{"id":2569,"date":"2018-06-19T20:36:18","date_gmt":"2018-06-19T20:36:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/1-22-how-substituents-affect-the-strength-of-an-acid\/"},"modified":"2018-08-06T12:53:16","modified_gmt":"2018-08-06T12:53:16","slug":"6-4-acid-strength-and-pka","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/6-4-acid-strength-and-pka\/","title":{"raw":"6.4. Acid strength and pKa","rendered":"6.4. Acid strength and pKa"},"content":{"raw":"<section class=\"mt-content-container\">\r\n<div id=\"s21358\" class=\"mt-include\">\r\n\r\nNow that we know how to quantify the strength of an acid or base, our next job is to gain an understanding of the fundamental reasons behind <em>why<\/em> one compound is more acidic or more basic than another.\u00a0 This is a big step: we are, for the first time, taking our knowledge of organic <em>structure<\/em> and applying it to a question of organic <em>reactivity<\/em>.\u00a0 Many of the ideas that we\u2019ll see for the first here will continue to apply throughout the book as we tackle many other organic reaction types.\r\n<div class=\"mt-section\">\r\n<h3 id=\"A:_Periodic_trends-21358\">A: Periodic trends<\/h3>\r\nFirst, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table.\u00a0 We\u2019ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules with the same functionalities, for example the side chains of the amino acids alanine (alkane), lysine (amine), and serine (alcohol).\r\n<div>\r\n<div class=\"textbox examples\">\r\n\r\n<strong>Horizontal periodic trend in acidity and basicity<\/strong>\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203548\/fig7-3-1.png\" alt=\"\" width=\"334\" height=\"280\" \/>\r\n\r\n<\/div>\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n\r\n\r\n<hr \/>\r\n\r\nWe can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen.\u00a0 The key to understanding this trend is to consider the hypothetical conjugate base in each case<em>: the more stable (weaker) the conjugate base, the stronger the acid<\/em>. Look at where the negative charge ends up in each conjugate base.\u00a0 In the conjugate base of ethane, the negative charge is borne by a carbon atom, while on the conjugate base of methylamine and ethanol the negative charge is located on a nitrogen and an oxygen, respectively.\u00a0 Remember that <a title=\"Section 2.3: Non-covalent interactions\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II\/2.4%3A_Non-covalent_interactions\" target=\"_blank\" rel=\"internal noopener\">electronegativity also increases as we move from left to right along a row of the periodic table<\/a>, meaning that oxygen is the most electronegative of the three atoms, and carbon the least.\r\n<div class=\"textbox key-takeaways\">\r\n\r\n<strong><em>The more electronegative an atom, the better able it is to bear a negative charge. Weaker bases have negative charges on more electronegative atoms; stronger bases have negative charges on less electronegative atoms.<\/em><\/strong>\r\n\r\n<\/div>\r\nThus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl carbanion anion is the least stable (highest energy, most basic).\u00a0 Conversely, ethanol is the strongest acid, and ethane the weakest acid.\r\n\r\nWhen moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity.\u00a0 This is best illustrated with the haloacids and halides: basicity, like electronegativity,\u00a0 increases as we move up the column.\r\n<div>\r\n<div>\r\n<div class=\"textbox shaded\">\r\n<p style=\"text-align: center\"><strong>Vertical periodic trend in acidity and basicity<\/strong><img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203551\/fig7-3-2.png\" alt=\"\" width=\"368\" height=\"179\" \/><\/p>\r\n\r\n<\/div>\r\nConversely, acidity in the haloacids increases as we move <em>down<\/em> the column.\r\n\r\n<\/div>\r\n<\/div>\r\nIn order to make sense of this trend, we will once again consider the stability of the conjugate bases.\u00a0 Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion.\u00a0 But in fact, it is the <em>least<\/em> stable, and the most basic!\u00a0\u00a0 It turns out that when moving vertically in the periodic table, the <em>size<\/em> of the atom trumps its electronegativity with regard to basicity.\u00a0 The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly\u00a0 larger volume:\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203553\/fig7-3-3.png\" alt=\"\" width=\"278\" height=\"144\" \/>\r\n\r\nThis illustrates a fundamental concept in organic chemistry:\r\n<div>\r\n<div class=\"textbox key-takeaways\"><em><strong>Electrostatic charges, whether positive or negative, are more stable when they are \u2018spread out\u2019 over a larger area.<\/strong><\/em><\/div>\r\n<\/div>\r\nWe will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts.\u00a0\u00a0 For now, we are applying the concept only to the influence of atomic radius on base strength.\u00a0 Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than a carboxylic acid.\u00a0 HI, with a pK<sub>a<\/sub> of about -9,\u00a0 is almost as strong as sulfuric acid.\r\n\r\nMore importantly to the study of biological organic chemistry, this trend tells us that <strong><em>thiols are more acidic than alcohols<\/em><\/strong>. The pK<sub>a<\/sub> of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pK<sub>a<\/sub> for the alcohol group on the serine side chain is on the order of 17.\r\n\r\nRemember the concept of 'driving force' that we learned about in chapter 6?\u00a0 Recall that the driving force for a reaction is usually based on two factors: relative charge stability, and relative total bond energy.\u00a0 Let's see how this applies to a simple acid-base reaction between hydrochloric acid and fluoride ion:\r\n\r\nHCl + F<sup>-<\/sup>\u00a0\u2192 HF + Cl-\r\n\r\nWe know that HCl (pK<sub>a<\/sub> -7) is a stronger acid than HF (pK<sub>a <\/sub>3.2), so the equilibrium for the reaction lies on the product side: the reaction is exergonic, and a 'driving force' pushes reactant to product.\r\n\r\nWhat explains this driving force?\u00a0 Consider first the charge factor: as we just learned, chloride ion (on the product side) is more stable than fluoride ion (on the reactant side).