Organic_Acids_and_bases

Organic_Acids_and_bases - Organic Acids and Bases This...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Organic Acids and Bases This document has been written to provide you with an overview of the fundamental concepts of organic acid‐base chemistry. Many organic compounds have not been included such as the basicity of pyrrole or the acidity of α hydrogens of carbonyl compounds. These topics are generally covered in the second and third organic chemistry courses. 1.0 2.0 2.1 2.2 2.3 2.4 2.5 2.6 3.0 3.1 3.2 4.0 Review of acids and bases Trends in acid‐base strengths Shifting of the acid‐base equilibrium Electronegativity Size Resonance Stabilization The proximity of electron‐withdrawing and electron‐donating groups: Inductive effects. Hybridization Base strength of Amines and Amides Amines Amides Some exceptions to the general trends of acidity and basicity pKa Tables 1 3 3 4 5 7 9 13 14 14 17 17 19 Dr. Steven Forsey 1 Organic Acids and Bases 1.0 Review of acids and bases Acid‐base reactions are the first step in understanding why reactions between organic molecules occur. This topic is studied in depth because acid‐base reactions are so fundamental, that the theory will be encountered in almost every topic in organic chemistry and when looking at the broadest sense of Lewis acids and bases, most organic reactions are acid‐base reactions. In this section you will learn why an acid or base is stronger or weaker than another. You will not be asked to memorize the dissociation constants but understand the trends that we see in acids and bases. This is a fundamental concept that will help you understand Organic Chemistry. This section can be difficult for some students because there is no set order to memorize, you must understand the concepts, look at the molecules of concern and understand the differences in the molecule and then make your decision. + Many organic reactions involve proton (H ) transfer or electron pair transfer. Thus, it will be helpful to review the Bronsted‐Lowry and Lewis theories of acids and bases. In the Bronsted‐Lowry Theory an acid is a proton donor and a base is a proton acceptor. In the Lewis Theory an acid is an electron pair acceptor and a base is an electron pair donor. In the Bronsted‐Lowry theory, an acid must contain an ionizable H atom and a base must contain a lone pair of electrons onto which a proton can bind. For this reason, an acid is often represented in this theory + by the general formula HA and a base is represented by :B. An acid HA will produce H3O ions in aqueous − solution and a base :B will produce OH ions because of the following ionization reactions. [H O + ] [A − ] H3O+(aq) + A−(aq) HA(aq) + H2O(l) Ka = 3 [HA] B(aq) + H2O(l) HB+(aq) + OH−(aq) Kb ≡ [HB+ ] [OH − ] [B] The equilibrium constants for the ionization reactions for HA and B: are called “ionization constants” and are denoted as Ka or Kb. The strength of an acid (or a base) is determined ultimately by the value of Ka (or Kb). A strong acid (or base) has a large ionization constant and a weak acid (or base) has a small ionization constant. Because the values of Ka and Kb can span an enormous range of values (1030 to 10−30), we often use pKa and pKb values instead. The following relationships are helpful for converting K values into pK values and vice versa. pK = − log10 K a or K = 10− pK (K= Ka or Kb) The expression K a = 10− pK shows that the larger the pKa value, the smaller the Ka value and the weaker the acid. Similarly, the larger the pKb value, the smaller the Kb value and the weaker the base. Strong acids have negative pKa values and show a strong tendency to lose a proton; strong bases have negative pKb values and show a strong tendency to gain a proton. We can classify acids or bases as being very strong, moderately strong, weak or very weak according to the extent to which the ionization reactions occur. At the end of this document there is a table of pKa values for some organic and inorganic acids. Notice that some protonated organic compounds are very strong acids, such as protonated, aldehydes and alcohols, and show a strong tendency to lose a proton while other protonated forms, such as protonated amines, are weak acids and show a small tendency to lose a proton. Similarly, some deprotanated organic compounds, such as deprotonated alkanes and alcohols, are very strong bases and are easily protonated while other deprotonated forms, such as deprotonated carboxylic acids, are weak bases. It will be Dr. Steven Forsey 2 extremely helpful for you to know these approximate pKa values. In general, protonated carbonyl compounds and protonated alcohols are strong acids with pKa values less than zero; carboxylic acids have pKa values around 5; protonated amines have pKa values around 10; alcohols have pKa values around 15; and alkanes, alkenes and alkynes greater than 41. 