycol" is most commonly applied to the 1,2-diols, also called vicinal diols. (p. 488) inorganic esters C ompounds derived from alcohols and inorganic acids with loss of water. (p. 49 1 ) Examples are o R-O-S-R II o s ulfonate esters Jones reagent II II R -O-S-O - R II o sulfate esters nitrate esters o II R -O-P-O-R o O-R phosphate esters I A s olution of dilute chromic acid dissolved in acetone, used for alcohol oxida tions. (p. 465). Lucas test A test used to determine whether an alcohol is primary, secondary, or tertiary. The test measures the rate of reaction with the Lucas reagent, ZnCl2 in concentrated HC!. Tertiary alcohols react fast, secondary alcohols react more slowly, and primary alcohols react very slow ly. (p. 474). nicotinamide adenine dinucleotide (NAD) A b iological oxidizing/reducing reagent that op erates in conjunction with enzymes such as alcohol dehydrogenase. (p. 467). oxidation Loss of H2; addition of 0 or O2; addition of X2 (halogens). Alternatively, an increase in the number of bonds to oxygen or halogens or a decrease in the number of bonds to hydrogen. (p. 460). pinacol rearrangement Dehydration of a glycol in which one of the groups migrates to give a ketone. (p. 488). pyridinium chlorochromate (PCC) A c omplex of chromium trioxide with pyridine and HC!. P CC oxidizes primary alcohols to aldehydes. (p. 464). reduction Addition of H2 (or H-); loss of 0 or O2; loss of X2 (halogens). Alternatively, a reduction in the number of bonds to oxygen or halogens or an increase in the number of bonds to hydrogen. (p. 460). Swern oxidation A m ild oxidation, using DMSO and oxalyl chloride, that can oxidize primary alcohols to aldehydes and secondary alcohols to ketones. (p. 465) tosylate ester An ester of an alcohol with para-toluenesulfonic acid. Like halide ions, the tosylate anion is an excellent leaving group. (p. 469) Williamson ether synthesis The SN2 reaction between an alkoxide ion and a primary alkyl halide or tosylate. The product is an ether. (p. 494) Study Problems 503 R - O : v R'- X .. v R -O-R' + X- I Essential Problem-Solving Skills in Chapter 1 1 1. 2. Identify whether oxidation or reduction is needed to interconvert alkanes, alcohols, aldehydes, ketones, and acids, and identify reagents that will accomplish the conversion. Predict the products of the reactions of alcohols with (a) O xidizing and reducing agents. ( b) Carboxylic acids and acid chlorides. ( c) D ehydrating reagents, especially H2S04 and H3P04. (d) Inorganic acids. ( e) S odium, potassium, and sodium hydride. Predict the products of reactions of alkoxide ions. Propose chemical tests to distinguish alcohols from the other types of compounds we have studied. Use your knowledge of alcohol and diol reactions to propose mechanisms and products of similar reactions you have never seen before. Show how to convert an alcohol to a related compound with a different functional group. Predict the products of pinacol realTangement and periodate cleavage of glycols. Use retrosynthetic analysis to propose effective single-step and multistep syntheses of compounds using alcohols as intermediates (especially those using Grignard and organolithium reagents to assemble the carbon skeletons). 3. 4. s. 6. 7. 8. Study Problems 11-39 11-40 11-41 Briefly define each term, and give an example. ( a) o xidation ( b) reduction (c) chromic acid oxidation (d) PCC oxidation ( e) ether (f) Williamson ether synthesis (g) alkoxide ion (h) carboxylic ester (i) F ischer esterification U) tosylate ester (k) alkyl phosphate ester (I) alkyl nitrate ester (m) Lucas test (n) p inacol realTangement In each case, show how you would synthesize the chloride, bromide, and iodide from the corresponding alcohol. (b) halocyclopentane (a) l -halobutane (halo = c hloro, bromo, iodo) (c) I -halo- l -methylcyclohexane ( d) I -halo-2-methylcyclohexane Predict the major products of the following reactions, including stereochemistry