Organic Chemistry Chapter 6 Alkyl Halides

Chapter 6 Alkyl Halides

Incomplete Causeway ( near Indiana, Peru) David Richardson

“Verily, knowledge is a lock and its key is the question.”

Imam Jafar Al-Sadiq

“Student: Dr. Einstein, Aren't these the same questions as last year's [physics] final exam?

Dr. Einstein: Yes; But this year the answers are different.”

Albert Einstein

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Alkyl halides can be named using the alkyl group and the attached functional group or as the halogen attached as a substituent. Also, alkyl halides can be named using the IUPAC system of nomenclature, i.e., identifying the longest continuous chain containing the alkyl group, and numbering the chain in such a manner that the alkyl group receives the lowest number.

Alkyl halides, RX, where X = Cl, Br, or I, readily undergo nucleophilic displacement reactions since X is a facile leaving group. The leaving group ability follows an expected pattern, i.e., iodine is a greater leaving group than bromide; and bromide is a better leaving group than chloride, and chloride is a better leaving group than fluoride.

RX + :Nu → RNu + XNucleophilic displacement reactions can occur via an SN2 (substitution nucleophilic bimolecular) mechanism or an SN1 (substitution nucleophilic unimolecular) mechanism.

Molecularity is the number of entities involved in an elementary step that leads to the formation of an intermediate or an activated complex if the reaction leads to a transition state before it goes to product.

In the case of a nucleophilic substitution reaction, the bimolecular process leads to the formation of an activated complex before forming the products. The following is an 3

energy profile diagram for this bimolecular reaction.

A unimolecular reaction requires one molecule that leads to the formation of an intermediate. A bimolecular reaction requires two molecules of the reactants to form the transition state. The bimolecular reaction takes place in one elementary step, but the unimolecular reaction is a two-step process. The first step is a slow step, followed by a rapid step that leads to the formation of product. Termolecular reactions are rare and generally involve processes using surface or wall phenomena or collision of molecules with their activated states.

An SN2 (substitution nucleophilic bimolecular) reaction can be explained by a single step mechanism involving a nucleophilic attack on an alkyl halide producing an inversion product.

Activated Complex

The activated complex is found at the transition state in the energy profile diagram.

4

The alkyl halide, RX, is found at the reactant stage and R-Nu is found at the product stage.

The SN2 reaction is second order, meaning that the rate of the reaction depends on the concentration of the alkyl halide to the first power and the concentration of the nucleophile to the first power as represented by equation 6.1.

Equation 6.1 Rate = k [Nu][RX] where k is the rate constant in molarity-1 time-1 [Nu] is the molarity, concentration in mol/L of the nucleophile [RX] is the molarity, concentration in mol/L of the alkyl halide If [Nu] = [RX] The rate of the reaction may be expressed by Equation 6.2.

Equation 6.2

where x = concentration of product formed [ao – x] = concentration of nucleophile and alkyl halide at some given point in time ao = the initial concentration of the alkyl halide and the initial concentration of the nucleophile.

The solution to Equation 6.2 is Equation 6.3, where the concentrations of the alkyl halide and the nucleophile are equal.

5

If ao – x = s Then, –dx = ds and

Integrating the equation between s1 and s2 gives

s = a0 -x and at t1 =0 , no product has yet formed x = 0 and s1 = a0 – 0 = ao at t2 = t and s2 = a0 – x

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Equation 6.3

This equation is analogous to the equation of a straight line, i.e., y = mx + b where

y=

x a 0 (a 0 -x)

and t = time and m, the slope, is equal to the rate constant k.

Figure 6.1 is a plot of concentration versus time. If a straight line is obtained, then the reaction is first order in both the nucleophile and the alkyl halide; therefore, the rate is second order overall.

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Figure 6.1 Plot of concentration versus time

If the initial concentrations are unequal, i.e., [Nu] ≠[RX], then the initial concentrations of the two substances can be represented by a0 and b0 respectively; therefore, the rate of the reaction can be described by Equation 6.4.

Equation 6.4

Equation 6.4 can be solved to give Equation 6.5

Equation 6.5 !

1 đ?‘?! (đ?‘Ž! − đ?‘Ľ) & đ?‘™đ?‘œđ?‘”" * / = đ?‘˜đ?‘Ą đ?‘Ž! − đ?‘?! đ?‘Ž! (đ?‘?! − đ?‘Ľ)

This is an equation for a straight line, i.e., similar to y = mx + b where

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1 đ?‘?! (đ?‘Ž! − đ?‘Ľ) đ?‘Ś = ! & đ?‘™đ?‘œđ?‘”" * / đ?‘Ž! − đ?‘?! đ?‘Ž! (đ?‘?! − đ?‘Ľ) and x=t and m , the slope, is equal to k A plot of y versus time would give a straight line (figure 6.2) indicates that the reaction is first order with respect to the starting materials, and second order overall where the two substances have different initial concentrations. The slope is equal to the rate constant.

Figure 6.2 Plot of concentration versus time for the alkyl halide and the nucleophile where the initial concentration of the alkyl halide is different from the initial concentration of the nucleophile

Kinetic information is important in establishing the series of elementary steps resulting in molecular interactions during chemical reactions. Kinetic data help to determine if a reaction follows a nucleophilic bimolecular pathway, a nucleophilic unimolecular pathway, an elimination bimolecular pathway, an elimination unimolecular pathway, or some other pathway.