2.0 Trends in acid‐base strengths 2.1 Shifting of the acid‐base equilibrium Why do acids have different acidities? As mentioned previously a strong acid has a weak conjugate base. Thus an acid is strong if its conjugate base is stable or not reactive. If the conjugate base is strong and not stable it will react with a proton and shift the equilibrium to the left. Most organic reactions are not performed in water and we are observing the acid‐base reactions between different organic compounds. However, the pKa values can be use to determine the reactivity between organic acids and bases. For example when sodium ethoxide is added to a solution containing methylamine an acid‐base reaction may occur CH3NH ˉ Na+ + CH3CH2OH CH3NH2 + CH3CH2Oˉ Na+ acid base conjugate base conjugate acid Notice after the chemical reaction the base becomes the conjugate acid and acid becomes the conjugate base. The conjugate base can now attack the conjugate acid to create the original acid‐base pair. CH3NH2 + CH3CH2Oˉ Na+ CH3NH ˉ Na+ + CH3CH2OH conjugate base conjugate acid acid base The original acid and base can then react again to regenerate the conjugate acid‐base pair. This is the physical process that is occurring in solution described by the equilibrium equation and when you look at this acid‐base reaction closely you will notice that there is a tug of war between two bases for the acidic proton. + CH3NH ˉ Na+ + CH3CH2OH CH3NH2 + CH3CH2Oˉ Na acid base base acid Who wins the fight for the proton? The strongest or most reactive base does. Which side does the equilibrium shift to? It will shift to generate the weaker or less reactive acid‐base pair. In this acid‐base reaction the equilibrium is shifted to the left because the sodium methylamide ion is a much stronger base than the sodium ethoxide ion. Remember the smaller the pKa or pKb value the stronger the acid or base. CH3NH ˉ Na+ + CH3CH2OH CH3NH2 + CH3CH2Oˉ Na+ pKb = ‐26 pKa = 16 pKa =8 pKb = ‐2 Thus when you are comparing the strength of different acids using pKa values you are looking at the ability of an acid to donate a proton to another base. If the conjugate base of the acid you are concerned with is stronger than the other bases in the system the proton will not be removed from the acid, as in the case with methylamine and sodium ethoxide. To remove a proton from methylamine you would have to use a stronger base such as a compound with a negatively changed carbon, for example CH3CH2ˉ. Acid‐base reactions are equilibriums and you must look at both sides of the equilibrium to decide which way the equilibrium will shift. The equilibrium will always shift to the weaker acid‐base pair or the less reactive species. One way of determining which way the equilibrium will shift is to look at the base strengths on both sides of the equilibrium equation. The stronger base is more reactive and will win the battle for the proton to produce the more stable less reactive and weaker base. But, what makes one base less basic than another? To determine this you must look at the stability or reactivity of the base. Dr. Steven Forsey 3 The stability of the base is dependent on the environment of the atom or atoms that donate the lone pair of electrons to the acid. You must consider what atom the electrons are on. Is the atom electronegative, is the atom small or large, what is the hybridization the atom and can the lone pair of electrons be delocalized throughout the molecule. All of these factors must be considered. 2.2 Electronegativity In the example given above we know from the pKa tables that the sodium methylamide ion is a much stronger base than the sodium ethoxide ion, or that ethanol is a stronger acid than methyl amine, but the big question is why? Consider the following molecules (CH3)3C‐H, (CH3)2N‐H, CH3O‐H and HF. How could you determine their relative acidity to each other without looking at the pKa tables? First write out the equilibriums and consider the stability of the conjugate bases. Which of the conjugates would be the weakest base (most stable) and shift the equilibrium towards the products to a greater extent? (CH3)3Cˉ + H+ (CH3)3C‐H (CH3)2N ˉ + H+ (CH3)2N‐H CH3O ˉ + H+ CH3O‐H H‐F F ˉ + H+ Notice that the charge is on different atoms and the most important difference B C N O F 2.0 2.5 3.0 3.5 4.0 between the atoms, when considering the stability of an anion is their electronegativity (Table 2.1). Remember electronegativity is the measure of the Al Si P S Cl 1.5 1.8 2.1 2.5 3.0 ability of an atom to attract electrons or charge density relative to another atom. Simply put electron density will shift to the atoms that have more Ga Ge As Se Br 1.8 2.0 2.2 2.6 2.8 protons in their nucleus. As you go across the periodic table from left to right the electronegativity of the atoms increase, (number of protons increases) thus Te I you would expect the stability of the conjugate bases to increase, (CH3)3Cˉ < 2.1 2.5 (CH3)2Nˉ < CH3Oˉ < Fˉ and the acidity to increase, (CH3)3CH < (CH3)2NH < CH3OH < HF. This is indeed what we find, as you go across the same row in the periodic 2.1 Electronegativity table the acidity of the compound increases. of selected atom Worked example Which of the following acid‐base reactions will not occur? A) B) C) Analyze: To determine if the reaction will proceed from left to right, first put lone pairs and charges on the atoms and identify the acids and bases on both side of the reaction. Arrows have been drawn to show the flow of electrons for the reactants side only. The red labeled electrons on the base show bond formation between the base and hydrogen, the base is donating its electrons to the acid. The blue electrons, illustrates the bond breaking that must occur for the base to form a new sigma bond with the hydrogen. Dr. Steven Forsey 4 The electrons from this hydrogen bond are transferred to what becomes the conjugate