Aliphatic Nucleophilic Substitution

Richard F. Daley and Sally J. Daley www.ochem4free.com

Organic Chemistry Chapter 12

12.1 Naming Single Bonded Heteroatom Functional Groups 579 12.2 Comparing Nucleophilic Substitution Reaction Mechanisms 586 12.3 The SN1 and SN2 Reaction Mechanisms 588 12.4 Stereochemistry of Nucleophilic Substitutions 592 12.5 The Substrate 595 12.6 Nucleophiles and Leaving Groups 601 12.7 Common Nucleophiles 606 12.8 The Reaction Medium 607 12.9 SN1 versus SN2 613 12.10 Halide Nucleophiles 613 Synthesis of 1-Bromobutane 616 12.11 Oxygen Nucleophiles 620 12.12 Nitrogen Nucleophiles 625 Synthesis of 2,5-Diaminoadipic Acid 628 12.13 Carbon Nucleophiles 631 12.14 Neighboring Group Participation 634 Special Topic - SN1 vs. SN2 637 Key Ideas from Chapter 12 641 Organic Chemistry - Ch 12 578 Daley & Daley

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Chapter 12

Aliphatic Nucleophilic Substitution

Chapter outline

12.1 Naming Single Bonded Heteroatom Functional Groups Naming alcohols, ethers, phenols, thiols, thioethers, sulfides, and amines 12.2 Comparing Nucleophilic Substitution Reaction Mechanisms A comparison of the mechanism for SN1 and SN2 reactions with the mechanism of nucleophilic substitution at carbonyl groups 12.3 The SN1 and SN2 Mechanisms More details for the SN1 and SN2 reaction mechanisms 12.4 Stereochemistry of Nucleophilic Substitutions The stereochemistry of the SN1 and SN2 reaction mechanisms 12.5 The Substrate Effect of substrate structure on the outcome of nucleophilic substitution reactions 12.6 Nucleophiles and Leaving Groups The factors that make good nucleophiles or leaving groups 12.7 Common Nucleophiles A listing of some of the common nucleophiles used in organic chemistry 12.8 The Reaction Medium Effect of the solvent on the outcome of nucleophilic substitution reactions 12.9 SN1 versus SN2 Factors that influence whether an E1 or an E2 mechanism will prevail 12.10 Halide Nucleophiles Substitution reactions with halide nucleophiles 12.11 Oxygen Nucleophiles Reactions using oxygen nucleophiles 12.12 Nitrogen Nucleophiles Using nitrogen nucleophiles in substitution reactions 12.13 Carbon Nucleophiles Substitution reactions with carbon nucleophiles derived from Grignard reagents and alkyllithium reagents 12.14 Neighboring Group Participation How a group adjacent to a functional group affects the outcome of reactions

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Objectives

✔ Apply the IUPAC rules of nomenclature to naming alcohols, thiols, ethers, sulfides, and amines

✔ Be able to write the mechanisms for both the SN1 and SN2 reaction types

✔ Know the effect of the SN1 and SN2 reaction mechanisms on the stereochemistry of reactions ✔ Recognize how the structure of the substrate affects the outcome of nucleophilic reactions ✔ Understand the factors that affect the stability of carbocations ✔ Recognize the factors that make good nucleophiles and good leaving groups ✔ Know how the solvent affects the outcome of nucleophilic substitution reactions ✔ Understand how halide, oxygen, nitrogen, and carbon nucleophiles react with various substrates ✔ Know how a functional group adjacent to the site of a nucleophilic substitution affects the course of reactions

If there was two birds sitting on a fence, he would bet you which one would fly first. - Mark Twain

liphatic nucleophilic substitution is the third A installment in the study of nucleophilic reactions in organic chemistry. The previous two installments, nucleophilic addition and nucleophilic substitution at the carbonyl group, describe the formation of new bonds to the carbonyl carbon of aldehydes, ketones, and the carboxylic acid family. Chapters 7 and 8 discuss the nature of the substituents bonded to the carbonyl carbon and how they determine whether a substitution or an addition reaction takes place. This chapter presents the study of aliphatic nucleophilic substitution reactions at sp3 hybridized carbon atoms. The outcome of

www.ochem4free.com 5 July 2005 Organic Chemistry - Ch 12 581 Daley & Daley aliphatic nucleophilic substitution reactions depends on such variables as nucleophilicity, stereochemistry, the reaction medium, and the intermediate's stability. Aliphatic nucleophilic substitution reactions have two, often competing, reaction mechanisms. These two mechanisms account for a large variety of reactions in organic chemistry. Although the two mechanisms are different, they both have a nucleophile, a leaving group, and a substrate; and all three play an important part in the outcome of each reaction. An understanding of the characteristics of these two mechanisms gives you a broader understanding of organic chemistry. What you learn about these two mechanisms will help you later with further organic syntheses and mechanisms. Through their work with aliphatic nucleophilic substitutions, chemists have perhaps learned more about organic reaction mechanisms than through their study of any other reaction type. However, these gains did not come easily. From the early years of the twentieth century until the 1970s, the debate raged over the issues surrounding aliphatic nucleophilic substitution. This debate included such things as experimental results that various investigators interpreted in opposite ways and experimental results that all agreed seemed mutually incompatible. Fortunately, most of these questions are now resolved.

12.1 Naming Single Bonded Heteroatom Functional Groups

This section covers the nomenclature of compounds that contain a singly bonded oxygen, sulfur, or nitrogen; that is, alcohols, ethers, phenols, thiols, thioethers (or sulfides), disulfides, and amines. Each of these functional groups replaces a hydrogen on the carbon skeleton of the organic compound. Each functional group also changes the suffix of the parent name in a way characteristic of that group. The parent compound, in most cases, includes the carbon in the main chain that bears the functional group. Alcohols replace a hydrogen on the parent carbon chain with an —OH group. The suffix for an alcohol is –ol in the IUPAC nomenclature. When naming an alcohol, follow these steps.

Step 1 Determine the parent chain. Step 2 Drop the final –e of the parent hydrocarbon name and add the –ol ending. Step 3 To number substituents, number the carbon chain to give the carbon bearing the —OH group the lowest possible number.

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CH3OH CH3CH2OH CH3CH2CHCH3 Methanol Ethanol 2-Butanol

OH OH

CH3

2-Methylcyclohexanol 3-Methyl-2-butanol

Frequently, chemists call the —OH group a hydroxy group and compounds that contain two or more hydroxy groups polyols. To A polyol is a molecule name polyols, use the IUPAC suffixes of –diol, –triol, and so forth. 1,2- that contains two or Ethanediol (HOCH CH OH) is an example of the proper naming of a more —OH groups. 2 2 polyol. Alcohols can be classified according to the number of carbon atoms bonded to the carbon that bears the alcohol group. For example, Primary, secondary, ethanol is a primary alcohol because the carbon bearing the hydroxy and tertiary alcohols have one, two, or three group is attached to only one other carbon. 2-Propanol is a secondary alkyl groups attached alcohol because the carbon bearing the —OH group is bonded to two to the carbon bearing other carbons, and 2-methyl-2-propanol is a tertiary alcohol because the —OH group. the —OH bearing carbon is bonded to three other carbons.

OH OH OH Ethanol 2-Propanol 2-Methyl-2-propanol A primary alcohol A secondary alcohol A tertiary alcohol

Benzyl alcohol is alcohol attached to a carbon that also bears Benzyl alcohol is a a phenyl group. carbon attached to a benzene ring and an —OH group. OH OH

I Benzyl alcohol 4-Iodobenzyl alcohol

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Chemists call a hydroxy group connected to a carbon atom of a A phenol is a benzene benzene ring a phenol after the simplest member of this group, ring substituted with hydroxybenzene or, more commonly, phenol. To name substituted an —OH group. phenols, number the ring so that the carbon bearing the —OH group is C—1. Because the common name for an alcohol is alkyl alcohol (i.e., ethanol is often called ethyl alcohol) many organic chemistry students are tempted to confuse benzyl alcohols with phenols. Be careful! Phenol nomenclature is complicated by the fact that many phenols have common names that are more widely used than the IUPAC names. For example, 2-methylphenol is known as o-cresol, 3- methylphenol is known as m-cresol and 4-methylphenol is known as p- cresol.

OH OH Cl OH

CH3 Phenol 3-Methylphenol 3-Chlorophenol (m-Cresol) (m-Chlorophenol)

A thiol, also called a Chemists call the sulfur analogs of the alcohols thiols, or mercaptan, contains an mercaptans. To name thiols, add the suffix –thiol to the parent —SH group. name. However, unlike most suffixes, the –thiol suffix does not replace the final –e of the hydrocarbon name.

CH3SH CH3CH2SH Methanethiol Ethanethiol

Thiols have a characteristic intensely unpleasant odor. Some common examples are the defensive odor of the skunk and the characteristic odors of garlic and onion. Replacing the oxygen in a phenol with a A thiophenol is the sulfur creates a class of compounds called thiophenols. Name sulfur analog of a thiophenols following the same rules for naming phenols. phenol. Exercise 12.1

Name the following compounds.

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a) b)

OH OH

c) d) Cl OH SH

e) f)

Sample solution

b) The common name for this compound is menthol. Menthol is extracted from the oil of peppermint. Its IUPAC name is 2-methyl-5- (1-methylethyl)cyclohexanol.

An ether contains two Ethers are compounds with two alkyl or aryl groups bonded to alkyl or aryl groups the same oxygen atom. Unfortunately, nomenclature for ethers is bonded to an oxygen. somewhat confusing because the IUPAC has not adopted a specific system for naming them. Currently, there are two systems that chemists usually use. One system names the least complicated of the alkyl groups bonded to the oxygen, plus the oxygen atom (RO—), as an An alkoxy group is alkoxy group, and the other alkyl group as the parent compound. The derived from an other system names the compound as an alkyl alkyl ether or as a alcohol with some substituent replacing dialkyl ether. This system calls both alkyl groups bonded to the the proton on the oxygen alkyl substituents of the ether. Here are some examples. oxygen.

CH3

CH3OCH CH CH OCH CH 3 2 2 3 CH3

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2-Methoxypropane Ethoxyethane or Or isopropyl methyl ether Diethyl ether

In general, when the two alkyl groups are of similar size, use the alkyl alkyl ether system; when one group is larger than the other, use the alkoxyalkane name. For cyclic ethers, the most common systematic name counts the oxygen as if it were a carbon in the ring and adds an oxa– prefix to the parent name. This prefix is rarely used with acyclic ethers. When numbering cyclic ethers, begin with the ether oxygen. An example of a cyclic compound is a three-membered ring with one oxygen and two carbons. It is named oxacyclopropane. The other examples shown below have their common names in parentheses.

O O Oxacyclopentane 1,4-Dioxacyclohexane (1,4-Dioxane) (Tetrahydrofuran)

Sulfides are the sulfur Sulfides, or alkylthio compounds, are sulfur analogs of analogs of an ether. ethers. The alkylthio name is comparable to the alkoxy name. A

Alkylthio groups are number of sulfides have two sulfur atoms bonded together with the sulfur analogs of formula RSSR. Chemists call these compounds disulfides. alkoxy groups. CH SCH(CH ) CH CH SCH CH Disulfides contain two 3 3 2 3 2 2 3 sulfurs bonded 2-Methylthiopropane Ethylthioethane together. or or Isopropyl methyl sulfide Diethyl sulfide

S S

Dicyclohexyl disulfide

Exercise 12.2

Using IUPAC nomenclature, name the following compounds.

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a) b) (CH ) COCH 3 3 3

c) d) O

SSCH3

(As a substituent, an ethenyl group is often called a vinyl group) e) f) S

Sample solution

e) The name of this cyclic ether is 3-(1-methylethyl)oxacyclopentane. Most chemists call this 3-isopropyltetrahydrofuran.

