Nomenclature: Functional group suffix = -ol Functional group prefix = hydroxy- Primary, secondary or tertiary? Alcohols are described as primary (1 o ), secondary (2 o ) or tertiary (3 o ) depending on how many alkyl substiutents are attached to the C-OH unit. t-butanol ethanol 2-propanol 2-methyl-2-propanol 1 o 2 o 3 o Check the designations of primary, secondary and tertiary by counting the number of C atoms attached to the C with the O attached. Physical Properties: The polar nature of the O-H bond (due to the electonegativity difference of the atoms ) results in the formation of hydrogen bonds with other alcohol molecules or other H-bonding systems (e.g. water). The implications of this are: o high melting and boiling points compared to analogous alkanes o high solubility in aqueous media Structure: The alcohol functional group consists of an O atom bonded to a C atom and a H atom via s bonds. Both thec-o and the O-H bonds are polar due to the high electronegativity of the O atom. Reactivity: The image shows the electrostatic potential for methanol. The more red an area is, the higher the electron density and the more blue an area is, the lower the electron density. The alcohol O atom are a region of high electron density (red) due to the lone pairs. (red) Alcohol oxygen atoms are Lewis
bases. So alcohols can react as either bases or nucleophiles at the oxygen. There is low electron density (blue) on H atom of the -OH group alcohol, i.e. H + character. So alcohols are acidic (pka ~ 16). Removal of the proton generates the alkoxide. The -OH group is a poor leaving group and needs to be converted to a better leaving group before substitution can occur. Acidity: Due to the electronegativity of the O atoms, alcohols are slightly acidic (pka 16-18) The anion derived by the deprotonation of an alcohol is the alkoxide. Alkoxides are important bases in organic chemistry. Alcohols react with Na (or K) like water to give the alkoxide: Nomenclature: Functional group suffix = -halide Functional group prefix = halo- Primary, secondary or tertiary? In a similar fashion to alcohols, alkyl halides are described as primary (1 o ), secondary (2 o ) or tertiary (3 o ) depending on how many alkyl substiutents are attached to the C-X unit. chloromethane bromoethane or or methyl chloride ethyl bromide 2-bromopropane or isopropyl bromide 2-bromo-2-methylpropane or tert-butyl bromide primary secondary tertiary
Physical Properties: The polar bond creates a molecular dipole that raises the melting points and boiling points compared to alkanes. Structure: The alkyl halide functional group consists of an sp 3 hybridized C atom bonded to a halogen, X, via a s bond. The carbon halogen bonds are typically quite polar due to the electronegativity and polarizability of the halogen. Reactivity: The halogens (Cl, Br and I) are good leaving groups. The polarity makes the C atom electrophilic and prone to attack by nucleophiles via S N 1 or S N 2 reactions. Bases can remove b-hydrogens and cause 1,2-elimination to form alkenes via E1 or E2 reactions. Insertion of a metal (esp. Mg) creates an organometalic species. Nucleophilic substitution reactions are an important class of reactions that allow the interconversion of functional groups, for example R-OH R-Br. Nucleophilic substitution will be explored in much more detail in chapter 8. What does the term "nucleophilic substitution" imply? A nucleophile is an the electron rich species that will react with an electron poor species. A substitution implies that one group replaces another There are two fundamental events in these substitution reactions: 1. formation of the new bond to the nucleophile 2. breaking of the bond to the leaving group
Depending on the relative timing of these events, two different mechanisms are possible: Bond breaking to form a carbocation preceeds the formation of the new bond : S N 1 reaction Simultaneous bond formation and bond breaking : S N 2 reaction S N 1 mechanism S N 1 indicates a substitution, nucleophilic, unimolecular reaction, described by the expression rate = k [R-LG] This pathway is a multi-step process with the following characteristics: step 1: rate determining (slow) loss of the leaving group, LG, to generate a carbocation intermediate, then step 2: rapid attack of a nucleophile on the electrophilic carbocation to form a new s bond Multi-step reactions have intermediates and several transition states (TS). In an S N 1 there is loss of the leaving group generating an intermediate carbocation which then undergoes a rapid reaction with the nucleophile. General case The reaction profiles shown here are simplified and do not include the equilibria for protonation of the -OH. SN1 reaction The following issues are relevant to the S N 1 reactions of alcohols: Effect of R- Reactivity order : (CH 3 ) 3 C- > (CH 3 ) 2 CH- > CH 3 CH 2 - > CH 3 - In an S N 1 reaction, the key step is the loss of the leaving group to form the intermediate carbocation. The more stable the carbocation is, the easier it is to form, and the faster the S N 1 reaction will be. Some students fall into the trap of thinking that the system with the less stable carbocation will react fastest, but they are forgetting that it is the generation of the carbocation that is rate determining. -LG The only event in the rate determining step of the S N 1 is breaking the C-LG
bond. For alcohols it is important to remember that -OH is a very poor leaving. In the reactions with HX, the -OH is protonated first to give an oxonium, providing the much better leaving group, a water molecule (see scheme below). Nu Since the nucleophile is not involved in the rate determining step of an S N 1 reaction, the nature of the nucleophile is unimportant. In the reactions of alcohols with HX, the reactivity trend of HI > HBr > HCl > HF is not due to the nucleophilicity of the halide ion but the acidity of HX which is involved in generating the leaving group prior to the rate determining step. Stereochemistry In an SN1, the nucleophile attacks the planar carbocation. Since there is an equally probability of attack on either face there will be a loss of stereochemistry at the reactive center and both possible products will be observed. Since a carbocation intermediate is formed, there is the possibility of rearrangements (e.g. 1,2-hydride or 1,2-alkyl shifts) to generate a more stable carbocation (see later). This is usually indicated by a change in the position of the halide compared to that of the original -OH group, or a change in the carbon skeleton of the product when compared to the starting material.