base. Then determine relative base strength. In these examples the least electronegative negatively charged atom will be the stronger base. The stronger base will react with the stronger acid to form the weaker acid‐base pair. H K O O A) HF K F + weaker weaker acid base base acid + + Note: Na and K are counter ion in these reaction and do not participate in the reaction. The alkoxide ion is a stronger base than the fluoride ion thus the above reaction will proceed from left to right. H H H Na N N + Na CH3 + H CH3 B) weaker base acid weaker acid base The methyl carbanion (methanide ion) is a stronger base than the cyclopentylamide ion thus the reaction will proceed from left to right H O Na + H O Na NH2 + N H C) stronger base stronger H acid base acid The 1‐propanolate ion (alkoxide ion) is a weaker base than the amide ion. The reaction will not occur. 2.3 Size Another important factor is the size of the atom that the charge is on. A larger atom can disperse the negative charge over a larger volume and thus increases the stability of the anion. Consider the following acids, HF, HCl, HBr, and HI and order the acids from strongest to weakest. Again consider the equilibrium and determine stability of the conjugate bases. HF H Cl H Br HI H H H H + + F Cl + Br + I Dr. Steven Forsey 5 As you go down the periodic table the size of the atoms increases. The iodide ion is much larger than the fluoride ion and will be much more stable with a negative charge even though fluorine is more electronegative than iodine. The order of increasing acid strength should be HI > HBr > HCl . HF and this is what we see. HF is a weak acid (pKa = 3.2) and HI is a very strong acid (pKa = ‐10) Thus, when comparing atoms in the same column in the periodic table size has a greater stabilizing effect than electronegativity and when comparing atoms in the same row electronegativity is more important. It should be noted here that electronegativity is a much stronger effect than size. As seen in Table 2.2 the pKa values decrease much more dramatically from methane to HF than HF to HI. ‐ increasing electronegativity ‐ increasing ability to stabilize a charge on conjugatebase NH3 H2O HF CH4 50 36 15.7 3.2 Increasing size H2 S HCl and ability to 7 ‐7 Increasing stabilize a charge H2Se HBr electronegativity on conjugate base 3.8 ‐9 H2Te HI 2.6 ‐10 Table 2.2 pKa values selected acids Worked example Order the following compounds from must acidic to least acidic. 2) CH3SH 3) CH3CH3 1) CH3OH Analyze: Draw the equilibriums for each acid and then determine the stability of the conjugate bases. In this example there are two factors to consider, size and electronegativity. CH3O H H + CH3O + CH3S H CH3S H H H + CH3C CH3C H H H H First compare the negatively charged carbon and oxygen since they are in the same row on the periodic table and are smaller than sulfur. Carbon is far less electronegative than oxygen thus a negatively charged carbon will be a much stronger base than oxygen with a negative charge. Ethane will be a much weaker acid than methanol. Now compare the negatively charged oxygen with the negatively charged sulfur. The sulfur atom is larger than the oxygen atom and will be more stable with a negative charge. Methanethiol will be a stronger acid because its conjugate base is more stable. The order of acid strength from strongest to weakest is CH3SH > CH3OH > CH3CH2 Note: In the above equilibriums the acids are not shown reacting with a base, the acids are shown by themselves. This is because we are not concerned with the acid reacting with a specific base, but with the ability of an acid to donate a proton to determine relative acid strength between the acids. The compound with the weakest conjugate base will be the most acidic. This is very important when designing chemical reactions. Dr. Steven Forsey 6 2.4 Resonance Stabilization In first year chemistry courses the concept of resonance hybrids was introduced. Resonance hybrids occur because more than one plausible Lewis structure can be written. A resonance hybrid is observed with the conjugate base of acetic acid, the acetate ion. Notice, that the arrow between the resonance hybrids is a double headed arrow and not an equilibrium arrow. This arrow indicates that the actual structure is a hybrid of the Lewis structures and the charge is spread over both oxygens. Experimentally this is what is observed. Carboxylates are symmetrical with carbon bond lengths of 1.26 A which is in between the typical carbon oxygen double (1.20 A ) and single (1.34 A ) bonds in the corresponding carboxylic acids. The negative charge is spread equally over both oxygens or we can say that the charge is delocalized. This resonance delocalizing of the negative charge plays a critical role in base stability. When the charge is spread over two or more atoms the base becomes more stable or less basic. Consider three different acids methanol, phenol and acetic acid, Again look at the stability of the conjugate bases to explain why one compound is more acidic than the other. The oxygen of the methoxide ion bears the full negative change alone whereas the phenoxide ion can delocalize or spread the negative charge over three carbons. The carbons are not as electronegative as oxygen and are not as stable with a negative charge as oxygen, but as you can see from the measured pKa values the equilibrium is shifted