Amines consist of a Amines are organic derivatives of ammonia (NH3) with one, nitrogen bonded to a two, or three alkyl groups to replace the hydrogens bonded to the combination of three carbon or hydrogen nitrogen. Chemists classify these derivatives as primary, secondary, or substituents. tertiary amines, respectively. The terms primary, secondary, and tertiary have somewhat different meanings for amines than they do for alcohols. Ammonia with one hydrogen replaced by an alkyl group (RNH2) is a primary amine; with two hydrogens replaced, it is a secondary amine (R2NH); and with all three replaced, it is a tertiary amine (R3N).

CH3NH2 (CH3)2NH (CH3)3N Methanamine N-Methylmethanamine N,N-Dimethylmethanamine or or or Methylamine Dimethylamine Trimethylamine (A primary amine) (A secondary amine) (A tertiary amine)

To name amines using IUPAC nomenclature, follow these steps.

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Step 1 Determine the parent compound. Step 2 Drop the final –e of the alkane name of the parent hydrocarbon and add the suffix –amine. A second system for naming amines uses the suffix –ylamine for naming amines with similar hydrocarbon substituents. For example, CH3NH2 is called methylamine. Step 3 Name the alkyl groups of secondary and tertiary amines as substituents to the parent compound name by adding the letter N to show that the group is bonded to the nitrogen atom. a) For secondary amines, add one N. b) For tertiary amines, add two Ns. Step 4 If the molecule contains a second nitrogen, denote any substituents on that nitrogen with the letter N'. Here are some examples of amines.

N(CH3)2

(CH3)2NCH2CH(CH3)2 N,N,2-Trimethylpropanamine N,N-Dimethylcyclohexanamine

CH3NHCH2CH2N(CH3)2 N,N,N'-Trimethylethanediamine

Step 5 Use the prefix amino–, or alkylamino–, when the amine is a substituent on a parent molecule with another functional group. For example, when an amine and an alcohol are in the same molecule, the alcohol group determines the parent name and either of the prefixes amino– or alkylamino– names the amine as a substituent group.

N(CH3)2

OH 2-(N,N-Dimethylamino)cyclopentanol

Step 6 For some amines, particularly the cyclic amines, chemists sometimes use the prefix aza–, which is similar to the oxa– prefix of ethers. The name for an amine group attached to a carbon atom of a benzene ring is aniline.

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NH2

N H Azacyclohexane Aniline (Piperidine)

Exercise 12.3

Using IUPAC nomenclature, name the following compounds.

a) b) (CH ) CCH NH CH CH 3 3 2 2 2 3 CH NCH CH CH 3 2 2 3 c) d) NH2

N(CH CH ) 2 3 2 e) f) NH CH CH CH CH CH NH H C CHCH CH NH 2 2 2 2 2 2 2 2 2 2 2

Sample solution b) N-Ethyl-N-methyl-1-propanamine

12.2 Comparing Nucleophilic Substitution Reaction Mechanisms

Recall from Chapter 8 that in a nucleophilic substitution reaction the nucleophile reacts with the substrate to displace the leaving group. The carbonyl carbon of a carboxylic acid, or one of its derivatives, is the obvious site for the reaction of a nucleophile because the carbonyl carbon is both positively polarized and unsaturated. Being positively polarized and unsaturated allows the carbonyl carbon to readily accept the bonding electron pair from the nucleophile. The sp3 hybridized carbon atoms also undergo nucleophilic substitution but for different reasons. Nucleophilic substitution reactions with sp3 hybridized carbon atoms take place because of the electronegativity difference between

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the leaving group and the carbon that it is bonded to. The leaving group is a heteroatom with a greater electronegativity than carbon. During a nucleophilic substitution reaction, the heteroatom departs with a pair of electrons formerly bonded to the carbon atom. Because the bond between an sp3 hybridized carbon and the leaving group is polar and because the leaving group is more electronegative than carbon, the carbon possesses a partial positive character. Thus, the carbon is an electrophile and is reactive towards a nucleophile. Although the fundamental reactions of a nucleophilic substitution of a carbonyl carbon and an sp3 hybridized carbon are the same, the mechanisms are not. With a carbonyl group, the π bond between the carbon and oxygen shifts making room for an incoming nucleophile to bond to the carbonyl carbon before the leaving group departs. An analogous nucleophilic substitution at an sp3 carbon would proceed as follows:

• Nu • + CL Nu C L Nu C + L•

However, this mechanism leads to an intermediate with five bonds to the carbon, which violates the octet rule. Thus, with an sp3 hybridized carbon, the nucleophile cannot first add to the hybridized carbon and then have the leaving group depart. It must follow some other mechanistic pathway. An aliphatic Aliphatic nucleophilic substitution reactions follow one of nucleophilic two mechanisms. Which mechanism a particular reaction takes substitution reaction occurs when an sp3 depends, in part, on the nature of the substrate. The difference hybridized substrate between the two mechanisms is when the leaving group departs. With reacts with a the first mechanism, the leaving group departs before the nucleophile nucleophile. arrives. In the second mechanism, the leaving group departs while the nucleophile is reacting. A carbocation is a The first of these two reaction mechanisms begins with a positively charged heterolytic cleavage of the bond between the sp3 carbon and the substrate with the leaving group in a heterolytic bond cleavage operation. The sp3 carbon charge on a carbon atom. is then electron-deficient. With the leaving group and a pair of electrons gone, the substrate is a positively charged carbocation. The The term SN1 means nucleophile then reacts with this carbocation in a heterogenic bond Substitution, forming the product. Chemists call a reaction that follows this Nucleophilic, mechanistic pathway an S 1 reaction. Unimolecular N

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CL C + L• Slow step

Nu • + C Nu C Fast step

The S 1 reaction mechanism N

In a reaction that follows the second reaction mechanism pathway, the leaving group departs as the nucleophile arrives. Simultaneously, the bond between the carbon and the leaving group breaks, and the bond between the carbon and the nucleophile forms. The term SN2 means Chemists call this type of reaction mechanism an S 2 reaction. An Substitution, N Nucleophilic, SN2 reaction mechanism describes a concerted reaction with a 1,3- Bimolecular. electron pair displacement operation.

A concerted reaction ‡ involves simultaneous • Nu C L Nu C + L• bond making and bond Nu • + CL breaking.

The SN2 Reaction Mechanism

12.3 The SN1 and SN2 Reaction Mechanisms

2-Chloro-2-methylpropane (tert-butyl chloride) reacts with water using acetone as the solvent to produce 2-methyl-2-propanol (tert-butyl alcohol). In this reaction, a hydroxy group substitutes itself for the chlorine of the substrate. The reaction follows the SN1 mechanistic pathway.

H3C H3C H2O C Cl COH H3C acetone H3C CH3 CH3 tert-Butyl chloride tert-Butyl alcohol

Chemists used this reaction to learn about the mechanism of the SN1 reaction. They chose this particular reaction because of the relative ease of using it to perform kinetic studies. A typical experiment involved the reaction of the chloride with water at a carefully controlled temperature. Chemists then followed the reaction progress by determining the amount of HCl produced.

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As they studied the results of this experiment, they learned that the rate of the reaction depended only on the concentration of the tert-butyl chloride. Thus, the rate-determining step of the reaction involved only the substrate and not the nucleophile. Chemists call The rate for a first such a reaction a first order reaction. Below is the rate equation for order reaction depends a first order reaction. only on the concentration of one reagent in the reaction rate = k1[(CH3)3CCl] mixture. This rate equation gives the mechanism its designation of SN1, or a In a unimolecular Substitution, Nucleophilic, unimolecular (1) reaction. reaction the rate The separation of charge that occurs in the heterolytic bond depends on the cleavage step makes it the rate-determining step in an SN1 pathway. concentration of only Thus, the two-step process fits better with first order kinetics. one chemical species. Applying the SN1 mechanism from Section 12.2 to the reaction of tert- butyl chloride with water gives the following mechanism:

H3C CH3 •• •• • C Cl• C + • Cl• •• •• Slow H3C CH3 H3C CH3

CH3 H3C H •• •• • C + H2O• C O Fast H3C H H3C CH3 CH3

H C H3C H 3 •• •• C O C OH + H Fast H C •• H3C 3 CH H CH3 3

Figure 6.4 is on page The reaction profile, which looks similar to the one in Figure 000. 6.4, shows this three-step process. Notice that the rate-determining step is the ionization of the carbon—chlorine bond and that the rate depends only on the concentration of the substrate. The carbocation appears at the energy minimum between the transition state coming from the substrate and the transition state leading to the product, making it an intermediate product.

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H C ‡ 3

C Cl ‡ H C H3C 3 H3C C OH2 H3C H3C

‡ H3C C O HCl CH3 H3C H C H3C H C 3 H3C CH3 C Cl + H O 2 H3C H3C H3C C OH + Cl H3C 2 H3C H C C OH + HCl 3 H3C H3C

Reaction Progress Figure 12.1. Reaction profile for the SN1 reaction.

Exercise 12.4

Explain these observations about the reaction of tert-butyl chloride with water to form tert-butyl alcohol. a) Adding hydroxide ion, a better nucleophile than water, to the reaction mixture makes little difference in the rate of reaction. b) However, adding chloride ion to the reaction mixture markedly slows down the rate of reaction.

The reaction of 1-chloropropane with water to form 1-propanol follows the SN2 reaction mechanism. When using acetone as the solvent, the rate of reaction is slow. However, adding hydroxide ion to the mixture increases the rate of reaction. These data are typical of an SN2 reaction mechanism.

OH CH3CH2CH2Cl CH3CH2CH2OH 1-Propanol

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The rate of this reaction depends on the concentration of both the alkyl chloride and the hydroxide ion. The kinetics make it a second The rate for a second order reaction, because changes in the concentrations of both order reaction depends reactants affect the rate of reaction. on the concentrations of two reagents in the c- reaction mixture. rate = k[CH3CH2CH2Cl][ OH]

The designation for this reaction is SN2, or a Substitution, In a bimolecular Nucleophilic, bimolecular (2) reaction. reaction the rate The mechanism for the reaction of 1-chloropropane with water depends on the concentrations of two is the concerted reaction shown in Section 12.2. With this mechanism, chemical species. the chloride ion leaves at the same time as the bond forms between the sp3 carbon atom and the hydroxide ion. At that moment, the sp3 carbon atom is fully bonded to three substituents and partially bonded to both the nucleophile and the leaving group.

‡ H H

CH3CH2CH2Cl + OH HO CCl

CH2CH3

CH CH CH OH + Cl 3 2 2

The reaction profile for the SN2 reaction looks similar to the Figure 6.3 is found on one in Figure 6.3. That profile shows only one transition state with no page 000. intermediates between the reactants and products. The reaction profile for the reaction of chloropropane with water, Figure 12.2, also shows only one transition state indicating that the reaction depends on the presence of both the substrate and the nucleophile.

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‡ H H HO CCl

CH2CH3

CH3CH2CH2Cl + OH

CH3CH2CH2OH + Cl

Reaction Progress

Figure 12.2. Reaction profile of the SN2 reaction.

Exercise 12.5

The reaction of ethyl bromide with the cyanide ion follows an SN2 pathway. a) Write the rate equation for this reaction. b) What happens to the rate of the reaction if the concentration of cyanide ion is doubled? What happens to the rate of the reaction if the concentrations of both cyanide and ethyl bromide are doubled?