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Evidence of the SN2 reaction Stereochemical evidence strongly suggests that primary alkyl halides follow an SN2 mechanism, secondary alkyl halides can follow an SN2, and tertiary alkyl halides do not follow an SN2 mechanism.

Walden, in 1893, observed that S (-) α –chlorosuccinic acid in the presence of potassium hydroxide forms R (+) malic acid.

The following chemical equation (where a chiral center, discussed in chapter 7, has undergone selective inversion suggestive of an SN2 reaction) illustrates this reaction.

The reaction process is frequently referred to as the Walden inversion.

(1)

S (-) α –chlorosuccinic acid

R (+) malic acid

In 1923, Phillips observed that R (+)-1-phenyl-2-propanol undergoes inversion of configuration when treated with p-toluenesulfonyl chloride followed by treatment with ethyl alcohol to produce S-2-ethoxy-3-phenylpropane. Chapter 7 introduces the concepts of R, S, (+), and (-). For now, observe that the reaction proceeds with inversion of configuration, i.e., the spatial arrangements of the atoms in S-2-ethoxy-3phenylpropane appear to be inverted compared to the spatial arrangements of the atoms in R (+)-1-phenyl-2-propanol. The same is true for R (+) malic acid and S (-) α – chlorosuccinic acid. The two steps in equations (2) illustrate these processes.

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(2)

S-2-ethoxy-3-phenylpropane

Chemical reactions (1) and (2) follow an SN2 pathway since rate studies indicate that the reactions are first order in the nucleophile, potassium hydroxide, for reaction (1) and ethyl alcohol for reaction (2), and first-order in the substrate, S (-) α –chlorosuccinic acid for reaction (1) and R (+)-1-phenyl-2-propyl p-methylbenzenesulfonate for reaction (2).

Therefore, the reactions are (overall) second order. If the concentration of the nucleophile is very much greater than the substrate, then the rate = k’ [Substrate] where k’ = k [Nu] and the concentration of the nucleophile varies little with time. This type of reaction is pseudo-first-order, and the nucleophile is the solvent. The reaction is a solvolysis reaction.

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Primary alkyl halides behave similarly as reactions (1) and (2), i.e., they are susceptible to SN2 reactions by a variety of chemical entities including -OH, -OR, - :C≥C-R, RMgX, - I-, - CN, RCOO--, - :NH2 3, :P(C6H5)3, --SH, --SR, sodium diethylmalonate (I) and sodium ethyl acetoacetate (II).

Primary alkyl halides substrates are more susceptible to SN2 reactions because the mechanism indicates that this one step process is very sensitive to steric inhibition; therefore, methyl halides can undergo SN2 reactions easier than primary halides that are more susceptible to SN2 reactions than secondary halides which are more sensitive to SN2 reactions than tertiary halides.

Substitution Nucleophilic Unimolecular Reactions Substitution nucleophilic unimolecular, SN1, reactions involve the slow dissociation of an alkyl halide followed by the rapid attack of a nucleophilic reagent on the resulting carbocation. Carbocation formation stabilized by the delocalization of the positive charge will preferably undergo SN1 reactions.

Two steps can represent the general reaction for an SN1 reaction. Step (1), the formation of a carbocation, is the slow step of the mechanism. Step (2), a fast step, is a nucleophilic attack on the carbocation intermediate formed in step (1).

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Step (1) is the rate determining step; therefore, the reaction is independent of the nature and concentration of the nucleophile. The rate of the reaction is dependent on the concentration of the alkyl halide and independent on the concentration of the nucleophile (Equation 6.6).

Equation 6.6

rate = k [RX]

The solution to a first order reaction may be treated in a similar manner as previously described for the solution for substitution nucleophilic bimolecular reactions. Equation 6.7 expresses the rate of a first order reaction.

Equation 6.7

dx = k(a 0 -x) dt where x = concentration of product and [ao – x] = concentration of alkyl halide at some given point in time ao = the initial concentration of the alkyl halide The solution to Equation 6.7 is Equation 6.8, where the rate of the reaction is independent of the concentration of the nucleophile.

dx ∫ a 0 -x = ∫ kdt If ao – x = s

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then –dx = ds and

âˆŤ

-ds = âˆŤ k dt s

Integrating the equation between s1 and s2 and t1 and t2 gives s2

t

ds 2 -âˆŤ = âˆŤ kdt s s1 t1 where s1 is a0 and s2 is a0-x and t1 = t and t2 = 0 Equation 6.8

⎥ a ⎤ kt log ⎢ 0 ⎼ = ⎣ a 0 -x ⎌ 2.303 or đ?‘™đ?‘œđ?‘”" 5

đ?‘Ž# 6 = đ?‘˜đ?‘Ą đ?‘Ž# − đ?‘Ľ

A plot of đ?‘™đ?‘œđ?‘”" 7%

%! ! &'

8 versus time would give Figure 6.3. The straight line indicates that

the reaction is first order in the alkyl halide and zero order in the nucleophile; therefore, first order overall. The slope is equal to k, and the unit for the first order reaction is time-1. Zero order in the nucleophile means that the reaction is independent of the concentration of the nucleophile; therefore, the rate of the reaction only depends on the concentration of the alkyl halide.

