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Reaction Mechanism

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Published in: Chemistry
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Reaction mechanism

Shilpi M / Rajkot

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  1. 03.1 03.2 03.3 03.4 03.5 03.6 03.7 03.8 03.9 03.10 03.11 03.1 6 ORGANIC CHEMISTRY: REACTION MECHANISMS Acids and Bases Carbanion Attack at a Carbonyl Group The Mechanism of Reduction Reactions Nucleophilic Attack by Water Nucleophilic Attack by an Alcohol Addition/Elimination Reactions of Carboxylic Acid Derivatives Free Radical Reactions Bimolecular Nucleophilic Substitution or SN2 Reactions Unimolecular Nucleophilic Substitution or SNI Reactions Elimination Reactions Substitution versus Elimination Reactions ACIDS AND BASES As noted in Section 11.1, for more than 300 years, chemists have classified substances that behave like vinegar as acids, while those that have properties like the ash from a wood fire have been called alkalies or bases. Today, when chemists use the words "acid" or "base" they refer to a model developed independently by Brønsted, Lowry, and Bjerrum. Since the most explicit statement of this theory was contained in the writings of Brønsted, it is most commonly known as the "Brønsted acid—base" theory. Brønsted Acid—Base Theory Brønsted argued that all acid—base reactions involve the transfer of an H ion, or proton. Water reacts with itself, for example, by transferring an H ion from one molecule to an- other to form an 1-130 ion and an OH ion. H20 + H20 H30+ + OH -H 1
  2. 2 ORGANIC REACTION MECHANISMS According to the theory, a Brønsted acid is a proton donor and a Brønsted base is a pro- ton acceptor. Acids are often divided into categories such as "strong" and "weak." One measure of the strength of an acid is the acid dissociation equilibrium constant, K a, for that acid. [H30 When Ka is relatively large, we have a strong acid. HCI 106 When it is small, we have a weak acid. —5 CH3C02H = 1.8 X 10 When it is very small, we have a very weak acid. 16 }-120 1.8 X 10 In 1909, S. P. L. Sørenson suggested that the enormous range of concentrations of the H30 and OH ions in aqueous solutions could be compressed into a more manageable set of data by taking advantage of logarithmic mathematics and calculating the pH or POH of the solution. pH POH -log[H30 ] -log[OH ] The "p" in pH and POH is an operator that indicates that the negative of the logarithm should be calculated for any quantity to which it is attached. Thus, PIC a is the negative of the logarithm of the acid dissociation equilibrium constant. pl
  3. ORGANIC REACTION MECHANISMS NH3 + }-120 NH4 + OH 3 Acid Base Acid Base Just as the magnitude of Ka is a measure of the strength of an acid, the value of Kb reflects the strength of its conjugate base. Consider what happens when we multiply the Ka ex- pression for a generic acid (HA) by the Kb expression for its conjugate base (A ). [H30 ] [H30 If we now replace each term in the equation by the appropriate equilibrium constant, we get the following result. — 14 = 1 X 10 Because the product of Ka times Kb is a relatively small number, either the acid or its con- jugate base can be "strong." But if one is strong, the other must be weak. Thus, a strong acid must have a weak conjugate base. HCI + H20 H30+ + Cl Strong acid Weak base A strong base, on the other hand, must have a weak conjugate acid. NH4 + OH NH3 + H20 Strong base Weak acid Brønsted Acids and Bases in Nonaqueous Solutions Water has a limiting effect on the strength of acids and bases. All strong acids behave the same in water—I M solutions of the strong acids all behave as 1 M solutions of the 1-130 ion—and very weak acids may not act as acids in water. Acid—base reactions don't have to occur in water, however. When other solvents are used, the full range of acid—base strength shown in Table 03.1 can be observed. The strongest acids are in the upper left corner of Table 03.1; the strongest bases are in the bottom right corner. Each base is strong enough to deprotonate the acid in any line above it. The hydride ion (H ), for example, can convert an alcohol into its conjugate base, CH30H+ H CH30 + 1-12 and the amide ion (NH2—) can deprotonate an alkyne. CH3C=CH + NH2 CH3C= + NH3
  4. 4 ORGANIC REACTION MECHANISMS TABLE 03.1 Typical Brønsted Acids and Their Conjugate Bases CH2—CH2 CH.-CH Compound HI HCI H2S04 H30 HN03 H3P04 CH3C02H H20 CH30H HC=CH NH3 CH4 3 x 109 1 x 106 1 x 103 55 28 —3 7.1 x 10 —5 1.8 x 10 1.0 x 10 —16 1.8 x 10 pl
  5. ORGANIC REACTION MECHANISMS 5 proton. The Lewis theory, however, vastly expanded the family of compounds that can be called "acids." Anything that has one or more empty valence-shell orbitals can act as an acid. The theory explains why BF3 reacts instantaneously with NH3. The nonbonding elec- trons on the nitrogen in ammonia are donated into an empty orbital on the boron to form a new covalent bond, as shown in Figure 03.1. It also explains why Cu2+ ions pick up ammonia to form the four-coordinate ion. H H H FIGURE 03.1 The reaction between BF3 acting as a Lewis acid and H NH3 acting as a Lewis base to form an acid—base complex. Cu2+(aq) + 4 NH3(aq) In this case, a pair of nonbonding electrons from each of the four NH3 molecules is do- nated into an empty orbital on the Cu ion to form a covalent Cu—N bond. NH3 NH3 Curved Arrow Symbolism The flow of electrons from a Lewis base to a Lewis acid is often indicated with a curved arrow. The arrow starts on a pair of nonbonding electrons on the Lewis base and points toward the Lewis acid with which it reacts. Because adding a pair of electrons to one point on a molecule often displaces electrons in the molecule, combinations of curved arrows are often used to describe even simple chemical reactions. Consider the following example, in which a pair of electrons on an NH2- ion are donated to the H ion formed when the elec- trons in one of the C—H bonds in acetylene are given to the carbon atom instead of be- ing shared by the C and H atoms in the bond. N: 03.2 CARBANION ATTACK AT A CARBONYL GROUP The discussion of acids and bases in the previous section helps us understand the chem- istry of the Grignard reagents introduced in Section 02.10. Grignard reagents are made by reacting an alkyl bromide with magnesium metal in diethyl ether.