where appropriate. ( b) ( S)-2-butyl tosylate + NaBr ( a) (R)-2-butanol + TsCI in pyridine (d) c yclopentylmethanol + Cr03 . pyridine HCl (c) c yclooctanol + Cr03/H2S04 ( e) c yclopentylmethanol + Na20'207/H2S04 (f) cyclopentanol + HCl/ZnCI2 (g) n-butanol + H Br (h) c yclooctylmethanol + CH3CH2MgBr (i) p otassium t-butoxide + methyl iodide U) sodium methoxide + I-butyl iodide (I) product from (k) + O s04/H202, then HI04 (k) c yclopentanol + H 2S04/heat (m) sodium ethoxide + l -bromobutane (n) sodium ethoxide + 2 -methyl-2-bromobutane 504 11-42 C hapter 1 1 : Reactions of Alcohols Show how you would accomplish the following synthetic conversions. (a) 11-43 11-44 Predict the major products of dehydration catalyzed by sulfuric acid. (a) I-hexanol (b) 2-hexanol (d) I -methylcyclopentanol (e) cyclopentylmethanol (c) o- oBr OH H -----'> H -----'> d d 0 H'CH' (b) 2 II H CH,CH,CH, (d) a I CH,B'. -----'> OH Predict the esterification products of the following acid/alcohol pairs. o(b) CH20H H OH (f) 2-methylcyclopentanol CH3 -C -OH (c) 3-pentanol 0 + CHCH,CH, I 0 II 11-45 Show how you would make the methanesulfonate ester of cyclohexanol, beginning with cyclohexanol and an appropriate acid chloride. o cyclohexyl methanesulfonate: 11-46 11-47 Show how you would convert (S)-2-hexanol to (b) ( R)-2-bromohexane (a) ( S)-2-chlorohexane When l -cyclohexylethanol is treated with concentrated aqueous HBr, the major product is I -bromo-l-ethylcyclohexane. S-CH3 ('f- O-II V H ( c) ( R)-2-hexanol (a) Give a mechanism for this reaction. (b) How would you convert l -cyclohexylethanol to ( l-bromoethyl)cyclohexane in good yield? ~ ------? HBr Hp 11-48 Show how you would make each compound, beginning with an alcohol of your choice. (a) d HO if ? if Br ( c ) (Y 0CH3 V C -OH LJ II (f) Study Problems 505 o (g) (i) H H ,r\; C H3 11-49 11-50 11-51 Predict the major products (including stereochemistry) when cis-3-methylcyclohexanol reacts with the following reagents, (c) Lucas reagent (b) SOCl2 (a) PBr3 (e) TsClIpyridine, then NaBr (d) concentrated HBr Show how you would use simple chemical tests to distinguish between the following pairs of compounds, In each case, describe what you would do and what you would observe, (a) I -butanol and 2-butanol (b) 2-butanol and 2-methyl-2-butanol (c) cyclohexanol and cyclohexene (d) cyclohexanol and cyclohexanone (e) cyclohexanone and l-methylcyclohexanol Write the important resonance structures of the following anions, H", ;" OTs I 11-52 Compound A is an optically active alcohoL Treatment with chromjc acid converts A into a ketone, B. In a separate reac tion, A is treated with PBr3, converting A into compound C. Compound C is purified, and then it is allowed to react with magnesium in ether to give a Grignard reagent, D, Compound B is added to the resulting solution of the Grignard reagent After hydrolysis of the initial product (E), this solution is found to contain 3,4-dimethyl-3-hexanoL Propose structures for compounds A, B, C, D, and K A C Mg, ether - - ,-;. D (Grignard reagent) ""--- ---- (a) H /C=C'" /H 0- (b) 0 c6 I (c) -O -S-CH3 o II o II 11-53 Give the structures of the intermediates and products V through Z, NCrp7' H2S04 ) B }----. E 3,4-dimethyl-3-hexanol Mg, ether , cyclopelltanol x y Z 11-54 Under acid catalysis, tetrahydrofurfuryl alcohol reacts to give surprisingly good yields of dihydropyran, Propose a mecha nism to explain this useful synthesis, V 11-55 Propose mechanisms for the following reactions, In most cases, more products are formed than are shown here, You only need to explain the formation of the products shown, however. tetrahydrofurfuryl alcohol dihydropyran HCI, ZnClz ) (a rrunor product) O CI o o 506 C hapter 1 1: Reactions of Alcohols 11-56 Show how you would synthesize the following compounds. As starting materials, you may use any alcohols containing four or fewer carbon atoms, cyclohexanol, and any necessary solvents and inorganic reagents. ) en OH OH H', CO CO + + u: OH (') U OH 11-57 Show how you would synthesize the following compound. As starting materials, you may use any alcohols containing five or fewer carbon atoms and any necessary solvents and inorganic reagents. 