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Figure 6.3 Plot of concentration of alkyl halide versus time

The first order calculations and the second order calculations can be applied to any reaction that follows first order kinetics or second order kinetics.

As previously indicated, the first step for nucleophilic unimolecular reactions involves the formation of a carbocation. The formation of the carbocation is the rate determining step or slow step for the reaction. The more stable the carbocation, the more likely the reaction will undergo an SN1 type reaction. Tertiary carbocations are more stable than secondary carbocations. Secondary carbocations are more stable than primary carbocations. Primary carbocations are more stable than the methyl carbocation.

A tertiary carbocation is stabilized by hyperconjugation as indicated in the following resonance structures for the tertiary butyl carbocation.

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The carbocation assumes a trigonal planar arrangement with a bond angle of approximately 120o.

The order of stability of carbocations is 3o > 2o > 1o > CH3+ and the order of alkyl halides toward SN1 reactions would be 3o alkyl halide > 2o alkyl halide > 1o alkyl halide > CH3X Evidence for an SN1 reaction Winstein provided evidence for an SN1 reaction when he added LiClO4 to a solvent separated ion pair and observed a salt effect. The salt effect is an increase in the rate of the SN1 reaction because the perchlorate anion, -ClO4, intercepts the solvent

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separated ion pair preventing the carbocation- X- ion-pair from undergoing internal return. Consequently, the carbocation is more susceptible to nucleophilic attack.

SN1 mechanisms also proceed with racemization or loss of optical activity (Chapter 7).

As an example, R (-)-2-bromooctane is converted into optically impure 2-octanol.

These concepts will be defined and explained in Chapter 7. Nevertheless, a visualization of what is occurring in this system can be understood by considering the following illustration. This illustration shows that if the reaction proceeds through a substitution nucleophilic unimolecular mechanism, there would be two products. One product would have the OH group with the same spatial arrangement as the original alkyl halide, and the other product would have the OH group with a spatial arrangement that would be opposite to that of the original alkyl halide.

Where x+ y= 1.00 and x > y

The alkyl halide ionizes to form the halide ion and the flat alkyl carbocation is attacked by the H2O nucleophile in the indicated directions, i.e., inversion of configuration or 17

retention of configuration. A random attack would produce a racemic mixture (an equal mixture of both isomers); however, the proximity of the halide ion at the reaction’s initiation prevents the attack from proceeding randomly; therefore, backside attack, the inversion product, would predominate. That is why x is greater than y.

Bridged Alkyl Halides Bridgehead alkyl halides such as 1-halobicyclo[2.2.2]octane will not undergo SN2 or SN1 reactions:

The bridge alkyl halide cannot undergo an SN1 reaction because the resulting carbocation cannot assume a trigonal planar or flat arrangement.

The bridge alkyl halide cannot undergo an SN2 reaction because the bridge alkyl halide cannot undergo inversion transformations expected of SN2 reactions.

Halogenation of Alkanes

Where X2 is F2 ; Cl2 ; or Br2.

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As indicated in Chapter 2, fluorine, chlorine, and bromine will undergo halogenation reactions, but iodine will not. The reactivity of halogenation is F2 > Cl2 > Br2.

The Mechanism of Halogenation Revisited Following is the mechanism for the halogenation of methane.

(1) X 2 (g) ⎯⎯ → 2 X⋅ hν (2) X ⋅

(g)

(3) ⋅ CH3

(g)

+ CH 4 (g) → HX (g) + ⋅ CH3 (g)

(g)

+ X 2 (g) → CH3X (g) + X ⋅(g)

The mechanism for the halogenation essentially involves three (3) steps. The first step, the initiation step, is the dissociation of a halogen, reaction (1), by light or heat. Steps (2) and (3) are propagation steps involving the continuation of the formation of free radicals. There are terminating steps that are not included in the mechanism. These

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steps are referred to as quenching steps.

Halogenation of alkanes follows a free radical mechanism. A free radical is an atom or group of atoms possessing an unpaired electron. The halogenation of methane follows a series of chain reactions since the reaction involves a series of steps where each step generates a reactive free radical intermediate that perpetuates the reaction. If Br2 is the halogen in question that interacts with light, then the following steps would explain the formation of the chloromethane.

The initiation step is an energy absorption step that generates a free radical. This step involves the homolytic cleavage of the bromine-bromine Ďƒ(4p+4p) bonding molecular orbital.

Once the free halogen free radical forms, it participates in propagating steps where one reactive species is consumed, and another is generated.

(2)

(3)

Halogenation concludes with a series of chain termination steps where the reactive species are consumed and not generated, i.e., they are quenched.

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There is evidence to support the halogenation mechanism. Tetraethyl lead decomposes at 140o C in the following manner.

As indicated in Chapter 2, the extremely reactive ethyl radical is formed at 140o C; therefore, a mixture of methane and bromine containing a small amount of tetraethyl lead should initiate the bromination of methane at 140o C instead of 250o C through the following series of steps:

The overall reaction would be:

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The structure of the methyl radical Experimental evidence suggests that the methyl radical is flat and the carbon atom (the central atom) is 2sp2 hybridized. The unpaired electron resides in a 2p atomic orbital perpendicular to the plane containing three (3) Ďƒ (2sp2 + 1s) bonding molecular orbitals.