more to the right for phenol verifying that the phenoxide ion is more stable or a weaker base than the methoxide ion. Dr. Steven Forsey 7 o o o A common misconception held by students is that if a compound has more resonance structures than another it will always be a weaker base. This is an incorrect statement. For example, acetic acid is more acidic than phenol even though there are only two resonance structures. The reason for this is that oxygen is more electronegative than carbon and resonance onto two electronegative oxygens is more stabilizing than resonance onto three carbons. Thus the acetate ion is a weaker base than the phenoxide ion and the equilibrium for acetic acid is shifter farther to the right making acetic acid a stronger acid than phenol. A fun way of understanding this concept is the hot potato analogy. If you hold onto a hot potato by yourself you will get burnt because the heat coming form the potato is localized in your hand. If you can share the hot potato with other people beside you, your hands will not have time to reach the full temperature of the hot potato. You are distributing the heat coming from the potato to other people and you will not get burnt. This is like the methoxide ion and the phenoxide ion. The methoxide ion does not have other atoms to share the charge with. The charge is localized on a single oxygen which makes it unstable or reactive, whereas the phenoxide ion has three carbon atoms to share the charge with and thus is more stable or less basic than the methoxide ion. Can we now take the analogy further with the acetate ion? The acetate ion is more stable or less basic than the phenoxide ion even though there are only two resonance structures. In this case we could say that the oxygens are wearing oven mitts compared to the carbon atoms. The oxygen atoms are more electronegative than carbon and can better stabilize the charge. It must be noted the acetic acid and phenol are weak acids and methanol is a very weak acid but when we are thinking about acid‐base reactions between organic molecules we need to know their relative strengths to determine if a reaction will proceed. Working example Which way is the equilibrium shifted in the following acid‐base reaction? Analyze: To determine which way the equilibrium is shifted first put lone pairs and charges onto the atoms and identify the acids and bases on both side of the reaction. Arrows have been drawn to show the flow of electrons for the reactants. The base donates a pair of electrons and the acid accepts the electrons. Then determine which base is the strongest or the weakest. Na H H H N N H + CH3CH2N + CH3CH2NH Na H acid base base acid When comparing the conjugate bases of the acids we do not have to consider electronegativity or size because the negative charge is on the same atom, nitrogen. However, the conjugate base of aniline is resonance stabilized, i.e. the negative change is delocalized through the aromatic system making it much more stable than the ethylamide ion. Dr. Steven Forsey 8 The equilibrium will shift to the right. 2.5 The proximity of electron‐withdrawing and electron‐donating groups: Inductive effects. Atoms bonded to each other with different electronegativities create a dipole through the bond because of the unequal sharing of electron density. The more electronegative atom becomes electron rich and the less electronegative atom becomes electron deficient. The electron rich anddeficient areas are denoted with δ‐ and δ+ symbols as shown in the figure below. The dipole that is created by the electronegative atom is called induction. Given below is a series of butanoic acids. pKa 4.83 4.52 4.05 2.86 Why is this acidity order observed? Again consider the conjugate bases and determine why the conjugate base of 2‐chlorobutanoic acid is the weakest or most stable base. O Cl O O O + Cl O + O O O + Cl Upon examination of the conjugate bases you will notice that as the electronegative chlorine and adjoining electron deficient carbon moves closer to the negatively charged oxygens the stability of the conjugate base increases. The electron deficient carbon helps stabilize the negatively charged region by pulling electron density inductively through the sigma bonds towards itself. As seen from the pKa values the magnitude of the inductive effect decreases rapidly as the electronegative atom moves further from the carboxylic acid group thus you will only observe major differences between acids when the electronegative atom is adjacent to the functional group. Dr. Steven Forsey 9 Worked example Order the following acids from strongest to weakest. O O Cl O HBr HCl Br Cl OH OH OH Analyze: First consider the conjugate bases of the acids O O δ- Cl O δδ Br Cl Cl Br δ+ O O O δ+ δ+ In this example there is a mixture of organic acids and inorganic acids. Do not panic think about what you know. What is a stronger acid, acetic acid (CH3COOH) or hydrochloric acid? Hydrochloric acid is. Both HCl and HBr are very strong acids and will be stronger than the carboxylic acids. The adjacent halogens next to the carboxylic acid groups will make these compound more acidic, but not as strong as HBr or HCl. Now consider which halogen is more stable with the negative charge. Since the bromide ion is larger and has a greater volume to stabilize the charge the bromide ion is more stable thus HBr is the strongest