\u00a0 This partially accounts for the driving force going from reactant to product in this reaction: we are going from less stable ion to a more stable ion.\r\n\r\nWhat about total bond energy, the other factor in driving force?\u00a0 If you consult a <a class=\"mt-disabled\" title=\"Organic_Chemistry\/Organic_Chemistry_With_a_Biological_Emphasis\/Chapter_07:_Organic_compounds_as_acids_and_bases\/Section_7.3:_Structural_effects_on_acidity_and_basicityhttp:\/\/chemwiki.ucdavis.edu\/Theoretical_Chemistry\/Chemical_Bonding\/General_Principles_of_Chemical_Bonding\/Bond_Energies\" target=\"_blank\" rel=\"broken\">table of bond energies<\/a>, you will see that the H-F bond on the product side is more energetic (stronger) than the H-Cl bond on the\u00a0 reactant side: 565 kJ\/mol <em>vs<\/em> 427 kJ\/mol, respectively).\u00a0 This also contributes to the driving force: we are moving from a weaker (less stable) bond to a stronger (more stable) bond.\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<h3 id=\"B:_Resonance_effects-21358\">B: Resonance effects<\/h3>\r\nIn the previous section we focused our attention on periodic trends - the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements.\u00a0 Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about <em>the same element acting as the proton donor\/acceptor<\/em>.\u00a0 The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups.\r\n\r\nDespite the fact that they are both oxygen acids, the pK<sub>a<\/sub> values of ethanol and acetic acid are strikingly different.\u00a0 What makes a carboxylic acid so much more acidic than an alcohol?\u00a0 As before, we begin by considering the stability of the conjugate bases, remembering that a\u00a0 more stable (weaker) conjugate base corresponds to a stronger acid.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203557\/fig7-3-4.png\" alt=\"\" width=\"539\" height=\"293\" \/>\r\n\r\nIn both species, the negative charge on the conjugate base is located on oxygen, so periodic trends cannot be invoked.\u00a0 For acetic acid, however, there is a key difference: two resonance contributors can be drawn for the conjugate base, and the negative charge can be delocalized (shared) over two oxygen atoms. In the ethoxide ion, by contrast, the negative charge is localized, or \u2018locked\u2019 on the single oxygen \u2013 it has nowhere else to go.\u00a0 This makes the ethoxide ion much less stable.\r\n\r\nRecall the important general statement that we made a little earlier: 'Electrostatic charges, whether positive or negative, are more stable when they are \u2018spread out\u2019 than when they are confined to one location.'\u00a0 Now, we are seeing this concept in another context, where a charge is being \u2018spread out\u2019 (in other words, delocalized) <em>by resonance<\/em>, rather than simply by the size of the atom involved.\r\n\r\nThe delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pK<sub>a<\/sub> units between ethanol and acetic acid (and remember, pK<sub>a<\/sub> is a log expression, so we are talking about a factor of 10<sup>12<\/sup> between the K<sub>a<\/sub> values for the two molecules!)\r\n\r\nThe resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but <em>not<\/em> basic when it is part of an amide group.\u00a0 Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a minor but still important resonance contributor in which the nitrogen lone pair is part of a pi bond.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203600\/fig7-3-5.png\" alt=\"\" width=\"581\" height=\"317\" \/>\r\n\r\nWhereas the lone pair of an amine nitrogen is \u2018stuck\u2019 in one place, the lone pair on an amide nitrogen is delocalized by resonance.\u00a0 Notice that in this case, we are extending our central statement to say that <em>electron density<\/em> \u2013 in the form of a lone pair \u2013 is stabilized by resonance delocalization, even though there is not a negative charge involved.\u00a0 Here\u2019s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton \u2013 these two electrons are too \u2018comfortable\u2019 being part of the delocalized pi bonding system.\u00a0 The lone pair on an amine nitrogen, by contrast, is not so comfortable - it is <em>not<\/em> part of a delocalized pi system, and is available to form a bond with any acidic proton that might be nearby.\r\n\r\nIf an amide group is protonated, it will be at the oxygen rather than the nitrogen.\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n1. a) Draw the Lewis structure of nitric acid, HNO<sub>3<\/sub>.\r\n\r\nb) Nitric acid is a strong acid - it has a pK<sub>a<\/sub> of -1.4.\u00a0 Make a structural argument to account for its strength. Your answer should involve the structure of nitrate, the conjugate base of nitric acid.\r\n\r\n[reveal-answer q=\"920451\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"920451\"]\r\n\r\nThe negative charge on the oxygen that results from deprotonation of the acid is delocalized by resonance.\r\n\r\n&nbsp;\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61509\/figES7-3-1.png?revision=1\" alt=\"\" width=\"312\" height=\"192\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n2. Rank the compounds below from most acidic to least acidic, and explain your reasoning.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203603\/figE7-3-1.png\" alt=\"\" width=\"516\" height=\"84\" \/>\r\n\r\n[reveal-answer q=\"581719\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"581719\"]\r\n\r\nA and B are ammonium groups, while C is an amine, so C is clearly the least acidic.\u00a0 Looking at the conjugate base of B, we see that the lone pair electrons can be delocalized by resonance, making this conjugate base more stable than the conjugate base of A, where the electrons cannot be stabilized by resonance.\u00a0\u00a0 Thus B is the most acidic.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61511\/figES7-3-2a.png?revision=1\" alt=\"\" width=\"379\" height=\"71\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n3. (challenging!) Often it requires some careful thought to predict the most acidic proton on a molecule.