The 2p atomic orbital has a lobe that lies above the plane containing the three (3) Ďƒ (2sp2 + 1s)

bonding molecular orbitals, and it has a lobe that lies below the plane containing

the three (3) Ďƒ (2sp2 + 1s) bonding molecular orbitals.

The following is an illustration of the methyl radical:

This picture suggests that the C-H bonds in the methyl radical lie in the same plane, and that the 2p atomic orbital containing one electron is perpendicular to the plane containing the three C-H molecular bonds.

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The Transition State When molecules react, bonds break, and bonds form.

The following three factors occur during a chemical process where bonds break, and bonds form.

(1)

The reactant molecules must collide.

(2)

The molecules must collide with the proper orientation so reactivity, i.e., bond

breaking and bond making can occur.

(3) Collisions between the reactant molecules must have sufficient kinetic energy so that bond breaking and making initiates; otherwise, the molecules will “bounce� off one another and no product(s) will form. The energy that the molecules possess upon collision must be sufficient to overcome an energy barrier to the reaction.

When these factors occur, they initiate bond breaking and bond making, and the molecules react. However, before bond formation and bond disruption, there is a nonisolable transition state where bond breaking and bond making are in a quasiintermediary state, i.e., bond formation and bond breaking are not complete. This quasi intermediate state is the activated complex and its energy is higher than both the energy of the reactants and the products. For example, in the propagation step where methane reacts with a halogen radical to form the methyl radical and hydrogen halide, the following represents the activated complex:

Activated Complex

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The following equation demonstrates the orientation of the reactants, the activated complex, and the product for the reaction between methane and a halogen radical:

Activated Complex

An energy profile diagram illustrates the progress of a reaction such as the one above, where the "top of the hill" represents the energy of the activated complex, and this “hilltop” position is the transition state.

The chemical species depicted at the “hilltop” at the transition state is the activated complex. The following picture is an illustration of the progress of a reaction (the reaction coordinates) showing the transition state that leads to the formation of the activated complex.

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The energy of activation, Eact, is the difference in energy between the reactants and the transition state. For a reaction to occur, the energy of the colliding molecules must exceed the Eact.

Nomenclature of Alcohols Chapter 12 will present a vigorous discussion about alcohols. Alcohols are a primary source for preparing alkyl halide. The nomenclature of simple alcohols includes the name of the alkyl group followed by the word alcohol, e.g., CH3CH2OH ethyl alcohol; (CH3)2CHOH isopropyl alcohol.

The IUPAC system of nomenclature is more complicated and includes selecting the longest continuous change containing the “OH” group. Then, replace the “e” of the

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alkane to develop the appropriate alkanol backbone and identify its location with the smallest possible number.

3-ethyl-3-heptanol

A 2004 IUPAC recommendation suggest the following nomenclature: 3-ethylheptan-3-ol The structure also could be named as a functional group: 1,1-diethylpentyl alcohol Alcohols can be classified in an analogous manner as alkyl halides, i.e., primary, secondary, and tertiary.

Tertiary alcohols do not have any hydrogen atoms attached to the carbon atom that the OH group is attached.

Secondary alcohols have one hydrogen atom attached to the carbon atom that the OH group is attached.

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Primary alcohols have two hydrogen atoms attached to the carbon atom that the OH group is attached.

Properties of Alcohols For compounds of the same general structural type, such as straight-chain hydrocarbons, the differences in boiling points correlate with differences in molecular masses. The higher the molecular mass, the greater the forces of attraction among the molecules, and the higher the boiling points. When comparing the boiling points of alcohols and alkanes (dissimilar molecules), the boiling points of the alcohols are higher.

Table 6.2 lists the boiling points of various alcohols compared to their molecular masses.

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compound

name

molecular mass,g/mol

boiling point, K

CH3CH2OH

ethyl alcohol

46.07

351.7

ethanol CH3CH3

ethane

44.11

231.1

CH3(CH2)2CH2OH

n-butyl alcohol

74.12

390.4

1-butanol Butan-1-ol CH3(CH2)3CH3

pentane

74.15

309.2

CH3(CH2)4CH2OH

n-hexyl alcohol

102.18

431.2

100.21

371.6

1-hexanol Hexan-1-ol CH3(CH2)5CH3

heptane

Table 6.2 Boiling points of selected alcohols

The higher boiling points of alcohols lie in their ability to undergo intermolecular hydrogen bonding or molecular association. Hydrogen bonding is related to the electrostatic nature of the alcohol resulting from an attraction between a hydrogen atom (bonded to an electronegative group) and another electronegative atom such as oxygen or nitrogen. Hydrogen bonds are weaker than ordinary chemical bonds, and the rupture of these bonds is necessary if alcohols are to boil; therefore, the additional energy requirement results in a higher boiling point.

Hydrogen bonding can be both intramolecular, within the molecule, and intermolecular, with other molecules.

Intermolecular Hydrogen Bonding

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Intramolecular Hydrogen Bonding

Water Water also exhibits hydrogen bonding. This is why water with a lower molecular mass than H2S is a liquid while H2S is a gas. The difference in the state of H2O and H2S doesn’t imply that there is no hydrogen bonding in H2S, but only that it is much weaker.