acid followed by HCl. Inductive effects drop off rapidly with distance thus you would expect 3‐chlorobutanoic acid to the be weakest acid. Now consider bromoethanoic acid and chloroethanoic acid. Which halogen will stabilize the carboxylate ion the most? Which halogen will create a greater delta positive charge on the adjacent carbon next to the anion? The more electronegative chlorine atom will. Therefore chloroethanoic acid is more acidic than bromoethanoic acid. Notice the difference and why HBr is a stronger acid than HCl and chloroethanoic acid is stronger than bromoethanoic acid. The difference is what atoms the charge is on. With HBr and HCl the negative charge is on the halogens thus size is the most important factor. With the carboxylic acids the charge is on the oxygens thus you must look at adjacent atoms to determine what inductive effects have on the stability of the anion. The pKa values are: HBr = ‐9, HCl = ‐7, ClCH2CO2H = 2.8, BrCH2CO2H = 2.9, CH3CH2ClCH2CO2H = 4.05 Worked example: nitroethanoic acid (pKa 1.68) is more acidic than fluoroethanoic acid (pKa = 2.59) why? Analyze: First consider the equilibriums and the conjugate bases of the acids O O O O2N N pKa = 1.68 +H OH O O O O δ F pKa = 2.59 F +H OH O δ+ The only difference between the conjugate bases is the inductive effects of the fluorine and the nitro group. The nitro group has a full positive charge on the nitrogen next to the anion and is therefore more stabilizing than the delta positive charge created by the fluorine. As seen from the worked examples electronegative atoms pull electron density towards their nucleus from adjacent atoms through σ bonds. These atoms or groups such as the nitro group given in the worked example are called electron‐withdrawing substituents and help stabilize the adjacent anion. There are also groups that are electron‐donating such as alkyl groups. Alkyl groups donate electron density Dr. Steven Forsey 10 inductively through σ bonds. This effect is called hyperconjugation and has the opposite effect of electron‐withdrawing groups. An important point to reemphasize is that inductive effects occurs through the sigma bond frame work of a molecule and drops off rapidly, where as resonance takes place through a π system and therefore has a longer range effect. The concept of electron donating and withdrawing groups and resonance effects is illustrated elegantly when we compare the pKa values of water (pka = 15.7), methanoic acid (pka = 3.8), chloroethanoic acid (pka = 2.8) and ethanoic acid (pka = 4.8). Comparing the acidities of water and methanoic acid we see a dramatic increase in the acidity of methanoic acid. This is because the conjugate base of methanoic acid is resonance stabilized by two oxygens equally sharing the negative charge where as the oxygen of the hydroxide ion bares the full negative charge and is considerably less stable. Methanoic acid is the simplest carboxylic acid and only has one hydrogen attached to the carboxylic acid functional group. The hydrogen atom is neither electron‐donating nor withdrawing. Chloroethanoic acid has one chlorine atom attached to the carbon adjacent to the carboxylic acid group. The chlorine atom is electron‐withdrawing and inductively induces a delta positive charge on the adjacent carbon that helps stabilize the carboxylate anion. Thus chloroethanoic acid is more acidic than methanoic acid, however the difference is not as significant as between methanoic acid and water. Contrastingly ethanoic acid has a methyl group attached to the carboxylic acid functional group. The alkyl group is electron donating and destabilizes the conjugate base making ethanoic acid less acidic than methanoic acid. Dr. Steven Forsey 11 Worked example Which way is the equilibrium shifted in the following acid‐base reaction? H2O + (CH3)3COˉ HOˉ + (CH3)3COH Analyze: To determine which way the equilibrium is shifted first put lone pairs and charges onto the atoms and identify the acids and bases on both side of the reaction. Arrows have been drawn to show the flow of electrons for the reactants. The base donates a pair of electrons and the acid accepts the electrons. Then determine which base is the strongest or the weakest. CH3 H CH3 CH 3 CH3 OC HO OC HO H + + CH3 base CH3 acid base acid The only difference in the bases is what the negatively charged oxygen is bonded to. The oxygen in the hydroxide ion is bonded to a hydrogen and the oxygen in the tert‐butoxide ion is bond to an alkyl group. Alkyl groups inductively donate electron density and destabilize the anion making it a stronger base than the hydroxide ion. The equilibrium is therefore shifted to the left to form the weaker acid‐base pair. The pKa values for water and tert‐butylalcohol is 15.7 and 18 respectively. Worked example Which cation is more stable? Analyze: Alkyl groups are electron‐donating and halogens are electron with‐drawing Structure 1 is more stable because the alkyl groups help stabilize the carbocation through hyperconjugation. Structure 2 is destabilized by the electron with‐drawing halogen because a delta positive charge is created next to positively charged carbon. Carbocation 1 is more stable. Worked example Which anion is more stable? Analyze: Alkyl groups are electron‐donating and halogens are electron with‐drawing Structure 2 will be stabilized by the adjacent electron with‐drawing chlorine