ALL REACTIONS FROM OCHEM (So far have completed chem 31-32, need to add chem 33)

Dehydration of alcohols

-Beta elimination reaction in which the elements of OH and H are removed from the alpha nad beta carbon atoms, respectively -This makes it possible to perform substitution and eliiniation reactions on alcohols -Typically carried out using acids scuh as H2SO4 or p-toluenesulfonic acid (TsOH) Dehydration of 2 degree and 3 degree alcohols -2 degree and 3 degree alcohols react with an E1 mechanism -The E1 dehydration of 2 degree and 3 degree alcohols with acid gives clean elimnation products without any by products form an S N 1 reaction Dehydration of 1 degree alcohols -Primary alcohols react by an E2 mechanism -1 degree carbocations are highly unstable, teh dehydration of 1 degree alchools cannot occur by an E1 mechanism involving a carbocation

Synthesis of ethers: Williamson ether synthesis

-First step is acid base, second step is SN2 -Use secondary or primary for second step Steric considerations of the group on the Br. Must be secondary or primary. If there is primary, chose that one

Acid-cataylzed tautomerization

Alkynes are also observed to undergo acid-catalyzed hydration (via a Markovnikov addition), but the reaction is slower than the corresponding reaction with alkenes. As noted earlier in this chapter, the difference in rate can be attributed to the high-energy, vinylic carbocation intermediate that is formed when an alkyne is protonated. The rate of alkyne hydration is markedly enhanced in the presence of mercuric sulfate (HgSO4), which catalyzes the reaction. The resulting compound is called an enol because it has a double bond (en) and an OH group (ol ). The enol cannot be isolated under these acidic conditions, because it rapidly converts into a ketone: The π bond of the enol is first protonated, generating a resonance-stabilized intermediate, which is then deprotonated to give the ketone. Notice that both steps of this mechanism are proton transfers. The result of this process is the migration of a proton from one location to another, accompanied by a change in location of the π bond

Summary of ring opening of epoxides

All of these nucleophiles are reagents that we have previously encountered, and they can all open epoxides. These reactions exhibit two important features that must be considered, regiochemistry and stereochemistry: 1. Regiochemistry. When the starting epoxide is unsymmetrical, the nucleophile attacks at the less substituted (less hindered) position. This steric effect is what we would expect from an SN2 process. 2. Stereochemistry. When the attack takes place at a chiral center, inversion of configuration is observed. This result is also expected for an SN2 process as a consequence of the requirement for backside attack of the nucleophile. Notice that the configuration of the other chiral center is not affected by the process. Only the center being attacked undergoes an inversion of configuration.

Hydrolysis of Amides under Basic Conditions

Amides are also hydrolyzed when heated in basic aqueous solutions, although the process is very slow. The accepted mechanism for this process (Mechanism 20.12) is directly analogous to the mecha- nism that we saw for the saponification of esters (Section 20.11). In the last step, formation of the carboxylate ion drives the reaction to completion and renders the process irreversible. After the reaction is complete, aqueous acidic workup of the carboxylate ion gives a carboxylic acid.

When is the mechanism E1 vs E2

Base strength and solvent type determines weather a reaction follows the E1 or E2 mechanism with 1 degree and 2 degree RX E2 mechanism: - much more common and usefull - favored by strong, negatively charged bases, especially -OH and -OR - The reaction occurs with 1 degree, 2 degree, and 3 degree alkyl halides. Order of reactivity: R3CX,R2CHX>RCH2X E1 mechanism: - much less useful because a mixture of S N 1 ad E1 products usually result -favored by weaker, natural bases, such as H2O and ROH -this mechanism doesn't occur with 1 degree RX because they form highly unstable 1 degree carbocations

Gringard reaction

C-C bond forming reaction R-OH as the product Notice that the proton source (H3O+ or H2O) is added to the reaction mixture in a separate step (similar to reduction with LiAlH4). Even a weak acid, such as H2O, cannot be present together with the Grignard reagent, because the Grignard reagent is also a strong base and will deprotonate water. The difference in pKa values is so vast that the reaction is essentially irreversible. Every water molecule present in the reaction flask will destroy one molecule of Grignard reagent. Grignard reagents will even react with the moisture in the air, so care must be taken to use conditions that scrupulously avoid the presence of water. After the Grignard reagent has attacked the ketone, an acid can be introduced into the reaction flask to protonate the alkoxide. This protonation step (also called workup) can be accomplished by introducing even a weak acid, such as H2O, into the reaction flask. Throughout the remainder of this text, we will always show H3O+ as the source of protons for the workup of a Grignard reaction. A Grignard reagent will react with a ketone or an aldehyde to produce an alcohol.

Mechanism for enamine formation (C=C-NR2) (secondary)

Enamines are compounds in which the nitrogen lone pair is delocalized by the presence of an adjacent CC double bond. A mechanism for enamine formation is shown in Mechanism 19.7. This mechanism of enamine formation is identical to the mechanism that was shown for imine formation except for the last step:

Reaction of epoxides: acid catalyzed ring opening

Epoxides are much more reactive than their ethers. Why? -They can be opened by both acids and bases -Works with HX, H3O+/H+/H2O, H+/ROH -In this section, we saw the reactions of epoxides with strong nucleophiles. The driving force for such reactions was the removal of ring strain associated with the threemembered ring of an epoxide. Ringopening reactions can also occur under acidic conditions. As an example, consider the reaction between ethylene oxide and HX. -The first step is a proton transfer, and the second step is nucleophilic attack (SN2) by a halide ion. This reaction can be accomplished with HCl, HBr, or HI. Other nucleophiles such as water or an alcohol can also open an epoxide ring under acidic conditions. A small amount of acid (often sulfuric acid) is used to catalyze the reaction. -The brackets around the H+ indicate that the acid functions as a catalyst. In each of the reactions above, the mechanism involves a proton transfer as the final step of the mechanism.

Acid-Catalyzed Hydrolysis of Esters

Esters can also be hydrolyzed under acidic conditions. -This process (Mechanism 20.8) is the reverse of a Fischer esterification.

Reactions of nitriles

For the last reaction, Notice that we use H2O (rather than H3O+) for the workup step, because the product of this reac- tion is an amine, and we want to avoid protonating the amine. Using H3O+ for the workup step would result in an ammonium ion (RNH3+).

Gilman reagent and LiAl(OR)3H

Gilman reagent: less reactive gingard (adds once) LiAl(OR)3H is a less reactive LiAH4 (adds once)

Gringard reactions with esters

Grignard reagents also react with esters to produce alcohols, with introduction of two R groups.

Hydrolysis of Nitriles

In aqueous acidic conditions, nitriles are hydrolyzed to afford amides, which are then further hydro- lyzed to yield carboxylic acids. Formation of the amide occurs via Mechanism 20.14, and conversion of the amide into the carbox- ylic acid was discussed earlier (Mechanism 20.11).

Mechanism for imine formation (C=N) (primary)

In mildly acidic conditions, an aldehyde or ketone will react with a primary amine to form an imine:

PCC

In order to produce the aldehyde without further oxidation, it is necessary to use a more selective oxidizing reagent—one that will react with the alcohol but will not react with the aldehyde. One such reagent is pyridinium chlorochromate (PCC), typically dissolved in methylene chloride (CH2Cl2): Under these conditions, the alcohol is oxidized to give an aldehyde which is not further oxidized to a carboxylic acid. When a secondary alcohol is treated with a chromium oxidizing agent (whether chromic acid or PCC), the alcohol is oxidized to give a ketone:

Synthesis and reactions of amides

Least reactive. The N is less reactive than O because it can hold a positive charge better -Acid halides are the most reactive of the carboxylic acid derivatives, so the yields are best when an acid chloride is used as a starting material.

Acid chlorides

Most reactive Can be converted to all other carboxylic acid derivatives

Alternatively, nitriles can also be hydrolyzed in aqueous base.

Once again, the nitrile is first converted to an amide (Mechanism 20.15), which is then converted to a carboxylic acid (see Mechanism 20.12).

Synthesis of nitriles

Reaction 1 is useful for preparing tertiary nitriles, which cannot be prepared via an SN2 process.