  6. 6 ORGANIC REACTION MECHANISMS Mg CH3Br CH3MgBr EGO An analogous reagent, known as an alkyllithium, can be prepared by reacting the alkyl bromide with lithium metal in diethyl ether. CH3Br + 2 Li —+ CH3Li + LiBr EGO In the course of the reactions the carbon atom is reduced from the —2 to the —4 oxidation state. Whereas the starting material contains a carbon atom with a partial positive charge, the carbon atom in the products of these reactions carries a partial negative charge. CH3Li and CH3MgBr can therefore be thought of as a source of the CH3 ion. CH3Li CHO + Li The CHO ion is the conjugate base of methane, which is the weakest Brønsted acid in Table 03.1. CH4 CHO + H —49 10 The CHO ion is therefore the strongest Brønsted base in the table. Exercise 03.1 A graduate student once tried to run the following reaction to prepare a Grignard reagent. Explain what he did wrong, state why the yield of the desired product was zero, and pre- dict the product he obtained. CH3CH2Br CH3CH2MgBr CH3CH20H Solution There is nothing wrong with the starting material (CH3CH2Br) or the product of the re- action (CH3CH2MgBr). Nor is there anything wrong with using magnesium metal to form a Grignard reagent. The only place where a mistake could have been made is in the choice of the solvent. The solvent that was used was ethanol (CH3CH20H), whereas the usual solvent for a Grignard reagent is diethyl ether. Ethanol is a protic solvent (see Section 02.4), which would react instantaneously with the carbanion in a Grignard reagent and thereby destroy the reagent to form ethane. CH3CH2- + H+ CH3CH3
  7. ORGANIC REACTION MECHANISMS 7 A subtle but important point must be made before we can extend our understanding of acid—base chemistry to the reaction between a Grignard or alkyllithium reagent and a carbonyl group. The data in Table 03.1 reflect the strengths of common acids and bases when they act as Brønsted acids or bases. The data predict that methyllithium should re- act with acetylene to form methane and an acetylide ion, for example. CH3Li + HC=CH CH4 + Li+ + The reaction should occur because it converts the stronger of a pair of Brønsted acids and the stronger of a pair of Brønsted bases into a weaker acid and a weaker base. :CH3 + HC—CH CH4 + Stronger base Stronger acid Weaker acid Weaker base The reaction between a carbonyl and CH3Li or CH3MgBr, on the other hand, involves at- tack by a CH3— ion acting as a Lewis base or nucleophile at the positive end of the car- bonyl group. CH3 CH3 H3C: C O CH3 C O CH3 CH3 This raises an interesting question: Is the stronger of a pair ofBrønsted bases always the stronger of a pair of Lewis bases? Unfortunately, the answer is no, it isn't. At times, the stronger of a pair of Brønsted bases is the weaker Lewis base or nucleophile. As a rule, however, strong 1 Brønsted bases are strong nucleophiles, and weak Brønsted bases are weak nucleophiles. Despite the enormous utility of Grignard reagents in organic chemistry, the exact mech- anism of the reaction between the reagents and a carbonyl is not known. There is reason to believe that two molecules of the Grignard reagent are involved in the reaction. The magnesium atom of one molecule of the reagent acts as a Lewis acid that interacts with the oxygen atom of the carbonyl group. The alkyl group of the other reagent molecule then acts as a Lewis base, attacking the positive end of the carbonyl. CH3 Mg CH3 Mg—Br In essence, the reaction involves the attack by a negatively charged CHO ion at the positively charged end of the carbonyl group. When this happens, the pair of nonbonding electrons on the CH3- ion are used to form a C—C bond. This, in turn, displaces the pair of electrons in the bond onto the other end of the carbonyl group. 1 See T. B. McMahon, T. Heinis, G. Nicol, J. K. Hovey, and P. Kebarle, J. Am. Chem. soc., 110, 7591 (1988).
