Alcohols and Alkyl Halides - Organic Chemistry - Lecture Notes, Study notes of Organic Chemistry

Download Alcohols and Alkyl Halides - Organic Chemistry - Lecture Notes and more Study notes Organic Chemistry in PDF only on Docsity! 1 Alcohols and Alkyl Halides 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 C and H 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, -OH. Alkyl halides – contain a halide. We use the symbol X to stand for any halogen (I, Br, Cl, F). “R” 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 ROH and a generalized alkyl halide is written as RX. Other functional groups that will be studied throughout the rest of the course include: Amines – contain a nitrogen, N. The amine may be primary (RNH2), secondary (R2NH), or tertiary (R3N) where the R groups can be the same or different. CH3CH2 N H H ethylamine (primary amine) CH3CH2 N H CH3 ethyl methylamine (secondary amine) CH3CH2 N CH2CH3 CH2CH3 triethylamine tertiary amine) Ethers – contain a C-O-C linkage, R-O-R’, where R and R’ can be the same or different. CH3CH2 O CH2CH3 diethyl ether Epoxides – these are special 3-membered ring ethers. O RR R R docsity.com 2 Nitriles – these contain the CN functional group, RCN. This functional group is also called a cyano-group. C NCH3CH2CH2 Nitroalkanes – these contain the nitro group, -NO2. CH3CH2 N O O Thiols – these contain an –SH group RSH. CH3CH2CH2 SH Sulfide – these are thio ethers, R-S-R’. CH3CH2CH2 S CH2CH2CH3 Carbonyl Derivatives – all contain the carbonyl group, C=O Aldehydes – have the carbonyl group at the end of the chain, -CHO. CH3CH2CH2 C O H Ketones – have the carbonyl group attached to two other carbons, one on each side, RCOR’. CH3CH2CH2 C O CH3 Carboxylic Acid – contain the functional group –CO2H. CH3CH2CH2 C O OH Carboxylic Acid Esters (or Esters) – contain the functional group –CO2R. CH3CH2CH2 C O O CH2CH3 Amides – contain the functional group –CONR’2 CH3CH2CH2 C O NH2 CH3CH2CH2 C O NH CH3 CH3CH2CH2 C O N CH2CH3 CH3 Carboxylic Acid Anhydrides – contain the functional group RCO2COR’ docsity.com 5 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. Carbon-hydrogen bonds are not polarized enough to engage in hydrogen bonding. H O CH2CH3 O HCH3CH2 O HCH3CH2 hydrogen bond O H O O H N N H O N H N For alkyl halides, the boiling point increases with the increasing size of the halogen. CH3 F boiling point CH3 Cl CH3 Br CH3 I -78°C -24 3 42 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 °C CH3Cl -24 CH2Cl2 CHCl3 CCl4 40 61 77 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 CF3CF3 still has a higher boiling point than ethane, CH3CH3, since it is more polar. boling point, °C CH3CH2F -32 CH3CHF2 CH3CF3 CF3CF3 -25 -47 -78 CH3CH3 -89 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. docsity.com 6 CH3CH2CH2CH2CH2CH2CH2CH2 OH non-polar polar Only 0.5 ml dissolves in one liter of water. Preparation of Alkyl Halides from Alcohols and HX General reaction: R OH + H X R X + H O H The reactivity order for the hydrogen halides parallels their acidity: H I > H Br > H Cl >> H F strongest acid weakest acid The reactivity order for the alcohols is: R C R R OH tertiary > R C R H OH > R C H H OH secondary primary > H C H H OH methyl Ex: CH3 C CH3 CH3 OH + H Cl room temp. (rt) CH3 C CH3 CH3 Cl + H2O 78-88%t-butanol t-butyl chloride Sometimes we write the reagents over the arrow in order to save space. CH3CH2CH2CH2 OH NaBr, H2SO4 heat CH3CH2CH2CH2 Br + H2O + Na+ -OSO3H Mechanism of the Reaction of t-Butanol and HCl 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 SN1 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 docsity.com 7 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 One: The SN1 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 2. CH3 C CH3 CH3 OH + H Cl acid base CH3 C CH3 CH3 O H H + Cl This is a simple acid-base reaction. It is a very fast reaction and therefore has a low activation energy (EACT). 