Hydrogen bonding is not only a characteristic of alcohols but also other compounds as well with hydrogen atoms attached to an electronegative atom. These compounds include but are not limited to, carboxylic acids, phenols, and primary and secondary amines.

Short chain alcohols, i.e., 2-5 carbon atoms, can undergo hydrogen bonding with water:

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Therefore, they are soluble in water.

Preparation of Alkyl Halides from Alcohols Reacting an alcohol with an appropriate haloacid, such as hydroiodic acid, HI; hydrobromic acid, HBr; hydrochloric acid, HCl; and hydrofluoric acid, HF, can result in the formation of alkyl halides.

ROH + HX → RX + H2O X = I, Br, Cl and F and HI > HBr > HCl > HF Tertiary alcohols can be converted into tertiary alkyl halides faster than secondary alcohols can be converted into secondary alkyl halides using haloacids. Secondary alcohols can be converted into secondary alkyl halides faster than primary alcohols can be converted to primary alkyl halides using haloacids. Primary alcohols can be converted to primary alkyl halides faster than methyl alcohol can be converted to methyl halides using haloacids. Following is the decrease in order of reactivity of alcohols with haloacids.

Also, the order of reactivity of the haloacids is HI reacts faster than HBr, and HBr reacts faster than HCl, and HCl reacts faster than HF. HI is a stronger acid than HBr; HBr is a stronger acid than HCl; and HCl is a stronger acid than HF.

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The ideal reaction conditions would be the reaction of a tertiary alcohol that’s more susceptible to substitution nucleophilic unimolecular reaction with hydroiodic acid, the strongest of the haloacids, that’s more susceptible to releasing hydrogen ions in solution (a measure of acid strength).

The mechanism for the reaction would involve protonation of the oxygen of the alcohol; loss of water, the loss of a facile leaving group; the generation of the stable tertiary carbocation; and nucleophilic attack by the weak base, the iodide ion-the conjugate base of the strong HI acid.

2-methyl-2-propanol or t-butyl alcohol or 2-methylpropan-2-ol reacts with hydroiodic acid to produce 2-iodo-2-methylpropane and water. The mechanism (the series of elementary steps) of the reaction is described as: (1)

HI

+

H2O

H3O

+

+

-

I

(2)

CH3

CH3 CH3

C CH3

..

OH ..

+

H3O

+

CH3

C

+ ..

OH ..

H

+

H2O

CH3

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(3)

CH3

CH3 CH3

C

+ ..

OH ..

CH3

C +

+

H2O

H

CH3

CH3

The generation of the carbocation is the slow step in the mechanism; therefore, it is the rate determining step.

(4)

CH3 CH3

C + CH3

CH3 +

-

I

CH3

C

I

CH3

It’s not surprising that secondary and primary alcohols do not react with HCl to produce appreciable amounts of the respective alkyl chlorides. This is not the case with HBr.

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Considering the reactivity of alcohols, i.e., tertiary alcohols react faster than secondary alcohols; and secondary alcohols react faster than primary alcohols and the order of reactivity of haloacids is HI reacts faster than HBr; HBr reacts faster than HCl, and HCl reacts faster than HF, HCl doesn’t form RCl with primary and secondary alcohols. The reaction uses heat because of thermodynamic considerations, and the reaction uses an in situ process with NaBr/H2SO4 to generate HBr. The mechanism of the reaction follows a similar format as the mechanism of HI with tert-butyl alcohol.

Steps (1) is a fast step and step (2) is a slow step, i.e., the rate determining step. Step (1) is the protonation of the oxygen atom of the OH group, and step (2) is the loss of water to form the carbocation.

The following is a mechanism that explains the reaction of tert-butyl alcohol with HBr.

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Again, the slow step of the mechanism is the formation of the carbocation.

Diagram 6.1 represents the energy profile diagram for the reaction.

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Diagram 6.1 Energy profile diagram of t-butyl alcohol reacting with HBr

Alcohols in the presence of a strong acid and heat will prefer to undergo elimination reactions rather than substitution reactions. Therefore, treating t-butyl alcohol with sulfuric acid at high temperatures would produce 2-methylpropene.

2,2-Dimethylpropanol in sulfuric acid with heat would produce 2-methyl-2-butene.

This product results from a rearrangement of a less stable carbocation.

Reactions of Alkyl Halides Primary alkyl halides will react with strong bases to displace the halide ion through substitution nucleophilic bimolecular (SN2) reactions.

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Iodide ion is a faster leaving group than the bromide ion. Bromide ion is a faster leaving group than the chloride ion. Chloride ion is a faster leaving group than the fluoride ion. Therefore, the leaving group ability is I- > Br - > Cl- > F-, and the iodide ion is a facile leaving group.

Methyl alkyl halides undergo SN2 reactions faster than primary alkyl halides. Primary alkyl halides undergo SN2 reactions faster than secondary alkyl halides. Secondary alkyl halides undergo SN2 reactions faster than tertiary alkyl halides. Therefore, the order of reactivity of alkyl halides toward SN2 reactions is methyl halides>primary alkyl halides>secondary alkyl halides>tertiary alkyl halides.

The following reaction shows how methoxide ion, --OCH3 displaces the “I�- in a substitution nucleophilic bimolecular reaction. The reaction is an example of an SN2 reaction where a strong base, sodium methoxide, displaces the iodide ion.