\u00a0 Ascorbic acid, also known as Vitamin C, has a pK<sub>a<\/sub> of 4.1 - the fact that this is in the range of carboxylic acids suggest to us that the negative charge on the conjugate base can be delocalized by resonance to <em>two<\/em> oxygen atoms.\u00a0 Which if the four OH protons on the molecule is most acidic?\u00a0 Draw the structure of ascorbate, the conjugate base of ascorbic acid, then draw a second resonance contributor showing how the negative charge is delocalized to a second oxygen atom.\u00a0 <em>Hint<\/em> - try removing each OH group in turn, then use your resonance drawing skills to figure out whether or not delocalization of charge can occur.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203605\/figE7-3-3.png\" alt=\"\" width=\"146\" height=\"172\" \/>\r\n\r\n[reveal-answer q=\"467594\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"467594\"]\r\n\r\nBelow is the structure of ascorbate, the conjugate base of ascorbic acid.\u00a0 Note that the negative charge can be delocalized by resonance to two oxygen atoms, which makes ascorbic acid similar in strength to carboxylic acids.\r\n\r\n&nbsp;\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61512\/figES7-3-3.png?revision=1\" alt=\"\" width=\"457\" height=\"129\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"mt-section\">\r\n\r\nResonance effects involving aromatic structures can have a dramatic influence on acidity and basicity.\u00a0 Notice, for example, the difference in acidity between phenol and cyclohexanol.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203619\/fig7-4-1.png\" alt=\"\" width=\"182\" height=\"130\" \/>\r\n\r\nLooking at the conjugate base of phenol, we see that the negative charge can be delocalized by resonance to three different carbons on the aromatic ring.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203622\/fig7-4-2.png\" alt=\"\" width=\"399\" height=\"261\" \/>\r\n\r\nAlthough these are all minor resonance contributors (negative charge is placed on a\u00a0 carbon rather than the more electronegative oxygen), they nonetheless have a significant effect on the acidity of the phenolic proton.\u00a0 Essentially, the benzene ring is acting as an electron-withdrawing group by resonance.\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercise<\/h3>\r\nDraw the conjugate base of 2-napthol (the major resonance contributor), and on your drawing indicate with arrows all of the atoms to which the negative charge can be delocalized by resonance.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203624\/figE7-4-1.png\" alt=\"\" width=\"114\" height=\"80\" \/>\r\n\r\n[reveal-answer q=\"114716\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"114716\"]\r\n\r\nThe negative charge can be delocalized by resonance to five carbons:\r\n\r\n<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61549\/figES7-4-1a.png?revision=1\" alt=\"\" width=\"272\" height=\"190\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nThe base-stabilizing effect of an aromatic ring can be accentuated by the presence of an additional electron-withdrawing substituent, such as a carbonyl.\u00a0 For the conjugate base of the phenol derivative below, an additional resonance contributor can be drawn in which the negative formal charge is placed on the carbonyl oxygen.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203627\/fig7-4-3.png\" alt=\"\" width=\"452\" height=\"177\" \/>\r\n\r\nNow the negative charge on the conjugate base can be spread out over <em>two<\/em> oxygens (in addition to three aromatic carbons). The phenol acid therefore has a pK<sub>a<\/sub> similar to that of a carboxylic acid, where the negative charge on the conjugate base is also delocalized to two oxygen atoms.\u00a0 The ketone group is acting as an electron withdrawing group - it is 'pulling' electron density towards itself, through both inductive and resonance effects.\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nThe position of the electron-withdrawing substituent relative to the phenol hydroxyl is very important in terms of its effect on acidity. Which of the two substituted phenols below is more acidic?\u00a0 Use resonance drawings to explain your answer.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203630\/figE7-4-2.png\" alt=\"\" width=\"251\" height=\"104\" \/>\r\n\r\n[reveal-answer q=\"981365\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"981365\"]\r\n\r\nWhen the aldehyde is in the 4 (<em>para<\/em>) position, the negative charge on the conjugate base can be delocalized to two oxygen atoms. In the other compound, the aldehyde is on the 3 (<em>meta<\/em>) position, and the negative charge cannot be delocalized to the aldehyde oxygen.\r\n\r\n<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61554\/figES7-4-2.png?revision=1\" alt=\"\" width=\"307\" height=\"153\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n2. Rank the four compounds below from most acidic to least.<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203632\/figE7-4-3.png\" alt=\"\" width=\"440\" height=\"93\" \/>\r\n\r\n[reveal-answer q=\"710916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"710916\"]\r\n\r\nThe order of acidity, going from left to right (with 1 being most acidic), is 2-1-4-3.\u00a0 The least acidic compound (second from the right) has no phenol group at all - aldehydes are not acidic.\u00a0 The most acidic compound (second from the left) is a phenol with an aldehyde in the 2 (<em>ortho<\/em>) position, and as a consequence the negative charge on the conjugate base can be delocalized to both oxygen atoms. \u00a0 In the compound with the aldehyde in the 3 (<em>meta<\/em>) position, there is an electron-withdrawing inductive effect, but NOT a resonance effect (the negative charge on the cannot be delocalized to the aldehyde oxygen).\r\n\r\n[\/hidden-answer]\r\n\r\n3. Nitro groups are very powerful electron-withdrawing groups.\u00a0 The phenol derivative picric acid (2,4,6 -trinitrophenol) has a pK<sub>a<\/sub> of 0.25, lower than that of trifluoroacetic acid. Use a resonance argument to explain why picric acid has such a low pKa.\r\n\r\n[reveal-answer q=\"475315\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"475315\"]\r\n\r\nThe negative charge on the conjugate base of picric acid can be delocalized to three different nitro oxygen atoms (in addition to the phenolate oxygen).