  8. 8 ORGANIC REACTION MECHANISMS q CH3 The second molecule of the Grignard reagent, which binds at the oxygen end of the car- bonyl, isn't consumed in the reaction. Its function is simple. When it acts as a Lewis acid, binding to the oxygen atom in the C=O double bond, it increases the polarity of the bond. By making the bond more polar, it increases the rate at which the CHO ion attacks the positive end of the C=O bond. 03.3 THE MECHANISM OF REDUCTION REACTIONS Two fundamentally different reducing agents have been used in the previous chapters on organic chemistry to add hydrogen across a double bond. In Section 01.6, a metal was used to catalyze the reaction between hydrogen gas and the C=C double bond in an alkene. H H —c—c H H A source of the hydride ion (H ), on the other hand, bonds in Section 02.6. o 1. LiAIH4 in ether was used to reduce C=O double CH3CH2CH CH3CH2CH20H 2.1120 The first step toward understanding the difference between these reactions involves noting that the first reaction uses a nonpolar reagent to reduce a nonpolar double bond. The atoms on the surface of a metal are different from those buried in the body of the solid because they cannot satisfy their tendency to form strong metal-metal bonds. Some metals can satisfy a portion of their combining power by binding hydrogen atoms and/or alkenes to the surface. c—c Transferring one of the hydrogen atoms on the metal surface to the alkene bound to the metal surface forms an alkyl group, which can bond to the metal until the second hydrogen atom can be added to form the alkane. H c c—
  9. Although the hydrogen atoms are transferred one at a time, the reaction is fast enough that both of the atoms usually end up on the same side of the C=C double bond. This can't be seen in most alkanes produced by the reaction because of the free rotation around C—C bonds. Reduction of a cycloalkene, however, gives a stereoselective product. ORGANIC REACTION MECHANISMS Cl OH Cl OH 9 112/Ni CH3 CH3 CH3 CH3 Reduction of an alkyne with hydrogen on a metal catalyst gives the corresponding alkane. By selectively "poisoning" the catalyst it is possible to reduce an alkyne to an alkene. Once again, the reaction is stereoselective, adding both hydrogen atoms from the same side of the CEC bond to form the cis-alkene. CH3 CH3 112 3 Pd on CaC03 c—c Because it is a polar reagent, LiAlH4 won't react with a C=C double bond. It acts as a source of the H ion, however, which is a strong Brønsted base and a strong nucleophile. The H ion can therefore attack the ö+ end of a polar double bond. C CH3 CH3 CH3 H: The neutral AlH3 molecule formed when an AlH4 ion acts as a hydride donor is a Lewis acid that coordinates to the negatively charged oxygen atom in the product of the reac- tion. When a protic solvent is added to the reaction in a second step, an alcohol is formed. ---AIH3 CFE —C — CH3 + H20 CH3 —C—CH3 + OH + AlH3 03.4 NUCLEOPHILIC ATTACK BY WATER In the 1930s and 1940s, Dashiell Hammett (1894—1961) created the genre of the "hard- boiled" detective novel with books such as The Maltese Falcon and The Thin Man. A com- mon occurrence in this literature was a character who "slipped someone a Mickey Finn" a dose of the sedative known as chloral hydrate dissolved in a drink that contains alcohol. Cl —C —C —H Chloral hydrate
  10. 10 ORGANIC REACTION MECHANISMS Chloral hydrate is a white solid formed by adding a molecule of water across the C— —o double bond in the corresponding aldehyde. Cl O Cl O H CI—C—C—H + H20 CI—C—C—H Cl Cl OH The equilibrium constant for the reaction is sensitive to the substituents on the carbonyl carbon. Electron-withdrawing substituents, such as the C13C group in chloral, drive the re- action toward the dialcohol, or diol (K >> 1). Electron-donating substituents, such as the pair of CH3 groups in acetone, pull the equilibrium back toward the aldehyde (K — 2 x 10 3). The rate of the reaction can be studied by following the incorporation of isotopically labeled water. The vast majority (99.76%) of water molecules contain 160, but some con- tain 170 (0.04%) or 180 (0.2%). When acetone is dissolved in a sample of water that has been enriched in 180, it gradually picks up the 180 isotope. O 180H 180 + H2180 + H20 OH The rate of the reaction is infinitesimally slow in a neutral solution (pH 7). But in the presence of a trace of acid (or base), the reaction occurs very rapidly. The role of the acid catalyst can be understood by noting that protonation of the oxygen atom increases the polarity of the carbonyl bond. •o This increases the rate at which a water molecule can act as a nucleophile toward the pos- itive end of the C=O double bond. Acid-catalyzed hydration, step 1 CH3 CH3 The product of the reaction then loses an H ion to form the diol.
  11. ORGANIC REACTION MECHANISMS 11 Acid-catalyzed hydration, step 2 C— CH3 CH3— C—CH3 + H The role of the base catalyst can be understood by noting that the OH ion is a much stronger nucleophile than water, strong enough to attack the carbonyl by itself. Base-catalyzed hydration, step 1 CH3 CH3 The product of the reaction then picks up a proton from a water molecule to form the diol and regenerate the OH ion. Base-catalyzed hydration, step 2 + H20 + OH There is a fundamental relationship between the mechanisms of the reactions at the carbonyl group introduced so far. In each case, a nucleophile or Lewis base attacks the positive end of the carbonyl group. And, in each case, the rate of reaction can be increased by coordinating a Lewis acid or electrophile at the other end of the carbonyl. Electrophiles Il (H , Mg2+, AlH3, etc.) Nucleophiles (CH3 , H ,H20, OH , etc.) There is a subtle difference between the reactions, however. Very strong nucleophiles, such as Grignard reagents and the hydride ion, add to the carbonyl in an irreversible reaction. — C — CH3 CH3 CH3 CH3 : CH3 Attack by a weaker nucleophile, such as water, is a reversible reaction that can occur in either direction. o OH + H20 CHEC CH 3 OH
  12. 12 ORGANIC REACTION MECHANISMS 03.5 NUCLEOPHILIC ATTACK BY AN ALCOHOL What would happen if we dissolved an aldehyde or ketone in an alcohol, instead of wa- ter? We would get a similar reaction, but now an ROH molecule is added across the C=O double bond. o + CH30H OCH3 Once again, the reaction is relatively slow in the absence of an acid or base catalyst. If we bubble HCI gas through the solution, or add a small quantity of concentrated H2S04, we get an acid-catalyzed reaction that occurs by a mechanism analogous to that described in 2 the previous section. Acid-catalyzed reaction of an alcohol with a carbonyl : O—CH3 CH3 CH3 CH3 CH3— C — CH3 CH3 The product of the reaction is known as a hemiacetal (literally, "half of an acetal"). If an anhydrous acid is added to a solution of the aldehyde in a large excess of alcohol, the re- action continues to form an acetal. o OCH3 HCI + 2 CH30H + H20 OCH3 Hemiacetals can be recognized by looking for a carbon atom that has both an and an —OR group. CH3 CH3CH20CHOH A hemiacetal Acetals, on the other hand, contain a carbon atom that has two —OR groups. CH3 CH3CH20COCH3 An CH3 —OH 2 Gaseous HCI and concentrated H2S04 give an acidic solution without contaminating the solution with wa- ter, which could react with the carbonyl.