Alcohol reactions

Reactions of alcohols -Whereas alkyl halides adn alkyl pseudohalides contain a good leaving group (X-) alcohols do not -For an alcohol to participate in nucleophilic subsitution reactions, the OH must be turned into a better leaving group, such as -OSO2R or -+OH2 Conversion of R-OH to R-X -The reaction of alcohols with HX (X=Cl, Br I) is a general method to prepare 1 degree, 2 degree, and 3 degree alkyl halides Reaction of 2 degree and 3 degree alcohols -React by an 2 N 1 mechanism -The S N 1 reaction of 2 degree and 3 degree alcohols with HC gives clean substitution products without any by products formed from an E1 reaction Reactions of methanol and 1 degree aclohols -Methanol and primary alcohols react by S N 1 mechanism

Syn dihydroxylation

Syn dihydroxylation results when an alkene is treated with KMnO4 or OsO4

Preparation of ethers

The carboxylic acid is first deprotonated to yield a carboxylate ion, which then functions as a nuc- leophile and attacks the alkyl halide in an SN2 process. The expected limitations of SN2 processes therefore apply. Specifically, tertiary alkyl halides cannot be used.

LAH carboxylic acid mechanic,

The first step of the process is likely a proton transfer, because LiAlH4 is not only a powerful nucleo- phile, but it can also function as a strong base, forming a carboxylate ion. -There are several possibilities for the rest of the mechanism. One possibility involves a reaction of the carboxylate ion with AlH3 followed by elimination to form an aldehyde: -Under these conditions, the aldehyde cannot be isolated. Instead, it is further attacked by LiAlH4 to form an alkoxide, which is then protonated when H3O+ is introduced into the reaction flask. An alternative method for reducing carboxylic acids involves the use of borane (BH3). -Reduction with borane is often preferred over reduction with LiAlH4, because borane reacts selec- tively with a carboxylic acid group in the presence of another carbonyl group. As an example, if the following reaction were performed with LiAlH4 instead of borane, both carbonyl groups would be reduced:

Example of reaction conversions

The reactions discussed thus far enable the interconversion between dihalides and terminal alkynes: -Put X at the most substituted carbon -Eliminates teh terminal alkyne

Acyl substitution reactions

The reactivity of carboxylic acid derivatives is similar to the reactivity of aldehydes and ketones in a number of ways. In both cases, the carbonyl group is electrophilic and subject to attack by a nucle- ophile. In both cases, the same rules and principles govern the proton transfers that accompany the reactions, as we will soon see. Nevertheless, there is one critical difference between carboxylic acid derivatives and aldehydes/ketones. Specifically, carboxylic acid derivatives possess a heteroatom that can function as a leaving group, while aldehydes and ketones do not. During the first step, in which a nucleophile attacks the carbonyl group, the hybridization state of the carbon atom changes. In both the starting material and product, the carbon atom is sp2 hybridized with trigonal planar geometry, but the same atom in the intermediate is sp3 hybridized with tetrahedral geometry. In recognition of this geometric change, the intermediate is often called a tetrahedral intermediate. In the second step, the carbonyl group is reformed via loss of a leaving group. Reformation of the CO double bond is a powerful driving force, and even poor leaving groups (such as RO ) can be expelled under certain conditions. Hydride ions (H−) and carbanions (C−) cannot function as leaving groups, so this type of reaction is not observed for ketones or aldehydes. When a nucleophile attacks a carbonyl group to form a tetrahedral intermediate, the carbonyl group will be re-formed, if possible, but H − and C − are generally not expelled as leaving groups.

Wolff-KIshner reaction

The reduction of an aldehyde or ketone into an alkane How? Reaction of the carbonyl group with hydrazine to for hydrazone followeed by treatment with KOH and heat -Second step is nonreversible -Notice that four of the five steps in the mechanism are proton transfers, the exception being the loss of N2 gas to generate a carbanion. The evolution of nitrogen gas renders this step irrevers- ible and forces the reaction to completion. As a result, the yields for this process are generally very good. -Carbanions are more stable and reactive than a carbocation

Preparation and reactions of acid chlorides

There is also R2CuLi and LiAl(OR)3H

Acid halide - mechanisms

These are the same three steps used in part 1 of the previous mechanism. This reaction produces HCl as a by-product. The HCl can often produce undesired reactions with other functional groups that might be present in the compound, so pyridine can be used to remove the HCl as it is produced. ' Pyridine is a base that reacts with HCl to form pyridinium chloride. This process effectively traps the HCl so that it is unavailable for any other side reactions.

Reactions of esters

Treating an ester with only one equivalent of LiAlH4 is not an efficient method for preparing an aldehyde, because aldehydes are more reactive than esters and will react with LiAlH4 immediately after being formed. If the desired product is an aldehyde, then DIBAH (diisobutylaluminum hydride) can be used as a reducing agent instead of LiAlH4. The reaction is performed at low temperature to prevent further reduction of the aldehyde.

Reduction of amides

When treated with excess LiAlH4, followed by aqueous workup, amides are converted into amines. -This is the first reaction we have seen that is somewhat different than the other reactions in this chapter. In this case, the carbonyl group is completely removed. Furthermore, the workup procedure is also different in this case. We use H2O (rather than H3O+), because the product of this reaction is an amine, and amines are protonated in the presence of H3O+ to give ammonium ions (RNH3+). To avoid protonating the amine, we use H2O for the workup of this reaction.

Addition of bromine or chlorine (Don't need to draw reactions)

alkynes have two π bonds rather than one and can, therefore, add two equivalents of the halogen to form a tetrahalide: CCl4 is a solvent In some cases, it is possible to add just one equivalent of halogen to produce a dihalide. Such a reac- tion generally proceeds via an anti addition (just as we saw with alkenes), producing the E isomer as the major product:

Epoxidation

the addition of a single oxygen atom to an alkene to form an epoxide -The first step of the process involves conversion of the alkene into an epoxide, and the second step involves opening the epoxide to form a trans diol (Mechanism 8.6). An epoxide is a three- membered, cyclic ether. Occurs via syn addition of an O atom from either side of the planar double bond, so both C-O bonds are inherently formed on the same side

Halogenation of alkenes: addition of Cl2 or Br2

-Carbocations are not intermediates in halogenation -Halogenation involves the addition of X2 (either Br2 or Cl2) across an alkene. -In this mechanism, an additional curved arrow has been introduced in the first step, forming a bridged intermediate rather than a free carbocation. This bridged intermediate, called a bromonium ion, is simi- lar in structure and reactivity to the mercurinium ion discussed in Section 8.6. Compare their structures: -In the second step of the proposed mechanism, the bromonium ion is attacked by the bromide ion that was produced in the first step. This step is an SN2 process and must therefore proceed via a back-side attack (as seen in Section 7.4). The requirement for back-side attack explains the observed stereochemical requirement for anti addition. The stereochemical outcome for halogenation reactions is dependent on the configuration of the starting alkene. For example, cis-2-butene will yield different products than trans-2-butene: -Anti addition across cis-2-butene leads to a pair of enantiomers, while anti addition across trans- 2-butene leads to a meso compound. These examples illustrate that the configuration of the starting alkene determines the configuration of the product for halogenation reactions.

(elimination followed by addition) can be used to change the posi- tion of an OH group, but we must remember that hydroxide is a very poor leaving group. Let's see what to do when dealing with an OH group. As an example, consider how the following transfor- mation might be achieved:

-It is possible to protonate an OH group using concentrated acid, which converts a bad leaving group into an excellent leaving group (see Section 7.10). How- ever, the acid-catalyzed dehydration of an alcohol cannot be used here because it is an E1 process, which always yields the Zaitsev product, not the Hofmann product. The regiochemical outcome of an E1 process cannot be controlled. The regiochemical outcome of an E2 process can be controlled, but an E2 process cannot be used in this example, because OH is a bad leaving group. It is not possible to protonate the OH group with a strong acid and then use a strong base to achieve an E2 reaction, because when mixed together, a strong base and a strong acid will simply neutralize each other. The question remains: How can an E2 process be performed when the desired leaving group is an OH group? -In Chapter 7, we explored a method that would allow an E2 process in this example, which main- tains control over the regiochemical outcome. The OH group can first be converted into a tosylate, which is a much better leaving group than OH (for a review of tosylates, see Section 7.10). After the OH group is converted into a tosylate, the strategy outlined above can be followed. Specifically, a strong, sterically hindered base is used for the elimination reaction, followed by anti-Markovnikov addition of H and OH:

Convert to -OTS, then do an SN2 reaction

-OH is a bad leaving group; -OTs is a good leaving group -The mechanism of adding TsCl to O (first step) is an SN2 reaction. The stereocenter doesn't change because the Ts attaches to O, not the carbon -As seen in Section 7.12, an alcohol can be converted into a tosylate, followed by nucleophilic attack. ---Using tosyl chloride and pyridine, the hydroxyl group is converted into a tosylate group (an excel- lent leaving group), which is susceptible to an SN2 process. Notice the stereochemical outcome of the previous reaction. The configuration of the chiral center is not inverted during formation of the tosylate, but it is inverted during the SN2 process. The net result is inversion of configuration.