12.4 Stereochemistry of Nucleophilic Substitutions

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Chemists often use the product of an sp3 hybridized carbon nucleophilic substitution reaction to identify which mechanism operated in that particular reaction. When the reactive site is a stereogenic center, a nucleophilic substitution reaction can yield An inversion reaction inversion, retention, or racemization of the configuration at the produces a product asymmetric carbon. See Figure 12.3. In an inversion, the product has with an opposite the opposite stereochemistry of the substrate. In a retention, the stereochemistry from product has the same stereochemistry of the substrate. Most the substrate. nucleophilic substitution reactions with retention products are beyond

Retention retains the the scope of an introductory organic course, so this book does not cover substrate any examples of them. With racemization, equal amounts of both stereochemistry. retention and inversion occur.

Racemization produces equal amounts of both. X X • Nu • +LCL Nu C + • Y Y Z Z

Inversion

X X Nu • +LCL CNu+ • Y Y Z Z

Retention

X X X

Nu • + CL CNu+ Nu C + L • Y Y Y Z Z Z

Equal amounts

Racemization

Figure 12.3. The three possible stereochemical outcomes for a nucleophilic substitution on an asymmetric, or chiral, substrate.

Because the asymmetric carbon atom in an SN1 reaction loses the leaving group before reacting with the nucleophile, the intermediate carbocation that forms at the loss of the leaving group For more on the has only three bonds to the carbon atom. According to the VSEPR VSEPR theory, see theory, these three bonds must lie in a plane with the carbon atom at Section 1.7, page 000. the center. With the carbon atom in that position, the intermediate is symmetrical as shown in Figure 12.4. That symmetry allows the nucleophile to react from either side of the carbon. Thus, an SN1 reaction produces a racemic product.

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X + C

Y Z

Nu•

Figure 12.4. The nucleophile can attack either side of the carbocation in an SN1 reaction.

In an SN2 reaction, the leaving group departs from the asymmetric carbon atom at the same time that the nucleophile arrives. Because the leaving group blocks access to the “front” of the carbon, the nucleophile must attack the “back” of the carbon. This attack from the “back” of the carbon causes the molecule to invert. Thus, SN2 reactions produce inversion products. To understand the stereochemistry of an SN2 reaction, reexamine the molecular orbital picture of an sp3 hybrid carbon For more on orbital. As you remember, an sp3 hybrid orbital has two lobes, one on hybridization, see either side of the carbon atom, 180o apart. The two lobes are of Section 1.7, page 000. different sizes: a small lobe, which chemists often neglect to draw, and a much larger lobe.

An sp3 hybrid orbital

The larger lobe is the lobe involved in bonding. It bonds the carbon to the leaving group and, after the leaving group leaves, it bonds the nucleophile to the carbon. However, the leaving group blocks the nucleophile from having access to the larger lobe of this orbital. To gain access to the larger lobe, the nucleophile attacks the back of the carbon and literally turns the tetrahedron of the carbon atom inside out—like an umbrella caught in the wind. Chemists call an attack on A backside attack is an the back of the carbon a backside attack. A backside attack, attack from the illustrated in Figure 12.5, leads to the inversion of the configuration of opposite side of the the carbon. Because the nucleophile reacts with the back of the carbon carbon from the bearing the leaving group, it is sensitive to the size of the other groups leaving group. on that carbon. Large groups tend to block access to the back of the carbon atom.

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X Nu • C L

Y Z

Figure 12.5. Attack by the nucleophile on the nonbonding lobe (backside) of the sp3 hybrid orbital.

When a reaction produces a greater quantity of one stereoisomer than it does of another, the reaction is either A stereospecific stereospecific or stereoselective. An SN2 reaction is stereospecific reaction has only one because its product is an inversion. A stereoselective reaction is stereochemical outcome. similar to a racemization except that it produces an unequal mixture of stereochemical products. A stereoselective reaction produces an Exercise 12.6 unequal mixture of stereochemical outcomes. Draw the structure of the product expected from the SN2 reaction of each of the following substrates with KOH.

a) (S)-2-Bromobutane b) Br

H H

CH 3 c) d) (R)-3-Chlorohexane H H3C C Cl

Sample solution

b) H

H OH

CH 3

12.5 The Substrate

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A number of variables affect how an aliphatic nucleophilic substitution reaction proceeds. These variables include the substrate, the nucleophile, the leaving group, and the reaction medium. This and the next several sections cover each of these variables in turn. After presenting the general aspects about each variable, the section then discusses how that variable particularly affects a nucleophilic substitution at an sp3 hybridized carbon atom. The first step in an SN1 reaction is the formation of a carbocation from the substrate. The ease with which a carbocation intermediate forms depends directly on the number of groups in the substrate available to stabilize the positive charge of that carbocation. When only alkyl groups are attached to the reactive carbon, the more highly substituted the carbon, the more stable the carbocation. Thus, the order for carbocation stability is 3o > 2o >> 1o > methyl. This order is the reactivity order for an SN1 reaction, as well. An alkyl group stabilizes the positive charge of a carbocation in two ways: 1) through the inductive effect, and 2) through the field effect. In the case of alkyl groups the inductive effect is simply the donation of electron density through the σ bonds from the alkyl groups to the positively charged carbon. This effect distributes the positive charge over several atoms. Thus, the more alkyl groups bonded to the carbon, the greater the amount of donation that takes place.

Electron density is donated

C from the carbons adjacent to the carbocation.

C C

The carbons in the alkyl groups bonded to the positively charged sp3 hybridized carbon have atoms bonded to them, as well. Through the field effect of the orbitals in those other atoms, the carbons that they are bonded to donate electrons to the empty p orbital of the positively charged carbon. This type of overlap, which is called hyperconjugation, spreads the positive charge over additional Hyperconjugation is the overlap of an empty atoms beyond those affected by the inductive effect. The effect of orbital on one atom hyperconjugation is much weaker than the bonding overlap of orbitals, with a filled, bonding but the combined inductive and field effects stabilize the positively orbital on an adjacent charged carbon of the carbocation. Figure 12.6 illustrates atom. hyperconjugation of a carbocation.

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Overlap H

+ CC

Figure 12.6. In this example, hyperconjugation comes from the partial overlap of the empty p orbital on the electron deficient carbon and the filled bonding molecular orbitals of an adjacent carbon—hydrogen bond.

A carbocation often undergoes structural changes, called A rearrangement rearrangements, to form a more stable carbocation. Rearrangements reaction is one where occur either after the formation of the carbocation or as the leaving the carbon skeleton group departs. If the rearrangement takes place as the leaving group changes or the site of the functional group departs, the reaction is a concerted reaction. Because it is so much moves on the carbon higher in energy than secondary or tertiary carbocations, the skeleton. formation of a primary carbocation is very unlikely. Almost any time a primary substrate is used, a rearrangement occurs in preference to carbocation formation. The driving force for the rearrangement of a carbocation is that the product carbocation has a lower energy, or is more stable, than the starting carbocation. For example, a primary or a secondary carbocation rearranges to form a lower energy tertiary carbocation. The rearrangement of the 3-methyl-2-butyl carbocation (secondary or 2o) to 2-methyl-2-butyl carbocation (tertiary or 3o) shows this.

In a hydride shift, a H H hydrogen on a carbon adjacent to the CH3CHCCH3 CH3CHCCH3 positively charged CH CH carbon moves with its 3 3 bonding electrons from its carbon to the positively charged The above reaction is also an example of a hydride shift. carbon. A second type of rearrangement is a methide shift. When there is no adjacent hydrogen, a methyl group with its pair of bonding A methide shift is electrons moves from its carbon to the adjacent positive carbon. An similar to a hydride shift. It occurs when example is the rearrangement of the 2,2-dimethylpropyl (neopentyl) there are no hydrogens carbocation to form the 2-methyl-2-butyl carbocation. to shift.

CH3

CH3CCH2 CH3CCH2CH3 CH CH 3 3

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In this example, the methide shift occurs while the leaving group is leaving because the primary carbocation requires too much energy to form directly. The energy difference between the primary and the tertiary carbocation is approximately 12 kcal/mole. Both the hydride and the methide shifts are part of a group of rearrangements called A [1,2]-shift means [1,2]-shifts. some group migrates from one atom to an adjacent atom. Exercise 12.7

Draw the structure of the product expected from an SN1 reaction of each of the following substrates with H2O. Justify your product with a mechanism.

a) 2-Bromo-3-methylbutane b)

CH2Br

c) d)

CH2Br Br

Sample solution

c) H

•• Br • ••

•• •OH •• • H2O • –H H

An SN2 reaction is very sensitive to crowding at the backside of the substrate's sp3 hybridized carbon atom. Thus, the more steric crowding present in a molecule, the slower is its rate of reaction. The rate of reaction for methyl groups and most primary alkyl groups is relatively rapid. For secondary groups the rate of reaction is generally much slower. And most tertiary groups fail to have a reasonable

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reaction rate at all. The reactivity order for the SN2 reaction is methyl o o o > 1 > 2 > 3 , which is the opposite of the reactivity order for the SN1 reaction. The reactivity order for the SN2 reaction is due to the steric hindrance of the larger groups. As the nucleophile moves to the back of the carbon bearing the leaving group, it must pass close to the other groups attached to that carbon. See Figure 12.7. (Please make models to help yourself clearly understand these ideas.) With small groups, such as hydrogens, the nucleophile reaches the back of the carbon with little interference. However, with larger groups, methyl groups in this example, the nucleophile must pass quite close to them to reach the carbon. With three methyl groups, the area between them is often smaller than the incoming nucleophile making it impossible for the nucleophile to reach the backside of the carbon atom.

H H - O C Cl H H

H H O- C Cl H

CH3

- O C Cl H

CH3

Figure 12.7. Comparison of the steric hindrance of an SN2 reaction of hydroxide with a methyl, primary, and tertiary halides.

Table 12.1 shows some examples of the effects of structure on the rate of reaction via SN1 and SN2 reaction mechanisms. These reaction rates are not absolute, but are based on the arbitrary assignment of 1 as the rate for a methyl group.

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Alkyl group Relative Rate for Relative Rate for SN1 Reaction SN2 Reaction CH3— 1 1 CH3CH2— 1 .03 (CH3)2CH— 12 .013 6 –6 (CH3)3C— 1.2 x 10 3 x 10

Table 12.1. Relative rates of reaction for SN1 and SN2 on different substrates with a chloride leaving group.

Solved Exercise 12.1

Each of the following alkyl halide substrates is chiral and is present as a single enantiomer. Show the stereochemistry of the product of the nucleophilic substitution reaction on these alkyl halides. a) CN H I KCN H O, CH OH 2 3

Solution The product of this reaction is a nitrile. Because iodide ion is a good leaving group and the substrate is reasonably open sterically, the reaction follows an SN2 mechanism. The product forms from an inversion reaction of the substrate.

H I NC H KCN H O, CH OH 2 3 b)

CH3OH

H3CBr H3COCH3

Solution The substrate in this reaction is tertiary and follows an SN1 mechanism. An SN1 reaction has a much higher rate than an SN2 reaction. The product formed is a racemic mixture.

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CH3OH

H3CBr H3COCH3

H CO CH 3 3 c) H H

Cl CH3CH2OH OCH2CH3

Solution This substrate easily forms a carbocation because the carbocation is resonance-stabilized.

Thus, the product formed is a racemic mixture.

H H

Cl CH3CH2OH OCH2CH3 +

OCH2CH3

H + +

H OCH CH H OCH CH 2 3 2 3

Exercise 12.8

Rank the following compounds in order of increasing reactivity towards SN1 substitution. Then rank them in order of increasing reactivity towards SN2 substitution.