  13. Hemiacetals and acetals play an important role in the chemistry of carbohydrates. Con- sider what would happen, for example, if the —OH group on the fifth carbon atom in a glucose molecule attacked the aldehyde at other end of the molecule. o CH OH CH20H The product of the reaction is a hemiacetal that contains a six-membered ring known as a pyranose. Two isomers of glucopyranose can be formed, depending on whether the —OH group attacks from above or below the C=O group. ORGANIC REACTION MECHANISMS CH20H 13 CH20H or OH HO H O OH OH HO HO a -D -Glucopyranose CH20H or OH HO H O OH OH HO HO CH20H O OH OH CH20H O OH OH ß -D -Glucopyranose An analogous intramolecular reaction can occur within a fructose molecule. CH20H c=o OH
  14. 14 ORGANIC REACTION MECHANISMS In this case, a hemiacetal is formed that contains a five-membered furanose ring. Once again, there are two isomers, depending on how the —OH group attacks the C=O group. OHCH O H HO OH H OH CH20H ß -D -Fructofuranose OHCH 0 CH20H H HO OH OH H a -D -Fructofuranose Sugars, such as glucose and fructose, can be linked to form complex carbohydrates by forming an acetal linkage between the —OH group on one sugar and the hemiacetal on the other. Sucrose, or cane sugar, for example, is an acetal formed by linking a-D- glucopyranose and ß-D-fructofuranose residues. CH20H O HO OH OH OH o Sucrose CH20H HO HOCH2 0 03.6 ADDITION/ELIMINATION REACTIONS OF CARBOXYLIC ACID DERIVATIVES The following reaction was used in Section 02.8 to illustrate the synthesis of an ester from a carboxylic acid. o o CH3COH + CH3CH20H CH3COCH2CH3 These reactions occur very slowly in the absence of a strong acid. When gaseous HCI is bubbled through the solution, or a small quantity of concentrated H2S04 is added, the re- actions reach equilibrium within a few hours. Once again, the acid protonates the oxygen of the C=O double bond, thereby increasing the polarity of the carbonyl group, which makes it more susceptible to attack by a nucleophile. As might be expected, the first step in the reaction involves attack by a nucleophile at the positively charged end of the C=O double bond. A pair of nonbonding electrons on the oxygen atom of the alcohol is donated to the carbon atom of the carbonyl to form a C—O bond. As the bond forms, the electrons in the bond of the carbonyl are displaced onto the oxygen atom. A proton is then transferred back to the solvent to give a tetrahe- dral addition intermediate. Nucleophilic addition CH3 —C— OH HOCH2CH3 : OH CH3— C— OH HOCH2CH3 : OH : OCH2CH3
  15. ORGANIC REACTION MECHANISMS 15 One of the —OH groups in the intermediate picks up a proton, loses a molecule of water, and then transfers a proton back to the solvent to give the ester. Nucleophilic elimination : OH -H20 OH -H CHEC—OH CHEC C—OCH2CH3 : OCH2CH3 : OCH2CH3 : OCH2CH3 The combination of addition and elimination reactions has the overall effect of substi- tuting one nucleophile for another—in this case, substituting an alcohol for water. The rate of these nucleophilic substitution reactions is determined by the ease with which the elim- ination step occurs. As a rule, the best leaving groups in nucleophilic substitution reactions are weak bases. The most reactive of the carboxylic acid derivatives are the acyl chlorides because the leaving group is a chloride ion, which is a very weak base (Kb 10 20). o o CH3CCl + CH30H CH3COCH3 + Cl Esters are less reactive because the leaving group is an alcohol, which is a slightly better base (Kb 10 14). O o CH3COCH2CH3 + CH30H CH3COCH3 + CH3CH20H Amides are even less reactive because the leaving group is ammonia or an amine, which are significantly more basic (Kb 10 5). o o CH3CONHCH3 + CH30H CH3COCH3 + CH3NH2 03.7 FREE RADICAL REACTIONS The starting point for reactions at a carbonyl involves attack by a nucleophile on the car- bon atom of the C=O double bond. CH3 CH3 Or it involves the heterolytic splitting of a bond to form a nucleophile that can attack the carbonyl group.