Markovnikov addition is observed when alkynes are treated with HX:

-Once again, the halogen is installed at the more substituted position. When the starting alkyne is treated with excess HX, two successive addition reactions occur, producing a geminal dihalide:

S N 1 vs E1 reactions S N 2 vs E2 reactions

-Reactions of 2 degree and 3 degree haloalkanes in polar protic solvent give mixtures of S N 1 and E1 products -For the majority of cases, S N 1 predominates over E1, however, the ratio of the two is difficult to predict' It is much easier to predict the ratio of S N 2 and E2 reactions. The guiding principles are: 1. Branchign near the carbon slows S N 2 reactions due to steric effects and increases the rate of E2 reaction due to greater stability of alkene products 2. When considering 2 alkyl halides halides, if the pKa of the conjugate acid of the nucleophile/base is: -<11 and is a good nucleophile, then an S N 2 reaction will dominate ->11, an E2 reaction will dominate For secondary alkyl halides

Alkylation of terminal alkynes

-Recall the terminal alkynes and acidic -Resulting conjugate base is a strong nucleophile -NH2/NaNH2 are the nucleophile The conjugate base of a terminal alkyne, called an alkynide ion, can only be formed with a sufficiently strong base. Sodium hydroxide (NaOH) is not a suitable base for this purpose, but sodium amide (NaNH2) can be used. There are several bases that can be used to deprotonate acetylene or terminal alkynes, as seen in Table 9.2. The three bases shown on the top left of the chart are all strong enough to deprotonate a terminal alkyne, and all three are commonly used to do so. Notice the position of the negative charge in each of these cases (N−, H−, or C−). In contrast, the three bases on the bottom left of the chart all have the negative charge on an oxygen atom, which is not strong enough to deprotonate a terminal alkyne.

Carbon nucleophiles: cyanohydrins

-The C≡N is an electron withdrawing group -C≡N is a cyano group -≡N is a strong nucleophile -HCN is a weak acid. CN- is a strong nucleophilex -KCN is potassium cyanide When treated with hydrogen cyanide (HCN), aldehydes and ketones are converted into cyanohy- drins, which are characterized by the presence of a cyano group and a hydroxyl group connected to the same carbon atom: -In the first step, a cyanide ion attacks the carbonyl group. The resulting intermediate then abstracts a proton from HCN, regenerating a cyanide ion. In this way, cyanide functions as a catalyst for the addition of HCN to the carbonyl group. Rather than using a catalytic amount of base to form cyanide ions, the reaction can simply be performed in a mixture of HCN and cyanide ions (from KCN). The process is reversible, and the yield of products is therefore determined by equilibrium concentrations. For most aldehydes and unhindered ketones, the equilibrium favors formation of the cyanohydrin:

Acetyl formation (addition of alcohols across C=O) -- Ketones, aldehydes, and cyclic acetals

-The equilibrium arrows in Mechanism 19.5 (acetal formation) indicate that the process is gov- erned by an equilibrium. For many simple aldehydes, the equilibrium favors formation of the acetal, so aldehydes are readily converted into acetals by treatment with two equivalents of alcohol in acidic conditions: -However, for most ketones, the equilibrium favors reactants rather than products: -In such cases, formation of the acetal can be accomplished by removing one of the products (water) via a special distillation technique. By removing water as it is formed, the reaction can be forced to completion. Notice that acetal formation requires two equivalents of the alcohol. That is, two molecules of ROH are required for every molecule of ketone. -Alternatively, a compound containing two OH groups can be used, forming a cyclic acetal. This reaction proceeds via the regular seven-step mech- anism for acetal formation: three steps for formation of the hemiacetal followed by four steps for formation of the cyclic acetal: -----This is more favorable because it has lower entropy

SN1

-involves more than one step -rate = k[substrate] If we carefully study the energy diagram in Figure 7.15, we can rationalize the observed first order kinetics for this process. For any process with more than one step, the step with the highest- energy transition state (the highest point on the curve) is called the ratedetermining step (RDS), because the rate of the entire process is most profoundly affected by any changes in the rate of that step. The rate-determining step is also generally the step with the highest energy of activation (for more on this, see the margin note). In Figure 7.15, which depicts an SN1 process, the first step (loss of a leaving group) is the rate-determining step of this process because the transition state for this step is the highest point on the curve and because this step has the highest energy of activation. The rates of the other steps are less significant in determining the overall rate of the entire process—if we could somehow lower the energy of the transition state of either the second or third step, the rate of the entire process would not be greatly affected. With this in mind, take notice that the solvent only functions as a nucleophile in the second step of our process (not in the rate-determining step). The rate-determining step is said to be unimolecular because it involves only one chemical entity (the substrate). The term SN1 is used to refer to unimolecular nucleophilic substitution reactions: Just as with teh S N 2 reaction, the best leaving groups are those that stabilize the negative charge in the transition state To promote an S N 1 reaction, use a polar protic solvent to stabilize the full and partial charges that form during the S N 1 mechanism

There are two common ways to deprotonate an alcohol, forming an alkoxide ion.

1. A strong base can be used to deprotonate the alcohol. A commonly used base is sodium hydride (NaH), because hydride (H−) deprotonates the alcohol to generate hydrogen gas, which bub- bles out of solution: 2. Alternatively, an alcohol can be deprotonated upon treatment with Li, Na, or K. These metals react with the alcohol to liberate hydrogen gas, producing the alkoxide ion.

Two important reductions of alkynes to alkenes...

1. Hydrogenation to a cis alkene (using a poisoned catalyst) Under these conditions, the cis alkene is difficult to isolate because it is even more reactive toward further hydrogenation than the starting alkyne. The obvious question then is whether it is possible to add just one equivalent of hydrogen to form the alkene. With the catalysts we have seen thus far (Pt, Pd, or Ni), this is difficult to achieve. However, with a partially deactivated catalyst, called a poisoned catalyst, it is possible to convert an alkyne into a cis alkene (without further reduction): -There are many poisoned catalysts. One common example is Lindlar's catalyst. -This process does not produce any trans alkene. The stereochemical outcome of alkyne hydrogenation can be rationalized in the same way that we rationalized the outcome of alkene hydrogenation (Section 8.9). Both hydrogen atoms are added to the same face of the alkyne (syn addition) to give the cis alkene as the major product. 2. Dissolving metal reduction to a trans alkene (involves radicals and a tougher mechanism So far, in this section, we have explored the conditions that enable the reduction of an alkyne to a cis alkene. Alkynes can also be reduced to trans alkenes via an entirely different reaction called dissolv- ing metal reduction: -The reagents employed are sodium metal (Na) in liquid ammonia (NH3 ) (NOT NaNH2). Ammonia has a very low boiling point (−33°C), so use of these reagents requires low temperature. When dissolved in liquid ammonia, sodium atoms serve as a source of electrons:

All ketone/aldehyde reactions will go one of two ways

1. Includes strong nucleophiles like LAH, RMgBR, NaBH4, OH, OR, SR. CN, H 2. Acidic conditions - Includes weak nucleophiles like H2O, ROH, NH2R (needs to to protonate first, creating a stronger electrophile). -In acidic conditions, the same two mechanistic steps are observed, but in reverse order—that is, the carbonyl group is first protonated and then undergoes a nucleophilic attack (Mechanism 19.2). -It is true that the carbonyl group is already a fairly strong electrophile; however, a protonated car- bonyl group bears a full positive charge, rendering the carbon atom even more electrophilic. This is especially important when weak nucleophiles, such as H2O or ROH, are employed -When a nucleophile attacks a carbonyl group under either acidic or basic conditions, the posi- tion of equilibrium is highly dependent on the ability of the nucleophile to function as a leaving group. A Grignard reagent is a very strong nucleophile, but it does not function as a leaving group (a carbanion

Synthesis of epoxides

1. Using peroxy acids (using mCPBA) -Concerted (one step -Retention of stereochemistry 2. From halohydrins (an SN2 reaction) Halohydrins can be converted into epoxides upon treatment with a strong base -The process is achieved via an intramolecular Williamson ether synthesis. An alkoxide ion is formed, which then functions as a nucleophile in an intramolecular SN2like process (Mechanism 13.4). This provides us with another way of forming an epoxide from an alkene: -The overall stereochemical outcome is the same as direct epoxidation with a peroxy acid, such as MCPBA. That is, substituents that are cis to each other in the starting alkene remain cis to each other in the epoxide, and substituents that are trans to each other in the starting alkene remain trans to each other in the epoxide:

Gringard reactions with carboxylic acids

A Grignard reagent will not attack the carbonyl group of a carboxylic acid. Instead, the Grignard reagent will simply function as a base to deprotonate the carboxylic acid. In other words, a Grignard reagent is incompatible with a carboxylic acid. For similar reasons, it is not possible to form a Grignard reagent in the presence of even mildly acidic protons, such as the proton of a hydroxyl group. This Grignard reagent cannot be formed, as it will simply attack itself to produce an alkoxide.

