Chapter 4 Alcohols and Alkyl Halides

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1 . 4 hapter 4 Alcohols and Alkyl alides Alcohols and alkyl halides are very important functional groups. A functional group is an atom or group of atoms that undergoes certain reactions that are typical of that functional group. It is important to recognize functional groups since it makes the organization and learning of organic chemistry much easier. There are several million organic compounds that are known and more are discovered or synthesized everyday but there is a limited number of functional groups. Functional groups we have already seen: Alkanes essentially have no functional group. They contain only and single bonds. Alkenes have the carbon-carbon double bond as the functional group. Alkynes have the carbon-carbon triple bond as the functional group. Arenes have the benzene ring as a functional group. New functional groups in this chapter: Alcohols contain the hydroxyl group, -. Alkyl halides contain a halide. We use the symbol X to stand for any halogen (I, Br,, F). is used to refer to any alkyl group regardless of size. It can be primary, secondary or tertiary, so a generalized alcohol is written as and a generalized alkyl halide is written as X. ther functional groups that will be studied throughout the rest of the course include: Amines contain a nitrogen, N. The amine may be primary (N 2 ), secondary ( 2 N), or tertiary ( 3 N) where the groups can be the same or different. 3 2 N ethylamine (primary amine) 3 2 N 3 ethyl methylamine (secondary amine) 3 2 N 2 3 triethylamine tertiary amine) 2 3 Ethers contain a -- linkage, --, where and can be the same or different diethyl ether Epoxides these are special 3-membered ring ethers. 1

2 . 4 Nitriles these contain the N functional group, N. This functional group is also called a cyano-group N Nitroalkanes these contain the nitro group, -N N Thiols these contain an S group S S Sulfide these are thio ethers, -S S arbonyl Derivatives all contain the carbonyl group, = Aldehydes have the carbonyl group at the end of the chain, Ketones have the carbonyl group attached to two other carbons, one on each side, arboxylic Acid contain the functional group arboxylic Acid Esters (or Esters) contain the functional group Amides contain the functional group N N N N arboxylic Acid Anhydrides contain the functional group 2 2

3 Nomenclature Alkyl halides: There are two ways to name alkyl halides. 1. Name the alkyl group first, then as a separate word name the halide. 3 F methyl fluoride Br butyl bromide 2. The systematic IUPA name is to treat the halide as a substituent on the longest alkyl chair that contains the halide. Alphabetize it along with any other substituents. Number the chain so that the substituents get the lowest number at the first point of difference between the two possible directions for numbering F 1-fluoropentane chloro-2-methylpentane Alcohols: For the systematic IUPA name, identify the parent alkane as the longest chain that contains the hydroxyl group. Drop the e from the alkane and add the suffix -ol. The alcohol group takes precedence over alkyl and halide substituents so number the parent chain in the direction that gives the alcohol the lowest number, giving the terminal carbon the number one. List the other substituents alphabetically, specifying their locations by number. In cyclic alcohol, the hydroxyl group always gets number one, though the number is not written. Number the other substituents so as to obtain the lowest number at the first pint of difference between the two possible numbering directions propanol propanol asses of Alcohols and Alkyl alides 3 Br bromo-5-methyl-2-hexanol chloro-3-ethylcyclohexanol A primary alcohol has a hydroxyl group attached to a primary carbon and a primary alkyl halide has a halogen attached to a primary carbon. 3

4 primary alcohol Br primary alkyl haldie A second alcohol has a hydroxyl group attached to a secondary carbon and a secondary alkyl halide has an alkyl group attached to a secondary carbon secondary alcohol secondary alkyl halide A tertiary alcohol has a hydroxyl group attached to a tertiary carbon and a tertiary alkyl halide has a halide attached to a tertiary carbon tertiary alcohol tertiary alkyl halide Both alcohols and alkyl halides have fairly large dipole moments. δ δ + δ δ + For alkyl halides, note the trend in bond lengths. The -F bond is the shortest and the -I bond is the longest as one would expect based on the much larger size of iodine versus fluorine. F < < Br < I arbon-halogen bond length Physical Properties: 1.4 A 1.79 A 1.97 A 2.16 A ompare boiling points: F b.p. -42 b.p b.p. 78 4