11-58 The following pseudo-syntheses (guaranteed not to work) exemplify a common conceptual error. heat 11-59 Two unknowns, X and Y, both having the molecular formula C4HsO, give the following results with four chemical tests. Propose structures for X and Y consistent with this information. Compound X Compound Y decolorizes no reaction Bromine (a) What is the conceptual error implicit in these syntheses? (b) Propose syntheses that are more likely to succeed. r-'H?S04 ) OH 11-60 The Williamson ether synthesis involves the displacement of an alkyl halide or tosylate by an alkoxide ion. Would the synthesis shown be possible by making a tosylate and displacing it? If so, show the sequence of reactions. If not, explain why not and show an alternative synthesis that would be more likely to work. bubbles no reaction Na Metal orange to green no reaction Chromic Acid Lucas Reagent no reaction no reaction OH make the tosylate and displace? ) OCH3 *11-61 C hromic acid oxidation of an alcohol (Section 1 1 -2A) occurs in two steps: formation of the chromate ester, followed by an elimination of H+ and chromium. Which step do you expect to be rate-limiting? Careful kinetic studies have shown that Compound A undergoes chromic acid oxidation over I a times as fast as Compound B. Explain this large difference in rates. S tudy Problems 5 07 *11-62 *11-63 Alcohols combine with ketones and aldehydes to form interesting derivatives, which we will discuss in Chapter 1 8. The following reactions show the hydrolysis of two such derivatives. Propose mechanisms for these reactions. Many alcohols undergo dehydration at aoc when treated with phosphorus oxychloride ( POCI 3 ) in the basic solvent pyri dine. (Phosphorus oxychloride is the acid chloride of phosphoric acid, with chlorine atoms in place of the hydroxyl groups of phosphoric acid.) ( Propose a mechanism for the dehydration of cyclopentanol using POCl3 and pyridine. The first half of the mecha a) nism, formation of a dichlorophosphate ester, is similar to the first half of the mechanism of reaction of an alcohol with thionyl chloride. Like a tosylate, the dichlorophosphate group is a good leaving group. The second half of the mechanism might be either first order or second order; draw both alternatives for now. (b) When trans-2-methylcyclopentanol undergoes dehydration using POCl3 in pyridine, the major product is 3-methyl cyclopentene, and not the Zaitsev product. What is the stereochemistry of the dehydration? What does this stereo chemistry imply about the correct mechanism in part (a)? Explain your reasoning. OH ~ H Compound A (H2Cr04 (slower) HO H Compound B ( a) (b) C) CXJ OCH, :, , HO f) H + C HpH

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Whitman - CHEM - 140
I nfra red S pectrosco py a n d M a ss S pectrometry12fixed mirrorsamplelaser calibration beam 12-1detectorI n trod ucti o nOne of the most important tasks of organic chemistry is the determination of organic structures. When an interesting compou
Whitman - CHEM - 140
13N uclea r M ag n etic Reso n a n ce S pectrosco pyH( \ \ Ji nduced magnetic fieldcirculationHNuclear magnetic resonance spectroscopy (NMR) i s the most powerful tool available for organic structure determination. Like IR spectroscopy, NMR can be
Whitman - CHEM - 140
E th ers, E poxi d es, a nd S u lfidesl S-crown-6 with K+ s olvated14or aryl (benzene ring) groups. Like alcohols, ethers are related to water, with alkyl groups replacing the hydrogen atoms. In an alcohol, one hydrogen atom of water is replaced by an
Whitman - CHEM - 140