The one step mechanism would involve the formation of an activated complex at the transition state where bond formation and bond breaking take place simultaneous followed by the displacement of the iodide ion and formation of the substitution product.

The mechanism is described in the following manner.

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The energy profile diagram for this reaction (Diagram 6.2) would represent the reactants going to the transition state (where the activated complex is formed), and then to the product or products.

transition state

potential energy reactants products reaction coordinates Diagram 6.2 The unimolecular energy profile diagram

Tertiary alkyl halides in the presence of strong bases would undergo elimination reactions rather than substitution reactions.

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The mechanism for this reaction involves the anti-elimination of iodide. The strong base abstracts a proton from one of the methyl groups, and the iodide ion leaves “anti� to the abstracted hydrogen atom. This reaction is an elimination bimolecular (E2) reaction because it occurs in one elementary step.

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An energy profile diagram, similar to Diagram 6.2, illustrates the reaction coordinates for this reaction. The reactants form an activated complex at the transition state, and then the activated complex forms the products.

An E-1 (elimination unimolecular) reaction will take place with tertiary alcohols if the reaction conditions involve a low concentration of a nucleophile and the nucleophile is a weak base.

The rates of SN2 reactions increase in polar aprotic solvents because the nucleophile, (as a charged species) becomes more available for nucleophilic attack at the reaction site of primary alkyl halides. The positive ions associated with nucleophiles complex with aprotic solvents and the negative nucleophiles are more available for nucleophilic attack. Just the opposite is true with protic solvents. Nucleophiles will molecularly associate with solvents containing OH groups (protic solvents); therefore, they will decrease the rate of SN2 reactions, because lone pair of electrons of the nucleophilic reagents would be less available for nucleophilic attack at the reaction site of primary alkyl halides.

Table 6.2 summarizes the conditions and parameters associated with SN1; SN2; E1; and E2 reactions.

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SN1 Reactions Reaction Conditions o Nature of RX Favors 3 RX Nucleophile weak Solvent protic formation of carbocation; nucleophilic Stereochemistry attack can take place on either side but not equally due to ion pairing Rate Expression k [3o RX]

Reaction Conditions Nature of RX Nucleophile Solvent Stereochemistry Rate Expression

SN2 Reactions o

Favors 1 RX strong aprotic with charged reactants Backside attack k[ Nu][ 1o RX]

E1 Reactions Reaction Conditions o Nature of RX Favors 3 RX Nucleophile weak Solvent protic Formation of a carbocation and base removes the proton resulting in formation Stereochemistry alkenes where the largest groups are trans to one another Rate Expression k [3o RX] E2 Reactions Reaction Conditions o Nature of RX Favors 3 RX Nucleophile strong Solvent aprotic in charged reactants abstraction of hydrogen takes place Stereochemistry opposite (anti) of the leaving Rate Expression k [base][ 3o RX] Table 6.2 Reaction Conditions for SN1; SN2; E1; and E2 Reactions

Reacting primary or secondary alcohols with phosphorus tribromide is a viable method for synthesizing primary alkyl bromides and secondary alkyl bromides. The reaction works best for primary alcohols, and the reaction follows a substitution nucleophilic 40

bimolecular type process.

3 ROH + PBr3 → 3 RBr + H3PO3

The following is a series of elementary steps that can rationale the formation of primary alkyl bromides by reacting phosphorus tribromide with primary and secondary alcohols.

The bromide attacks the carbon from the rear following an SN2 mechanism. This occurs also in steps (4) and (6) of the mechanism; therefore, if stereochemistry (Chapter 7) is possible, then the reaction would take place with inversion of configuration.

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Primary and secondary alkyl chlorides can be prepared from primary and secondary alcohols and thionyl chloride, and the reaction would proceed in the following manner.

ROH + SOCl2 → RCl + SO2 + HCl where ROH is a primary or secondary alcohol For example, 1-pentanol will react with thionyl chloride to produce 1-chloropentane.

CH3CH2CH2CH2CH2OH + SOCl2 → CH3CH2CH2CH2CH2I + SO2 + HCl 42

As previously indicated, the reaction works better with primary alcohols. Following are the series of elementary steps that rationalize the formation of alkyl chlorides from the reaction of primary and secondary alcohols with thionyl chloride.

43

Problems 1.

Suggest syntheses for the following from the indicated starting material and any necessary inorganic and/or organic materials. i. 2-bromopropane from 1-propanol ii. 1-bomopropane from 1-propanol iii.

iv

v. methyl propyl ether from propene vi.

2-methylpentane from propene as the only source of organic material

2.

The reaction of 3-bromo-2,2-dimethylbutane with ethanol exhibits the properties of a first order reaction in which the rate of the reaction depends on the concentration of 3bromo-2,2-dimethylbutane and is independent of the concentration of ethanol. The major product of the reaction is 2-ethoxy-2,3-dimethylbutane. In addition, 2,344

dimethyl-2-butene, 2,3-dimethyl-1-butene, and 3,3-dimethyl-1-butene are produced as minor products. Suggest a mechanism that would be compatible with these observed data.

3.

If 3-bromo-2,2-dimethylbutane is treated with sodium ethoxide, the only product isolated is 3,3-dimethyl-1-butene. Suggest a rationale for this observation. Write a rate expression that would account for this observation.

4.