SN2

A concerted reaction: bond breaking and bond making occur at the same time -the mechanism must exhibit a step in which the alkyl halide and the nucleophile collide with each other. Because that step involves two chemical entities, it is said to be bimolecular. Ingold and Hughes coined the term SN2 to refer to bimolecular nucleophilic substitution reactions: -Rate=k[electrophile][nucleophile] All S N 2 reactions proceed with backside attack of the nucleophile, resulting in inversion of configuriation at a stereogenic center There are two ways to explain why the reaction proceeds through back-side attack: 1. The lone pairs of the leaving group create regions of high electron density that effectively block the front side of the substrate, so the nucleophile can only approach from the back side. 2.Molecular orbital (MO) theory provides a more sophisticated answer. Recall that molecular orbitals are associated with the entire molecule (as opposed to atomic orbitals, which are asso- ciated with individual atoms). According to MO theory, the electron density flows from the HOMO of the nucleophile into the LUMO of the electrophile. As an example let's focus our attention on the LUMO of methyl bromide (Figure 7.3). - If a nucleophile attacks methyl bromide from the front side, the nucleophile will encounter a node and as a result, no net bonding will result from the overlap between the HOMO of the nucleophile and the LUMO of the electrophile. In contrast, nucleophilic attack from the back side allows for efficient over- lap between the HOMO of the nucleophile and the LUMO of the electrophile. - The observed stereochemical outcome for an SN2 process (inversion of configuration) is consistent with a concerted mecha- nism. The nucleophile attacks with simultaneous loss of the leaving group.

Electrophilic addition of HX (adds two times because there are two pi bonds)

A similar mechanism can be proposed for addition of HX to a triple bond: (1) protonation to form a carbocation followed by (2) nucleophilic attack: -This proposed mechanism invokes an intermediate vinylic carbocation (vinyl = a carbon atom bearing a double bond) and can successfully explain the observed regioselectivity. Specifically, the reaction is expected to proceed via the more stable, secondary vinylic carbocation, rather than via the less stable, primary vinylic carbocation:

Ozonolysis

Addition of ozone to the pi bond of the alkene forms an unstable intermediate called a molozonide, which rearranges to an ozonide by a stepwise process There are many reagents tthat will add across an alkene and complete cleave the C-C bond. In this section, we will explore one such reaction, called ozonolysis. Consider the following example -Notice that the CC bond is completely split apart to form two CO bonds. Therefore, issues of stereochemistry and regiochemistry become irrelevant. In order to understand how this reaction occurs, we must first explore the structure of ozone. -Ozone is a compound with the following resonance structures: Ozone is formed primarily in the upper atmosphere where oxygen gas (O2) is bombarded with ultraviolet light. The ozone layer in our atmosphere serves to protect us from harmful UV radiation from the sun. Ozone can also be prepared in the laboratory, where it can serve a useful purpose. Ozone will react with an alkene to produce an initial, primary ozonide (or molozonide), which undergoes rearrangement to produce a more stable ozonide: -When treated with a mild reducing agent, the ozonide is converted into products:

Acid catalyzed hydration

Addition of water across a double bond in the presence of an acid is called acid-catalyzed hydra- tion. For most simple alkenes, this reaction proceeds via a Markovnikov addition, as shown. The net result is an addition of H and OH across the π bond, with the OH group positioned at the more substituted carbon: The reagent, H3O+, represents the presence of both water (H2O) and an acid source, such as sulfuric acid. These conditions can be shown in the following way: -where the brackets indicate that the proton source is not consumed in the reaction. It is a catalyst, and, therefore, this reaction is said to be an acid-catalyzed hydration. -The rate of acid-catalyzed hydration is very much dependent on the structure of the starting alkene. Compare the relative rates of the following three reactions and analyze the effects of an alkyl substituent on the relative rate of each reaction: With each additional alkyl group, the reaction rate increases by many orders of magnitude. Also notice that the OH group is installed at the more substituted position

Nucleophilic addition of H2O: hydration

Addition of water can give a 1,1-diol in equilibrium with the carbonyl compound. Equilibrium favors carbonyl -The position of equilibrium generally favors the carbonyl group rather than the hydrate, except in the case of very simple aldehydes, such as formaldehyde: -The rate of reaction is relatively slow under neutral conditions but is readily enhanced in the presence of either acid or base. That is, the reaction can be either acid-catalyzed or base-catalyzed, allowing the equilibrium to be achieved much more rapidly. Base catalyzed hydration: -In the first step, a hydroxide ion (rather than water) functions as a nucleophile. Then, in the second step, the intermediate is protonated with water, regenerating a hydroxide ion. In this way, hydroxide serves as a catalyst for the addition of water across the carbonyl group. -Under basic conditions, a mechanism will only be reasonable if it avoids the use or formation of strong acids (only weak acids may be present). Acid catalyzed hydration: -Under acid-catalyzed conditions, the carbonyl group is first protonated, generating a positively charged intermediate that is extremely electrophilic (it bears a full positive charge). This intermedi- ate is then attacked by water to form an oxonium ion (a cation in which the positive charge resides on an oxygen atom), which is deprotonated to give the product. -Under acidic conditions, a mechanism will only be reasonable if it avoids the use or formation of strong bases (only weak bases may be present).

Catalytic Hydrogenation

Alkynes undergo many of the same addition reactions as alkenes. For example, alkynes will undergo catalytic hydrogenation just as alkenes do: In the process, the alkyne consumes two equivalents of molecular hydrogen:

Wilkinson's catalyst

As seen earlier in this section, when hydrogenation involves the creation of either one or two chiral centers, a pair of enantiomers is expected: In both reactions a racemic mixture is formed. This raises the obvious question: Is it possible to create only one enantiomer rather than a pair of enantiomers? In other words, is it possible to perform an asymmetric hydrogenation? asymmetric induction might be possible through the use of a chiral catalyst. He reasoned that a chiral catalyst should be capable of lower- ing the energy of activation for formation of one enantiomer more dramatically than the other enantio- mer (Figure 8.9). In this way, a chiral catalyst could theoretically favor the production of one enantiomer over another, leading to an observed enantiomeric excess (ee). Knowles succeeded in developing a chiral catalyst by preparing a cleverly modified version of Wilkinson's catalyst. Recall that Wilkinson's catalyst has three triphenylphosphine ligands: Knowles' idea was to use chiral phosphine ligands, rather than symmetrical phosphine ligands:

Two reducing agents

Both LAH and NaBH4 reduce ketones and aldehydes. LAH will also reduce carboxylic acids and esters. LAH is more reactive

Anti dihydroxylation

Dihydroxylation is the addition of two hydroxyl groups to a double bond, forming a 1,2-diol -We have seen that bromonium and mercurinium ions can be attacked by water from the back side. In much the same way, a protonated epoxide can also be attacked by water from the back side, as seen in Mechanism 8.6. The necessity for back-side attack (SN2) explains the observed stereochemi- cal preference for anti addition. -In the final step of the mechanism, the oxonium ion is deprotonated to yield a trans diol. Once again, notice that water is shown as the base (rather than hydroxide) in order to stay consistent with the conditions. In acidic conditions, hydroxide ions are not present in sufficient quantity to partici- pate in the reaction, and therefore, they cannot be involved when drawing the mechanism. -First part of pic is epoxidation and second part is Anti dihydroxylation