Conjugated Systems, Orbital Symmetry, and Ultraviolet SpectroscopyDouble bonds can interact with each other if they are separated by just one single bond. Such interacting double bonds are said to be conjugated. D ouble bonds with two or more single bond
Whitman - CHEM - 140
Aromatic Compounds16H -,. .In 1 825, Michael Faraday isolated a pure compound of boiling point 80C from the oily mixture that condensed from illuminating gas, the fuel burned in gaslights. Elemental analysis showed an unusually small hydrogen-to-carbon
Whitman - CHEM - 140
Reactions of Aromatic CompoundsEPM17of anisoleWith an understanding of the properties that make a compound aromatic, we now consider the reactions of aromatic compounds. A large part of this chapter is devoted to e lectrophilic aromatic substitution,
Whitman - CHEM - 140
18Ketones and AldehydesWheywiarle stofucentcompoundsance taoinorganithce chemistry, biochemistry, andin biololgy.because dy i cont ing C tofethe common rtyalpesmportcarbonyl compounds are l isted in Table detai Some of Csarbonylare constituentareofevery
Whitman - CHEM - 140
A m i n eselectrostatic potential map of trimethylamine191 9- 1 Introd uction( S)-coni inebonded toarecompounds. atom. As serve many functions orsome ofalkyl mostarylsuch as theorganic derivatives of class, arnines includein more organisms,important
Whitman - CHEM - 140
20C a rboxyl ic Aci d sThe combination of a carbonyl group and a hydroxyl on the same carbon atom is called a carboxyl group. Compounds containing the carboxyl group are distinctly acidic and are called c arboxylic acids. -C -O- Hcarboxyl group2 0- 1
Whitman - CHEM - 140
Whitman - CHEM - 140
nH aJHUa1d-NOS \f3d9-2Lh -E -O N8SI
Whitman - CHEM - 140
LUT ONS MANUALCalifornia Polytechnic State UniversityJan William SimekORC CHEMISTRYSIXTH EDITIONL. G. Wade, Jr.&quot;.JL.-':'Prentice HallUpper Saddle River, NJ07458
Whitman - CHEM - 140
Assistant Editor: Carole Snyder Project Manager: Kristen Kaiser Executive Editor: Nicole Folchetti Executive Managing Editor: Kathleen Schiaparelli Assistant Managing Editor: Becca Richter Production Editor: Kathryn O'Neill Supplement Cover Manager: Paul
Whitman - CHEM - 140
PrefaceTABLE OF CONTENTS.vSymbols and Abbreviations . viiChapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chap
Whitman - CHEM - 140
PREFACE Hints for Passing Organic ChemistryDo you want to pass your course in organic chemistry? Here is my best advice, based on over thirty years of observing students learning organic chemistry: Hint #1: Do the problems. It seems straightforward, but
Whitman - CHEM - 140
Some Web Stuff Prentice-Hall maintains a w eb site dedicated to the Wade text: try www.prenhall.com/wade. Two essential web sites providing spectra are listed on the bottom of p. 270. Acknowledgments No project of this scope is ever done alone. These are
Whitman - CHEM - 140
SYMBOLS AND ABBREVIA TIONSB elow is a list of symbols and abbreviations used in this Solutions Manual, consistent with those used in the textbook by Wade. (Do not expect all of these to make sense to you now. You will learn them throughout your study of
Whitman - CHEM - 140
Symbols and Abbreviations, continued SUBSTITUENT GROUPS, continued Ac Cy Ts an acetyl group: a cyclohexyl group: CH3 - C oII0-o II B oc a t-butoxycarbonyl group (amino acid and peptide chemistry): (CH3)3C -0 - C IItosyl, or p-toluenesulfonyl group: CH
Whitman - CHEM - 140
Symbols and Abbreviations, c ontinued REAGENTS AND SOLVENTS, c ontinued pyridinium chlorochromate, Cr03 HCI N PCCS ia2BHdisiamylboraneC H3 H H H CH 3 I I I I I H-C-C-B-C-C-H I I I I C H3 CH 3 CH3 CH3 ')THFtetrahydrofuranS PECTROSCOPY infrared spect
Whitman - CHEM - 140
C HAPTER I-INTRODUCTION A ND REVIEW1-11-2P 1 s 2 2s 2 2p6 3s 2 3px 1 3py 1 3pz 1 1 s 2 2s2 2p6 3s 1 S 1 s 2 2s 2 2p6 3s 2 3px 2 3py 13pz 1 Mg 1 s 2 2s2 2p6 3s 2 Is 2 2s2 2p6 3s 2 3px 1 CI 1 s 2 2s2 2p6 3s 2 3px2 3py2 3pz 1 AI Ar 1 s 2 2s 2 2p6 3s 2 3px
Whitman - CHEM - 140