  16. 16 ORGANIC REACTION MECHANISMS c. CH3_c CH3 CH3 CH3 In either case, the reaction is carried out by a reagent that donates a pair of electrons to a carbon atom to form a new covalent bond. Section 02.3 introduced a reaction that occurs by a very different mechanism: free rad- ical halogenation of an alkane. The first step in these reactions is the homolytic splitting of a bond to give a pair of free radicals. Chain initiation A series of reactions then occurs that involves a chain reaction. Consider the chlorination of propane, for example. A Cl• atom can attack the CH3 group at one end of the molecule. Chain propagation H H H H H H : CI—H + H H H H H H Or it can attack the CH2 group in the center of the molecule. Chain propagation : CI—H + The free radicals generated in the reactions then react with chlorine to form either I-chloro- propane or 2-chloropropane and regenerate a Cl• radical. Chain propagation H H H H H H H H H : CI—CI: H : CI—CI: H H —c—c H —c—c H H H H H H —C— H
  17. ORGANIC REACTION MECHANISMS 17 There are six hydrogen atoms in the two CH3 groups and two hydrogens in the CH2 group in propane. If attack occurred randomly, six-eighths (or three-quarters) of the prod- uct of the reaction would be I-chloropropane. The distribution of products of the reaction, however, suggests that I-chloropropane is formed slightly less often than 2-chloropropane. Cl CH3CH2CH3 + C12 —+ CICH2CH2CH3 + CH3CHCH3 I-Chloropropane (45%) 2-Chloropropane (55%) This can be explained by noting that the 20 radical formed by removing a hydrogen atom from the CH2 group in the center of the molecule is slightly more stable than the 10 radi- cal produced when a hydrogen atom is removed from one of the CH3 groups at either end of the molecule. The difference between the radicals can be appreciated by considering the energy it takes to break the C—H bond in the following compounds. CH3 CH3 CH3 CH3 CH3 CH3—C• + H• CH3 CH3 CH3—C• + H• CH3 AHO = -381 kJ/molrxn AHO = -395 kJ/molrxn AHO - -410 kJ/molrxn These data suggest that it takes less energy to break a C—H bond as the number of alkyl groups on the carbon atom that contains the bond increases. This can be explained by as- suming that the products of the bond-breaking reaction become more stable as the num- ber of alkyl groups increases. In other words, 30 radicals are more stable than 20 radicals, which are more stable than 10 radicals. CH3 CH3—C• > CH3—C. > H—C• CH3 30 CH3 20 CH3 10 The activation energy for the chain propagation steps in free radical bromination re- actions is significantly larger than the activation energy for the steps during chlorination. As a result, free radical bromination reactions are more selective than chlorination reac- tions. Bromination reactions are far more likely to give the product predicted from the relative stability of the free radical intermediate. Bromination of 2-methylpropane, for ex- ample, gives almost exclusively 2-bromo-2-methylpropane, not the statistically more likely I-bromo-2-methylpropane.
  18. 18 ORGANIC REACTION MECHANISMS CH3 CH3 CH3CHCH3 + Br2 CH3CCH3 + HBr 990/0 03.8 BIMOLECULAR NUCLEOPHILIC SUBSTITUTION OR SN2 REACTIONS Most of our knowledge of the mechanisms of chemical reactions has come from the study of the factors that influence the rate of these reactions. The type of reaction that has been studied more than any other involves attack by a nucleophile on a saturated carbon atom. Consider the following reaction, for example, which converts an alkyl bromide into an al- cohol. CH3Br + OH CH30H + Br In the course of the reaction, one nucleophile (the OH ion) is substituted for another (the Br ion). This is therefore a nucleophilic substitution reaction. The rate of the reaction is first-order in both CH3Br and the OH ion, and second- order overall. Rate = ) In the 1930s, Sir Christopher Ingold proposed a mechanism for the reaction in which both the alkyl halide and the hydroxyl ion are involved in the rate-limiting or slowest step of the reaction. The OH ion attacks the "backside" of the CH3Br molecule. (It attacks the carbon atom at a point directly opposite the Br substituent or leaving group.) When this happens, a pair of nonbonding electrons on the OH ion are used to form a covalent bond to the carbon atom at the same time that the carbon—bromine is broken, as shown in Fig- ure 03.2. Because the rate-limiting step in the reaction involves both the CH3Br and OH molecules, it is called a bimolecular nucleophilic substitution reaction, or SN2 reaction. C—Br HO----C----Br HO HO FIGURE 03.2 The mechanism for nucleophilic attack on CH3Br. The most important point to remember about the mechanism of SN2 reactions is that they occur in a single step. The species in the middle of Figure 03.2 is known as a transi- tion state. If you envision the reaction as an endless series of snapshots that capture the infinitesimally small changes which occur as one bond forms and the other bond breaks, the transition state is the snapshot in the series that has the highest energy—and is there- fore the least stable. The transition state has an infinitesimally short lifetime, on the order —12 of 10 second. In the course of an SN2 reaction, the other three substituents on the carbon atom are "flipped" from one side of the atom to the other. This inevitably leads to inversion of the
  19. ORGANIC REACTION MECHANISMS 19 configuration at a stereocenter. Consider the following reaction, for example, in which cis- I-bromo-3-methylcyclopentane is converted to trans-3-methylcyclopentanol. CH3 CH3 OH OH Or consider the reaction in which the R isomer of 2-bromobutane is transformed into the S isomer of 2-butanol. CH3CH2 HO — + c Cl--13i'7 HO—C CH2CH3 CH3 03.9 UNIMOLECULAR NUCLEOPHILIC SUBSTITUTION OR SNI REACTIONS The kinetics of nucleophilic substitution reactions have been studied in greater detail than any other type of reaction because they don't always proceed through the same mecha- nism. Consider the reaction between the OH ion and t-butyl bromide, for example. (CH3)3CBr + OH (CH3)3COH + Br The rate of the reaction depends only on the concentration of the alkyl bromide. (Adding more OH ion to the solution has no effect on the rate of reaction.) Rate = Ingold and co-workers argued that this rate law is consistent with a mechanism in which the rate-limiting or slowest step involves the breaking of the carbon—bromine bond to form a pair of ions. As one might expect, the pair of electrons in the C—Br bond end up on the more electronegative bromine atom. Rate-limiting step CH3 CH3 CH3 CH3—C+ + Br CH3 Because the bromine atom has formally gained an electron from the carbon atom, it is now a negatively charged Br ion. Because the carbon atom has formally lost an electron, it is now a "carbocation.' The first step in the mechanism is a relatively slow reaction because the activation energy for this step is roughly 80 kJ/molrxn. If the reaction is done in water, the next step is extremely fast. The (CH3)3C ion is a Lewis acid because it has an empty orbital that can be used to accept a pair of electrons. Water, on the other hand, is a reasonably good Lewis base. A Lewis acid-base reaction therefore rapidly occurs in which a pair of nonbonding electrons on a water molecule are donated to the carbocation to form a covalent C—O bond.