Mechanism for dissolving metal reductions (WONT NEED TO DRAW THIS) (ONLY NEED TO UNDERSTAND THE MOVEMENT OF ELECTRONS AND WHY TRANS IS FORMED)

In the first step of the mechanism, a single electron is transferred to the alkyne, generating an inter- mediate that is called a radical anion. It is an anion because of the charge associated with the lone pair, and it is a radical because of the unpaired electron: -intermediate is a highly reactive radical anion. The anion quickly protonated from the solvent. -Notice the use of single-barbed curved arrows in the first step of the mechanism to form the intermediate radical anion. The nature of this intermediate explains the stereochemical preference for formation of a trans alkene. Specifically, the intermediate achieves a lower energy state when the paired and unpaired electrons are positioned as far apart as possible, minimizing their repulsion: -The reaction proceeds more rapidly through the lower energy intermediate, which is then pro- tonated by ammonia under these conditions. In the remaining two steps of the mechanism, another electron is transferred, followed by one more proton transfer. That is, the net addition of molecular hydrogen (H2) is achieved via the installation of two electrons and two protons in the following order: e−, H+, e−, H+. The net result is an anti addition of two hydrogen atoms across the alkyne

Preparation of sulfides from thiols Predict the product

In the first step, hydroxide is used to deprotonate the thiol and produce a thiolate ion. The thiolate ion then functions as a nucleophile and attacks an alkyl halide, producing the sulfide. The process follows an SN2 pathway, so the regular restrictions apply. The reaction works well with methyl and primary alkyl halides, can often be accomplished with secondary alkyl halides, and does not work for tertiary alkyl halides. Since sulfides are structurally similar to ethers, we might expect sulfides to be as unreactive as ethers, but this is not the case. Sulfides undergo several important reactions. Sulfides will attack alkyl halides in an SN2 process. -The product of this step is a powerful alkylating agent, because it is capable of transferring a methyl group to a nucleophile.

How to make a gringard reagent

In this section, we will discuss formation of alcohols using Grignard reagents. A Grignard reagent is formed by the reaction between an alkyl halide and magnesium. A Grignard reagent is characterized by the presence of a C-Mg bond. Carbon is more electronegative than magnesium, so the carbon atom withdraws electron density from magnesium via induction. This gives rise to a partial negative charge (δ−) on the carbon atom. In fact, the difference in electronegativity between C and Mg is so large that the bond can be treated as ionic.

This method can be used to convert alkenes into diols with fairly high yields, but there are several disadvantages. In particular, OsO4 is expensive and toxic. To deal with these issues, several methods have been developed that use a co-oxidant that serves to regenerate OsO4 as it is consumed during the reaction. (conceptual checkpoint 8.31)

In this way, OsO4 functions as a catalyst so that even small quantities can produce large quantities of the diol. Typical co-oxidants include N-methylmorpholine N-oxide (NMO) and tert- butyl hydroperoxide: A different, although mechanistically similar, method for achieving syn dihydroxylation involves treatment of alkenes with cold potassium permanganate under basic conditions: Once again, a concerted process adds both oxygen atoms simultaneously across the double bond. Notice the similarity between the mechanisms of these two methods (OsO4 vs. KMnO4).

anti-Markovnikov addition

Interestingly, attempts to repeat Markovnikov's observations occasionally met with failure. In many cases involving the addition of HBr, the observed regioselectivity was, in fact, the opposite of what was expected—that is, the bromine atom would be installed at the less sub- stituted carbon, which came to be known as anti-Markovnikov addition. These curious observations fueled much speculation over the underlying cause, with some researchers even suggesting that the phases of the moon had an impact on the course of the reaction. Over time, it was realized that purity of reagents was the critical feature. Specifically, Markovnikov addition was observed whenever purified reagents were used, while the use of impure reagents sometimes led to anti-Markovnikov addition. Further investigation revealed the identity of the impurity that most greatly affected the regioselectivity of the reaction. It was found that alkyl peroxides (ROOR), even in trace amounts, would cause HBr to add across an alkene in an anti-Markovnikov fashion: Ianti-Markovnikov addition of HBr is known to proceed via an entirely different mechanism, one involving radical intermediates. This radical pathway (anti- Markovnikov addition) is efficient for the addition of HBr but not HCl or HI, and the details of that process will be discussed in Section 10.10. For now, we will simply note that the regiochemical out- come of HBr addition can be controlled by choosing whether or not to use peroxides:

Mechanism for oxidation by chromic acids

Know step 2 (E2) -Chromic acid oxidations are believed to proceed via a mechanism comprised of two stages (Mech- anism 12.8). The first stage involves formation of a chromate ester, and the second stage is an E2 process to form a carbon-oxygen π bond (rather than a carbon-carbon π bond). -When a primary alcohol is oxidized with chromic acid, a carboxylic acid is obtained. It is generally difficult to control the reaction to produce the aldehyde.

LAH as a reducing agent

LAH is also a delivery agent of H−, but it is a much stronger reagent. It reacts violently with protic solvents (such as water), and therefore, a protic solvent cannot be present together with LiAlH4 in the reaction flask. First the ketone or aldehyde is treated with LiAlH4, and then, in a separate step (called workup), the pro- ton source is added to the reaction flask. For the workup step, H3O+ can be used as the proton source, although H2O can also be used as a proton source: -When LiAlH4 is used to reduce a ketone or aldehyde, the workup step must be shown as a separate step, because LiAlH4 cannot be in the reaction flask together with H3O+(or H2O). Both NaBH4 and LiAlH4 will reduce ketones or aldehydes. These hydride delivery agents offer one significant advantage over catalytic hydrogenation in that they can selectively reduce a carbonyl group in the presence of a C=C bond. Consider the following example: When treated with a hydride reducing agent, only the carbonyl group is reduced. In contrast, catalytic hydrogenation will also reduce the C=C bond under the conditions required to reduce the carbonyl group (high temperature and pressure). For this reason, hydride reducing agents such as NaBH4 and LiAlH4 are generally preferred over catalytic hydrogenation.

E2

Like Sn2 reactions, E2 reactions are concerted - all bonds are broken and formed in a single step -When an alkyl halide is treated with a strong base, the rate of reaction is generally found to be dependent on the concentrations of both the alkyl halide and the base. Specifically, doubling the concentration of the alkyl halide causes the reaction rate to double, and similarly, doubling the concentration of the base also causes the rate to double. rate=k[substrate][base] -Based on these observations, we conclude that the mechanism must exhibit a step in which the alkyl halide and the base collide with each other. Because that step involves two chemical entities, it is said to be bimolecular -The experimental observations for E2 reactions are consistent with a concerted process, involving both the base and the alkyl halide in a single step (Mechanism 7.2). Such a process is expected to exhibit second-order kinetics, exactly as observed. If the observations had been otherwise, a concerted mechanism could not have been proposed. In Section 7.9, we will explore other observa- tions that are consistent with a concerted process. tertiary alkyl halides undergo E2 reactions quite rapidly. A substitution reaction occurs when the reagent functions as a nucleophile and attacks the α position, while an elimination reaction occurs when the reagent functions as a base and abstracts a proton from a β position. With a tertiary sub- strate, steric hindrance prevents the reagent from functioning as a nucleophile at an appreciable rate, but the reagent can still function as a base without encountering much steric hindrance (Figure 7.9). In general, a cis alkene will be less stable than its stereoisomeric trans alkene.