1-5The symbols &quot; 8+&quot; and &quot; 8-&quot; indicate bond polarity by showing partial charge. (In the arrow symbolism, the arrow should point to the partial negative charge.) (a) C -C 1(b) (g)C- O8+(c) (h)C-N N-S8-N-O8-8+(i) N - B8-8+1-6Non-zero forma
Whitman - CHEM - 140
1-7continued. .(c) :O-N=O (d) (e)I I+O=N-O: .+.-H-C=C-C-HHHHI. H-C-C=C-HHHHIIIH - C = C - C - H . . H - C - C=C - H I I I I I I HHH HHH (f) Sulfur can have up to 12 e lectrons around it because it has d orbitals accessible . :0 : : 0: :
Whitman - CHEM - 140
1 -8 continued H H H + + + \+ \ \ C - C = N - O: C = C - N - O: C=C - N=O (b) / / / 1 1 1 11 1 1 H H H H :0 : H :0 : H :0 : minor major major These two forms have equivalent energy and are major because they have full octets, more bonds, and less charge s
Whitman - CHEM - 140
1 -8 continued :0 : II (h) H - C - N - HI.II-I . . .. :0 : + I H - C =N H-I(no charge separation)majorHm inorH1 -9 Your Lewis structures may appear different from these. As long as the atoms are connected in the same order and by the same type
Whitman - CHEM - 140
1 -10 continued(e )(g)H H :0 : H H H (h) :0 : H I II I I II I I I H ' /C:-., /,C-C - H H-C - C-C-C-C-H c &quot;c I I I I I I H II H HHH C C, H/ , C 'i H I H 1 -1 1 There is often more than one correct way to write condensed structural formulas. You must oft
Whitman - CHEM - 140
1 - 1 2 continued (b) 32.0 g C 1 2.0 g/mole 2 .67 moles C 1.34 moles 1.99 2 C 6.67 g H 5H 4 .93 6 .60 m oles H -:- 1.34 m oles l.01 g/mole IS.7 g N 1 4.0 g/mole 1 .34 moles N -:- 1.34 moles 1 N 4 2.6 g a . 1 6.0 g/mole - 2.66 moles a -;- 1.34 moles 1 .99
Whitman - CHEM - 140
1 -13 1 mole HEr (a) 5 .00 g HBr x 80.9 g HBr 0.06 1 8 moles HEr0.0618 moles=0 .06 1 8 moles HBr0.06 1 8 moles H 30 + (100% dissociated) x=1 00 mL-H30 +10001LmL=0.6 1 8 moles H30 + 1 L solution=pH=10gIO [H30+]-logl o (0.6 1 8)(b)0 .03
Whitman - CHEM - 140
1-15 (a) HCOOH stronger acid pKa 3.76 (b) CH3COOweaker base (c) CH30H stronger acid pKa 1 5.5+-CN stronger base+-.-HCOOweaker base+H CN FAVORS weaker PRODUCTS acid pK a 9.22+C H30H weaker acid pK a 1 5 .5 NaNH2 stronger base.-CH 3 COOH strong
Whitman - CHEM - 140
1 -l7 In Solved Problem 1-4, the structures of ethanol and methylamine are shown to be similar to methanol and ammonia, respectively. We must infer that their acid-base properties are also similar. (a) This problem can be viewed in two ways. 1 ) Quantitat
Whitman - CHEM - 140
1 -1 8 continued (d) - . 0 Na+ :O - H + H-S-H . stronger acid stronger base (e) H /' I CH3 - H + C H3-O: stronger base H stronger acid+ &quot; _ equilibrium favors PRODUCTS _-. H - O - H + Na+ : S-H conjugate acid conjugate base weaker base weaker acid CH3
Whitman - CHEM - 140
1 - 19 continued (e) Bronsted-Lowry-c-proton transferC H - C - C - H +/.- H = H -. - C - H .- H - b = b - H :O - ) k base acid. . . :(f)C H3 - - H H base.+C H3 - S=!: acid()i:H :0:+:Cl :..-_+H - O-H.1 -20 Learning organic chemistry
Whitman - CHEM - 140
1 -24Cl: N (b) P (a) N . . . p . : C l I C l: I I c : . .!'&quot; I C l: :Cl: :Cl: :Cl: :CI: C ANNOT EXIST NCls violates the octet rule; nitrogen can have no more than eight electrons (or four atoms) around it. Phosphorus, a third-row element, can have more t
Whitman - CHEM - 140
1-26 continued H - O: H :0:. .( c)H-C = C-C-C-C-O-H HIIIIIHIHHII1 -27 In each set below, the second structure is a more correct line formula. Since chemists are human (surprise!), they will take shortcuts where possible; the first structure i
Whitman - CHEM - 140
1 -28 continued (c) There are several other possibilities as well. Your answer may be correct even if it does not appear here. Check with others in your study group. HH H HHH HHH I I I I I I I I I :O - C-C - O - C - H :O - C - C-C - O: :O-C-C - C - H I I
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HH (b) H . _ _H \ HI H . C ,0 0/ C &quot; / N . H H I HH H H :0: III&quot; H - N - C-C-C-C-O - H IIII HHHH(e)H H (g) I \ C-C \ :0: H, &quot; II \ H - C-C C-S=O / \ / I H C = C :0: I \ I HH HHH \I /C, ,.H C, :N II H I H H\ C I H / H - C , C - H H HH I I,. I \ CH H -