\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61555\/figES7-4-4.png?revision=1\" alt=\"\" width=\"645\" height=\"196\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<h3 id=\"C:_Inductive_effects-21358\">C: Inductive effects<\/h3>\r\nCompare the pK<sub>a<\/sub> values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203608\/fig7-3-6.png\" alt=\"\" width=\"463\" height=\"95\" \/>\r\n\r\nThe presence of the chlorine atoms clearly increases the acidity of the carboxylic acid group, but the argument here does not have to do with resonance delocalization, because no additional resonance contributors can be drawn for the chlorinated molecules.\u00a0 Rather, the explanation for this phenomenon involves something called the <strong>inductive effect<\/strong>.\u00a0 A chlorine atom is more electronegative than a hydrogen, and thus is able to \u2018induce\u2019, or \u2018pull\u2019 electron density towards itself, away from the carboxylate group.\u00a0 In effect, the chlorine atoms are helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect.\u00a0 In this context, the chlorine substituent can be referred to as an <strong>electron-withdrawing group<\/strong>. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pK<sub>a<\/sub> values between an alcohol and a carboxylic acid.\u00a0 In general, <em>resonance effects are more powerful than inductive effects<\/em>.\r\n\r\nBecause the inductive effect depends on electronegativity, fluorine substituents have a more pronounced\u00a0 pKa-lowered effect than chlorine substituents.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203610\/fig7-3-7.png\" alt=\"\" width=\"288\" height=\"122\" \/>\r\n\r\nIn addition, the\u00a0 inductive takes place through covalent bonds, and its influence decreases markedly with distance \u2013 thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203613\/fig7-3-8.png\" alt=\"\" width=\"297\" height=\"98\" \/>\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercise<\/h3>\r\nRank the compounds below from most acidic to least acidic, and explain your reasoning.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203617\/figE7-3-4.png\" alt=\"\" width=\"275\" height=\"129\" \/>\r\n\r\n[reveal-answer q=\"891926\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"891926\"]\r\n\r\nWe must consider the electronegativity and the position of the halogen substituent in terms of inductive effects.\u00a0 A is the strongest acid, as <a title=\"Section 2.3: Non-covalent interactions\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II\/2.4%3A_Non-covalent_interactions\" target=\"_blank\" rel=\"internal noopener\">chlorine is more electronegative than bromine.<\/a>\u00a0 B is more acidic than C, as the bromine is closer (in terms of the number of bonds) to the site of acidity.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<h3>Combinations of effects<\/h3>\r\nConsider the acidity of 4-methoxyphenol, compared to phenol:\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203635\/fig7-4-4.png\" alt=\"\" width=\"210\" height=\"166\" \/>\r\n\r\nNotice that the methoxy group increases the pK<sub>a<\/sub> of the phenol group - it makes it <em>less<\/em> acidic. Why is this? At first inspection, you might assume that the methoxy substituent, with its electronegative oxygen, would be an electron-withdrawing group by induction.\u00a0 That is correct, but only to a point.\u00a0 The oxygen atom does indeed exert an electron-withdrawing inductive effect, but the lone pairs on the oxygen cause the exact opposite effect \u2013 <em>the methoxy group is an <strong>electron-donating group by resonance<\/strong><\/em>.\u00a0 A resonance contributor can be drawn in which a formal negative charge is placed on the carbon adjacent to the negatively-charged phenolate oxygen.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203637\/fig7-4-5.png\" alt=\"\" width=\"237\" height=\"162\" \/>\r\n\r\nBecause of like-charge repulsion, this <em>destabilizes<\/em> the negative charge on the phenolate oxygen, making it more basic. It may help to visualize the methoxy group \u2018pushing\u2019 electrons towards the lone pair electrons of the phenolate oxygen, causing them to be less 'comfortable' and more reactive. The example above is a somewhat confusing but quite common situation in organic chemistry - a functional group, in this case a methoxy group, is exerting both an inductive effect and a resonance effect, <em>but in opposite directions<\/em> (the inductive effect is electron-withdrawing, the resonance effect is electron-donating).\u00a0 As a general rule a resonance effect is more powerful than an inductive effect - so overall, the methoxy group is acting as an electron donating group. A good rule of thumb to remember:\r\n<div class=\"textbox key-takeaways\">\r\n<p style=\"text-align: center\"><strong>When resonance and induction compete, resonance usually wins!<\/strong><\/p>\r\n\r\n<\/div>\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercise<\/h3>\r\nRank the three compounds below from lowest pKa to highest, and explain your reasoning.\u00a0 <em>Hint<\/em> - think about both resonance and inductive effects!\r\n\r\n<img class=\"internal default\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203639\/figE7-4-5.png\" alt=\"\" width=\"339\" height=\"117\" \/>\r\n\r\n[reveal-answer q=\"951666\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"951666\"]\r\n\r\nCompound C has the lowest pK<sub>a<\/sub> (most acidic): the oxygen acts as an electron withdrawing group by induction.\u00a0 Compound A has the highest pK<sub>a<\/sub> (the oxygen is in a position to act as an electron donating group by resonance, thus destabilizing the negative charge of the conjugate base).