Carbonyl groups can be converted into alcohols

More bonds in oxygen → more bonds in hydrogen The conversion of a ketone (or aldehyde) to an alcohol is a reduction. The reaction requires a reducing agent, which is itself oxidized as a result of the reaction. In this section, we will explore three reducing agents that can be used to convert a ketone or aldehyde to an alcohol: 1. In Chapter 8, we learned that an alkene can undergo hydrogenation in the presence of a metal catalyst such as platinum, palladium, or nickel. A similar reaction can occur for ketones or aldehydes, although more forcing conditions are generally required (higher temperature and pressure), so this method is rarely used by synthetic chemists. 2. Sodium borohydride (NaBH4) is a common reducing agent that can be used to reduce ketones or aldehydes. -Sodium borohydride functions as a source of hydride (H:−) and the solvent functions as the source of a proton (H+). The solvent can be ethanol, methanol, or water. The precise mecha- nism of action has been heavily investigated and is somewhat complex. Nevertheless, Mecha- nism 12.1 presents a simplified version that will be sufficient for our purposes. The first step involves the transfer of a hydride to the carbonyl group (C==O bond) and the second step is a proton transfer

Hydration of terminal alkynes, 2 ways

Note that the product is either a ketone or aldehyde enol is initially forms and is converted into a ketone (or aldehyde) -These are called tautomers. They are constitutional isomers. -The enol and ketone are said to be tautomers, which are constitutional isomers that rapidly inter- convert via the migration of a proton. The interconversion between an enol and a ketone is called keto-enol tautomerization. Tautomerization is an equilibrium process, which means that the equi- librium will establish specific concentrations for both the enol and the ketone. Generally, the ketone is highly favored, and the concentration of enol will be quite small. Be very careful not to confuse tau- tomers with resonance structures. Tautomers are constitutional isomers that exist in equilibrium with one another. Once the equilibrium has been reached, the concentrations of ketone and enol can be measured. In contrast, resonance structures are not different compounds and they are not in equilib- rium with one another. Resonance structures simply represent different drawings of one compound. -Keto-enol tautomerization is an equilibrium process that is catalyzed by even trace amounts of acid (or base). Glassware that is scrupulously cleaned will still have trace amounts of acid or base adsorbed to its surface. As a result, it is extremely difficult to prevent a keto-enol tautomerization from reaching equilibrium, which generally occurs very rapidly.

demercu- ration

Notice that the attack takes place at the more substituted position, ultimately leading to Markovnikov addition. After attack of the nucleophile, the mercury can be removed through a process called demercu- ration, which can be accomplished with sodium borohydride. There is much evidence that demercura- tion occurs via a radical process. The net result is the addition of H and a nucleophile across an alkene: Many nucleophiles can be used, including water. This reaction sequence provides for a two-step process that enables the hydration of an alkene with- out carbocation rearrangements:

add and then eliminate:(skill builder 8.11, 8.12)

Once again, a retrosynthetic analysis of the alkene target molecule reveals this synthetic strategy: -The target molecule (a monosubstituted alkene) can be prepared by an elimination reaction, so we delete the π bond and place H and Br atoms on those two carbon atoms to draw a suitable alkyl halide starting material. The Br atom is placed on the more-substituted carbon atom, because that allows us to continue the retrosynthesis back to the given starting material (which has a π bond at that position). -This two-step sequence (addition followed by elimination) makes it possible to change the posi- tion of a π bond. When using this strategy, the regioselectivity of each step can be carefully controlled. In the first step (addition), Markovnikov addition is achieved by using HBr, while anti-Markovnikov addition is achieved by using HBr with peroxides. In the second step (elimination), the Zaitsev prod- uct can be obtained by using a strong base, while the Hofmann product can be obtained by using a strong, sterically hindered base.

Conversion into R-X

Primary and secondary alcohols can react with SOCl2 or PBr3 via an SN2 process: (pic) -Both go in SN2 reaction -Better to use these reactants than HCl or HBr because HCl and HBr are strong acids and create carbocations, which could be rearranged. -These two reactions are believed to proceed via similar mechanisms. In the reaction with SOCl2, the first few steps convert a bad leaving group into a good leaving group, then the halide attacks in an SN2 process (as shown in Mechanism 12.6, although the exact mechanism with SOCl2 depends on the experimental conditions). Mechanisms of ROH→ RBr (pic) - Same idea for SOCl2 - Notice the similarity among all of the SN2 processes that we have seen in this section. All involve the conversion of the hydroxyl group into a better leaving group followed by nucleop- hilic attack. If any of these reactions occurs at a chiral center, inversion of configuration is to be expected. tertiary: (pic)

SN2 Reactions with Alcohols

Primary and secondary alcohols will undergo substitution reactions with a variety of reagents, all of which proceed via an SN2 process. In this section, we will explore three such reactions that all employ an SN2 process. -Primary alcohols will react with HBr via an SN2 process. -The hydroxyl group is first protonated, converting it into an excellent leaving group, followed by an SN2 process. This reaction works well for HBr but does not work well for HCl. To replace the hydroxyl group with chloride, ZnCl2 can be used as a catalyst. -Because of its ionic nature, the use of ZnCl2 is limited to alcohols that are water-soluble. For such alcohols (Section 12.1), ZnCl2 converts the hydroxyl group into a better leaving group, thereby enabling the subsequent SN2 process

Wittig reaction mechanism (making ylide + carbonyl group_

Produces a phosphine oxide, which helps drive the reaction. Don't need to know the reaction. The ylide carbon is the nucleophile, the carbonyl is the electrophile,

Acetyl formation (addition of alcohols across C=O)

Produces a protected carbonyl, creating a tertiary carbon that isn't as reactive and less electrophilic -Common acids used for this purpose include para-toluenesulfonic acid (TsOH) and sulfuric acid (H2SO4) As mentioned earlier, the acid catalyst serves an important role in this reaction. Specifically, in the pres- ence of an acid, the carbonyl group is protonated, rendering the carbon atom even more electrophilic. This is necessary because the nucleophile (an alcohol) is weak; it reacts with the carbonyl group more rapidly if the carbonyl group is first protonated.

E1 reaction of alcohols

Recall from Section 7.10 that alcohols undergo elimination reactions in acidic conditions. For tertiary alcohols, this transformation follows an E1 mechanism: Recall that the two core steps of an E1 mechanism are loss of a leaving group followed by a proton transfer. However, when the starting material is an alcohol, the mechanism must begin with an addi- tional step, in which the hydroxyl group is protonated. Recall that elimination generally favors the more substituted alkene.

SN1 Reactions with Alcohols

Recall that an SN1 mechanism has two core steps (loss of leaving group and nucleophilic attack). When the starting material is an alcohol, the mechanism must begin with an additional step, in which the hydroxyl group is protonated. This reaction proceeds via a carbocation intermediate and is therefore most appropriate for tertiary alcohols. When dealing with a primary or sec

Mechanism for LAH reducing an ester to an alcohol

Reduction of the ester involves the transfer of hydride two times, as seen in Mechanism 12.3. LiAlH4 delivers hydride to the carbonyl group, but then loss of a leaving group causes the carbonyl group to re-form. In the presence of LiAlH4, the newly formed carbonyl group can be attacked again by hydride. The leaving group in the second step of the mechanism is a methoxide ion, which is generally not a good leaving group. For example, methoxide does not function as a leaving group in E2 or SN2 reactions. The reason that it can function as a leaving group in this case stems from the nature of the intermediate after the first attack of hydride. That intermediate is high in energy and already exhibits a negatively charged oxygen atom. The intermediate is therefore capable of ejecting methoxide, because such a step is not uphill in energy.

Synthesis of ROHs

SN1/SN2 -A primary substrate will require SN2 conditions (a strong nucleophile), while a tertiary substrate will require SN1 conditions (a weak nucleophile). -With a secondary substrate, neither SN2 nor SN1 is particularly effective for preparing a secondary alcohol. Under SN1 conditions, the reaction is generally too slow, while SN2 conditions (use of hydroxide as the nucleophile) will generally favor elimination over substitution. Hydration reactions and acid catalyzed reactions -Acid-catalyzed hydration proceeds with Markovnikov addition (Section 8.6). That is, the hydroxyl group is installed at the more substituted position. It is a useful method if the substrate is not sus- ceptible to carbocation rearrangements (Section 6.11). In a case where the substrate can possibly rearrange, oxymercuration-demercuration can be employed. This approach also proceeds via Mar- kovnikov addition, but it does not involve carbocation rearrangements. Hydroboration-oxidation is used to achieve an anti-Markovnikov addition of water.

Catalytic hydrogenation

The addition of H2 only occurs in the presence of a metal catalyst involves the addition of molecular hydrogen (H2) across a double bond in the presence of a metal catalyst; for example: The net result of this process is to reduce an alkene to an alkane. there are no chiral centers in the product, so the stereospecificity of the pro- cess is irrelevant. In order to explore whether a reaction displays stereospecificity, we must examine a case where two new chiral centers are being formed. For example, consider the following case: With two chiral centers, there are four possible stereoisomeric products (two pairs of enantiomers): However, the reaction does not produce all four products. Only one pair of enantiomers is observed, the pair that results from a syn addition. To understand the reason for the observed stereospecificity, we must take a close look at the reagents and their proposed interactions. In any given case, the stereochemical outcome is dependent on the number of chiral centers formed in the process, as summarized in Table 8.2.