5 . 4 Fluoroethane is more polar than propane and has a slightly higher boiling point due to increases in dipole-induced dipole interactions but the alcohol has a much higher boiling point due to intermolecular hydrogen bonding. In order for hydrogen bonding to occur a hydrogen must be attached to an electronegative atom like oxygen or nitrogen or fluorine that has a lone pair. Each hydrogen bond is worth about 20 KJ/mol or ~ 5 Kcal/mol in energy. arbon-hydrogen bonds are not polarized enough to engage in hydrogen bonding hydrogen bond N N N N For alkyl halides, the boiling point increases with the increasing size of the halogen. 3 F 3 3 Br 3 I boiling point This is because the larger halogens are more polarizable. In the very large iodine, the electrons are much farther from the nucleus and more easily distorted. Therefore, there are stronger induced dipole-induced dipole interactions. The boiling point also increases with the increasing number of halogens boiling point Again, this is due to increased induced dipole-induced dipole interactions. But fluorine is an exception due to the fact that fluorine is not very polarizable, though F 3 F 3 still has a higher boiling point than ethane, 3 3, since it is more polar. boling point, 3 2 F F 2 3 F 3 F 3 F Water solubility: All alkyl halides are insoluble in water, like alkanes. Low molecular weight alcohols (methanol, ethanol, 1-propanol, 2-propanol) are miscible with water. Miscible means they are soluble in all proportions and form one layer with water. This is due to intermolecular hydrogen bonding. As the hydrocarbon chain gets longer, the solubility decreases

6 nly 0.5 ml dissolves in one liter of water. non-polar polar Preparation of Alkyl alides from Alcohols and X General reaction: + X X + The reactivity order for the hydrogen halides parallels their acidity: I > Br > >> F weakest acid strongest acid The reactivity order for the alcohols is: > > tertiary secondary primary > methyl Ex: room temp. + 3 (rt) t-butanol t-butyl chloride 78-88% Sometimes we write the reagents over the arrow in order to save space NaBr, 2S 4 heat Br Na + - S 3 Mechanism of the eaction of t-butanol and When t-butanol reacts with hydrochloric acid to form t-butyl chloride, this is a substitution reaction. We substitute the hydroxyl group with the chlorine. The mechanism is a Nucleophilic Substitution Unimolecular or S N 1 mechanism. The mechanism of a chemical reaction is a detailed, step-by-step description of how the reaction occurs. As chemists, we can never actually observe a reaction occurring and so we have to infer the actual 6

7 . 4 mechanism based on what intermediates are formed and based on our understanding of chemical principles. Generally most chemists will agree on a mechanism and it is critical in the study of organic chemistry to learn the proper mechanism and to keep track of how the electrons are moving in a given reaction by means of the curved arrow formalism. Step ne: The S N 1 mechanism under consideration here involves three separate steps. Step one is proton transfer from the hydrochloric acid to the oxygen lone pair of the alcohol. We protonate the hydroxyl group to make it into a good leaving group in step acid base This is a simple acid-base reaction. It is a very fast reaction and therefore has a low activation energy (E AT ). The rate of the reaction is determined by the height of energy barrier. In general, proton transfer reactions are the fastest reactions in chemistry. The reaction is exothermic, which means heat is given off and it is favorable. The - bond is weak and easily broken. The hydrogen forms a stronger bond with oxygen. If we make a graph of what happens to the potential energy as the reaction proceeds we see that initially the potential energy increases slightly. This is the activation energy or energy barrier to the reaction. The potential energy then reaches a maximum. We define this maximum as the transition state or TS. This literally is the point of transition between starting materials and the product. Potential Energy E AT Transition State (TS) Exothermic eaction, early transition state eaction Progress We cannot actually isolate the transition state to study its structure. It has a very short lifetime and is an unstable structure that is literally making the transition from starting material to product. It is very important to understanding the full mechanism to have some sense of the structure of the transition state because it is the height of the maximum point of the transition state, the activation energy, that determines the rate and the more stable the transition state the lower it is in energy. 7