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 H-Cl 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. EACT Reaction Progress Po te nt ia l E ne rg y CH3 C CH3 CH3 OH H Cl Transition State (TS) CH3 C CH3 CH3 O H H Exothermic Reaction, early transition state 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. docsity.com 10 Reaction Progress Po te nt ia l E ne rg y CH3 C CH3 CH3 EACT3 CH3 C CH3 CH3 Cl The overall potential energy diagram for all three steps is as follows: Reaction Progress Po te nt ia l E ne rg y EACT1 EACT2 EACT3 CH3 C CH3 CH3 Cl CH3 C CH3 CH3 CH3 C CH3 CH3 O H H 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. Carbocation Stability Alkyl groups stabilize a carbocation by donating electrons and helping to spread out the positive charge. docsity.com 11 CH3 C CH3 CH3 > most stable CH3 C CH3 H > CH3 C H H > H C H H tertiary carbocation secondary carbocation primary carbocation methyl carbocation 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 C-C bond are more polarizable than the electrons in a C-H bond. So replacing H’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. CH3 C CH3 CH3 (2) Hyperconjugation: This is a resonance effect due to overlap of the six C-H bonds with the empty p orbital of the carbocation. For this overlap to be effective the C-H bonds must be on the carbon directly connected to the carbocation. CH3 CH3H H The two electrons from the C-H bond on the alkyl group attached to the carbocation can donate into the empty p orbital of the carbocation. The two C-H 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: C CH3 CH3 H H H C CH3 CH3 H H H 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. docsity.com 12 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 (EACT3°) so that it forms much faster. po te nt ia l e ne rg y R C R R R C R R O HH 3° R C R R R C R H O HH 2° 1° R C H H O HH methyl H C H H O HH R C H H H C H H EACT3° EACT2° EACT1° EACT 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. SN2 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. CH3CH2CH2 O H + H Cl CH3CH2CH2 O H H + fast Cl 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. docsity.com 15 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, NO2, is also a stable radical and the important cell- cell signaling molecule, nitrogen monoxide, NO, is a radical as well. O O O N O molecular oxygen nitrogen dioxide N O nitrogen monoxide Carbon 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. Carbon 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. R C R R > R C R H > R C H H > H C H H tertiary radical most stable methyl radical least stable The methyl radical is planar, trigonal with the hybridization very nearly sp2. The t-butyl radical is slightly pyramidal but still flattened and closer to sp2 than sp3 hybridization. C H H H 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, ΔH, 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 ΔH = 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. Recall that in heterolytic cleavage one atom gets both electrons. docsity.com 16 X Y X + Yheterolytic cleavage One 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. Compare the bond dissociation energy of a primary versus a secondary carbon. CH3CH2CH2 H CH3CH2CH2 + H ΔH = +410 KJ/mol CH3 CH CH3 H CH3 H C CH3 + H ΔH = +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. CH3 C CH3 CH3 CH3 C CH3 + H ΔH = +380 KJ/mol CH3 CH CH2 CH3 H CH3 CH CH2 CH3 + H ΔH = +410 KJ/mol H CH3 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. Halogenation of alkanes occurs by means of a radical chain reaction. This is shown below for the chlorination of methane. docsity.com 17 Inititation: Cl Cl heat or UV light 2 Cl Propagation: (1) Cl + H C H H H Cl H + C H H H(2) + Cl Cl H C H H Cl + Cl 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. H C H Cl H Cl+ H C H Cl H Clstep 1 step 2 H C H Cl Cl Cl H C H Cl Cl + Cl+ 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: docsity.com 20 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. CH3 C H CH3 CH3 Br2 CH3 C Br CH3 CH3 CH3 C H CH3 CH2 Br <1%>99% UV 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. CH3 C H CH3 CH3 + Cl CH3 C H CH3 CH2 + H Cl break 1° C-H +422 form H-Cl -431 ΔH = -9 KJ/mol CH3 C H CH3 CH3 + Cl CH3 C CH3 CH3 + H Cl break 3° C-H +400 form H-Cl -431 ΔH = -31 KJ/mol CH3 C H CH3 CH3 + Br CH3 C H CH3 CH2 + H Br break 1° C-H +422 form HBr = -366 ΔH = +56 KJ/mol CH3 C H CH3 CH3 + Br CH3 C CH3 CH3 + H Br break 3° C-H +400 ΔH = +34 KJ/mol form HBr = -366 Hammond’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. docsity.com 21 Br + H(CH3)3 CH3 C H CH3 CH2 CH3 C CH3 CH3 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. Reaction Progress Po te nt ia l E ne rg y 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. Cl + H(CH3)3 CH3 C H CH3 CH2 CH3 C CH3 CH3 Reaction Progress Po te nt ia l E ne rg y 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. docsity.com