  20. 20 ORGANIC REACTION MECHANISMS CH3 CH3 CH3 C +H20 CH3 C OH2 CH3 CH3 The product of the reaction is a stronger acid than water. As a result, it transfers a proton to water. CH3 CH3 —C— OH2+ + H 20 CH3 CH3 CH3 —C—OH + H30 CH3 Because the slowest step of the reaction only involves t-butyl bromide, the overall rate of reaction depends only on the concentration of that species. The reaction is therefore a uni- molecular nucleophilic substitution reaction, or SNI reaction. The central carbon atom in the t-butyl carbocation formed in the first step of the re- action is planar, as shown in Figure 03.3. This means that water can attack the carbo- cation in the second step with equal probability from either side of the carbon atom. This has no effect on the products of the reaction, because the starting material is not opti- cally active. But what would happen if we started with an optically active halide, such as —bromobutane? CH3—C CH3 FIGURE 03.3 The geometry of the (CH3)3C ion. Regardless of whether we start with the R or S isomer of 2-bromobutane, we get the same intermediate when the C—Br bond breaks. CH3CH2 CH3 CH2CH3 c CH3 'H CH2CH3 c CH3 The intermediate formed in the first step in the SNI mechanism is therefore achiral. Mixtures of equal quantities of the +/— or R/S stereoisomers of a compound are said to be racemic. The term traces back to the Latin word racemus, which means "a cluster of grapes." Just as there is an equal probability of finding grapes on either side of the stem in a cluster of grapes, there is an equal probability of finding the R and S enantiomers in a racemic mixture. SNI reactions are therefore said to proceed with racemization. If we start with a pure sample of (R)-2-bromobutane, for example, we expect the product of the SNI reaction with the OH ion to be a racemic mixture of the two enantiomers of 2-butanol. We are now ready to address a pair of important questions. First, why does CH3Br re- act with the OH ion by the SN2 mechanism if (CH3)3CBr does not? The SN2 mechanism requires direct attack by the OH ion on the carbon atom that carries the C—Br bond. It is much easier for the OH ion to get past the small hydrogen atoms in CH3Br than it is for the ion to get past the bulkier CH3 groups in (CH3)3CBr.
  21. ORGANIC REACTION MECHANISMS 21 c HO ---C---Br HO HO Thus, SN2 reactions at the 10 carbon atom in CH3Br are much faster than the analogous reaction at the 30 carbon atom in (CH3)3CBr. Why, then, does (CH3)3CBr react with the OH ion by the SNI mechanism if CH3Br does not? The SNI reaction proceeds through a carbocation intermediate, and the stabil- ity of such ions decreases in the following order. CH3 CH3—C+ > CH3—C+ > H—C+ > Organic chemists explain this by noting that alkyl groups are slightly "electron releasing." They can donate electron density to a neighboring group. This tends to delocalize the charge over a larger volume of the molecule, which stabilizes the carbocation. When we encountered a similar phenomenon in the chemistry of free radicals in Sec- tion 03.7 we noted that 30 radicals are roughly 30 kJ/mol more stable than 10 radicals. In this case, the difference is much larger. A 30 carbocation is 340 kJ/mol more stable than a 10 carbocation!3 As a result, it is much easier for (CH3)3CBr to form a carbocation inter- mediate than it is for CH3Br to undergo a similar reaction. In theory, both starting materials could undergo both reaction mechanisms. But the rate of SN2 reactions for CH3Br are much faster than the corresponding SNI reactions, whereas the rate of SNI reactions for (CH3)3CBr are very much faster than SN2 reactions. 03.10 ELIMINATION REACTIONS Why do we need to worry about whether a nucleophilic substitution reaction occurs by an SNI or SN2 mechanism? At first glance, it would appear that the same product is obtained regardless of the mechanism of the reaction. Consider the following substitution reaction, CH3 CH3 CH3 for example. OCH3 CH3CHCH2CH3 + CH30 CH3CHCH2CH3 + Br The only apparent difference between the two mechanisms is the stereochemistry of the product. If the reaction proceeds through an SN2 mechanism, it gives inversion of config- uration—conversion of an R starting material into an S product, or vice versa. If the reac- tion proceeds through a carbocation intermediate via an SNI mechanism, we get a racemic mixture. The importance of understanding the mechanism of nucleophilic substitution reactions can best be appreciated by studying the distribution of products of the example given above. 3J. E. Bartmess, Mass Spectrometry Review, 8, 297-343 (1989).