8 . 4 ammond s Postulate: this is a theory or postulate that predicts the structure of the transition state based on whether a reaction is exothermic or endothermic. It states that if two species are similar in energy, they are similar in structure. In general: For an exothermic reaction, the transition state is similar in energy to the starting materials so it resembles the starting materials in structure. Bond breaking and bond formation have not proceeded very far. We call this an early transition state and show it using dotted lines for the bonds that are breaking and forming δ + δ+ δ symbol for transition state The oxygen-hydrogen bond is still much longer than the - bond; bond formation has not proceeded very far. For an endothermic reaction, we have a late transition state. Bond breaking and bond formation is nearly complete and the transition state resembles the product. We will see an example of an endothermic reaction and a late transition state in step two of our S N 1 mechanism. Step Two: Step two is the slow, rate-determining step. It is a unimolecular reaction in which the protonated alcohol from step one leaves as the neutral water to form a carbocation This step is endothermic since a bond is being broken but no bond is being formed. It requires energy to break a bond. The carbocation product of this reaction is a high energy intermediate. The potential energy versus reaction progress diagram shows that the transition state is similar in energy to the carbocation product. So there is lots of positive charge on the carbocation intermediate. 8

9 . 4 TS - has lots of carbcation character because bond cleaving is almost complete. Potential Energy E AT 3 3 δ + 3 Bond breaking is almost complete so there is lots of positive charge on the carbon. δ + eaction Progress arbocations are planar, trigonal structures with an sp 2 carbon that has an empty p-orbital perpendicular to the plane of the carbon atom and its three substituents. The carbocation has only six electrons around it so it is a Lewis Acid. Another term for a Lewis acid is electrophile. An electrophile is a species that loves (from the Greek philos) electrons sp 2 carbon empty p orbital The empty p orbital is perpendicular to the plane of the three atoms attached to the sp 2 carbon. Step 3: The chlorine attacks the empty p orbital on the carbocation to form a new - bond. This step is very fast and thermodynamically very favorable since a bond is being formed and no bond is being broken. The chlorine is a nucleophile. Nucleophiles are Lewis bases that have a lone pair and/or a minus charge. They donate electrons to the electrophile. We always show the arrow for the electrons as moving from the nucleophile lone pair to the electrophile fast This step is fast and has a low activation energy. It is exothermic. 9

10 . 4 E AT3 Potential Energy eaction Progress The overall potential energy diagram for all three steps is as follows: E AT E AT2 Potential Energy 3 E AT eaction Progress 3 The second step, the formation of the carbocation has the highest activation energy and therefore it is the slowest step and determines the overall rate of the reaction. arbocation Stability Alkyl groups stabilize a carbocation by donating electrons and helping to spread out the positive charge. 10