Base catalyzed tautomerization

The initial product of this reaction is an enol that cannot be isolated because it is rapidly converted into an aldehyde via tautomerization. As we saw earlier in this section, tautomerization cannot be prevented, and it is catalyzed by either acid or base. In this case, basic conditions are employed, so the tautomerization process occurs via a base-catalyzed mechanism (Mechanism 9.3). Notice that the order of events under base-catalyzed conditions is the reverse of the order under acid-catalyzed conditions. That is, the enol is first deprotonated (in contrast, under acid-catalyzed conditions, the first step is to protonate the enol). Deprotonation of the enol leads to a resonance- stabilized anion called an enolate ion, which is then protonated to generate the aldehyde.

Summary of S N 1 and S N 2 reactions

The mechanism that is favored is hye one that has the fastest reaction rate fora given set of reactants. This is what dictates the major product formed SN2 Rate equation:rate=k[substrate][nucleophile] Rate of reaction: Methyl > 1 degree > 2 degree > 3 degree Stereochemistry: Inversion of configuration Alkyl halide: CH3X, RCH2X (1 degree) -favored by: strong nucleophiles (usually a net negative charge), polar parotid solvents SN1 Rate equation: k[substrate] Rate of reaction: 3 degree> 2 degree > 1 degree > methyl Stereochemistry: racemization (with slight preference for inversion due to ion pairs) Alkyl halide: R3X (3 degree) -favored by: weak nucleophiles (usually neutral), polar protic solvents SN1 and SN2 Alkyl halide: R2CHx (2 degree). The mechanism depends on the condiitons -strong nucleophiles favor the SN2 mechanism over the SN1 mechanism. For example, RO- is a stronger nucleophile than ROH, so RO- favors the SN2 reaction and ROH favors the SN1 reaction -protic solvents favor the SN1 mechanism and parotid solvents favor the SN2 mechanism. For example, H2O and CH3OH are polar protic solvents that favor the SN1 mechanism, whereas acetone ((CH3)2C=O) and DMSO are polar parotid solvents that favor the SN2 mechanism

one-step method for moving the location of a bromine atom (skill builder 8.10)

The net result is the change in position of the Br atom. How can this type of transformation be achieved? This chapter did not present a one-step method for moving the location of a bromine atom. However, this transformation can be accomplished in two steps: an elimination reaction fol- lowed by an addition reaction: This same strategy could be revealed by performing a retrosynthetic analysis of the target molecule. There is more than one potential starting material to synthesize the tertiary alkyl bromide, but only the trisubstituted alkene is an intermediate compound that can also be readily prepared from the given starting material. An alkene is an ideal intermediate to change the position of a bromine atom, because elimination of an alkyl bromide generates a π bond, and addition of HBr to the π bond generates an alkyl bromide. When performing this two-step sequence, there are a few important issues to keep in mind. In the first step (elimination), the product can be the more substituted alkene (Zaitsev product) or the less substituted alkene (Hofmann product): Note that a careful choice of reagents makes it possible to control the regiochemical outcome, as we first saw in Section 7.7. With a strong base, such as sodium methoxide (NaOMe) or sodium ethoxide (NaOEt), the major product is the more substituted alkene. With a strong, sterically hindered base, such as potassium tert-butoxide (t-BuOK), the major product is the less substituted alkene. After forming the double bond, the regiochemical outcome of the next step (addition of HBr) can also be controlled by the reagents used. HBr produces a Markovnikov addition while HBr/ROOR produces an anti-Markovnikov addition.

S N 1 or S N 2 mechanism?

The nucleophile: -Strong nucleohiles favor a S N 2 mechanism -Weak nucleophiles favor an S N 1 mechanism by decreasing the rate of any competing S N 2 reaction The leaving group: -A better leaving roup increases the rate of both SN1 and SN2 reactions -A better leaving gorup is more able to accept hte negative charge The solvent: -Polar potic solvents favor S N 1 reactions because the ionic intermediates (both cations and anions) are stabilized by solvation -Polar aprotic solvents favor S N 2 reactions because nucelophiles are not well solvated, adn therefore are more nucleophilic What about sp2 hybriidzed organohalides? -S N 1 and S N 2 reactions can only happen at sp3 hybdirized carbon centers

S N 1 and E1

The rate equation is said to be first order, because the rate is linearly dependent on the concen- tration of only one compound (the substrate). This observation demonstrates that the products must be formed via processes other than SN2 and E2 reactions, which would be expected to exhibit second-order rate equations. In this section, we will explore new mechanisms, called SN1 and E1, that account for these and other observations. Both mechanisms begin with the same first step—loss of a leaving group to give a carbocation intermediate: The tertiary alkyl halide is said to undergo ionization, because it dissociates into a pair of ions (a car- bocation and a halide ion). The fate of the carbocation intermediate then determines the resulting product. There are two possibilities, depending on the role played by the solvent in the next step. Either the solvent can function as a nucleophile, giving a substitution reaction, or the solvent can function as a base, giving an elimination reaction.

Hydration by HOH and hydroalkoxylation

The truth is that most reactions represent equilibrium processes; however, organic chemists generally draw equilibrium arrows only in situations where the equilibrium can be easily manipulated (allowing for control over the product distribution). Acid- catalyzed dehydration is an excellent example of such a reaction. -For acid-catalyzed dehy- dration, there is yet another way to control the equilibrium. The equilibrium is sensitive not only to temperature but also to the concentration of water that is present. By controlling the amount of water present (using either concentrated acid or dilute acid), one side of the equilib- rium can be favored over the other -Control over this equilibrium derives from an understanding of Le Châtelier's principle. To understand how this principle applies, consider the above process after equi- librium has been established. Notice that water is on the left side of the reaction. How would the introduction of more water affect the system? The concentrations would no longer be at equilibrium, and the system would have to adjust to reestablish new equilibrium concentra- tions. The introduction of more water will cause the position of equilibrium to move in such a way that more of the alcohol is produced. Therefore, dilute acid is used to convert an alkene into an alcohol. On the flip side, removing water from the system would cause the equilibrium to favor the alkene. Therefore, concentrated acid is used to favor formation of the alkene. Alternatively, water can also be removed from the system via a distillation or chemical process, which would also favor the alkene. In summary, the outcome of a reaction can be greatly affected by carefully choosing the reaction conditions and concentrations of reagents. -Hydroalkoxylation results in addition of an alcohol to form an ether

Make thiols by SN2 reactions

Thiols can be prepared via an SN2 reaction between sodium hydrosulfide (NaSH) and a suitable alkyl halide This reaction can occur even at secondary substrates without competing E2 reactions, because the hydrosulfide ion (HS−) is an excellent nucleophile and a poor base. When this nucleophile attacks a chiral center, inversion of configuration is observed.

Mechanism for reversing acetal - principle of microscopic reversibility

This process is called a hydrolysis reaction, because bonds are cleaved (shown with red wavy lines) by treatment with water. Acetal hydrolysis generally requires acid catalysis. That is, acetals do not undergo hydrolysis under aqueous basic conditions: Acetal hydrolysis is believed to proceed via Mechanism 19.9. Since acidic conditions are employed, the mechanism does not involve any reagents or intermediates that are strong bases. For example, in the third step of the mechanism, water is used as a nucleophile, rather than hydroxide, because the latter is not present in substantial quantities. Similarly, H3O+ is used in all protonation steps (such as the first step), because the use of water as a proton source would generate a hydroxide ion, which is unlikely to form in acidic conditions. Always remember that a mechanism must be consistent with the conditions employed. In acidic conditions, strong bases should not be invoked as reagents or intermediates.