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1 -34 Non-zero formal charges are shown by the atoms. + + (a) H - C = N = N : H - C - N - N:H ,H/ H 1 (b) H C 1 1 1 , H H H - C -+N - O: / H H I H H C I 1 + + 1 (c) H - C = C-C - H (d) H - C-N=O (e) H - C-O - C - H H /1' H H 1 I +1 111 11 HCH HHH H : 0:
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1 -37 continued (c)&lt;&gt;C H24(d) (e)0+ . 0-0 &lt;&gt;+0 /; :4+ O. .CH,-+C H2(f)(g) (h)(i) C H 3 - C = C - C = C - C - CH3 I I I I I HHHHHQ:0 :&lt; &gt; : -o + (&gt; CN- - CN+-H + 0 . 0 0+0. .0/ +CH2+()C H,0 .-/--:0=0 .-H4H.0+. o ._
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1-38 . (a) O=S-O:+.:O-S=O++...O=S=O.(b) 0=0-0:+.(c) The last resonance form of S0 2 has no equivalent form in 03. Sulfur, a third row element, can have more than eight electrons around it because of d orbitals, whereas oxygen, a second row e
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1 -40 continued (c) :0 : :0: II -. . II C H3 - C - ? - C - CH3 minorH. : 0:. .1-1. . .C H3 - C = C - C - CH3I:0:\ -majorHIII:0:.1-1. . ..C H3 - C - C = C - CH3II. :0:n egative charge on electronegative atoms-equal energy (d)+ + -+ C H3
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1 -4 1 continued (c) H - C - CH3 \ Hno resonance stabi li zati onH - C - C - N: \ H..H - C = C = N: \ H ./ CH c' 2 \H C .:/+more stable-reson ance stabi l i zedmore stable-resonance stabilized (e) CH3 - N - CH 3 1 CH3 - C - CH 3+ .I-I . . - .CC
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1 -45 The newly formed bond is shown in bold. . ('. . CH 3 - - CH3 + (a) CH 3 - .O: CH 3-CI. : . . electrophile nucleophile Lewis acid Lewis base (b) CH3 - O - CH 3 + H - O - H +1 ) H 3 c ucleoPhile Lewis base e I ectrophIO Ie Lewis acid (c) CH 3 - ? - CH
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(h)1 -45 continu F-B - F I F e lectrophile Lewis acidCH2 = CH2 nucleophile Lewis baseF + -I F - f - CH2 - CH2 F+( i) B F3 - CH2 - CH2 electrophile Lewis acid 1 -46 (a) H2S04++C H2 = CH2 nucleophile Lewis base. B F3 - CH2 - CH2 - CH2 - CH2+C H3C
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1 -48 From the amounts of CO2 and H20 generated, the milligrams of C and H in the original sample can be determined, thus giving by difference the amount of oxygen in the 5.00 mg sample. From these values, the empirical formula and empirical weight can be
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C HAPTER 2-STRUCTURE AND PROPERTIES OF ORGANIC MOLECULES2 - 1 The fundamental principle of organic chemistry is that a molecule's chemical and physical properties depend on the molecule's structure: the structure-function or structure-reactivity correlat
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2 -5 (a) linear, bond angle 1 80 (b) all atoms are sp3 ; tetrahedral geometry and bond angles of 1 09 around each atom not a bqnd-;-shows not a bond-s how s '. H H lone Rair coming \ , lbon pair gomg ehmd paper &quot;'- / out 01 paper I I H , . 0 , . H H-C-O
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2 -6 Carbon-2 is sp hybridized. If the p orbitals making the pi bond between C-l and C-2 are in the plane of the paper (putting the hydrogens in front of and behind the paper), then the other p orbital on C-2 must be perpendicular to the plane of the pape
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2-7 continued (c) the nitrogen and the carbon bonded to it are sp hybridized; the left carbon is sp2H \ 1 200 c /H( .N.:1 800HII H /c-cN:oil(d) the boron and the oxygens bonded to it are sp2 hybridizedH - O: H \ I B-O: I H - O. : .II.PH-
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2-8 Very commonly in organic chemistry, we have to determine whether two structures are the same or different, and if they are different, what structural features are different. In order for two structures to be the same, all bonding connections have to b