11 > 3 > 3 > 3 3 tertiary carbocation secondary carbocation primary carbocation methyl carbocation most stable least stable There are two ways that the alkyl groups release electrons. (1) The Inductive Effect: This is due to the polarization of sigma bonds. The electrons in a - bond are more polarizable than the electrons in a - bond. So replacing s with alkyl groups has the effect of increasing the polarizability of the bonds to the carbocation and allows for better electron donation through the sigma bonds. This has the result of making the carbocation less electron deficient and therefore more stable and lower in energy (2) yperconjugation: This is a resonance effect due to overlap of the six - bonds with the empty p orbital of the carbocation. For this overlap to be effective the - bonds must be on the carbon directly connected to the carbocation. 3 3 The two electrons from the - bond on the alkyl group attached to the carbocation can donate into the empty p orbital of the carbocation. The two - electrons are partially delocalized onto the p orbital. The more alkyl groups that are attached to the carbocation, the more electron donation there is. Another representation of this effect that is sometimes shown is as follows: Now we can understand why tertiary alcohols react faster than secondary alcohols and why secondary alcohols react faster than primary ones. It is due to the fact that the tertiary alcohols, upon protonation and loss of water, form the more stable tertiary carbocations. Tertiary carbocations are lower in energy than secondary and primary and methyl carbocations and form faster. 11

12 . 4 A graphical comparison of the relative energies of the various carbocations shows that the tertiary is much lower in energy and furthermore has a lower activation energy (E AT3 ) so that it forms much faster. potential energy E AT2 E AT1 E AT methyl 3 E AT3 2 1 methyl Generally speaking, the primary and methyl carbocations are too high in energy to be formed and are never observed. So with primary and methyl alcohols there is a different mechanism. S N 2 Mechanism: Nucleophilic Substitution Bimolecular With methyl alcohol and primary alcohol we have a different mechanism called the SN2 mechanism, where the 2 refers to the fact that the rate determining step has two molecules coming together (i.e. bi-molecular). The first step is the same as in the tertiary alcohol case: the alcohol oxygen is protonated so as to make it into a good leaving group fast In the second step, the chlorine anion produced in the first step directly attacks the carbon bearing the protonated hydroxyl group to replace it. The carbon bearing the oxygen has a partial positive charge since the oxygen is more electronegative than carbon. 12

13 slow δ + The negatively charged chlorine is attracted to the partial (+) charge on the carbon bearing the oxygen. As will be discussed in much more detail in hapter 8, the attack by the nucleophilic chlorine occurs from the backside. The transition state is pictured below. δ δ + In the transition state breaking of the - 2 bond is more or less sequenced with formation of the - bond. Think of the chlorine electrons as pushing out the electrons of the - bond. The 2 group leaves with these two electrons. The potential energy versus reaction progress diagram shows two transition states corresponding to the two steps. The second step has the higher activation energy and is therefore the slow step and overall rate-determining step. Potential Energy EAT E AT2 δ δ eaction Progress ther ways to convert alcohols to alky halides (1) Thionyl hloride: We can treat alcohols with boiling thionyl chloride (S 2 ) to convert the alcohol to an alkyl chloride. This method is considerably milder than using concentrated hydrochloric acid and is useful when the molecule contains sensitive functional groups that would react with the strong acid. This is usually used with primary and secondary alcohols since, as we have seen, tertiary alcohols react easily at low temperatures with hydrochloric acid. ften a weak base such as potassium carbonate (K 2 3) or pyridine ( 5 5 N) is used to neutralize the that is produced. 13

14 . 4 + S heat + S 2 + For example: S heat K (2) Phosphorus Tribromide: A mild way to convert alcohols to alkyl bromides involves the treatment of the alcohol with phosphorus tribromide, PBr 3. ne equivalent of phosphorus tribromide will convert three molecules of the alcohol. 3 + PBr 3 3 Br + 3 P 3 phosphorus acid alogenation of Alkanes A hydrogen atom on a simple alkane can be replaced with a halogen atom. + X 2 X + X X = Br, The order of reactivity for the halogens is: F 2 > 2 > Br 2 >I 2. Fluorine is too reactive. It is a strong oxidizing reagent and the reaction is strongly exothermic and difficult to control. Iodine is unreactive. The reaction is endothermic and not useful. Therefore, the only useful halogenations occur with bromine and chlorine. This is not a nucleophilic substitution reaction like the reactions we just studied with alcohols and halogen acids. The - bond is not polarized and : - is not a leaving group. This is an example of a free radical reaction. hlorination of methane: The reaction can give up to four products due to over reaction. n a large industrial scale the products can be separated by distillation. heat + uv light