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Further reading<\/h3>\r\n<ul>\r\n \t<li><a href=\"https:\/\/chem.libretexts.org\/LibreTexts\/Purdue\/Purdue_Chem_26100%3A_Organic_Chemistry_I_(Wenthold)\/Chapter_01%3A_Introduction_and_Review\/1.8_Acids_and_Bases\/Overview_of_Acids_and_Bases\">Overview of Acids and Bases<\/a><\/li>\r\n<\/ul>\r\n<img class=\" wp-image-4634 alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/01151246\/static_qr_code_without_logo1-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/>\r\n<ul>\r\n \t<li><a href=\"https:\/\/chem.libretexts.org\/Homework_Exercises\/Exercises%3A_Organic_Chemistry\/Organic%3A_Acid%2F%2FBase_Practice_Problems\">Base Practice problems<\/a><\/li>\r\n<\/ul>\r\n<img class=\"size-thumbnail wp-image-4635 alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/01151706\/static_qr_code_without_logo2-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>","rendered":"<section class=\"mt-content-container\">\n<div id=\"s21358\" class=\"mt-include\">\n<p>Now that we know how to quantify the strength of an acid or base, our next job is to gain an understanding of the fundamental reasons behind <em>why<\/em> one compound is more acidic or more basic than another.\u00a0 This is a big step: we are, for the first time, taking our knowledge of organic <em>structure<\/em> and applying it to a question of organic <em>reactivity<\/em>.\u00a0 Many of the ideas that we\u2019ll see for the first here will continue to apply throughout the book as we tackle many other organic reaction types.<\/p>\n<div class=\"mt-section\">\n<h3 id=\"A:_Periodic_trends-21358\">A: Periodic trends<\/h3>\n<p>First, we will focus on individual atoms, and think about trends associated with the position of an element on the periodic table.\u00a0 We\u2019ll use as our first models the simple organic compounds ethane, methylamine, and ethanol, but the concepts apply equally to more complex biomolecules with the same functionalities, for example the side chains of the amino acids alanine (alkane), lysine (amine), and serine (alcohol).<\/p>\n<div>\n<div class=\"textbox examples\">\n<p><strong>Horizontal periodic trend in acidity and basicity<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203548\/fig7-3-1.png\" alt=\"\" width=\"334\" height=\"280\" \/><\/p>\n<\/div>\n<\/div>\n<hr \/>\n<hr \/>\n<p>We can see a clear trend in acidity as we move from left to right along the second row of the periodic table from carbon to nitrogen to oxygen.\u00a0 The key to understanding this trend is to consider the hypothetical conjugate base in each case<em>: the more stable (weaker) the conjugate base, the stronger the acid<\/em>. Look at where the negative charge ends up in each conjugate base.\u00a0 In the conjugate base of ethane, the negative charge is borne by a carbon atom, while on the conjugate base of methylamine and ethanol the negative charge is located on a nitrogen and an oxygen, respectively.\u00a0 Remember that <a title=\"Section 2.3: Non-covalent interactions\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II\/2.4%3A_Non-covalent_interactions\" target=\"_blank\" rel=\"internal noopener\">electronegativity also increases as we move from left to right along a row of the periodic table<\/a>, meaning that oxygen is the most electronegative of the three atoms, and carbon the least.<\/p>\n<div class=\"textbox key-takeaways\">\n<p><strong><em>The more electronegative an atom, the better able it is to bear a negative charge. Weaker bases have negative charges on more electronegative atoms; stronger bases have negative charges on less electronegative atoms.<\/em><\/strong><\/p>\n<\/div>\n<p>Thus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl carbanion anion is the least stable (highest energy, most basic).\u00a0 Conversely, ethanol is the strongest acid, and ethane the weakest acid.<\/p>\n<p>When moving vertically within a given column of the periodic table, we again observe a clear periodic trend in acidity.\u00a0 This is best illustrated with the haloacids and halides: basicity, like electronegativity,\u00a0 increases as we move up the column.<\/p>\n<div>\n<div>\n<div class=\"textbox shaded\">\n<p style=\"text-align: center\"><strong>Vertical periodic trend in acidity and basicity<\/strong><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203551\/fig7-3-2.png\" alt=\"\" width=\"368\" height=\"179\" \/><\/p>\n<\/div>\n<p>Conversely, acidity in the haloacids increases as we move <em>down<\/em> the column.<\/p>\n<\/div>\n<\/div>\n<p>In order to make sense of this trend, we will once again consider the stability of the conjugate bases.\u00a0 Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion.\u00a0 But in fact, it is the <em>least<\/em> stable, and the most basic!\u00a0\u00a0 It turns out that when moving vertically in the periodic table, the <em>size<\/em> of the atom trumps its electronegativity with regard to basicity.\u00a0 The atomic radius of iodine is approximately twice that of fluorine, so in an iodide ion, the negative charge is spread out over a significantly\u00a0 larger volume:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203553\/fig7-3-3.png\" alt=\"\" width=\"278\" height=\"144\" \/><\/p>\n<p>This illustrates a fundamental concept in organic chemistry:<\/p>\n<div>\n<div class=\"textbox key-takeaways\"><em><strong>Electrostatic charges, whether positive or negative, are more stable when they are \u2018spread out\u2019 over a larger area.<\/strong><\/em><\/div>\n<\/div>\n<p>We will see this idea expressed again and again throughout our study of organic reactivity, in many different contexts.\u00a0\u00a0 For now, we are applying the concept only to the influence of atomic radius on base strength.\u00a0 Because fluoride is the least stable (most basic) of the halide conjugate bases, HF is the least acidic of the haloacids, only slightly stronger than a carboxylic acid.\u00a0 HI, with a pK<sub>a<\/sub> of about -9,\u00a0 is almost as strong as sulfuric acid.<\/p>\n<p>More importantly to the study of biological organic chemistry, this trend tells us that <strong><em>thiols are more acidic than alcohols<\/em><\/strong>. The pK<sub>a<\/sub> of the thiol group on the cysteine side chain, for example, is approximately 8.3, while the pK<sub>a<\/sub> for the alcohol group on the serine side chain is on the order of 17.<\/p>\n<p>Remember the concept of &#8216;driving force&#8217; that we learned about in chapter 6?\u00a0 Recall that the driving force for a reaction is usually based on two factors: relative charge stability, and relative total bond energy.