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(CH3)2 CHCH2Br CH3CH2Br (CH3CH2)3CBr

12.6 Nucleophiles and Leaving Groups

3 In an SN1 reaction, the addition of the nucleophile to the sp hybridized carbon takes place after the leaving group departs. Therefore, the nucleophile has no effect on the rate of the reaction because the rate of addition of the nucleophile is much faster than the rate of ionization. In an SN2 reaction, on the other hand, the addition of the nucleophile to the sp3 hybridized carbon takes place as the leaving group departs. So the nucleophile plays an important part in the rate of reaction. Understanding nucleophiles will help you understand how these reactions work. Both nucleophiles and leaving groups are bases. In an aliphatic nucleophilic substitution reaction the nucleophile is generally a stronger base than the leaving group. Nucleophiles are good electron donors and, in actuality, are Lewis bases. Chemists frequently differentiate between nucleophiles and bases—although those differences are often fuzzy. Use the following as a working definition to determine the differences between nucleophiles and bases: a base is To abstract an atom a compound that abstracts a proton, and a nucleophile is a compound means to remove it. that reacts with any other electrophilic atom. A stronger base abstracts the proton more rapidly, or more completely, than a weaker base. A stronger nucleophile reacts with an electrophile more rapidly, or more completely, than a weaker nucleophile. Under the appropriate conditions, a Lewis base acts as either a base or a nucleophile. When comparing similar species across the periodic table, basicity, and nucleophilicity correlate. NH3 is both a stronger base and a stronger nucleophile than H2O. And H2O is a stronger base and a stronger nucleophile than HF. However, this generalization does not apply when comparing atoms down the periodic table. As atomic number increases within a family of elements, basicity decreases and nucleophilicity increases. For example, H2S is a better nucleophile but a weaker base than H2O. Figure 12.8 summarizes these trends.

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Basicity Nucleophilicity

Figure 12.8. Trends of nucleophilicity and basicity for similar species in polar solvents, like water, with a proton attached to a heteroatom.

The concept of hard and soft acids and bases also helps Recall from Section differentiate bases from nucleophiles because nucleophilicity increases 5.3, page 000, that a as a base becomes softer. For example, the ability of the halide ions to hard base is small and act as leaving groups increases from fluoride (the hardest base in the nonpolarizable halogen family) to iodide (the softest base). whereas a soft base is The sulfite ion (SO 2c- ) demonstrates the differences between large and polarizible. 3 bases and nucleophiles. The sulfite ion has a central sulfur with three oxygen atoms bonded to it. Each oxygen atom bears a single formal negative charge, and the sulfur bears a single formal positive charge:

•• •• •• •O S O• •• ••

• O• ••

In the presence of an acid, the sulfite ion acts as a base; in the presence of an electrophile, it acts as a nucleophile. The oxygens of the sulfite ion are more basic than the sulfur, so one of the oxygens will c- react with a proton in acid to form the hydrogen sulfite (HSO3 ), or bisulfite, ion.

•• •• •• •OHS O •• ••

• O• ••

On the other hand, the sulfur is more nucleophilic than the oxygens; 2c- thus, SO3 reacts with methyl iodide to produce methyl sulfonate.

CH3 •• •• • S • ••O ••O

• • ••O Methyl sulfonate

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Even though the oxygens are negatively charged and are stronger bases than the sulfur, the sulfur is a better nucleophile when reacting with methyl iodide. The size and shape of a compound also help determine whether that compound is a nucleophile or a base. When the compound acts as a base to pick up a small proton, its size is less important. However, when the compound must act as a nucleophile and bond with a tetrahedral carbon in an SN2 reaction, the size of the compound determines whether or not it faces steric constraints. The ethoxide ion and the tert-butoxide ion help illustrate how the size and shape of a compound affect the basicity and nucleophilicity of a compound. Both the ethoxide ion and the tert- butoxide are conjugate bases of alcohols and, of the two, the tert- butoxide ion is a slightly stronger base. However, the two compounds are very different in size and shape. The ethoxide ion is much smaller, and its shape leaves the oxygen much more available for bonding than does the shape of the tert-butoxide ion. Therefore, the ethoxide ion readily reacts with a carbon bearing a leaving group, but the tert- butoxide ion is too large for the backside attack, so it cannot function as an effective nucleophile.

CH3

CH3CH2O CH3 CO

CH3

Ethoxide ion tert-Butoxide ion

Table 12.2 lists some relative rates for the SN2 reaction of various nucleophiles with methyl iodide in DMF. Note that methyl iodide does not have any steric crowding, thus, even large nucleophiles readily react with it.

Nucleophile Relative Nucleophile Relative Rate Rate c- 6 CH3OH 1 CH3O 2.0 x 10 c- 2 c- 6 F 5.0 x 10 CH3 CH2O 4.3 x 10 c- 4 c- 6 CH3COO 2.0 x 10 CN 5.0 x 10 c- 4 c- 6 Cl 2.4 x 10 (CH3) 3CO 7.0 x 10 5 7 NH3 3.1 x 10 (CH3)2NH 1.1 x 10 5 c- 7 (CH3)2S 3.6 x 10 CH3S 1.6 x 10 c- 5 c- 7 N3 5.9 x 10 I 2.1 x 10 c 5 9 Br - 6.1 x 10 (CH3CH2)3P 3.0 x 10

Table 12.2. Relative rates for SN2 reaction in DMF with methyl iodide as the substrate.

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Exercise 12.9

In each of the following pairs of reagents, which is a better nucleophile in an SN2 reaction with CH3Cl in a methanol solution?

⊕ a) CH3OH or CH3SH b) NH3 or NH4 c- c- c- c) I or Br d) CH3O or CH3OH e) (CH3)3N or (CH3)3B f) (CH3)3P or (CH3)3N

Sample solution c) Ic- is a better nucleophile than Brc- because Ic- is a softer base than Brc- .

Recall from Chapter 8 that the best leaving groups are weaker bases than the incoming nucleophile. To facilitate some reactions, chemists use methods that convert poor leaving groups into good leaving groups. For example, the leaving group of an alcohol is a hydroxide ion. Because the hydroxide ion is a stronger base than many nucleophiles, it is a relatively poor leaving group, which prevents alcohols from readily undergoing nucleophilic substitutions. However, when chemists add some acid to the alcohol, the hydroxyl group protonates and becomes a better leaving group.

HA •• •• ROH RO•• H H

Chemists also use protonation to make ethers and some amines better leaving groups. In each of these cases, the leaving group becomes a neutral molecule, which is a much weaker base than the original group. Water, alcohols, and some amines make very good leaving groups. A second method that chemists use to improve a group's ability to leave is to convert it to some other functional group that is a better leaving group than the original group. For example, an alcohol can react with p-toluenesulfonyl chloride (usually called tosyl chloride) to form a p-toluenesulfonate. p-Toluenesulfonate is a very good leaving group because it is a weaker base than c- OH.

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O O

ROH+ CH3 SCl CH3 SOR + HCl O O

Tosyl chloride An alkyl tosylate

Usually, chemists abbreviate tosyl chloride as TsCl. They also write the above reaction in the following way.

TsCl ROH ROTs

Solved Exercise 12.2

1-Propyl methanesulfonate is dissolved in acetone containing equivalent amounts of both sodium chloride and sodium bromide. One product predominates in the reaction. What is the product? Explain.

CH3CH2CH2 O S CH3 O

1-Propyl methanesulfonate

Solution 1-Bromopropane is the predominant product because bromide ion is a better nucleophile than chloride ion (see Table 12.2) in an SN2 reaction on a primary substrate.

12.7 Common Nucleophiles

Organic chemists use a variety of common nucleophiles. Table 12.3 lists some of these nucleophiles. This chapter discusses many of these nucleophiles. You will see many of the others in nearly every chapter in the rest of the text.

Nucleophile Product Nucleophile Product Clc- R—Cl R'Sc- R—SR' c Br - R—Br NH3 R—NH2 c I - R—I R'NH2 R—NHR' H2O R—OH R'2NH R—NR'2 c ⊕ HO - R—OH R'3N R— NR'3 c- R'OH R—OR' N3 R—N3 (Azide ion) (Alkyl azide)

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Nucleophile Product Nucleophile Product c ⊕ R'O - R—OR' R'3P R—P R'3 R'COOH R—OOCR' Hc- R—H c- c- R'COO R—OOCR' R'3C R—CR'3 H2S R—SH R C CR' R'C C c- R—SH HS C RNC N O O R'SH R—SR' C C CCR

Table 12.3. The most common nucleophiles used in organic chemistry.

12.8 The Reaction Medium

Only rarely do chemists simply mix a substrate and reagent together to run a reaction. Instead, they dissolve them in a solvent, perhaps adding a catalyst, and then carefully adjust the temperature. All these components make up the reaction medium, or environment, in which a chemical transformation takes place. Of these various components, the one that overwhelmingly influences the course of the reaction is the solvent. The solvent has so much influence because it is present in a much higher concentration than any other component. As the reaction progresses, the solvent interacts with the substrate, the nucleophile, the transition states, and the intermediates, thereby influencing the rate of the reaction. Understanding the influence of the solvent in a reaction helps you to know which solvent to choose when you want to run a reaction. Understanding the influence of the solvent also helps you to analyze the course of a reaction as you study a reaction mechanism. Solvent polarity, usually measured by the dielectric constant, is a major factor in how a reaction proceeds. The dielectric constant is a A solvated ion has measure of the ability of the solvent to solvate, or stabilize, ions in solvent molecules solution. Remember that the rate-determining step in an SN1 reaction associated with it in is the formation of the carbocation. More polar solvents accelerate the the solution. rate of formation of a carbocation, thereby increasing the SN1 reaction rates. The data in Table 12.4 show the ability of the more polar solvents to accelerate the ionization of tert-butyl chloride.

Solvent Structure Dielectric Relative rate constant (ε) (at 25oC) 3 Water H2O 80 8 x 10 3 Methanol CH3OH 33 1 x 10

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Solvent Structure Dielectric Relative rate constant (ε) (at 25oC) 2 Ethanol CH3CH2OH 24 2 x 10 Acetone CH3COCH3 21 1 –3 Ethyl ether (CH3CH2)2O 4 1 x 10 –4 Pentane CH3CH2CH2CH2CH3 2 < 10

Table 12.4. Dielectric constants for some representative solvents and their effect on the rate of ionization for tert-butyl chloride.

A polar solvent strongly solvates the ions formed in the SN1 With a solvent cage, the reaction by building a solvent cage around each ion. In a solvent positive ends of the cage, each molecule of solvent donates a little negative charge from its solvent molecules nonbonded pair of electrons to the cation of the substrate. The higher associate around a negative ion, or the the dielectric constant of a solvent, the stronger its ability to solvate negative end of the ions in a reaction. Solvation lowers the energy of the ions and of the solvent molecules transition state leading to them. For example, when the oxygen of associate around a water donates its electron density to a cation, that cation becomes positive ion. solvated.

H H O

H H O RO H H

O H H Solvated cation in water

Water can also solvate an anion. The anion donates electron density to the partially positive hydrogens of the H—O bond.

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H O H

H O H XHH O

H O H Solvated anion in water

Solvent polarity also affects SN2 reaction mechanisms. However, because the mechanism does not involve ionization, solvation of any intermediates is unimportant. The important factor is the solvation of starting materials and products. Solvation of SN2 reactions fits into four categories. The difference between each category depends on the charges of the starting materials. The first category of SN2 reactions is the Type I SN2 reaction. In a Type I SN2 reaction, the substrate and nucleophile are neutral, and the initial product has a positive charge. The solvent builds a solvent cage around the product; thus, lowering its energy level. With a Type I SN2 reaction, a high polarity solvent actually causes a large increase in the rate of the reaction.

Nu • + RL R Nu + L•

SN2 Reaction - Type I

In the second category of SN2 reactions, the Type II SN2 reaction, the substrate is neutral and the nucleophile has a negative charge. With a Type II SN2 reaction, the high solvent polarity increases the solvation of the nucleophile. The solvent cage formed around the nucleophile is so stable that it must disperse before the nucleophile can react with the substrate. Thus, with the Type II SN2 reaction, increasing solvent polarity causes a small decrease in the rate of reaction.