15 . 4 This is an example of a free radical reaction. A radical is a species that has a free unpaired electron. There are several examples of stable radicals, the most common of which is molecular oxygen. Nitrogen dioxide, N 2, is also a stable radical and the important cellcell signaling molecule, nitrogen monoxide, N, is a radical as well. N N molecular oxygen nitrogen dioxide nitrogen monoxide arbon free radicals are usually much less stable. They are missing one electron and are generally very reactive intermediates that exist only for a short time. arbon radical stability parallels that of carbocation stability. Since carbon radicals are missing one electron and so are electron deficient species, they are stabilized by electron donating substituents such as alkyl groups. Therefore, tertiary radicals are the most stable and methyl radicals are the least stable. Note also that the carbon radical has no charge. tertiary radical most stable > > > methyl radical least stable The methyl radical is planar, trigonal with the hybridization very nearly sp 2. The t-butyl radical is slightly pyramidal but still flattened and closer to sp 2 than sp 3 hybridization. Alkyl groups stabilize the radical just like they stabilize carbocations. Evidence for this stabilization can be seen in terms of the Bond Dissociation Enthalpy (BDE). For homolytic cleavage each atom gets one of the two electrons in the bond between the two atoms. The heat change, Δ, in this reaction is the bond dissociation energy. Note that by convention in radical reactions, we use a single-headed arrow and that the product(s) of hemolytic cleavage are uncharged species. X Y X + Y Δ = BDE In homolytic cleavage, each atom gets one electron. Note that no charge develops and that the arrows are single headed arrows, showing that only one electron is moving. ecall that in heterolytic cleavage one atom gets both electrons. 15

16 . 4 heterolytic cleavage X Y X + Y ne atom gets both electrons and charges develop in the product(s). It is generally the case that hemolytic cleavage requires less energy than heterolytic cleavage. For carbon atoms, the bond dissociation enthalpies are known and can be found in tables. ompare the bond dissociation energy of a primary versus a secondary carbon Δ = +410 KJ/mol Δ = +397 KJ/mol We see the cleavage of a secondary carbon is 13 KJ/mol more favorable, implying that the secondary carbon radical is more stable than the primary by 13 KJ/mol. We can use the bond dissociation enthalpies to compare the stabilities of a primary versus a tertiary carbon radical Δ = +410 KJ/mol Δ = +380 KJ/mol 3 We see that the tertiary radical is more stable than the primary radical by 30 KJ/mol and more stable than the secondary radical by 17 KJ/mol. alogenation of alkanes occurs by means of a radical chain reaction. This is shown below for the chlorination of methane. 16

17 . 4 Inititation: heat or UV light Propagation: (1) (2) + + The first step is the initiation step. This is the hemolytic cleavage of the chlorine molecule into two chlorine radicals. This bond cleavage occurs due to heating or to ultra-violet (UV) light. Generally, this step occurs only once. The next two steps occur over and over again. These are the propagation steps. In each propagation step a new radical is formed which then goes on to react to generate a new radical, which continues the chain reaction. We can also get dichloromethane, tricholormethane and even tetrachloromethane. This occurs when the initial chloromethane continues to react and is more likely to occur toward the end of the reaction as the concentration of starting methane begins to decrease and the concentration of chloromethane increases. step 1 + step The chain reaction can be broken by termination steps. Termination steps are those in which two radicals come together to react with each other. This does not generate a new radical and so the chain is broken. Some termination steps for the chlorination of methane are: 17