atom where as the alkyl groups donate electron density and destabilize the carbanion. Structure 2 is the more stable carbanion. Dr. Steven Forsey 12 2.6 Hybridization The last factor to look at is the hybrization of the atom. Consider the follow acids and their conjugate bases. pKa > 70 (CH3)3CH (CH3)3Cˉ + H+ CH2=CH2 pKa = 26 All of these acids are very weak but when comparing the pKa values we see that the alkyne is the most acidic followed by the alkene and then the alkane. The trend in the stability of the conjugated bases is not due to electronegativity because they are all carbanions; nor is it due to delocalization. The only difference is the hybridization of the carbons baring the negative charge. The alkyne is sp hybridized and has 50% s and 50% p character. The alkene is sp2 hybridized and has 1/3 s and 2/3 p character. The alkane is sp3 hybridized and has ¼ s and ¾ p character. Thus as the hybrization of the carbon changes from sp to sp3 the amount of p character increases and the farther the electrons in the hybridized orbital are from the nucleus. What is more stabilizing, electrons close to the nucleus or farther away? To answer this question think about the difference in stability of a 2s orbital versus a 2p orbital; the 2s electrons are more stable because they penetrate closer to the nucleus and have a smaller charge separation. Because the lone pair electrons on the sp hybridized carbon are closer to the nucleus the alkynes conjugate base is the most stable making it the most acidic. Worked example Does the following equilibrium shift to the left or right? H HC C H + HC C + Analyze: To determine which way the equilibrium is shifted first put lone pairs and charges onto the atoms and identify the acids and bases on both side of the reaction. Arrows have been drawn to show the flow of electrons for the reactants. The base donates a pair of electrons and the acid accepts the electrons. Then determine which base is the strongest or the weakest. H HC C H + HC C + When comparing the bases on both sides of the equation we do not have to consider electronegativity of the charged atom or inductive effects. The only difference is the hybridization of the carbon atom. The sp 2 carbanion is a weaker base than the sp hybridized carbanion therefore the equilibrium will be shifted to the left. HC≡CH HC≡Cˉ + H+ CH2=CHˉ + H+ pKa = 44 Dr. Steven Forsey 13 3.0 Base strength of Amines and Amides 3.1 Amines Amines are weak bases and will react with water to form a hydroxide ion. R3NH+ + HO ˉ R3N + H2O From the above equilibrium we can see that anything stabilizes the free amine (the base) relative to the conjugate acid makes the base less basic (shifts the equilibrium to the left). Anything that destabilizes the base the more reactive or more basic the amine will be and the equilibrium shift to the right. Electron withdrawing groups stabilize the amine by withdrawing electron density from the nitrogen making them weaker and electron donating groups make the amine more basic. CH3 Worked example N N NH O NH N Order the bases from strongest to weakest. 1) 2) 3) 4) 5) Cl CH3 piperidine morpholine pyridine 2,6-dimethylpyridine 3-chloropyridine Analysis: Lets consider the equilibriums; which equilibrium would be shifted more to conjugate acid. To answer this you must consider the previous topics, electronegativity, size, inductive effects, resonance stabilization and hybridization. (Remember: The smaller the pkb the stronger the base) First lets consider the hybridization of the of the nitrogen atoms. N N H Compounds 1 and 2 are sp3 hybridized and 3, 4 and 5 are sp2 sp2 3 sp hybridized. We would expect that 3, 4 and 5 to be less basic than 1 and 2 5) 1) 2) 3) 4) because the nitrogen’s lone pair of electrons are in sp2 hybridized orbitals, have pKb = 4.7 CH3 pKb = 11.2 pKb = 8.75 pKb = 2.9 pKb = 7.3 δ+ more s character and are closer nucleus 3 (more stable) than the sp hybridized lone N N NH O NH N pair. Note the electrons in the sp2 δ+ δ hybridized orbitals do not resonate into the δ+ δo Cl CH3 aromatic ring because they are at 90 to the H2 O H 2O H2O H2O H2 O π bonds and π bonds are created by the sideways overlap of p orbitals not CH3 hybridized orbitals. Next look at the δ+ inductive effects of 1 and 2; the oxygen in 1 NH NH NH2 O NH2 NH is electron withdrawing and will pull δδ+ electron density from the nitrogen atom. δ+ δstabilizing This brings the lone of pair of electrons Cl destabilizing destabilizing CH3 pKa = 11.1 pKa = 9.3 closer to the nucleus and makes them more pKa = 5.25 pKa = 6.7 pKa = 2.8 stable or less basic than 1. We can also look + HO + HO + HO + HO + HO at the conjugate acids. The conjugate acid of 2 is less stable than for 1 because of the inductive effects of the oxygen. The oxygen creates delta positively charged carbons that destabilize the positively charged nitrogen of the conjugate acid and shifts the equilibrium towards the base. Thus 1 is more basic than 2. For 3, 4 and 5 the only difference between the compounds is the electron withdrawing and donating groups next to the nitrogen atom. Compound 4 has two electron donating methyl groups attached to the aromatic ring. These groups donate electron density into the ring and push the lone pair of electrons on the nitrogen farther from the nucleus, making them less stable and more basic than 3 and 5. Also the positively charged