Nu • + RL R Nu + L•

SN2 Reaction - Type II

In the third category of SN2 reactions, the Type III SN2 reaction, the leaving group has a positive charge and the nucleophile

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has a negative charge. High solvent polarity increases the solvation of the substrate and the nucleophile. The solvent builds a solvent cage around both the substrate and the nucleophile, and both solvent cages must disperse before the reaction can take place. With the Type III SN2 reaction, increasing solvent polarity causes a large decrease in the rate of reaction.

• Nu • + R L R Nu + L•

SN2 Reaction - Type III

The fourth category of SN2 reactions is the Type IV SN2 reaction. In a Type IV SN2 reaction, the leaving group has a positive charge and the nucleophile is neutral. A high polarity solvent builds a solvent cage around the substrate. Before the reaction can proceed beyond this point, the nucleophile must penetrate this solvent cage. With a Type IV SN2 reaction, increasing solvent polarity causes a small decrease in the rate of reaction.

Nu• + R L R Nu + L•

SN2 Reaction - Type IV

So far, all the solvents covered in this book are polar-protic A polar-protic solvent is a polar solvent solvents. Polar-protic solvents have a heteroatom that bears a proton, containing a and they easily solvate both positive and negative ions. With polar- heteroatom bearing an protic solvents, the anion becomes solvated by the hydrogen bonding acidic proton. with the solvent. Polar-protic solvents work well as solvents for all four types of SN2 reactions as well as SN1 reactions. Two common examples of this type of solvent are water and methanol. A second category of solvents commonly used in organic A polar-aprotic solvent chemistry is the polar-aprotic solvents. Even though polar-aprotic is a polar solvent with solvents lack protons on a heteroatom, they can be quite polar. Thus, no acidic protons. they have the ability to solvate ions, even without the polar hydrogens. Polar-aprotic solvents will solvate a nucleophile less than does a polar-protic solvent, which gives the nucleophiles more freedom to react. The rate for SN2 reactions in polar-aprotic solvents is highest for types II and III. SN1 reactions generally do not proceed as well with these solvents. Table 12.5 lists a few representative polar-aprotic solvents.

Solvent Structure Dielectric (Common abbreviation) constant (ε)

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Solvent Structure Dielectric (Common abbreviation) constant (ε) Dimethyl sulfoxide O (DMSO) CH SCH 45 3 3

CH CN Acetonitrile 3 38

Dimethylformamide O (DMF) HC N(CH ) 37 3 2

Hexamethylphosphoramide O (HMPA) 30 (CH3)2N P N(CH3)2

N(CH3)2

O Acetone CH CCH 21 3 3

Tetrahydrofuran (THF) 7 O

Table 12.5. Some representative polar-aprotic solvents and their dielectric constants.

Polar-aprotic solvents promote a much faster rate of reaction in SN2 reactions than do polar-protic solvents because polar-aprotic solvents solvate only the positive ions. Solvating only the positive ion leaves the anion unsolvated. These unsolvated, or “naked,” anions are much more reactive than when solvated. For example, compare the c- rate of reaction for the Type II reaction of azide ion (N3 ) with methyl iodide using methanol, a polar-protic solvent, versus dimethylformamide (DMF), a polar-aprotic solvent.

CH I + N CH N + I 3 3 3 3

Methanol and DMF have similar dielectric constants. The rate of reaction in DMF solution, however, is 10,000 times larger than the rate of reaction in the methanol solution.

Solved Exercise 12.3

When dissolved in hot ethanol, compound A is completely stable, but compound B is rapidly converted to another compound. What is the product

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of the reaction of ethanol with compound B? Explain the differences in reactivity between compound A and compound B.

Br Br

Solution Compound B reacts with ethanol forming an ether.

CH3CH2OH Br OCH2CH3

This reaction occurs because compound B is a tertiary alkyl halide and ethanol is a polar-protic solvent that promotes ionization of the substrate. On the other hand, compound A is a primary substrate that is sterically blocked for SN2 reaction. In addition, ethanol is not a strong enough nucleophile to react with compound A.

Exercise 12.10

In water or methanol the order of halide nucleophile reactivity is Ic- > Brc- > Clc- . However, with a solvent such as DMSO or DMF, the order of reactivity reverses to Clc- > Brc- > Ic- . Explain this difference.

A crown ether is a The presence of a crown ether in a reaction essentially cyclic polyether having cancels the effects of the solvent by complexing the positive counter an oxygen every third atom around the ring. ion. Crown ethers, discovered in the 1960s, are cyclic polyethers. They get their name from their three-dimensional crown shapes. Chemists use crown ethers as complexing agents because of their high affinity for various metal cations. As a complexing agent, a crown ether forms a structure with the metal cation similar to a solvent cage. For example, 18-crown-6 (18 is the ring size and 6 is the number of oxygen atoms in the ring) binds strongly to a potassium ion.

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O O O O O O K K

O O O O O O

18-Crown-6 Potassium 18-crown-6 complex

Another word for the complex formed by a crown ether and a Chelation is the metal ion is a chelation. A chelate (Greek chele, “claw”) is a binding formation of a complex between a metal ion and an electron-donating molecule. 18-Crown-6 is between a metal ion quite specific for the chelation of potassium whereas 15-crown-5 (the guest) and an chelates sodium ions, and 12-crown-4 chelates lithium ions. In these electron-donating molecule (the host). complexes, chemists call the crown ether the host and the metal ion the guest. Chelation makes ionic substances soluble in nonpolar organic solvents where they are not usually soluble. In fact, after chelation, most ionic substrates are much more soluble in nonpolar organic solvents than they are in the solvents that they are soluble in without the chelation. Because the crown ether does not complex with the anion during a chelation, it leaves the anion readily available for reaction. These anions, sometimes called “naked” anions because they are not solvated, are much more reactive than they are when solvated. Even “poor” nucleophiles become quite reactive. For example, the reaction of benzyl bromide with potassium acetate yields less than 10% of the product, benzyl acetate. After adding 18-crown-6, however, the yield jumps to 93%. The solvent used for the reaction is acetonitrile (CH3CN), a polar-aprotic solvent that does not readily dissolve ionic compounds.

O O

CH2Br + CH3CO K CH3COCH2

Benzyl acetate Benzyl bromide Potassium acetate (93%)

12.9 SN1 Versus SN2

So far, Chapter 12 has presented a number of factors that help decide whether an SN1 or an SN2 substitution reaction takes place. This section summarizes these factors in a series of tables. Table 12.6

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compares the effects of various substrates on the outcome of the reaction. Table 12.7 shows how the nucleophile/base, the leaving group, and the solvent determine which mechanism a substrate will follow.

R R CH L R CH L 3 2 R CH L R C L R Methyl 1o 2o 3o SN2 only. Mainly SN2 with weak No SN2. In a bases. Also SN1 in solvolysis reaction acid, especially if the or under acidic substrate can conditions, SN1. rearrange.

Table 12.6. Summary of reaction pathways of substitution reactions by substrate type.

Factor SN1 SN2 Nucleophile Poor nucleophile. Nucleophile is soft (base). base. Leaving group. Better if a weaker base than the nucleophile. Solvent. Polar-protic solvent. Polar-aprotic solvent.

Table 12.7. Summary of effects of nucleophile, leaving group, and solvent on the mechanism of a particular substrate.

12.10 Halide Nucleophiles

In the previous discussions of aliphatic nucleophilic substitution reactions, many of the examples used alkyl halides as the substrate with the halide ion acting as the leaving group. Under the appropriate circumstances, halides play a double role in aliphatic nucleophilic substitutions: they act as the leaving group or they act as the nucleophile. Chemists easily synthesize alkyl halides from alcohols. They use these alkyl halides as reactive intermediates in organic syntheses. Chemists usually synthesize alkyl bromides via the reaction of an alcohol with hydrogen bromide. Because the c- OH ion is a strong base it is a very poor leaving group. However, the use of hydrogen An oxonium ion is an bromide allows protonation of the alcohol to form the oxonium ion. ether or alcohol with The oxonium ion is an excellent leaving group because it is a relatively the oxygen protonated. weak base.

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•• HA •• RO•• H ROH H Oxonium ion

An SN2 reaction occurs with a primary or secondary substrate because c- ⊕ the Br nucleophile easily displaces the — OH2 group. An SN1 ⊕ reaction occurs with a tertiary substrate. First, the — OH2 group leaves to form the carbocation. Then the carbocation reacts with the c- Br nucleophile in an SN1 reaction.

•• •• • RCH2 OH RCH2 Br• ••

•• H •Br• •• SN2 synthesis of an alkyl bromide

R R R •• •• R C OH R C R C Br• •• R H R R •• •Br• •• SN1 synthesis of an alkyl bromide

Chemists prepare alkyl chlorides in a manner similar to the way that they prepare alkyl bromides. However, hydrogen chloride is a weaker acid than hydrogen bromide so formation of the oxonium ion is less favorable with hydrogen chloride than it is with hydrogen bromide. The chloride ion is also a weaker nucleophile than the bromide ion, so the formation of the alkyl chloride occurs with a poorer yield than does the alkyl bromide. Hydrogen chloride reacts well with tertiary alcohols but much less satisfactorily with primary and many The Lucas Reagent is a secondary alcohols. Chemists often use the Lucas Reagent for solution of ZnCl2 dissolved in preparing primary and secondary halides because it adds the strong concentrated HCl. Lewis acid ZnCl2 to the hydrogen chloride solution and enhances the reactivity of the alcohols.

•• ZnCl2 •• ROH ROH •• ZnCl 2

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The resulting zinc chloride-alcohol complex is much more reactive towards aliphatic nucleophilic substitutions than hydrogen chloride alone. With the Lucas Reagent, the complex reacts with the chloride ion to produce the alkyl chloride.

•• • Cl• •• •• •• R Cl• ROH •• ZnCl 2

Synthesizing iodoalkanes is much more difficult than synthesizing either alkyl chlorides or alkyl bromides. Generally, iodoalkanes are too expensive and too reactive for widespread use. They also often promote undesirable side reactions. When chemists do use iodoalkanes, they synthesize them by reaction of a chloroalkane with sodium iodide in acetone.

NaI R Cl RI acetone

This reaction is driven to completion because sodium chloride is essentially insoluble in acetone, whereas sodium iodide is soluble in acetone. The reaction follows SN2 kinetics. In addition to the above methods for preparing alkyl halides, chemists use thionyl chloride and phosphorus chloride to make alkyl chlorides from alcohols. Because phosphorus bromide and phosphorus The term in situ means iodide is so unstable, chemists usually synthesize them in situ using in place. A synthesis phosphorus and bromine and iodine. In the mechanism for this done in situ completes method, an ester of an inorganic acid forms, then the ester group a reaction in the converts the alcohol to a good leaving group. The leaving group leaves reaction flask where the product is to be following an attack by the halide nucleophile. used without isolation. SOCl2 ROH R Cl

PCl3 ROH R Cl

P, Br2 ROH R Br

P, I2 ROH R I

Below are some examples of alkyl halide synthesis.

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HCl OH Cl ZnCl2 1-Chloropentane (81%)

Cl NaI I Acetone

3-Methyl-1-iodobutane (97%)

CH2OH CH2Cl SOCl2 Benzene Pyridine

CH3 CH3 3-Methylbenzyl chloride (89%)

Synthesis of 1-Bromobutane

NaBr OH Br H2SO4, 1-Bromobutane (78%)

Place 1.8 mL (24 mmol) of 1-butanol in a flask. Add 2.5 g (24 mmol) of sodium bromide and 2 mL of water. Cool the flask in an ice bath and slowly add 1.6 mL of concentrated sulfuric acid. Reflux this mixture for 45 minutes. Cool the apparatus to room temperature. Remove the sulfuric acid layer and wash the organic layer with 5 mL of water followed by 2 mL of saturated sodium bicarbonate. After separating the layers, dry the organic layer with anhydrous sodium sulfate. Distill the product. Yield of 1- bromobutane is 2.5 g (78%), bp 101-104oC.