\u00a0 Let&#8217;s see how this applies to a simple acid-base reaction between hydrochloric acid and fluoride ion:<\/p>\n<p>HCl + F<sup>&#8211;<\/sup>\u00a0\u2192 HF + Cl-<\/p>\n<p>We know that HCl (pK<sub>a<\/sub> -7) is a stronger acid than HF (pK<sub>a <\/sub>3.2), so the equilibrium for the reaction lies on the product side: the reaction is exergonic, and a &#8216;driving force&#8217; pushes reactant to product.<\/p>\n<p>What explains this driving force?\u00a0 Consider first the charge factor: as we just learned, chloride ion (on the product side) is more stable than fluoride ion (on the reactant side).\u00a0 This partially accounts for the driving force going from reactant to product in this reaction: we are going from less stable ion to a more stable ion.<\/p>\n<p>What about total bond energy, the other factor in driving force?\u00a0 If you consult a <a class=\"mt-disabled\" title=\"Organic_Chemistry\/Organic_Chemistry_With_a_Biological_Emphasis\/Chapter_07:_Organic_compounds_as_acids_and_bases\/Section_7.3:_Structural_effects_on_acidity_and_basicityhttp:\/\/chemwiki.ucdavis.edu\/Theoretical_Chemistry\/Chemical_Bonding\/General_Principles_of_Chemical_Bonding\/Bond_Energies\" target=\"_blank\" rel=\"broken\">table of bond energies<\/a>, you will see that the H-F bond on the product side is more energetic (stronger) than the H-Cl bond on the\u00a0 reactant side: 565 kJ\/mol <em>vs<\/em> 427 kJ\/mol, respectively).\u00a0 This also contributes to the driving force: we are moving from a weaker (less stable) bond to a stronger (more stable) bond.<\/p>\n<\/div>\n<div class=\"mt-section\">\n<h3 id=\"B:_Resonance_effects-21358\">B: Resonance effects<\/h3>\n<p>In the previous section we focused our attention on periodic trends &#8211; the differences in acidity and basicity between groups where the exchangeable proton was bound to different elements.\u00a0 Now, it is time to think about how the structure of different organic groups contributes to their relative acidity or basicity, even when we are talking about <em>the same element acting as the proton donor\/acceptor<\/em>.\u00a0 The first model pair we will consider is ethanol and acetic acid, but the conclusions we reach will be equally valid for all alcohol and carboxylic acid groups.<\/p>\n<p>Despite the fact that they are both oxygen acids, the pK<sub>a<\/sub> values of ethanol and acetic acid are strikingly different.\u00a0 What makes a carboxylic acid so much more acidic than an alcohol?\u00a0 As before, we begin by considering the stability of the conjugate bases, remembering that a\u00a0 more stable (weaker) conjugate base corresponds to a stronger acid.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203557\/fig7-3-4.png\" alt=\"\" width=\"539\" height=\"293\" \/><\/p>\n<p>In both species, the negative charge on the conjugate base is located on oxygen, so periodic trends cannot be invoked.\u00a0 For acetic acid, however, there is a key difference: two resonance contributors can be drawn for the conjugate base, and the negative charge can be delocalized (shared) over two oxygen atoms. In the ethoxide ion, by contrast, the negative charge is localized, or \u2018locked\u2019 on the single oxygen \u2013 it has nowhere else to go.\u00a0 This makes the ethoxide ion much less stable.<\/p>\n<p>Recall the important general statement that we made a little earlier: &#8216;Electrostatic charges, whether positive or negative, are more stable when they are \u2018spread out\u2019 than when they are confined to one location.&#8217;\u00a0 Now, we are seeing this concept in another context, where a charge is being \u2018spread out\u2019 (in other words, delocalized) <em>by resonance<\/em>, rather than simply by the size of the atom involved.<\/p>\n<p>The delocalization of charge by resonance has a very powerful effect on the reactivity of organic molecules, enough to account for the difference of over 12 pK<sub>a<\/sub> units between ethanol and acetic acid (and remember, pK<sub>a<\/sub> is a log expression, so we are talking about a factor of 10<sup>12<\/sup> between the K<sub>a<\/sub> values for the two molecules!)<\/p>\n<p>The resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but <em>not<\/em> basic when it is part of an amide group.\u00a0 Recall that in an amide, there is significant double-bond character to the carbon-nitrogen bond, due to a minor but still important resonance contributor in which the nitrogen lone pair is part of a pi bond.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203600\/fig7-3-5.png\" alt=\"\" width=\"581\" height=\"317\" \/><\/p>\n<p>Whereas the lone pair of an amine nitrogen is \u2018stuck\u2019 in one place, the lone pair on an amide nitrogen is delocalized by resonance.\u00a0 Notice that in this case, we are extending our central statement to say that <em>electron density<\/em> \u2013 in the form of a lone pair \u2013 is stabilized by resonance delocalization, even though there is not a negative charge involved.\u00a0 Here\u2019s another way to think about it: the lone pair on an amide nitrogen is not available for bonding with a proton \u2013 these two electrons are too \u2018comfortable\u2019 being part of the delocalized pi bonding system.\u00a0 The lone pair on an amine nitrogen, by contrast, is not so comfortable &#8211; it is <em>not<\/em> part of a delocalized pi system, and is available to form a bond with any acidic proton that might be nearby.<\/p>\n<p>If an amide group is protonated, it will be at the oxygen rather than the nitrogen.<\/p>\n<div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>1. a) Draw the Lewis structure of nitric acid, HNO<sub>3<\/sub>.<\/p>\n<p>b) Nitric acid is a strong acid &#8211; it has a pK<sub>a<\/sub> of -1.4.\u00a0 Make a structural argument to account for its strength. Your answer should involve the structure of nitrate, the conjugate base of nitric acid.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q920451\">Show Solution<\/span><\/p>\n<div id=\"q920451\" class=\"hidden-answer\" style=\"display: none\">\n<p>The negative charge on the oxygen that results from deprotonation of the acid is delocalized by resonance.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61509\/figES7-3-1.png?revision=1\" alt=\"\" width=\"312\" height=\"192\" \/><\/p>\n<\/div>\n<\/div>\n<p>2. Rank the compounds below from most acidic to least acidic, and explain your reasoning.