18 We can use bond dissociation enthalpies to calculate the overall in enthalpy, Δ, for the reaction. In general, the overall change in enthalpy, Δ, is equal to the sum of the bond dissociation enthalpies of the bonds broken plus the sum of the bond dissociation enthalpies of the bonds formed. We will use the convention that a bond broken is a positive number, since energy is required to break a bond, and a bond formed is a negative number, since heat is given off when a bond is formed. Δ = ΣBDE bonds broken + ΣBDE bonds broken Where bonds broken = (+) bonds formed = (-) Typically, we ignore the initiation step since it happens only once, where as the propagation steps can happen thousands of times. step break methyl - bond +435 KJ/mol form - bond -431 KJ/mol Δ = = +4 KJ/mol step 2 + break - bond +242 KJ/mol + form - bond -349 KJ/mol Δ = -107 KJ/mol Δ rxn = = -103 KJ/mol To find the overall enthalpy of reaction, add step 1 and step 2 to get (-107 KJ/mol + 4 KJ/mol) -103 KJ/mol. The reaction is exothermic and therefore favorable. If the same calculation is done for fluorination, we find that Δ is -426 KJ/mol. This is a very large amount of energy that is released for each step. As more molecules react, more energy is released, the reaction mixture heats up and reacts even faster. It quickly becomes uncontrollable. Free radical fluorinations are dangerously explosive. 18

19 . 4 If we do the same calculation for bromination, we find that the Δ = -30 KJ/mol, considerably less than that for chlorination and much less than that for fluorination. As we will see, free radical bromination reactions are much more selective than free radical chlorinations. For iodination, Δ = +54 KJ/mol. In the case of iodination, the reaction is unfavorable and free radical iodination reactions are not feasible. alogenation of igher Alkanes If the molecule is symmetrical, we get only one monochloro- product. 2, UV light But if the molecule is not symmetrical, a mixture of products can form. We see that the major product is the one that results from the formation of the secondary radical , UV light % 72% Since there are six primary hydrogens and only four secondary hydrogens, we would expect a mixture of 60% (6 of 10) 1-chlorobutane and 40% (4 of 10) of the 2-chlorobutane. But we see 72% of 2-chlorobutane. This tells us that the secondary radical is formed faster than the primary radical by a factor of 72/28 x 6/4 = 3.9/1. Bromination is even more selective. Tertiary hydrogens are abstracted selectively. For example, compare the product rations of free radical chlorination and free radical bromination of 2-methylpropane UV % 63% ere we see that there are nine primary hydrogens and only one tertiary hydrogen, so if primary hydrogens and tertiary hydrogens reacted at the same rate we would expect a product ratio of 10% 2-chloro-2-methyl propane and 90% 1-chloro-2-methylpropane. So clearly, the tertiary position reacts faster by a ratio of 37/63 x 9/1 =

20 . 4 For bromination the results are very striking: bromination is much more selective for removal of the tertiary hydrogen than chlorination. Doing the same calculation that we did before, we see that in the bromination reaction, the tertiary hydrogen reacts 99/1 x 9/1 = 891 times faster Br Br Br UV 3 3 >99% <1% The reason for this is that the first step in the chlorination reaction is quite exothermic (early transition state) for both primary and tertiary products while the first step in the bromination reaction is endothermic (late transition state) for both products form break form break Δ = -9 KJ/mol Δ = -31 KJ/mol Br Br 3 3 form Br = -366 break Br Br 3 3 form Br = -366 break Δ = +56 KJ/mol Δ = +34 KJ/mol ammond s postulate tells us that we have a late transition state in an endothermic reaction so that there is a large difference in energy between the transition states leading to the primary radical and the tertiary radical. 20

21 Potential Energy Br + ( 3 ) Endothermic reaction with a late transition state so that there is a lot of radical character in the transitions state and the difference in stability of a primary versus tertiary radical is important. eaction Progress The chlorination reaction is exothermic. Therefore there is an early transition state with little radical character and the difference in energy between the transition states is small. Potential Energy + ( 3 ) In the exothermic reation there is a relatively small difference in energy in the transition states since bond breaking is not very far advanced and there is little radical character. 3 eaction Progress

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