conjugate acid is stabilized by the electron donating alkyl groups. These two factors make 2 more basic than 3 and 5. The chlorine atom in 5 inductively withdraws electron density from the aromatic ring and consequently the lone pair of electrons on the nitrogen are pulled closer to Dr. Steven Forsey 14 the nucleus, making them less basic. The chlorine also destabilizes the conjugate base of 5 because the positively charged nitrogen atom is destabilized by the adjacent delta positively charged carbon atom. The combination of the sp2 hybridized nitrogen and the adjacent electronegative atom makes 5 the least basic compound in this group. Thus base strength from strongest to weakest is: 1 > 2 > 4 > 3 > 5 Now consider the following trend: CH3NH2 CH3NHCH3 (CH3)3N Amine NH3 4.75 3.34 3.27 4.19 pKb The first three amines follow the expected trend. Alkyl groups are electron donating and push the lone pair of electrons farther away from the nitrogen’s nucleus and make them more reactive. Also the alkyl groups help stabilize the positively charged conjugate acid. These two factors shift the equilibrium to the right as the number of alkyl groups increases. But why does the pkb increase for trimethylamine? A very important factor that we have not touched on is solvent effects and how the equilibriums can be changed by using different solvents or even by just changing the ionic strength of the solution. We will only briefly touch on solvent effects here. As discussed in the physical properties sections of the course, to dissolve a solute the solvent molecules must surround the solute molecules and suspend the solute in the solvent. To dissolve a salt strong ion‐dipole bonds are needed between the solvent and the ions. Thus good solvents for ionic compounds must be highly polar and have the ability to surround the ions and form strong bonds to dissolve the ions. Water is such a solvent. The water molecule has an electronegative oxygen with two small hydrogens attached to it. The electron density is pulled away from the hydrogens nucleus to create a very polar molecule. Water solvates cations through the donation of unshared electron pairs of the oxygen atom to the vacant orbitals of the cation and because the hydrogen atoms are tiny, water molecules are able to surround the cation tightly. Water molecules will also form strong ion‐dipole electrostatic forces between the hydrogen atoms and the anion through hydrogen bonding. Again because the hydrogen atoms are small a number of water molecules can surround the anion without crowding. The figure below illustrates the solvation of a salt with water. The water molecules oxygen and hydrogens create electrostatic bonds with the ions to form a cage around the ions to solubilize them. This is called a solvent shell. The water molecules in the bulk solution are freer to move around while the solvent shell molecules are bound to the ions. H H H O H H O H H H O O H H O H H O H H H O H H H H O H O O H H H O O H H H H H H δ+ O H H O O O δ- H H H O O H H + H δ+ O δ+ H H O H H δ+ H H H O O O H δ O O H H O H H H H H H HH O O H O H O H O H H H O H H H O H H H O O H H H Dr. Steven Forsey 15 Compounds like methanol are also able to solubilize salts because of the –OH group. However because of the hydrophobic alkyl group and the steric hindrance created by the methyl group, methanol is less polar than water and is inferior to water in its ability to solvate salts. How does this relate to the equilibrium of the amines? When the amine becomes protonated the cation is solvated by water through ion‐dipole interactions. The 1o 2 o and 3 o amines all have hydrophobic alkyl groups attached to the amine which have nonbonding interactions with water. Thus water is forced to create a very ordered cage around the alkyl groups. This is an unfavorable interaction since systems like to maximize disorder (entropy). As the number of alkyl groups increases the ability of the water molecules to solvate the positively changed conjugate acid decreases. Thus trimethylamine is a weaker base than diethylamine because of the inability of the water molecules to solvate the cation and the equilibrium is shifted to the free amine. Another factor to consider is the steric interaction of the alkyl groups around the nitrogen. As the number of alkyl groups increase the ability of the water molecules to have close contact decreases because of steric interaction. This also explains why cyclic amines with their alkyl groups “tied back” are more basic than comparable open‐chained secondary amines. Dr. Steven Forsey 16 3.2 Amides Amides are very weak bases and are considered to be a neutral functional group. It requires a very strong acid to protonate an amide. Notice below that when protonation occurs the oxygen is protonated and not the nitrogen. This can be explained by the amides resonance structure that generates a negative charge on the oxygen and a positive charge on the nitrogen. This resonance structure even though charged is important. Amides have a planar structure and the nitrogen possesses a trigonal planar geometry as shown in the bond angles of formamide. The shape of the nitrogen allows for the lone pair electrons on the nitrogen to occupy a p orbital and permit π overlap with the carbonyl carbons p orbital. The C‐N bond has double‐bond character like an alkene and for formamide the barrier of rotation