Arrange the following alcohols in order of reactivity toward HBr. 2-butanol, 1-bromo-2-butanol, 1,1-dibromo-2-butanol, 1,1,1-tribromo-2-butanol 5.

Suggest the major product anticipated in each of the following reactions. (a)

(b)

45

(c)

(d)

(e)

(f)

6.

The rate of the reaction between 2-methyl-2-butanol with HCl is unimolecular. Predict the product or products of the reaction and write the series of elementary steps that rationalize the formation of product.

7.

The rate of the reaction of 1-pentanol with HI is bimolecular. Predict the product of the 46

reaction and write the series of elementary steps that rationalize the formation of product.

8.

For the following reaction:

Suggest structures for A and B Suggest the series of elementary steps that would rationalize the formation of A and B. 9.

Write a series of elementary steps that rationalize the formation of the designated products. (a)

(b)

47

(c)

10. (a) Describe the conditions that would favor substitution nucleophilic bimolecular reactions. (b) Describe the conditions that would favor substitution nucleophilic unimolecular reactions. (c) Describe the conditions that would favor elimination bimolecular reactions. (d) Describe the conditions that would favor elimination unimolecular reactions.

11.

Predict the products for the following chemical reactions; (a)

48

(b)

12.

The following reaction was observed:

In addition, other alkenes are produced.

Suggest a mechanism for the formation of the two indicated alkenes 13.

Suggest a synthesis for bromocyclohexane from 1,3-butadiene and any other necessary inorganic and organic material.

14.

Suggest syntheses for the following from butane as the only organic starting material and any necessary inorganic material. (a)

n-octane

(b)

1-bromobutane

(c)

2-bromobutane

(d)

3-methylheptane

49

15.

Suggest the product expected in the following reaction and suggest a mechanism for the reaction.

16.

Suggest the major product for the reaction of the indicated materials with 1iodobutane. a. Concentrated NaOH b. Cold sulfuric acid c. hot concentrated sulfuric acid d. Li, then copper (I) iodide, then ethyl bromide e. Mg/ether, then deuterium oxide f. Sodium bromide in acetone 17.

3,3-Dimethyl-1-butene reacts with hydrogen iodide to produce 3-iodo-2,2dimethylbutane and 2-iodo-2,3-dimethylbutane. Give a rational for the formation of these two products.

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18.

The following reaction proceeds slowly. Give a rationale for this observation.

19.

Write structural formulas for (a) 2,6-dibromo-4-methyl-4-octanol (b) 1-bromo-2-cyclobutylethane (c) trans-4-tert-butylbromocyclohexane (d) 2-bromo-2-cyclopentylpropane (e) 5-bromo-2,3-dimethylhexane 20.

Methyl bromide depletes the ozone in the atmosphere; therefore, its use in agriculture as a soil fumigant is no longer desirable. Methyl bromide can be prepared industrially from methanol and hydrogen bromide. Following are hypothetical kinetic data for this reaction at fifty degrees centigrade.

[Methanol], moles/liters 0.10 0.050 0.025 0.0125

[HBr], moles/liters

Time, seconds

0.10 0.050 0.025 0.0125

0 60 177 412

(a) Use Microsoft Excel to determine if the reaction follows a unimolecular nucleophilic

51

substitution reaction or a bimolecular nucleophilic substitution reaction. (b) Determine the rate constant for the reaction. (c) Write the series of elementary steps that would account for the formation of the soil fumigant, methyl bromide as a product.

52

Solutions to Problems Chapter 6 1.

Suggest syntheses for the following from the indicated starting material and any necessary inorganic and/or organic materials. (i)

2-bromopropane from 1-propanol

(ii)

1-bomopropane from 1-propanol

(iii)

53

(iv)

54

(v) methyl propyl ether from propene

(vi) 2-methylpentane from propene as the only source of organic material

55

2.

The reaction of 3-bromo-2,2-dimethylbutane with ethanol exhibits the properties of a first order reaction in which the rate of the reaction depends on the concentration of 3bromo-2,2-dimethylbutane and is independent of the concentration of ethanol. The major product of the reaction is 2-ethoxy-2,3-dimethylbutane. In addition, 2,3dimethyl-2-butene, 2,3-dimethyl-1-butene, and 3,3-dimethyl-1-butene are produced as minor products. Suggest a mechanism that would be compatible with these observed data.

1.

56

2.

3.

2-ethoxy-2,3-dimethylbutane Minor products

3,3-dimethyl-1-butene

2,3-dimethyl-2-butene

57

2,3-dimethyl-1-butene

3.

If 3-bromo-2,2-dimethylbutane is treated with sodium ethoxide, the only product isolated is 3,3-dimethyl-1-butene. Suggest a rationale for this observation. Write a rate expression that would account for this observation.

The only hydrogen atoms that are available for abstraction are located on the methyl group of the carbon atom to the right of the carbon atom containing the bromine atom.

Therefore, elimination must take place on that carbon atom anti to the bromine atom:

Let

then

rate = k [ RBr ] ⎡⎣CH 3CH 2O- Na + ⎤⎦

58

4.

Arrange the following alcohols in order of reactivity toward HBr. 2-butanol, 1-bromo-2-butanol, 1,1-dibromo-2-butanol, 1,1,1-tribromo-2-butanol 1,1,1-tribromo-2-butanol>1,1-dibromo-2-butanol>1-bromo-2-butanol>2-butanol 5.