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2-1 1 M odels will be helpful here. (a) cis-trans isomers-the first is trans, the second is cis (b) constitutional isomers-the carbon skeleton is different (c) constitutional isomers-the bromines are on different carbons in the first structure, on the sam
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2 -14 continued (c)L &quot; c / n et dipole 0 F =F(d)( e)0, . 0 'x o &quot; &quot; o ( g)( i)ro or 0; 13 1 1 C /H ' CH3 n etn et/ N ,,/Q1 oeach end oxygen has one-half negative charge as it is the composite of two resonance forms; see n et solution to 1 -
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( hydrogen bonds shown as wavy bond)2- 1 7 ( a) (CH3hCHCH2CH2CH(CH3h has less branching and boils at a higher temperature than (CH3hCC(CH3h . (b) CH3(CH2)sCH2 0 H can form hydrogen bonds and will boil at a much higher temperature than CH3(CH2) 6CH3 which
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( a)2-19HHHHH I I I I I H-C-C-C-C-C-H I I I I I HHHHH( d)a lkane (Usually, we use the tenn &quot;alkane&quot; only when no other groups are present.)H I H-C-C H-C H'( b)H HH I I I H-C-C=C-C-C-H I I I I I HHHHHa lkene( c)HHH I I I H - C - C : C - C - C -
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2 -2 1( a)H1 1H1 1011 1H1 1HH-C-C-C-N-C-H H H H H( b)H-C-C-N-C-C-H H11H1 1 1H1 1H1 1)H1 1H1 1011H1 1H-C-C-C-O-C-H H H'HHHHa mideHa mineHI IHH1 1C'HHe ster( e)H-C-C-O-C-C-H HIIHI 1HI I(f)H1 1H1 1
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2-22(d)c ontinuedII H C - NHa mide( this also looks like an aldeh yde, b u t an amide h as higher &quot;priority &quot; as y o u wi l l s e e l ater)g( e)CH6 c:3)Ha mineS uggested by student R i c h ard King: R i s the symbol that organic chemists use t
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2 -24 c ontinued(g) aldehyde: contains a carbonyl group with a hydrogen on one side HHa H-C-C-C-H H HIIIIII( h) aromatic hydrocarbon : a cyclic hydrocarbon with alternating double and single bonds( i ) carboxylic acid: contains a c arbonyl group w
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2 -27 c ontinued (h)i n front o f the p lane of the p aperbeh i nd the p l ane of the pape;(i) (j cfw_)H/a&quot;'&quot;Cangles around sp 3 atom angles around sp 2 atoms1 09 1 20HI, ,HH.-/2 -28For cl ari ty in these picture s , b o n d s between
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( i)s p 3 _sp 3I OJ - &lt; o, f&quot;' ;H_s p 3 _sp 2sp 2 -ssp 2 _ sp 3 2 -29 The second resonance form of formamide is a minor but significant resonance contri butor. It shows that the nitrogen-carbon bond h as some double bond character, requiring that
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2 -32( a) cis (b) The cop l an ar atoms i n the structures to the left and below are marked w i th asterisks.(c)HH JC*C - C ,&quot; ,&quot;T here are sti l l six copl anar atoms.2 -33QD M(d), ,eH,CH]H*transC ollinear atoms are marked with asteri s
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2-35 (a) consti tutional i somers-the c arbon skeleton s are different (b) constitutional i somers-the position of the c h l orine atom has changed (c) c i s-trans i somers-the first is c is, t he second i s trans ( d) constitutional i somers-the c arbon
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2 -38 Diethyl ether and I -butanol each have one oxygen , so each c an form hydrogen bonds with w ater (water supplies the H for hydrogen bonding with dieth yl ether) ; their water solubi lities should be similar. The boiling point of I -butanol is much h
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2-42( a) ether ether alkene (c) aldehyde (d) ketone (b)(e)ester (cyclic)( f)aromaticamide (cyclic)alkene(g)( h)ammeester2-43C H3..:0: /C'II:0: CH3 C H3 / /t&quot; S .:0 : C H3- . .f-J. . .CH 3/ /I S .+2 sp -planartp3 -tetrahedral