To do an E2, must make -OH a good leaving group (-OTs)

This transformation can also be accomplished via an E2 pathway if the hydroxyl group is first con- verted into a better leaving group, such as a tosylate. A strong base can then be employed to accom- plish an E2 reaction. As seen in Section 7.10, when more than one alkene product is possible, an E2 process will generally favor formation of the more substituted alkene (unless a sterically hindered base is used), and no carbocation rearrangements are observed in E2 processes. -Avoids N2SO4 -No rearrangement -Controls alkene position by base

synthesis of Sulfonates

Tosylates (ROTs) are more commonly used and can be prepared by treating the corresponding alcohol (ROH) with tosyl chloride (TsCl) in the presence of pyridine. Under these conditions, the alcohol serves as a nucleophile and displaces the chloride ion in TsCl, and the resulting oxonium ion is deprotonated by pyridine (a base) to give an alkyl tosylate: The bond between R and O (in ROH) is not broken during the process, and as a result, a chiral alcohol will retain its configuration when converted to its corresponding tosylate: The configuration (S ) does not change during formation of the tosylate. A common abbreviation for pyridine is "py," as shown in the reaction above. Alkyl sulfonates undergo substitution and elimination reactions, much like the corresponding alkyl halides. For example, when treated with a strong nucleophile, an alkyl sulfonate will undergo an SN2 reaction, characterized by inversion of configuration:

Under these strongly acidic conditions, the OH group can be protonated, thereby converting it into a good leaving group. This renders the substrate susceptible to an E1 process, giving an alkene as the product:

Under these strongly acidic conditions, the OH group can be protonated, thereby converting it into a good leaving group. This renders the substrate susceptible to an E1 process, giving an alkene as the product: The overall process involves formation of an alkene via the removal of water (H and OH) and is therefore called a dehydration reaction. As shown in the mechanism above, dehydration of an alcohol (under acidic conditions) occurs via a three-step mechanism. The first step is protonation of the alcohol, followed by the two steps of an E1 process (loss of leaving group and deproton- ation). When drawing the first step of this mechanism, notice that a hydronium ion (H3O+) is shown as the source of the proton, rather than H2SO4. Recall that the leveling effect (Sec- tion 3.7) prevents the presence of acids stronger than H3O+ in an aqueous environment. Con- centrated sulfuric acid is generally a mixture of H2SO4 and water, so H3O+ will be the strongest acid present in substantial quantities.

Protecting groups - process

Unlike a hydroxyl group, this protecting group is compatible with a Grignard reagent because it has no acidic protons. In other words, the OH group is "protected" from undergoing a proton transfer reaction after being converted to an OTMS group. The OH group of an alcohol is converted into an OTMS group upon treatment with tri- methylsilyl chloride, abbreviated TMSCl. This reaction is believed to proceed via an SN2-like process (called SN2-Si), in which the hydroxyl group functions as a nucleophile to attack the silicon atom and a chloride ion is expelled as a leaving group. A base, such as triethylamine, is then used to remove the proton connected to the oxygen atom. Notice that the first step involves an SN2-like process occurring at a tertiary substrate. This should seem surprising because in Chapter 7 we learned that a nucleophile cannot effectively attack a sterically hindered substrate. This case is different because the electrophilic center is a silicon atom rather than a carbon atom. Bonds to silicon atoms are typically much longer than bonds to carbon atoms, and this longer bond length opens up the back side for attack. After the desired Grignard reaction has been performed, the trimethylsilyl group can be removed easily with either H3O+ or fluoride ion. A commonly used source of fluoride ion is tetrabutylammonium fluoride (TBAF).

Solvolysis

When a tertiary alkyl halide undergoes ionization in a polar solvent, such as EtOH, the solvent can func- tion as a nucleophile and attack the intermediate carbocation, resulting in a two-step substitution process: -Since a solvent molecule functions as the attacking nucleophile, the process is called a solvolysis. Notice that the result of this two-step substitution process is an oxonium ion (an intermediate with a positive charge on an oxygen atom). This oxonium ion can then lose its proton to a solvent mole- cule, giving the observed substitution product: -The first region (blue) shows the two steps of the substitution process (loss of a leaving group to give a carbocation intermediate, followed by nucleophilic attack). The next region (purple) shows the proton transfer step that removes the proton from the oxonium ion, giving the observed product. If we focus our attention on the first two steps of this process (the substitution steps), we will find that these are exactly the same two steps of an S 2 process, but the steps do not occur at the same time. In an S 2 process, these two steps (loss of leaving group and nucleophilic attack) occur in a concerted fashion (simultaneously), and there is no carbocation intermediate. But in this case, unlike in an SN2 process, the two steps occur separately and the reaction proceeds via a carbocation intermediate. The geometry of the carbocation intermediate allows nucleophilic attack from both sides of the planar species

Nucleophilic opening of epoxides

When an epoxide is subjected to attack by a strong nucleophile, a ring-opening reaction occurs. For example, consider the opening of ethylene oxide by a hydroxide ion.

Cleavage by ozonolysis + problems

When treated with ozone followed by water, alkynes undergo oxidative cleavage to produce carboxylic acids: -When a terminal alkyne undergoes oxidative cleavage, the terminal carbon atom is converted into carbon dioxide:

Halohydrin formation

With unsymmetrical alkenes the preferred product has the electrophile X+ bonded to the less substituted carbon and OH bonded to the more substituted carbon -In most cases, halohydrin formation is observed to be a regioselective process. Specifically, an OH group is generally installed at the more substituted position: The proposed mechanism for halohydrin formation can justify the observed regioselectivity. Recall that in the second step of the mechanism, the bromonium ion is captured by a water molecule: Focus carefully on the position of the positive charge throughout the reaction. Think of the positive charge as a hole (or more accurately, a site of electron deficiency) that is passed from one place to another. It begins on the bromine atom and is transferred to the oxygen atom. In order to do so, the positive charge must pass through a carbon atom in the transition state: In other words, the transition state for this step will bear partial carbocationic character. This explains why the water molecule is observed to attack the more substituted carbon. The more sub- stituted carbon is more capable of stabilizing the partial positive charge in the transition state. As a result, the transition state will be lower in energy when nucleophilic attack occurs at the more substituted carbon atom. The proposed mechanism is therefore consistent with the observed regi- oselectivity of halohydrin formation.

Crown ethers

Without the crown ether, KF would simply not dissolve in benzene. The presence of 18crown6 generates a complex that dissolves in benzene. The result is a solution containing fluoride ions, which enables us to perform substitution reactions with F− as a nucleophile. Generally, it is too difficult to use F− as a nucleophile, because it will usually interact too strongly with the polar solvents in which it dissolves. The strong interaction between fluoride ions and polar solvents makes it difficult for F− to become "free" to serve as a nucleophile. However, the use of 18crown6 allows the creation of free fluoride ions in a nonpolar solvent, making substitution reactions possible. For example: Another example is the ability of 18crown6 to dissolve potassium permanganate (KMnO4) in ben zene. Such a solution is very useful for performing a wide variety of oxidation reactions. Other metal cations can be solvated by other crown ethers. For example, a lithium ion is solvated by 12crown4, and a sodium ion is solvated by 15crown5.

Hydroboration oxidation

a two step reaction sequence that converts an alkene to alcohol -a method for achieving an anti-Markovnikov addition of water. This process, called hydroboration-oxidation, installs the OH group at the less substituted position: The stereochemical outcome of this reaction is also of particular interest. Specifically, when two new chiral centers are formed, the addition of water (H and OH) is observed to occur in a way that places the H and OH on the same face of the π bond: -Hydroboration-oxidation fractions proceed via syn addition Mechanism involves a concerted addition of H and BH2 to the same face of a double bond. Stereochemistry is retained in the second step

Oxymercuration-demercuration

acid-catalyzed hydration can be used to achieve a Markovnikov addition of water across an alkene. The utility of that process is somewhat diminished by the fact that carbocation rearrangements can produce a mixture of products : In cases where protonation of the alkene ultimately leads to carbocation rearrangements, acid- catalyzed hydration is an inefficient method for adding water across the alkene. Many other methods can achieve a Markovnikov addition of water across an alkene without carbocation rearrangements. One of the oldest known methods is called oxymercuration-demercuration: To understand this process, we must explore the reagents employed. The process begins when mer- curic acetate, Hg(OAc)2, dissociates to form a mercuric cation:

Hydrohalogenation

addition of H-X across a double bond - opposite of dehydrohalogenation The treatment of alkenes with HX (where X = Cl, Br, or I) results in an addition reaction called hydrohalogenation, in which H and X are added across the π bond: In the first step, the π bond of the alkene is protonated, generating a carbocation interme- diate. In the second step, this intermediate is attacked by a bromide ion. Figure 8.2 shows an energy diagram for this two-step process. The observed regioselectivity for this process can be attributed to the first step of the mechanism (proton transfer), which is the rate-determining step because it exhibits a higher transition state energy than the second step of the mechanism. -After the nucleophilic attack, it can either form a syn or anti

dihydroxylation reactions

character- ized by the addition of OH and OH across an alkene. As an example, consider the dihydroxylation of ethylene to produce ethylene glycol: There are a number of reagents well suited to carry out this transformation. Some reagents pro- vide for an anti dihydroxylation, while others provide for a syn dihydroxylation. In this section, we will explore a two-step procedure for achieving anti dihydroxylation