SN1 MECHANISM FOR REACTION OF ALCOHOLS WITH HBr Step 1: An acid/base reaction. Protonation of the alcoholic oxygen to make a better leaving group. This step is very fast and reversible. The lone pairs on the oxygen make it a Lewis base. Step 2: Cleavage of the C-O bond allows the loss of the good leaving group, a neutral water molecule, to give a carbocation intermediate. This is the rate determining step (bond breaking is endothermic) Step 3: Attack of the nucleophilic bromide ion on the electrophilic carbocation creates the alkyl bromide. Carbocations Stability: The general stability order of simple alkyl carbocations is: (most stable) 3 o > 2 o > 1 o > methyl (least stable)
This is because alkyl groups are weakly electron donating due to hyperconjugation and inductive effects. Resonance effects can further stabilize carbocations when present. Structure: Alkyl carbocations are sp 2 hybridized, planar systems at the cationic C center. The p-orbital that is not utilized in the hybrids is empty and is often shown bearing the positive charge since it represents the orbital available to accept electrons. Reactivity: As they have an incomplete octet, carbocation s are excellent electrophil es and react readily with nucleophile s. Alternative ly, loss of H + can generate a
π bond. The electrostatic potential diagrams clearly show the cationic center in blue, this is where the nucleophile will attack. Rearrangements: Carbocations are prone to rearrangement via 1,2-hyride or 1,2-alkyl shifts if it generates a more stable carbocation Reactions involving carbocations: 1. Substitutions via the S N 1 2. Eliminations via the E1 3. Additions to alkenes and alkynes (HX, H 3 O + ) S N 2 mechanism S N 2 indicates a substitution, nucleophilic, bimolecular reaction, described by the expression rate = k [Nu][R-LG] This pathway is a concerted process (single step) as shown by the following reaction coordinate diagrams, where there is simultaneous attack of the nucleophile and displacement of the leaving group.
Single step reactions have no intermedia tes and single transition state (TS). In an S N 2 there is simultaneous formation of the carbonnucleophile bond and breaking of the carbonleaving group bond, hence the reaction proceeds via a TS in which the central C is partially bonded to five groups. The reaction profiles shown here are simplified and do not include the equilibria for protonation of the -OH. General case SN2 reaction The following issues are relevant to the S N 2 reactions of alcohols: Effects of R- Reactivity order : CH 3 - > CH 3 CH 2 - > (CH 3 ) 2 CH- > (CH 3 ) 3 C- For alcohols reacting with HX, methyl and 1 o systems are more likely to react via an S N 2 reaction since the carbocations are too high energy for the S N 1 pathway to occur.
-LG Once again the leaving group is a water molecule formed by protonation of the - OH group. -OH on its own is a poor leaving group. Nu Since the nucleophile is involved in the rate determining step, the nature of the nucleophile is very important in an S N 2 reaction. More reactive nucleophiles will favor an S N 2 reaction. Stereochemistry When the nucleophile attacks in an S N 2 reaction, it is on the opposite side to the position of the leaving group. As a result, the reaction will proceed with an inversion of configuration. S N 2 MECHANISM FOR REACTION OF ALCOHOLS WITH HBr Step 1: An acid/base reaction. Protonation of the alcoholic oxygen to make a better leaving group. This step is very fast and reversible. The lone pairs on the oxygen make it a Lewis base. Step 2: Simultaneous formation of C-Br bond and cleavage of the C-O bond allows the loss of the good leaving group, a neutral water molecule, to give a the alkyl bromide. This is the rate determining step.