Discussion Questions

1. Which layer would you expect to contain the 1-bromobutane, the upper or lower layer? If, instead of 1-bromobutane, you prepared 1-chlorobutane, would the organic layer be the same one (top or bottom)? 2. Does this reaction follow the SN1 or SN2 mechanism? 3. How would you expect this procedure to change if the reaction were run with 2- methyl-2-propanol instead of 1-butanol? Would this reaction follow the SN1 or SN2 mechanism?

Exercise 12.11

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Predict the major products of each of the following reactions.

a) HBr CH3CH2CH2CH2CHCH3

b) OH HCl

ZnCl2

PCl3

HO H

d) Cl

CHCH3 NaI acetone

CH2OH SOCl2

f) OH P, I2

Sample Solution e)

CH2OH CH2Cl SOCl2

Ethers are usually quite unreactive. To react, they require an acid to protonate the oxygen and a good nucleophile. Two commonly

www.ochem4free.com 5 July 2005 Organic Chemistry - Ch 12 621 Daley & Daley used examples are hot, concentrated solutions of HBr or HI. For sterically open ethers, the reaction mechanism is SN2. For more hindered ethers, the reaction pathway is SN1. The first step in either pathway is the protonation of the ether oxygen.

H HBr CH OCH CH OCH 3 3 3 3

The second step in an SN2 reaction is a nucleophilic substitution by the halide ion.

H Br CH Br + CH OH CH3OCH3 3 3

The reaction of HBr with an ether forms an alkyl bromide and an alcohol. The alcohol reacts with more HBr to form a second alkyl bromide. However, this reaction is normally slow enough that the alcohol can be isolated. In a reaction of an ether with HI, the iodide ion is a better nucleophile than the bromide ion so the alcohol cannot be isolated because it reacts so quickly with more HI. The following reactions are examples of ether reacting with HBr and HI. HBr CH3CH2CH2OCH2CH2CH3 CH3CH2CH2Br + CH3CH2CH2OH 1-Bromopropane 1-Propanol

OCH2CH3 I HI + CH3CH2I

Iodocyclohexane Iodoethane

Exercise 12.12

In practice, an ether cleavage uses a large excess of HBr or HI. However, the chemist observes no alcohol product and isolates only the haloalkane. Write a mechanism to explain this observation.

An exception to the unreactivity of ethers is the oxirane ring. This 3-membered ring is highly strained and very susceptible to reaction with a variety of ring-opening reactions. For example, oxirane ring ethers are among the few ethers that react with HCl directly to cleave the ring.

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O Cl HCl CH3CH CH2 CH3CHCH2OH 2-Chloropropanol (98%)

The opening of the oxirane ring using HCl follows an SN2-like reaction mechanism.

• •• • •• • •• • • • O • O • Cl• H Cl• •• •• •• CH3CHCH2OH CH3CH CH2 CH3CH CH2 ••

• Cl• ••

Unlike a typical SN2 reaction, the incoming nucleophile reacts at the secondary carbon because it has a larger partial positive charge than the primary carbon.

• O •

CH3CH CH2

Larger partial positive charge on this carbon.

Please note that under basic conditions in a ring opening reaction of an oxirane, the nucleophile reacts with the least substituted carbon atom.

12.11 Oxygen Nucleophiles

Aliphatic nucleophilic substitution reactions that involve an oxygen atom nucleophile are very important in organic chemistry. Oxygen nucleophiles fall into three major categories: water and alcohols, hydroxide and alkoxide ions, and carboxylate ions. The members of each category have about the same base strength, but each category has a different base strength; thus, the particular category of nucleophiles chosen for a reaction depends on the base strength needed to accomplish that reaction. Water and hydroxide

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ions form alcohols, alcohols and alkoxide ions form ethers, and the carboxylate ions form esters. When studying the reaction mechanism of aliphatic nucleophilic substitution, chemists used the reaction of alkyl halides with water or hydroxide ion to form alcohols. However, this method is not useful for the synthesis of alcohols because, as you studied in Section 12.10, the best substrate to use when synthesizing an alkyl halide is an alcohol. A hydride reduction of a carbonyl group or the reaction of an organometallic with a carbonyl are better ways to synthesize alcohols. The SN2 reaction of water with primary alkyl halides and most secondary alkyl halides is very slow and follows the reverse pathway of the reaction forming alkyl halides from primary and secondary alcohols. The SN1 reaction of tertiary alkyl halides and some secondary alkyl halides with water is a more rapid reaction. This increase in rate occurs because water is a polar-protic solvent and promotes the ionization of tertiary and many secondary substrates. The SN1 reaction of tertiary, and some secondary, halides with water In a solvolysis reaction, is an example of a solvolysis reaction. The mechanism of a the substrate reacts with the solvent. solvolysis reaction follows the reverse pathway of the reaction of a halide nucleophile with an alcohol. The hydroxide ion is a better nucleophile than water, so adding it to the reaction mixture increases the rate of reaction. Adding hydroxide ion also increases the rate of side reactions. The most significant of which is a competing elimination reaction. In an elimination reaction, the nucleophile reacts with a hydrogen on a carbon adjacent to the leaving group taking it away. However, because the hydroxide ion is a strong base a double bond forms rather than another group substituting itself for the leaving group. Elimination occurs because the hydroxide ion is not only a better nucleophile than water but is also a moderately strong base. This basicity causes the competing elimination reaction. Elimination of alkyl An elimination reaction is often a competing reaction for an halides is discussed in SN2 substitution reaction and even more frequently for an SN1 Section 13.7, page 000. substitution reaction. An elimination reaction is an especially significant competing reaction for reactions involving hydroxide ion with alkyl halides. With these reactions, the proportion of competing elimination reaction rises with the increase of steric crowding on the substrate. For example, compare the percentages of the product mixture in the reaction of the following alkyl bromide isomers with hydroxide ion.

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OH CH3CH2CH2CH2CH2Br CH3CH2CH2CH2CH2OH + CH3CH2CH2CH CH2 H2O 1-Pentanol 1-Pentene 90% 10%

CH3 CH 3 CH3 OH CH3CH2CBr CH3CH2COH + CH3CH C + CH3CH2C CH2 H2O CH3 CH3 CH3 CH3

2-Methyl-2-butanol 2-Methyl-2-butene 1-Methyl-1-butene 1% 94% 5%

Exercise 12.13

Write a mechanism for the solvolysis reaction of tert-butyl chloride with water.

Alcohols are somewhat better nucleophiles than water. A solvolysis reaction of alcohols with tertiary alkyl halides and some secondary alkyl halides is a useful method for the synthesis of ethers.

CH3 CH3 CH3CH2OH CH3 C Cl CH3 C OCH2CH3

CH3 CH3 Ethyl tert-butyl ether (89%)

The solvolysis of primary alkyl halides and most secondary alkyl halides with alcohols does not generally work well. However, the use of alkoxide ion, a stronger nucleophile, on these substrates does make the reaction go. This reaction is an irreversible preparation for In the Williamson ether ethers and is called the Williamson ether synthesis. The synthesis, an alkyl Williamson ether synthesis works because an alcohol reacts with halide reacts, in an SN2 sodium metal or some stronger base to form the alkoxide ion. reaction, with an alkoxide ion forming an ether. 2 ROH + 2 Na 2 RO Na + H 2

The alkoxide ion then reacts with an alkyl halide to form an ether. This reaction follows an SN2 reaction mechanism, as follows.

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•• CH3CH2O• •• •• •• CH3CH2 Br• CH3CH2 OCH2CH3 •• ••

Here is an example of the Williamson ether synthesis:

CH3CH2CH2O Na CH3CH2Br CH3CH2OCH2CH2CH3 Ethyl propyl ether (70%)

This method produces cyclic ethers as well as acyclic ethers. To make cyclic ethers, the hydroxide ion reacts with the alcohol to form an alkoxide anion. The alkoxide anion then reacts with the alkyl halide portion of the molecule forming the cyclic ether. Note that the alkoxide ion is about the same base strength as the hydroxide ion, thus establishing an equilibrium.

OH NaOH O Cl 1,2-Epoxycyclohexane (Cyclohexene oxide) (77%)

NaOH Cl OH O Oxacyclohexane (Tetrahydropyran) (89%)

Exercise 12.14

Write a mechanism for the synthesis of oxacyclohexane shown in the previous example.

Similar to the synthesis of alcohols, the Williamson ether synthesis is also susceptible to a competing elimination reaction. The reaction of a primary alcohol with a primary alkoxide favors a substitution reaction. However, as steric hindrance increases, the amount of competition from elimination reactions also increases. For

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example, if one of the reactants has a tertiary structure, the amount of elimination reaction is often 30-40% of the amount of substitution reaction. When both reagents have a tertiary structure, virtually no substitution reaction occurs. The acidity of phenol is Phenols (pKa ~ 10) are much more acidic than are alcohols discussed in section (pK ~ 17). Thus, the preparation of phenyl alkyl ethers requires a 5.4, page 000. a much weaker base than does the preparation of alkyl ethers. Sodium carbonate readily forms the phenoxide ion but is much too weak to form an alkoxide ion. Thus, the synthesis of a phenyl alkyl ether occurs with no side reaction. For example, chemists easily accomplish the preparation of anisole by using sodium carbonate as a base.

OH OCH3 CH3I

Na2CO3 Anisole (92%)

The use of a polar-aprotic solvent increases the nucleophilicity of the nucleophile. Often, this increase in nucleophilicity in turn increases the yield of the substitution reaction product. For example, the synthesis of dipropyl ether from sodium propoxide and 1- chloropropane using 1-propanol as the solvent gives nearly a 70% yield of product. Running the same reaction using dimethyl sulfoxide (DMSO), a polar-aprotic solvent as the solvent increases the yield to nearly 100%.

CH3CH2CH2O Na + CH3CH2CH2Cl (CH3CH2CH2)2O DMSO Dipropyl ether (99%)

Carboxylate ions are relatively poor nucleophiles. Their alkylation with an alkyl halide, though not widely used, is a method for the formation of esters. This reaction follows an SN2 substitution reaction mechanism. When using carboxylate ions as the nucleophile, chemists enhance their nucleophilicity by using a polar-aprotic solvent. The example shown below uses dimethylformamide (DMF) as the polar-aprotic solvent.

O O CH3CO K CH CH CH I CH3COCH2CH2CH3 3 2 2 DMF Propyl acetate (96%)

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Solvolysis is a very good method for the preparation of esters using a tertiary alkyl group as the substrate. For this reaction, chemists dissolve the alkyl halide in the carboxylic acid, then heat the reaction mixture. In the example below, a good yield of the tertiary alkyl ester forms.

O CH3 O CH3 CH3COH CH3 CCl CH COCCH 3 3 CH3 CH3 tert-Butyl acetate (81%)

Exercise 12.15

Predict the major products of each of the following reactions.

CH2CH2Br CH3O

b) Cl NaOH CH3CHCHCH3

c) CH 3 CH3OH

d) OH CH3CH2I

Na2CO3

CH OH 2

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1) Na

2) CH3I OH

f) Br CH3CH2OH

CH Br 2

Sample Solution b) Cl NaOH CH3CHCHCH3 CH3C H CHCH3 O OH

12.12 Nitrogen Nucleophiles

Chemists often synthesize amines via an SN2 reaction of ammonia, or some other amine, with an alkyl halide substrate. The reaction doesn't work well with tertiary halides because they are too sterically hindered. Poor yields are often achieved with secondary halides, too, because of significant amounts of competing elimination reaction. The reaction does work well, however, with primary halides. Although the example here shows chlorine, bromine and iodine are also excellent leaving groups.