  22. 22 ORGANIC REACTION MECHANISMS When 2-bromobutane is allowed to react with the methoxide ion in methanol, less than half of the starting material is converted into methyl isopropyl ether; the rest is transformed into 2-propene. OCH3 CH3CHCH2CH3 + CHO - CH3CHCH2CH3 + CH2=CHCH2CH3 ca. 60% The reaction that produces the alkene involves the loss of an HBr molecule to form a C=C double bond. It is therefore an example of an elimination reaction. Starting materials that are likely to undergo a bimolecular SN2 reaction undergo elim- ination reactions by a bimolecular elimination mechanism, or E2 reaction. This is a one- step reaction in which the nucleophile attacks a C—H bond on the carbon atom adjacent to the site of SN2 reaction. CH30 C1130H H—CH2CHCH2CH3 CH2=CHCH2CH3 + Br Starting materials that are likely to undergo a unimolecular SNI reaction undergo elim- ination reactions by a unimolecular elimination mechanism, or El reaction. As might be expected, the rate-limiting step is the formation of the carbocation. Rate-limiting step CH3 CH3—C—Br CH3— CH3 CH3 CH3 The solvent then acts as a base, removing an H ion from one of the alkyl groups adja- cent to the carbocation. The electrons in the C—H bond that is broken are donated to the empty orbital on the carbocation to form a double bond. CH30HÄ) H CH3 CH3 CH3 C+ CH2=C CH3 CH3 CH3 03.11 SUBSTITUTION VERSUS ELIMINATION REACTIONS There are three ways of pushing the reaction between an alkyl halide and a nucleophile toward elimination instead of substitution. Start with a highly substituted substrate, which is more likely to undergo elimination. Only 10% of a primary alkyl bromide undergoes elimination to form an alkene, for example, when it reacts with an alkoxide ion dissolved in alcohol. The vast majority of the starting material goes on to form the product expected for an SN2 reaction.
  23. ORGANIC REACTION MECHANISMS C1130 CH3CH2CH2Br CH3CH2CH20CH3 CH30H 23 More than half of a secondary alkyl bromide undergoes elimination under the same conditions, as we have already seen. OCH3 CH3CHCH2CH3 + CH30- CH3CHCH2CH3 + CH2=CHCH2CH3 When the starting material is a tertiary alkyl halide, more than 90% of the prod- uct is formed by an El elimination reaction. CH3 C1130 CH3— C—Br CH30H CH3 CH3 CH3 Use a very strong base as the nucleophile. When we use a relatively weak base, such as ethyl alcohol, only about 20% of t-butyl bromide undergoes elimination. CH3 CH3CH20H CH3 CH3 CHEC—OCH2CH3 + CH2=C CH3 CH3 -800/0 CH3 -200/0 In the presence of the ethoxide ion, which is a much stronger base, the product of the reaction is predominantly the alkene. CH3 CH3CH20 CH3CH20H CH3 CH3 CH3 •50/0 Increase the temperature at which the reaction is run. Because both El and E2 reactions lead to an increase in the number of particles in the system, they are as- sociated with a positive entropy term. Thus, increasing the temperature of the re- action makes the overall free energy of reaction more negative, and the reaction becomes more favorable. Summary of Substitution/Elimination Reactions Methyl halides and primary alkyl halides such as CH3CH2Br undergo nucleophilic substitution reactions. CH3CH2Br + CN- CH3CH2CN + Br
  24. 24 ORGANIC REACTION MECHANISMS Secondary alkyl halides undergo SN2 reactions when handled gently—at low tem- peratures and with moderately strong nucleophiles. SH CH3CHCH3 + HS- CH3CHCH3 + Br At high temperatures, or in the presence of a strong base, secondary alkyl halides undergo E2 elimination reactions. (CH3)3CO CH3CHCH3 CH2=CHCH3 heat Tertiary alkyl halides undergo a combination of SNI and El reactions. If the reac- tion is kept cool, and the nucleophile is a relatively weak base, it is possible to get nucleophilic substitution. At high temperatures, or with strong bases, elimination reactions predominate. KEY TERMS Acetal Acid Acid dissociation equilibrium constant, Alkali Base Bimolecular nucleophilic substitution reaction Brønsted acid PROBLEMS Acids and Bases CH3 C1130 CH3 Brønsted base Diol El reaction E2 reaction Elimination reaction Hemiacetal Heterolytic Homolytic Leaving group CH3 CH3 Lewis acid Lewis base Nucleophilic substitution reaction SNI reaction SN2 reaction Transition state Unimolecular nucleophilic substitution reaction 1. Predict the products of the reaction between ethanol (CH3CH20H) and sodium hy- dride (NaH). 2. Write the Lewis structure for the conjugate base of ethanol (CH3CH20H). 3. Which of the following is the conjugate acid of ethanol (CH3CH20H)? (a) CH3CH20H (b) CH3CH20 (c) CH3CH20H2 (d) CH3CH2 (e) f-130 4. Some beginning chemistry students get confused when acetic acid is described as a weak acid because they see four hydrogen atoms in a CH3C02H molecule. Explain why only one of the H atoms dissociates when acetic acid is dissolved in water.