Radical Substitution Summary Substitution of R-H by -X provides the alkyl halide, R-X and HX. Alkane R-H relative reactivity order : 3 o > 2 o > 1 o > methyl Halogen reactivity F 2 > Cl 2 > Br 2 > I 2 Only chlorination and bromination are useful in the laboratory. Reaction proceeds via a radical chain mechanism Radicals A radical is a species that contains unpaired electrons. Typically formed by a homolytic bond cleavage as represented by the fishhook curved arrows: Stability: The general stability order of simple alkyl radicals is: (most stable) 3 o > 2 o > 1 o > methyl (least stable) This is because alkyl groups are weakly electron donating due to hyperconjugation and inductive effects. Note that this is the same order as for carbocations. Resonance effects can further stabilize radicals when present.
Structure: Alkyl radicals are sp 2 hybridized, planar systems at the radical C center. The p-orbital that is not utilized in the hybrids contains the single electron. Reactivity: As they have an incomplete octet, radicals are excellent electrophiles and react readily with nucleophiles. Alternatively, loss of H. can generate a p bond Rearrangements: Unlike carbocations, radicals do not tend to undergo rearrangements Reactions involving radicals: Radical Substitution Mechanism Unlike the large majority of reactions that you will see in your organic chemistry course, radical mechanism require that fishhook curly arrows that represent the motion of a single electron are used. These can be a little more confusing and more difficult to master. We suggest you get to grips with normal curved arrows first. However, the mechanism for the bromination of methane is shown below, but the mechanism for chlorination or higher alkanes in the same. Note that it contains three distinct type of steps, depending on the net change in the number of radicals that are present. RADICAL CHAIN MECHANIS M FOR REACTION OF METHANE WITH Br 2
Step 1 (Initiation) Heat or uv light cause the weak halogen bond to undergo homolytic cleavage to generate two bromine radicals and starting the chain process. Step 2 (Propagatio n) (a) A bromine radical abstracts a hydrogen to form HBr and a methyl radical, then (b) The methyl radical abstracts a bromine atom from another molecule of Br 2 to form the methyl bromide product and another bromine radical, whic
h can then itself undergo reaction 2(a) creating a cycle that can repeat. Step 3 (Terminatio n) Various reactions between the possible pairs of radicals allow for the formation of ethane, Br 2 or the product, methyl bromide. These reactions remove radicals and do not perpetuate the cycle. Selectivity There are two components to understanding the selectivity of radical halogenations of alkanes: reactivity of R-H system reactivity of X. R-H The strength of the R-H varies slightly depending on whether the H is 1 o, 2 o or 3 o.
The following table shows the bond dissociation energy, that is the energy required to break the bond in a homolytic fashion, generating R. and H. Type R-H kj/mol kcal/mol CH 3 -H 435 104 1 o CH 3 CH 2 -H 410 98 2 o (CH 3 ) 2 CH-H 397 95 3 o (CH 3 ) 3 C-H 380 91 Halogen radical, X. Note how the bonds get weaker as we move down the table, so the R. also get easier to form, with 3 o being the easiest. Bromine radicals are less reactive than chlorine radicals Br. tends to be more selective in its reactions, and prefers to react with the weaker R-H bonds. The more reactive chlorine radical is less discriminating in what it reacts with. The selectivity of the radical reactions can be predicted mathematically based on a combination of an experimentally determined reactivity factor, R i, and a statistical factor, nh i. In order to use the equation shown below we need to look at our original alkane and look at each H in turn to see what product it would give if it were to be susbtituted. This is an exercise in recognizing different types of hydrogen, something that will be important later. %Pi nhi Ri Σi = % yield of product "i" = number of H of type "i" = reactivity factor for type "i" = sum for all types Reactivity factors, R i Br Cl 1 o 1 1 2 o 82 3.9 3 o 1640 5.2 What do the reactivity factors indicate? Well as an example of the conclusions we could make: Bromination is 1640 times more likely to occur at a 3 o position than 1 o Chlorination is 5.2 times more likely to occur at a 3 o position than 1 o Lets work an example, say propane, CH 3 CH 2 CH 3 How many different monochlorides can be produced by radical chlorination? 1-chloropropane and 2 -chloropropane, so we have two types of H, the 6 x 1 o in -
CH 3 and the 2 x 2 o in-ch 2 -. Use the diagrams below to highlight this if you are unsure. 1 o CH 2 o CH Reset propane 1-chloropropane 2-chloropropane Now for the calculations, so plugging the values into the equations we get (the reactivity factors R i are in the table above): % 1-chloropropane = 100 x (6 x 1) / (6 x 1 + 2 x 3.9) = 100 x 6 / 13.8 = 43.5 % (experimental = 44 %) % 2-chloropropane = 100 x (2 x 3.9) / (6 x 1 + 2 x 3.9) = 100 x 7.8 / 13.8 = 56.5 % (experimental = 56 %) What about bromination of propane? Most of the process is the same, all we have to do is change the reactivty factors. % 1-bromopropane = 100 x (6 x 1) / (6 x 1 + 2 x 82) = 100 x 6 / 170 = 3.5 % (experimental = 4 %) % 2-bromoopropane = 100 x (2 x 82) / (6 x 1 + 2 x 82) = 100 x 164 / 170 = 96.5 % (experimental = 96 %) Note that the results match well with experimental values and that they illustrate the high regioselectivity of the bromination reaction for the 2 o radical, whereas in the chlorination the number of 1 o H dictates the regioslectivity. There are other examples in the Self Assessment. Radical Halogenation of Alkanes Reaction type: Radical Substitution Summary: When treated with Br 2 or Cl 2, radical substitution of R-H generates the alkyl halide and HX.