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203603\/figE7-3-1.png\" alt=\"\" width=\"516\" height=\"84\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q581719\">Show Solution<\/span><\/p>\n<div id=\"q581719\" class=\"hidden-answer\" style=\"display: none\">\n<p>A and B are ammonium groups, while C is an amine, so C is clearly the least acidic.\u00a0 Looking at the conjugate base of B, we see that the lone pair electrons can be delocalized by resonance, making this conjugate base more stable than the conjugate base of A, where the electrons cannot be stabilized by resonance.\u00a0\u00a0 Thus B is the most acidic.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61511\/figES7-3-2a.png?revision=1\" alt=\"\" width=\"379\" height=\"71\" \/><\/p>\n<\/div>\n<\/div>\n<p>3. (challenging!) Often it requires some careful thought to predict the most acidic proton on a molecule.\u00a0 Ascorbic acid, also known as Vitamin C, has a pK<sub>a<\/sub> of 4.1 &#8211; the fact that this is in the range of carboxylic acids suggest to us that the negative charge on the conjugate base can be delocalized by resonance to <em>two<\/em> oxygen atoms.\u00a0 Which if the four OH protons on the molecule is most acidic?\u00a0 Draw the structure of ascorbate, the conjugate base of ascorbic acid, then draw a second resonance contributor showing how the negative charge is delocalized to a second oxygen atom.\u00a0 <em>Hint<\/em> &#8211; try removing each OH group in turn, then use your resonance drawing skills to figure out whether or not delocalization of charge can occur.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203605\/figE7-3-3.png\" alt=\"\" width=\"146\" height=\"172\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q467594\">Show Solution<\/span><\/p>\n<div id=\"q467594\" class=\"hidden-answer\" style=\"display: none\">\n<p>Below is the structure of ascorbate, the conjugate base of ascorbic acid.\u00a0 Note that the negative charge can be delocalized by resonance to two oxygen atoms, which makes ascorbic acid similar in strength to carboxylic acids.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61512\/figES7-3-3.png?revision=1\" alt=\"\" width=\"457\" height=\"129\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"mt-section\">\n<p>Resonance effects involving aromatic structures can have a dramatic influence on acidity and basicity.\u00a0 Notice, for example, the difference in acidity between phenol and cyclohexanol.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203619\/fig7-4-1.png\" alt=\"\" width=\"182\" height=\"130\" \/><\/p>\n<p>Looking at the conjugate base of phenol, we see that the negative charge can be delocalized by resonance to three different carbons on the aromatic ring.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203622\/fig7-4-2.png\" alt=\"\" width=\"399\" height=\"261\" \/><\/p>\n<p>Although these are all minor resonance contributors (negative charge is placed on a\u00a0 carbon rather than the more electronegative oxygen), they nonetheless have a significant effect on the acidity of the phenolic proton.\u00a0 Essentially, the benzene ring is acting as an electron-withdrawing group by resonance.<\/p>\n<div>\n<div class=\"textbox exercises\">\n<h3>Exercise<\/h3>\n<p>Draw the conjugate base of 2-napthol (the major resonance contributor), and on your drawing indicate with arrows all of the atoms to which the negative charge can be delocalized by resonance.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203624\/figE7-4-1.png\" alt=\"\" width=\"114\" height=\"80\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q114716\">Show Solution<\/span><\/p>\n<div id=\"q114716\" class=\"hidden-answer\" style=\"display: none\">\n<p>The negative charge can be delocalized by resonance to five carbons:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61549\/figES7-4-1a.png?revision=1\" alt=\"\" width=\"272\" height=\"190\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The base-stabilizing effect of an aromatic ring can be accentuated by the presence of an additional electron-withdrawing substituent, such as a carbonyl.\u00a0 For the conjugate base of the phenol derivative below, an additional resonance contributor can be drawn in which the negative formal charge is placed on the carbonyl oxygen.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203627\/fig7-4-3.png\" alt=\"\" width=\"452\" height=\"177\" \/><\/p>\n<p>Now the negative charge on the conjugate base can be spread out over <em>two<\/em> oxygens (in addition to three aromatic carbons). The phenol acid therefore has a pK<sub>a<\/sub> similar to that of a carboxylic acid, where the negative charge on the conjugate base is also delocalized to two oxygen atoms.\u00a0 The ketone group is acting as an electron withdrawing group &#8211; it is &#8216;pulling&#8217; electron density towards itself, through both inductive and resonance effects.<\/p>\n<div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>The position of the electron-withdrawing substituent relative to the phenol hydroxyl is very important in terms of its effect on acidity. Which of the two substituted phenols below is more acidic?\u00a0 Use resonance drawings to explain your answer.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203630\/figE7-4-2.png\" alt=\"\" width=\"251\" height=\"104\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q981365\">Show Solution<\/span><\/p>\n<div id=\"q981365\" class=\"hidden-answer\" style=\"display: none\">\n<p>When the aldehyde is in the 4 (<em>para<\/em>) position, the negative charge on the conjugate base can be delocalized to two oxygen atoms. In the other compound, the aldehyde is on the 3 (<em>meta<\/em>) position, and the negative charge cannot be delocalized to the aldehyde oxygen.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61554\/figES7-4-2.png?revision=1\" alt=\"\" width=\"307\" height=\"153\" \/><\/p>\n<\/div>\n<\/div>\n<p>2. Rank the four compounds below from most acidic to least.