  25. ORGANIC REACTION MECHANISMS 5. Which of the following are Lewis acids but not a Brønsted acid? (a) H (b) NH4 (c) BF3 (d) CH3CH20H (e) Mg2 + 6. Arrange the following compounds in order of increasing basicity. (a) NH3 (b) NH2 (c) NH4 (d) N3 7. Identify the Brønsted acids in the following reaction. HC=CH + CH3Li HC=C + CH4 + Li 8. Arrange the following hydrocarbons in order of increasing acidity. (a) C2H2 (b) C2H4 (c) 25 9. Which of the following reagents is a strong enough base to generate the ethoxide ion (CH3CH20 ) from ethanol (CH3CH20H)? (a) NaOH (b) NaH (c) NaNH2 (d) CH3MgBr (e) HC=CNa 10. Which of the following acid—base reactions should occur as written? (a) CH3CH20H + NaOH Na + CH3CH20 + H20 (b) CH4 + NaNH2 CH3 + Na + NH3 (c) CH3Li + HC=CH CH4 + HC=CLi (d) CH3CH20H + HS CH3CH20 + Attack at a Carbonyl 11. Use Lewis structures and the concept of oxidation—reduction reactions to explain why it takes two moles of lithium metal to reduce a mole of methyl bromide to form methyl- lithium. CH3Br + 2 Li CH3Li + LiBr 12. Explain why ethers, but not alcohols, are used as solvents for reactions that involve Grignard or alkyllithium reagents. 13. Explain why the bonding of a Lewis acid at the oxygen atom of a carbonyl group in- creases the rate at which nucleophilic attack occurs at the carbon atom. 14. Predict the products of the following reactions. o Li -l- CH3CCH2CH3 A B 15. Predict the products of the following reactions. o C—CH3 + CH3MgBr A B 16. Design a two-step reaction sequence using either a Grignard or alkyllithium reagent that could be used to produce the following compound. OH C—CH2CH3 CH3
  26. 26 ORGANIC REACTION MECHANISMS The Mechanism of Reduction Reactions 17. Both H2 on a metal catalyst and LiAlH4, can be used to reduce a carbonyl to an al- cohol. Explain why H2 on a metal catalyst can be used to reduce a C=C double bond but not LiAlH4. 18. Explain why the yield of reactions that try to reduce a carbonyl with LiAlH4 in ethanol is effectively zero. 19. Reduction of a C=C double bond with H2 gas on a metal catalyst gives a product in which the two hydrogen atoms are added to the same side of the double bond. What does this tell us about the relative rates of the first and second steps in the reaction? 20. Why is it important to "poison" the metal catalyst before trying to reduce an alkyne to an alkene with 1-12? 21. The reagent that attacks the carbon atom is the same when either LiAlH4 or NaBH4 is used to reduce a C=O double bond. (In each case, it is the H ion.) What does the fact that LiAlH4 is significantly more reactive than NaBH4 tell us about the relative acidity of AlH3 and BI--13 as Lewis acids? 22. Predict the product of the following reaction. o CH3 CH3 23. Predict the product of the following reaction. 1. LiAIH4 in ether 2.1130 CH3CæCCH3 Pd on CaC03 Nucleophilic Attack by Water or an Alcohol 24. Explain why the rate of nucleophilic attack on a carbonyl by either water or an alco- hol is relatively slow in the absence of an acid or base catalyst. 25. Identify the starting materials that would give the following product. OH C—OCH2CH3 CH3 26. Identify the starting materials that would give the following acetal, which has a de- lightful odor of 'Bing' cherries. OCH2CH3 CH OCH2CH3 27. Maltose, or malt sugar, is an important component of the barley malt used to brew beer. This disaccharide is formed when the —OH group on C-4 of one a-D-glucopyranose forms an acetal by reacting with C-1 of a second a-D-glucopyranose residue. Draw the structure of the compound.
  27. ORGANIC REACTION MECHANISMS Addition/Elimination Reactions 27 28. Explain why acyl chlorides are more reactive than carboxylic acid esters toward addi- tion/elimination reactions. 29. Use the fact that the best leaving groups for addition/elimination reactions are weak bases to explain why attack on a carbonyl by the H ion or a source of the CHO ion are not reversible reactions. 30. Which of the following would be the best leaving group? (a) NH2- (pl Br > CE > F . Is this consistent with what we know about the relative acidities ofH1 (Ka = 3 X 109), HBr (Ka = 1 X 109), HCI (Ka = 106), and HF (Ka = 7.2 X 10 4)?
  28. 28 ORGANIC REACTION MECHANISMS Or does it suggest that there is a difference between trends in nucleophilicity and ba- sicity as we go down a column of the periodic table? (In other words, the X ion be- comes less basic and more nucleophilic as we go down the column.) 39. Write the mechanism for the following reaction. CH3 CH3 CH3 + OCHO + Br CH3 40. Identify a set of starting materials that would give the following compound as the prod- uct of a nucleophilic substitution reaction. CH3CH2NH2 41. Assume that you had a choice of two reagents to react with the following starting ma- terial, CH30H and the CH30 ion. CH3 C—Br Which nucleophile would be more likely to give a racemized product? Elimination Reactions 42. Write the mechanism for the following reaction. CH3 + OCHO CH2=C CH3 CH3 CH3 43. Which of the following would be the most likely to undergo an El elimination reac- tion with a very strong base, such as the (CH3)3CO ion? c 44. Predict the product of the following elimination reaction. + (CH3)3CO 45. Elimination reactions can often give two different products, depending on which car- bon atom adjacent to the C—X bond is attacked. As a rule, the dominant product of
  29. the reactions is the most highly substituted alkene. Predict the product of the follow- ORGANIC REACTION MECHANISMS CH—CH 29 ing elimination reaction. CH3 CH3 CH3 + (CH3)3CO Substitution versus Elimination Reactions 46. Assume that 2-bromopropane undergoes a combination of E2 and SN2 reactions when it reacts with the CH30 ion in methanol. Predict the products of the reactions. 47. Predict the most likely product or products of the following reactions. (a) (b) (c) CH30H CH3NH2 C1130 heat 48. Predict whether the following reactions are more likely to undergo elimination or sub- stitution. Identify the mechanism of the dominant reaction (El versus E2', SNI versus (a) (b) (c) NH2 —C—Br 1120 49. Explain why the following reaction will not give the indicated product. Predict the prod- uct that would form. CH3 CH3 CH3 CH3 3 + CHEC—O CH3 CH3 CH3 CH3 50. Explain why the following reaction will not give the indicated product. Predict the prod- uct that would form. 1120
  30. 30 ORGANIC REACTION MECHANISMS Integrated Problems 51. The following Lewis structures were drawn correctly by an organic chemistry student who forgot to indicate whether the molecules carry a positive or negative charge. Cor- rect the work by specifying whether each molecule is negatively charged, positively charged, or neutral. H H H H H H

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