Alkane R-H relative reactivity order : tertiary > secondary > primary > methyl. Halogen reactivity F 2 > Cl 2 > Br 2 > I 2 Only chlorination and bromination are useful in the laboratory. Bromination is selective for the R-H that gives the most stable radical. Chlorination is less selective Reaction proceeds via an radical chain mechanism which involves radical intermediates. The termination steps are of low probability due to the low concentration of the radical species meaning that the chances of them colliding is very low. RADICAL CHAIN MECHANIS M FOR REACTION OF METHANE WITH Br 2 Step 1 (Initiation) Heat or uv light cause the weak halogen bond to undergo homolytic cleavage to generate two bromine radicals and starting the chain process.
Step 2 (Propagatio n) (a) A bromine radical abstracts a hydrogen to form HBr and a methyl radical, then (b) The methyl radical abstracts a bromine atom from another molecule of Br 2 to form the methyl bromide product and another bromine radical, whic h can then itself undergo reaction 2(a) creating a cycle that can repeat.
Step 3 (Terminatio n) Various reactions between the possible pairs of radicals allow for the formation of ethane, Br 2 or the product, methyl bromide. These reactions remove radicals and do not perpetuate the cycle. More highly brominated by-products are possible if methyl bromide reacts with a bromine radical in the same fashion as methane does. Can you draw the cycle that leads to the formation of dibromomethane? Reaction of Alcohols with Hydrogen Halides Reaction type: Nucleophilic Substitution (S N 1 or S N 2) Summary:
When treated with HBr or HCl alcohols typically undergo a nucleophilic substitution reaction to generate an alkyl halide and water. Alcohol relative reactivity order : 3 o > 2 o > 1 o > methyl. Hydrogen halide reactivity order : HI > HBr > HCl > HF (paralleling acidity order). Reaction usually proceeds via an S N 1 mechanism which proceeds via a carbocation intermediate, that can also undergo rearrangement. Methanol and primary alcohols will proceed via an S N 2 mechanism since these have highly unfavorable carbocations. The reaction of alcohols with HCl in the presence of ZnCl 2 (catalyst) forms the basis of the Lucas test for alcohols.
S N 1 MECHANISM FOR REACTION OF ALCOHOLS WITH HBr Step 1: An acid/base reaction. Protonation of the alcoholic oxygen to make a better leaving group. This step is very fast and reversible. The lone pairs on the oxygen make it a Lewis base. Step 2: Cleavage of the C-O bond allows the loss of the good leaving group, a neutral water molecule, to give a carbocation intermediate. This is the rate determining step (bond breaking is endothermic) Step 3: Attack of the nucleophilic bromide ion on the electrophilic carbcation creates the alkyl bromide.
Reaction of Alcohols with other Halogenating agents (SOCl 2, PX 3 ) Reaction type: Nucleophilic Substitution (S N 1 or S N 2) Summary: Alcohols can also be converted to alkyl chlorides using thionyl chloride, SOCl 2, or phosphorous trichloride, PCl 3. Alkyl bromides can be prepared in a similar reaction using PBr 3. Used mostly for 1 o and 2 o ROH In each case a base is used to "mop-up" the acidic by-product. Common bases are triethylamine, Et 3 N, or pyridine, C 6 H 5 N. In each case the -OH reacts first as a nucleophile, attacking the electrophilic center of the halogenating agent. A displaced halide ion then completes the substitution displacing the leaving group. Note that it is not -OH that leaves, but a much better leaving group. The advantage of these reagents is in that the reaction is not under the strongly acidic conditions like using HCl or HBr.