for the C‐N bond is 75 kJ/mol. pKb = 13.37 O 117o O O C H C H N C R NH2 R NH2 121o 119o H H formamide H O O HC C R NH2 NH H 4.0 Some exceptions to the general trends of acidity and basicity The proceeding sections having given you the ability to qualitatively determine relative acid‐base strength by looking at the stability of the bases. However, it impossible to know the exact acidity or basicity of a compound just by looking at it, this has to be done experimentally. For example which is more basic 2 pyridine of aniline? The pyridine’s nitrogen is sp hybridized which should make it less basic than aniline. However, the lone pair on the aniline’s nitrogen atom resonate into the aromatic ring and are resonance stabilized. Which has the greater stabilizing effect? The only way to know this is to do the experiment. Experimentally we find that aniline (pKb = 9.4) is less basic than pyridine (pKb = 8.5) thus resonance in this example has a greater stabilizing effect than hybridization. NH2 NH2 NH2 NH2 N pyridine aniline Another interesting pKb value that is not obvious is the amide ion ion (NH2ˉ, pKb = ‐24) versus a sp hybridized carbanion (i.e HC≡Cˉ, pKb = ‐12). Let’s look at the trends of the acids, examine their pka values given below and try to explain what we see experimentally. (note: all of these are very weak acids) HC≡CH (pKa = 26) > NH3 (pKa = 38) > H2C=CH2 (pKa = 44) > CH3CH3 (pKa = 50) From previous examples we expect electronegativity to have a greater stabilizing effect on the stability of the conjugate base than hybridization. And indeed it does. The sp3 negatively charged nitrogen atom is a weaker base than a negatively charged sp3 carbon because it is more electronegative. Thus NH3 is a stronger acid than CH3CH3. The amide ion is also a weaker base than a sp2 hybridized carbanion. Thus NH3 is stronger acid than H2C=CH2 and CH3CH3. However NH2ˉ is a stronger base than a sp hybridized Dr. Steven Forsey 17 carbanion. How can we explain this? The explanation must be that sp hybridization of the carbanion has a greater stabilizingeffect than increased electronegativity of the sp3 hybridized nitrogen atom. We can visualize what is happing with the following simplified orbital diagram. 3 3 2 sp sp sp sp C¯ N¯ C¯ C¯ The change in hybridization from sp3. sp2 to sp brings the electrons closer to the nucleus of the carbon atom and increases the stability of the carbanion. Similarly increasing electronegativity brings the electrons closer to the nucleus and helps stabilize the amide ion compared to the sp3 and sp2 carbanions. It just so happens that the sp hybridization of the carbanion has a greater stabilizing effect than the increase in electronegativity of the sp3 nitrogen atom. The only way you would know this is to do the experiments or look up the pka values. There are other examples in which one factor has a greater effect on the acidity of a molecule than what you may expect. For example 2,4‐dinitrophenol (pka = 4.08) is more acidic than acetic acid (pka = 4.80) because of the enhance stabilization of the conjugate base by the nitro‐groups. 1,4,6‐trintrophenol (pka = 0.38) is even more acidic than trichloroacetic acid (pka = 0.64) OH OH NO2 NO2 O2N O O CH3COH CF3COH pKa=0.64 pKa=4.80 NO2 NO2 pKa=0.38 pKa=4.08 For Chem 266 you need to know that R‐NHˉ is a stronger base than R‐C≡Cˉ because this reaction is commonly used to form the acetylnide anion which is then used as a nucleophile. As long as I am teaching Chem 266 (Dr. Steven Forsey) this will be the only trend breaker you need to know. You also need to remember the difference in basicity between aniline and pyridine. However if you are given pKa or pKb values you should able to rationalize why you see the observed values. Such as why a phenol like 2,4‐ dinitrophenol (pka = 4.08) is more acidic than acetic acid (pka = 4.80). Dr. Steven Forsey 18 Acid or Conjugate acid HI HBr pKa -10.1 -10 -9 -8 Base or Conjugate Base I¯ Br¯ Very strong acids Very weak bases -7.3 HCl -7 -6.8 -6.1 H2SO4 -5 to -3.0 -3.6 -2.4 -2.5 H3O HNO3 CH3 SO3H + Cl¯ HSO4¯ -1.7 -1.3 -1.2 -0.6 H2O NO3¯ SO3 NO 2 NO2 O2N OH NO2 0.38 O2N O NO2 Strong acids Weak bases 0.0 0.2 0.64 O2N + NH3 HSO4¯ H3PO4 1.0 2.0 2.1 1.3 O2N NH2 SO4 H2PO4¯ 2- H N+ Cl N 2.8 2.7 2.8 2.9 Cl Dr. Steven Forsey 19 Acid or Conjugate acid pKa 3.2 Base or Conjugate Base HF HNO2 3.2 3.4 3.4 3.8 3.9 4.0 F¯ NO2¯ Cl + NH3 4.15 4.2 4.3 Cl O CO NH2 CH3O O COH 4.5 4.6 Weal bases Weak acids 4.8 4.8 CH3 NH2 5.1 5.2 6.2 H2CO3 6.4 6.7 HCO3¯ H2S 7.0 7.1 HS¯ 7.2 7.8 8.4 HCN 9.1 9.3 CN¯ Dr. Steven Forsey 20 Acid or Conjugate acid NH4+ OH pKa 9.4 9.4 9.51 10 10.2 Base or Conjugate Base NH3 O HCO3¯ CH3NO2 10.2 10.2 10.3 CO3-2¯ ¯CH2NO2 CH3CH2SH (CH3)3NH+ CH3NH3+ + (CH3)2NH2 CH3CH2NH3+ + NH2 10.5 10.6 10.7 10.7 10.7 11.1 12.4 13.7 13.7 15.0 CH3CH2S¯ (CH3)3N CH3NH2 (CH3)2NH CH3CH2NH2 NH CF3CH2OH CF3CH2O¯ NH C NH2 NH2 Very strong bases Very weak acids CH3OH H2O CH3CH2OH 15.5 15.7 16.0 16 17 CH3O¯ HO¯ CH3CH2O¯ (CH3)3COH 18.0 20 (CH3)3CO¯ CHCl3 HC/CH H2 NH3 CH3NH2 25 25 35 36 40 41 ¯CCl3 HC/C¯ H¯ NH2¯ CH3NH¯ 43 CH2=CHCH3 CH2=CH2 CH4 CH3CH3 (CH3)3CH 43 44 50 50 >70 CH2=CHCH3¯ CH2=CH2¯ CH3¯ CH3CH2¯ (CH3)3C¯ Dr. Steven Forsey 21 ...
View Full Document

This note was uploaded on 11/14/2010 for the course CHEM CHEM266 taught by Professor Forsey during the Fall '10 term at Waterloo.

Ask a homework question - tutors are online