•• RNH2 + R'CH2 Cl RNH2CH2R' Cl Primary amine Salt of a secondary amine

As the reaction proceeds, it initially forms a non-nucleophilic salt. This salt is in equilibrium with the original amine.

•• •• RNH + RNH2CH2R' Cl RNH Cl + RNHCH R' 2 3 2

Thus, the secondary amine becomes a competitive nucleophile with the original amine.

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•• •• RNHCH2R' + R'CH2Cl RNCH2R' CH R' 2

The product of this step is also a nucleophile. It alkylates forming a tetraalkyl ammonium salt.

CH2R' •• RNCH2R' + R'CH2Cl RNCH2R' Cl

CH2R' CH2R' A tetraalkyl ammonium salt

The above reaction is useful under two conditions: 1) when there is a large excess of the original nucleophilic amine (or ammonia) to reduce the amount of competitive alkylation, and 2) when there is an excess of the alkyl halide to “exhaustively alkylate” the nitrogen when forming a tetraalkyl ammonium salt. Under the first condition of adding a large excess of nucleophile to the reaction mixture, chemists typically use ammonia (NH3) as the nucleophile. Ammonia is cheap and relatively easy to handle. To run the reaction, they add a primary alkyl halide to at least a ten-fold excess of ammonia. The excess amount of ammonia effectively eliminates dialkylation and higher alkylation products. At the end of the reaction, they simply distill the ammonia from the reaction mixture, leaving only the pure product behind.

NH CH Br 3 CH NH 2 (Excess) 2 2

Aminomethylcyclopentane (67%)

When doing an exhaustive alkylation to form the tetraalkyl ammonium salt, chemists go to the other extreme. They add an excess of the alkyl halide to the amine or ammonia. Often, they also add a mild base, such as Na2CO3, to neutralize the large amount of HX formed during the reaction. Chemists have devised a variety of methods to overcome the limitations of producing amines, particularly primary amines. Recall the three methods presented earlier: the reaction of nitriles or amides with a hydride nucleophile and the reaction of a ketone or aldehyde The Gabriel synthesis with an amine and NaCNBH3. A fourth method is the Gabriel involves reaction of the synthesis. In the Gabriel synthesis, the nitrogen nucleophile is an anion of phthalimide with an alkyl halide in anion that readily forms from phthalimide in a reaction with KOH in an SN2 reaction.

www.ochem4free.com 5 July 2005 Organic Chemistry - Ch 12 630 Daley & Daley ethanol or sometimes water. In either case, the reaction forms a resonance-stabilized anion.

O O

KOH NH N Ethanol

O O

N N

O O

This resonance-stabilized anion, which is a good nucleophile, reacts with a primary alkyl halide (usually a bromide) in an SN2 substitution reaction.

O O

RCH2Br N NCH2R

O O

The resulting imide then reacts with hydrazine to form the primary amine and phthalimidehydrazide. This reaction, a nucleophilic substitution at the carbonyl group, involves an amide exchange. An amide exchange is the reaction of an amide with an amine to form another amide and amine.

O O

NH2NH2 NH NCH2R RCH2NH2 + NH

O O

Below is an example of the Gabriel synthesis.

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1) KOH, H2O NH2NH2 NH CH3CH2CHCH2CH2NH2 2) CH3CH2CHCH2CH2Br CH3 CH3 O 3-Methyl-1-pentanamine (88%)

Synthesis of 2,5-Diaminoadipic Acid

N Br O O N O O

CH3O O CH3O OCH3 OCH3 O Br O N O O

NH2NH2

NH2 O NH2 O HCl HO CH3O OH H2O OCH3 O NH2 O NH2 2,5-Diaminoadipic acid (80%)

Dimethyl 1,4-diphthalimidoadipate

Heat a mixture of 1.45 g (5 mmol) of dimethyl 2,5-dibromoadipate, 1.85 g (10 mmol) of potassium phthalimide, and 5.2 mL of dimethylformamide. At 50oC a slightly exothermic reaction begins. Maintain the temperature for 40 min at 90oC. Cool the reaction mixture and add 6 mL of chloroform. Pour the mixture into 25 mL of water. Separate the layers and wash the aqueous layer twice with 5 mL of chloroform. Wash the combined chloroform extracts with 5 mL of 0.1M sodium hydroxide and 5 mL of water. Dry the solution with anhydrous sodium sulfate. Remove the chloroform on a rotary evaporator until crystals just begin to appear. Add 6 mL of ether and collect the crystals. Wash the crystals with two 5 mL portions of cold ether. The product is sufficiently pure for the next step. If you wish, you may recrystallize twice: first from ethyl acetate and then from benzene.

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2,5-Diaminoadipic acid

To 1.75 g of crude product from the above step, add 20 mL of methanol and 0.45 mL of 85% aqueous hydrazine solution. Reflux for 1 hour. Cool the reaction mixture and add 7.5 mL of water. Remove the methanol on a rotary evaporator. Add 7.5 mL of concentrated hydrochloric acid. Reflux for 1 hour. Cool to 0oC and filter the solution. On a rotary evaporator concentrate the aqueous solution until a moist solid remains. Dissolve the solid in 15 mL of water and neutralize (pH ~ 6) the solution with 2M aqueous sodium hydroxide. Filter out any insoluble material. Cool the resulting solution for 12 hours at 0oC, then collect the crystals. Yield 0.53 g (60%) of product, mp decomposes.

Discussion Questions

1. After refluxing with hydrazine, you remove and discard a solid material. What is the structure of that solid material? 2. Why must you adjust the pH of the solution of 2,5-diaminoadipic acid to 6 to crystallize the product?

c- Another common nitrogen nucleophile is the azide (N3 ) ion. An azide ion, the conjugate base of the acid HN3, is a resonance hybrid of the following structures:

2 2 NNN NNN NNN

The anion is relatively non-basic (the pKa of HN3 is 11), but it is a good nucleophile and reacts readily in an SN2 reaction to form alkyl azides.

NaN3 Br N CH3OH, H2O 3 1-Butyl azide (86%)

Alkyl azides are reduced with LiAlH4 to produce the corresponding amines. This reaction provides a convenient synthesis for amines.

N3 NH2 1) LiAlH4

2) H3O

Cyclohexanamine (94%)

Exercise 12.16

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Predict the major products of each of the following reactions.

NH3 CH3CHCH2CH2Br Large excess CH 3

NH2 CH3I Excess

c) NaN3

d) O

1) KOH NH CH2Cl

O 2)

3) NH NH 2 2

CH3CH2Br

N CH 3

f) O

1) KOH NH 2) BrCH2CH2CH2Br 3) NH NH O 2 2

Sample Solution a)

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NH3 CH3CHCH2CH2Br CH3CHCH2CH2NH2 Large excess CH CH 3 3

12.13 Carbon Nucleophiles

The study of stabilized carbanions in Chapter 19 considers carbon nucleophiles in greater detail than this section does. However, several important reactions also fit in this chapter. These reactions use the cyanide ion nucleophile, the acetylide ion, and the organometallics. Because the cyanide ion is similar to the iodide ion in reactivity, the cyanide ion is a good nucleophile. Cyanide ion substitutions proceed through the SN2 mechanism, so the cyanide ion is useful with primary substrates, as well as many secondary substrates.

KCN Cl CN H2O Hexanenitrile (81%)

Br CN NaCN

H2O

Cyclohexanecarbonitrile (67%)

A cyanide ion substitution reaction often proceeds with difficulty because both sodium and potassium cyanides are insoluble in most organic solvents and the substrates are insoluble in water. Using crown ethers, or some other member of this group of compounds A phase-transfer called phase-transfer catalysts, helps to overcome these problems. catalyst is a compound For example, crown ethers carry a variety of negatively charged that carries some chemical species into a nucleophiles into nonpolar solvents. solvent in which it is Other widely used phase-transfer catalysts are quaternary ⊕ c- otherwise insoluble. ammonium salts (R4N X ). Quaternary ammonium salts often show varying solubilities in different solvents with changes in the anion. For example, tetrabutyl ammonium chloride is quite soluble in water but shows low solubility in many organic solvents. On the other hand, tetrabutyl ammonium cyanide has a higher solubility in organic solvents. In water that contains the cyanide ion, the tetrabutyl

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ammonium chloride and the tetrabutyl ammonium cyanide are in equilibrium.

Bu N Cl ++CN Bu N CN Cl 4 4

Once in the organic solvent, the cyanide ion reacts with the substrate, releasing the halide ion. The tetrabutyl ammonium halide then migrates to the aqueous phase ready to carry another cyanide ion to the organic phase.

Bu N CN ++RCH Cl Bu N Cl RCH CN 4 2 4 2

When they react with a strong base such as sodium amide (NaNH2) or Grignard reagents, terminal alkynes (RCCH) have An acetylide ion is the sufficient acidity to form carbanions, also called acetylide ions. The conjugate base formed ease of formation of the acetylide ion is due to the sp hybrid orbital of by reaction of a strong the triple bond carbon. Because this bond is shorter than the sp2 or base with a terminal sp3 carbon to hydrogen bonds, it is also weaker. Acetylide ions are alkyne. good nucleophiles and react readily in SN2 substitution reactions.

NaNH2 RC CH RC C Na NH (–33oC) 3

R'MgBr RC CH RC C MgBr ether

Chemists prepare sodium amide by adding sodium to liquid ammonia in the presence of catalytic quantities of the iron(III) cation. This process is colorful. The solution of sodium in liquid ammonia is an intense deep blue color. As the sodium reacts with the mixture, it changes to a suspension of a gray solid in a colorless liquid. At this point, the chemists add the terminal alkyne, followed by the alkylating agent. The reaction mixture makes no further appreciable color change.

R'CH2Br RC C Na R'CH2CCR NH (–33oC) 3

Grignard reagents and organolithium reagents are not usually good sources of carbon nucleophiles in SN2 substitution reactions. They are such strong bases that they tend to promote elimination reactions instead. However, they do substitute on some particularly

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reactive substrates. For example, epoxides are strained cyclic ethers that react with both Grignard reagents and organolithium reagents. This reaction is notable in that it results in the extension of the carbon chain with a reactive functional group on the second carbon.

MgCl O CH CH OH 1) 2 2

2) H3O 2-Cyclohexylethanol (69%)

Exercise 12.17

Propose a synthesis for the following compounds.

a) 4-Nonyne b) 3-Methylhexanol c) 2-Cyclohexyl-1-pentanol d) 1-Decyne e) 2-Phenylethanol

Sample Solution a) There are two possible ways to synthesize this compound. 1)

1) NaNH2/NH3 CH3CH2CH2CH2CCH CH3CH2CH2C CCH2CH2CH2CH3 2) CH CH CH Br 3 2 2 2) 1) NaNH2/NH3 CH3CH2CH2CCH CH3CH2CH2C CCH2CH2CH2CH3 2) CH3CH2CH2CH2Br

12.14 Neighboring Group Participation

With the exception of the formation of cyclic ethers, the substitution reactions covered to this point have involved separate substrates and nucleophiles. With the formation of cyclic ethers, nucleophilic substitution can also be an intramolecular process. In an intramolecular reaction, called neighboring group participation, With neighboring the presence of functional groups in the substrate other than the group participation, a leaving group affects the rate and/or the stereochemistry of the functional group in the substrate, other than reaction. Thus, when the nucleophile and leaving groups are a part of the leaving group, the same molecule, the result is a cyclic compound. affects the outcome of the reaction.