Suggest the major product anticipated in each of the following reactions. (a)

(b)

59

(c)

No reaction (d)

(e)

(f)

60

6.

The rate of the reaction between 2-methyl-2-butanol with HCl is unimolecular. Predict the product or products of the reaction, and write the series of elementary steps that rationalize the formation of product.

1.

2.

3.

7.

The rate of the reaction of 1-pentanol with HI is bimolecular. Predict the product of the reaction, and write the series of elementary steps that rationalize the formation of product.

61

where X = I 1. For the following reaction:

Where x + y = 1 Suggest structures for A and B Suggest the series of elementary steps that would rationalize the

formation of

A and B. 1.

62

2.

3.

63

8.

Write a series of elementary steps that rationalize the formation of the designated products. (a)

1.

2.

3.

secondary carbocation

tertiary carbocation

64

4.

(b)

1.

2.

65

3.

4.

(c)

1.

66

2.

3.

4.

Formation of the five-member ring structure: 1.

67

2.

3.

4.

10. (a)

Describe the conditions that would favor substitution nucleophilic bimolecular

reactions.

Primary alkyl halides, strong nucleophiles, and protic solvents favor substitution bimolecular reactions. (b)

Describe the conditions that would favor substitution nucleophilic unimolecular

reactions.

68

Tertiary alkyl halides, weak nucleophiles, and protic solvents with charged reactants favor substitution nucleophilic unimolecular reactions. (c)

Describe the conditions that would favor elimination bimolecular reactions.

Tertiary alkyl halides, strong nucleophiles, and aprotic solvents in charged reactants favor elimination bimolecular reactions. (d)

Describe the conditions that would favor elimination unimolecular reactions.

Tertiary alkyl halides, weak nucleopliles, and protic solvents favor substitution nucleophilic unimolecular reactions.

11. Predict the products for the following chemical reactions; (a)

(b)

69

12.

The following reaction was observed:

In addition, other alkenes are produced.

Suggest a mechanism for the formation of the two indicated alkenes 1.

2.

3.

70

4.

or

13.

Suggest a synthesis for bromocyclohexane from 1,3-butadiene and any other necessary inorganic and organic material.

14.

Suggest syntheses for the following from butane as the only organic starting material and any necessary inorganic material. (a) n-octane

71

(b) 1-bromobutane

(c)

2-bromobutane

72

(d) 3-methylheptane

73

15.

Suggest the product expected in the following reaction, and suggest a mechanism for the reaction.

74

16.

Suggest the major product for the reaction of the indicated materials with 1iodobutane. (a) NaOH

(b) Cold sulfuric acid

(c) hot concentrated sulfuric acid

75

(d)

Li, then copper (I) iodide, then ethyl bromide

(e)

Mg/ether, then deuterium oxide

(f)

Sodium bromide in acetone

76

17.

3,3-Dimethyl-1-butene reacts with hydrogen iodide to produce 3-iodo-2,2dimethylbutane and 2-iodo-2,3-dimethylbutane. Give a rational for the formation of these two products.

3-iodo-2,2-dimethylbutane 1,2-methide shift:

2-iodo-2,3-dimethylbutane

77

18.

The following reaction proceeds slowly. Give a rationale for this observation.

19.

Write structural formulas for (a) 2,6-dibromo-4-methyl-4-octanol

(b) 1-bromo-2-cyclobutylethane

(c) trans-4-tert-butylbromocyclohexane

(d) 2-bromo-2-cyclopentylpropane

78

(e) 5-bromo-2,3-dimethylhexane

20.

Methyl bromide depletes the ozone in the atmosphere; therefore, its use in agriculture as a soil fumigant is no longer desirable. Methyl bromide can be prepared industrially from methanol and hydrogen bromide. Following are hypothetical kinetic data for this reaction at fifty degrees centigrade.

[Methanol], moles/liters 0.10 0.050 0.025 0.0125

[HBr], moles/liters

Time, seconds

0.10 0.050 0.025 0.0125

0 60 177 412

(a) Use Microsoft Excel to determine if the reaction follows a unimolecular nucleophilic substitution reaction or a bimolecular nucleophilic substitution reaction.

A plot of x/[ao(ao -x)] versus times gives a straight line. This indicates that the reaction is second order, i.e., first order in methanol and first order in HBr.

79

x/[ao(ao -x)] 0 10 30 66

time (second) 0 60 177 412

70" 60"

x/[a(a&x)])

50" 40" y"="0.1617x"

30" 20" 10" 0" 0"

50"

100"

150"

200"

250"

300"

350"

400"

450"

*me,)second)

The equation for the line is

x = kt a o (a o -x) Therefore, a plot of

x versus t a o (a o -x) gives a straight line with a slope equal to 0.16 L mole-1 s-1, and the intercept equals 0.

Note that the units are 1/Ms or M-1 s-1 or L mol-1 s-1. The units are obtained from the change in concentration (M/M2 = 1/M) divided by the change in time (seconds).

80

(b) Determine the rate constant for the reaction.

The rate constant is the slope of the line; therefore, k is 0.162 mole-1 s-1

(c) Write the series of elementary steps that would account for the formation of the soil fumigant, methyl bromide as a product.

Where X is Br

81