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NaOH Cl OH O Tetrahydropyran (89%)

NH2 HN Cl NaOH

CH3CH2OH

2,3-Diphenylaziridine (94%)

Intramolecular processes have a more favorable entropy than do analogous intermolecular processes because the two separate molecules are not required to come together. Five- or six-membered rings form with reaction rates that are typically about 106 faster than comparable acyclic reactions. Even a three-membered ring forms about 103 faster than a similar acyclic process. Formation of a three-membered ring is due more to proximity effects than entropy. For example, the formation of an epoxide from a chlorohydrin requires only a small movement of the atoms for bond formation. The oxygen anion is about 250 pm away from the carbon bearing the chlorine. The length of a C—O bond is about 150 pm.

250 pm 150 pm

•• •• •• • • • • O• CH2 Cl• • O •• CH2 C C H H H H

Neighboring group participation in a reaction is not always as obvious as in the above reactions. Often the only obvious effect is a marked rate enhancement over similar reactions with no neighboring group participation. Chemists call such rate enhancements Anchemeric assistance anchemeric assistance (Greek anchi + meros meaning neighboring results in a rate parts). For example, the hydrolysis reaction of β-chloroethyl ethyl enhancement in a sulfide is 3,000 times faster than the hydrolysis of 1-chloropentane. reaction involving neighboring group assistance.

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CH3CH2SCH2CH2Cl +HH2OCH3CH2SCH2CH2OH + Cl

-Chloroethyl ethyl sulfide -Hydroxyethyl ethyl sulfide

CH3CH2CH2CH2CH2Cl +HH2O CH3CH2CH2CH2CH2OH + Cl 1-Pentanol 1-Chloropentane

The first reaction is a first order reaction. Its rate depends only on the concentration of the substrate. The concentration of the nucleophile has no effect on the rate. This is surprising because the second reaction follows second order kinetics. The difference between the two reactions above must have something to do with the sulfur atom, but the sulfur is too far from the reaction site to electronically affect the rate of the reaction. However, sulfur is a good nucleophile and the chloride ion is a good leaving group, so the sulfur forms a cyclic sulfonium salt by displacing the chloride ion.

•• • Cl• •• •• CH CH Cl• CH2 •• 2 2 •• •• CH CH S CH3CH2 S 3 2 •• CH2 Sulfonium ion

The intermediate sulfonium ion is very reactive with an incoming nucleophile because the positive sulfur is a good leaving group. Thus, the nucleophile reacts with the sulfonium ion, which is not an ordinary primary alkyl halide.

•• CH • •• •• 2 H2O• •• CH3CH2 S CH3CH2 SCH2CH2OH CH (-H ) •• •• 2

Solved Exercise 12.4

Often chemists label a compound in an experiment with an uncommon isotope of one of its atoms to use that atom as a probe to understand the mechanism of the reaction. The accelerated rate of the nucleophilic substitution reaction of β-chloroethyl ethyl sulfide is an interesting example of the use of labeling to probe the reaction. If the carbon bearing the chlorine is labeled with 14C, what labeling pattern would be observed in the product if a) there is no neighboring group participation or b) neighboring group participation occurs?

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14 CH3CH2SCH2 CH2Cl Labeled -chloroethyl ethyl sulfide

Solution a) If there is no neighboring group effect, then the product will be labeled on the carbon bearing the —OH group.

14 CH CH SCH CH OH 3 2 2 2 b) If neighboring group participation is important in this reaction, then there will be a 1:1 mixture with the labeled carbon either bearing the —OH group or attached to the sulfur.

•• • Cl• •• 14 Water can react equally •• CH2 well with either carbon. CH3CH2 S CH2

14 14 CH CH SCH CH OH CH CH S CH CH OH 3 2 2 2 3 2 2 2

The actual result is a 50:50 mixture of the two label positions produced as in part b).

Exercise 12.18

The rate of hydrolysis of β-chloroethyl ethyl ether is much slower than the rate of hydrolysis of β-chloroethyl ethyl sulfide. Explain this observation.

Special Topic–SN1 Versus SN2

The SN1 and SN2 reaction mechanisms represent the two extremes of nucleophilic substitution reactions at an sp3 saturated carbon. With an SN1 reaction, the reaction rate depends only on the substrate because the ionization of the bond between the electrophilic carbon and the leaving group is a slower step than the reaction with the nucleophile. With an SN2 reaction, the rate of reaction depends on the concentrations of both the substrate and the nucleophile because the bond breaking and bond formation occur simultaneously.

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Most aliphatic nucleophilic substitution reactions fall somewhere between these two extremes. Some unimolecular processes Borderline behavior occur with a partial amount of inversion, but some of the bimolecular occurs when a processes exhibit less than 100% inversion. These “in-between” nucleophilic reactions exhibit what chemists call borderline behavior. The substitution reaction does not follow either reaction is not strictly an SN1 or an SN2 reaction but shows some components of both mechanisms. Where a particular reaction falls in SN1 or SN2 kinetics. the continuum between SN1 and SN2 mechanisms depends largely on the substrate; although the solvent, nucleophile, and leaving group also play a part. Richard A. Sneen of Purdue University suggested a unified theory that explains the “pure” SN1 and SN2 mechanisms as well as the borderline behavior (Acc Chem. Res., 6, 46-53 (1973)). His theory, simply put, says that every substrate first ionizes to form an intermediate ion pair. That ion pair then converts to products.

Nu RL R L RNu

According to his theory, the rate-determining step of the two equilibria above is the factor that determines whether a particular reaction follows an SN1 or an SN2 mechanism. If the formation of the ion pair determines the rate, then the reaction follows SN1 kinetics. If the reaction of the ion pair determines the rate, then the reaction follows SN2 kinetics. Reactions following borderline behavior involve both steps in determining the reaction rate. An important consideration for both SN1 and SN2 reaction mechanisms is the actual steps in ionization. As ionization occurs, a A caged ion pair is a cage of solvent molecules surrounds the pair of ions. Chemists often pair of ions surrounded call this formation a caged ion pair. A caged ion pair is essentially by solvent molecules. the same as a solvent cage, except a caged ion pair involves a pair of ions, whereas a solvent cage involves only one ion. Shortly after ionization occurs, only a few solvent molecules separate the two ions, A solvent-separated ion and the ion pair is called a solvent-separated ion pair. Finally, the pair has a few ions become completely separated and exist as free solvated ions. molecules of solvent Figure 12.9 depicts the steps of this process. between a pair of ions.

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H H O H H O O H H O H H H H H2O H2O RL O R L H O R O H L H H H H H O H H H O O H O H H O H O H H

Caged ion pair Solvent-separated ion pair

H2O

H H O O H H H H H O O R O + H L H H H H H O O H H O H Free ion pair

Figure 12.9. Schematic of the ionization process.

With a high polarity solvent, the substrate tends to ionize further than with a low polarity solvent. Ionization that occurs with a high polarity solvent in combination with a substrate that forms a more stable carbocation, or has a very good leaving group, tends to show SN1 kinetics. A low polarity solvent discourages ionization. A substrate that does not form a stable carbocation, or that has a poor leaving group, shows SN2 kinetics. The most important factor in the kinetics of an aliphatic nucleophilic substitution is the substrate's structure and its ability to ionize. Nearly as important is the ability of the solvent to promote ionization. Of somewhat less importance is the ability of the leaving group to leave. In relation to the leaving group, the more SN2 character that a reaction possesses, the greater is the importance of the strength of the nucleophile. Only when the nucleophile is a stronger base than the leaving group does the equilibrium favor the product. When a reaction mixture consists of a primary substrate in a low polarity solvent that does not promote ionization and a good nucleophile, then the reaction favors the SN2 reaction mechanism. Methyl compounds react the fastest. Primary compounds react nearly as fast and secondary compounds less so. Tertiary compounds do not react via the SN2 mechanism.

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Solvents that promote the formation of free ions from the substrate, tertiary substrates, or other compounds that ionize readily, the SN1 reaction mechanism is favored. The intermediate carbocation in an SN1 reaction is a strong Lewis acid. Acidic reaction conditions encourage the formation of the intermediate carbocation, but in a basic reaction medium, carbocations do not form. If the reaction medium is basic and the nucleophile is a strong enough base, the reaction prefers an elimination mechanism.

Exercise 12.19

Predict the major products of each of the following reactions. Determine whether the reaction is primarily SN1 or SN2.

CH3 HCl COH ZnCl2 CH 3

CH3CCNa Br NH 3

CH2OTs NaN3 DMF

Br CH3CH2OH

CH 3

CH3 CH3CH2I CH3CNH2 (excess) CH 3

Sample Solution c) This is an SN2 reaction.

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CH2OTs CH2N3 NaN3 DMF

Key Ideas from Chapter 12

❏ An sp3 hybridized carbon undergoes nucleophilic substitution if it has a bond to an atom more electronegative than itself.

❏ Nucleophilic substitution at a saturated carbon atom follows one of two mechanisms. In the first mechanism, the leaving group departs before the nucleophile arrives. This is the SN1 mechanism. In the second mechanism, the leaving group departs as the nucleophile arrives. This is the SN2 mechanism.

❏ The rate for an SN1 mechanism depends only on the concentration of the substrate. Thus, an SN1 reaction follows first order kinetics and proceeds through a carbocation intermediate.

❏ The rate for an SN2 mechanism depends on the concentrations of both the nucleophile and the substrate. Thus, an SN2 reaction follows second order kinetics and is a concerted reaction.

❏ Because the SN1 mechanism has a symmetrical carbocation intermediate, it loses all stereochemical information in the reaction. The reaction proceeds with racemization of configuration.

❏ In the SN2 reaction, the nucleophile approaches from “behind” the leaving group resulting in an inversion of the configuration of the carbon in the substrate. In the inversion of configuration of the SN2 mechanism, the product has the opposite configuration of the starting material.

❏ The SN1 mechanism works best in reactions that produce the most stable carbocations. Tertiary carbocations are more stable than either secondary or primary carbocations because of hyperconjugation.

❏ When a primary or secondary carbocation forms, it rearranges, if possible, to form a tertiary carbocation.

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❏ The SN2 mechanism works best in reactions where the structure of the substrate gives the nucleophile access to the backside of the electrophilic carbon atom. Thus, the highest rate of reaction is with methyl, or primary, substrates.

❏ Nucleophiles and leaving groups are both bases. Usually, the leaving group is a weaker base than the nucleophile. Most good nucleophiles are soft bases.

❏ Reactions with both high solvent polarity and the ability of the solvent to solvate both the carbocation and the nucleophile promote SN1 reaction pathways.

❏ Low solvent polarity that does not stabilize carbocation formation promotes SN2 reaction pathways.

❏ Crown ethers and quaternary ammonium salts are examples of phase-transfer catalysts. A phase-transfer catalyst changes an ionic species that is otherwise insoluble in a low polarity solvent to one that is soluble. This change makes a nucleophile much more reactive.

❏ Halide nucleophiles react via either the SN1 or the SN2 mechanism depending on the substrate and reaction conditions.

❏ Water and alcohols are good nucleophiles for solvolysis reactions. Their conjugate bases are good nucleophiles as well, but they tend to promote elimination reactions as side reactions to substitution reactions.

❏ Ammonia and amines are excellent nucleophiles in SN2 reactions.

❏ Cyanide and acetylide ions are also good nucleophiles in SN2 substitution reactions.

❏ Most nucleophilic substitutions are neither pure SN1 nor SN2 reactions but fall somewhere in between. An ionization process has been proposed to explain the SN1-SN2 continuum. The closer together an ion pair stays, the more SN2 characteristics that reaction possesses. The further they move apart, the more SN1 character the reaction shows.