<img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203632\/figE7-4-3.png\" alt=\"\" width=\"440\" height=\"93\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q710916\">Show Solution<\/span><\/p>\n<div id=\"q710916\" class=\"hidden-answer\" style=\"display: none\">\n<p>The order of acidity, going from left to right (with 1 being most acidic), is 2-1-4-3.\u00a0 The least acidic compound (second from the right) has no phenol group at all &#8211; aldehydes are not acidic.\u00a0 The most acidic compound (second from the left) is a phenol with an aldehyde in the 2 (<em>ortho<\/em>) position, and as a consequence the negative charge on the conjugate base can be delocalized to both oxygen atoms. \u00a0 In the compound with the aldehyde in the 3 (<em>meta<\/em>) position, there is an electron-withdrawing inductive effect, but NOT a resonance effect (the negative charge on the cannot be delocalized to the aldehyde oxygen).<\/p>\n<\/div>\n<\/div>\n<p>3. Nitro groups are very powerful electron-withdrawing groups.\u00a0 The phenol derivative picric acid (2,4,6 -trinitrophenol) has a pK<sub>a<\/sub> of 0.25, lower than that of trifluoroacetic acid. Use a resonance argument to explain why picric acid has such a low pKa.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q475315\">Show Solution<\/span><\/p>\n<div id=\"q475315\" class=\"hidden-answer\" style=\"display: none\">\n<p>The negative charge on the conjugate base of picric acid can be delocalized to three different nitro oxygen atoms (in addition to the phenolate oxygen).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/61555\/figES7-4-4.png?revision=1\" alt=\"\" width=\"645\" height=\"196\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3 id=\"C:_Inductive_effects-21358\">C: Inductive effects<\/h3>\n<p>Compare the pK<sub>a<\/sub> values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203608\/fig7-3-6.png\" alt=\"\" width=\"463\" height=\"95\" \/><\/p>\n<p>The presence of the chlorine atoms clearly increases the acidity of the carboxylic acid group, but the argument here does not have to do with resonance delocalization, because no additional resonance contributors can be drawn for the chlorinated molecules.\u00a0 Rather, the explanation for this phenomenon involves something called the <strong>inductive effect<\/strong>.\u00a0 A chlorine atom is more electronegative than a hydrogen, and thus is able to \u2018induce\u2019, or \u2018pull\u2019 electron density towards itself, away from the carboxylate group.\u00a0 In effect, the chlorine atoms are helping to further spread out the electron density of the conjugate base, which as we know has a stabilizing effect.\u00a0 In this context, the chlorine substituent can be referred to as an <strong>electron-withdrawing group<\/strong>. Notice that the pKa-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pK<sub>a<\/sub> values between an alcohol and a carboxylic acid.\u00a0 In general, <em>resonance effects are more powerful than inductive effects<\/em>.<\/p>\n<p>Because the inductive effect depends on electronegativity, fluorine substituents have a more pronounced\u00a0 pKa-lowered effect than chlorine substituents.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203610\/fig7-3-7.png\" alt=\"\" width=\"288\" height=\"122\" \/><\/p>\n<p>In addition, the\u00a0 inductive takes place through covalent bonds, and its influence decreases markedly with distance \u2013 thus a chlorine two carbons away from a carboxylic acid group has a decreased effect compared to a chlorine just one carbon away.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203613\/fig7-3-8.png\" alt=\"\" width=\"297\" height=\"98\" \/><\/p>\n<div>\n<div class=\"textbox exercises\">\n<h3>Exercise<\/h3>\n<p>Rank the compounds below from most acidic to least acidic, and explain your reasoning.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203617\/figE7-3-4.png\" alt=\"\" width=\"275\" height=\"129\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q891926\">Show Solution<\/span><\/p>\n<div id=\"q891926\" class=\"hidden-answer\" style=\"display: none\">\n<p>We must consider the electronegativity and the position of the halogen substituent in terms of inductive effects.\u00a0 A is the strongest acid, as <a title=\"Section 2.3: Non-covalent interactions\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II\/2.4%3A_Non-covalent_interactions\" target=\"_blank\" rel=\"internal noopener\">chlorine is more electronegative than bromine.<\/a>\u00a0 B is more acidic than C, as the bromine is closer (in terms of the number of bonds) to the site of acidity.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h3>Combinations of effects<\/h3>\n<p>Consider the acidity of 4-methoxyphenol, compared to phenol:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203635\/fig7-4-4.png\" alt=\"\" width=\"210\" height=\"166\" \/><\/p>\n<p>Notice that the methoxy group increases the pK<sub>a<\/sub> of the phenol group &#8211; it makes it <em>less<\/em> acidic. Why is this? At first inspection, you might assume that the methoxy substituent, with its electronegative oxygen, would be an electron-withdrawing group by induction.\u00a0 That is correct, but only to a point.\u00a0 The oxygen atom does indeed exert an electron-withdrawing inductive effect, but the lone pairs on the oxygen cause the exact opposite effect \u2013 <em>the methoxy group is an <strong>electron-donating group by resonance<\/strong><\/em>.\u00a0 A resonance contributor can be drawn in which a formal negative charge is placed on the carbon adjacent to the negatively-charged phenolate oxygen.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203637\/fig7-4-5.png\" alt=\"\" width=\"237\" height=\"162\" \/><\/p>\n<p>Because of like-charge repulsion, this <em>destabilizes<\/em> the negative charge on the phenolate oxygen, making it more basic. It may help to visualize the methoxy group \u2018pushing\u2019 electrons towards the lone pair electrons of the phenolate oxygen, causing them to be less &#8216;comfortable&#8217; and more reactive. The example above is a somewhat confusing but quite common situation in organic chemistry &#8211; a functional group, in this case a methoxy group, is exerting both an inductive effect and a resonance effect, <em>but in opposite directions<\/em> (the inductive effect is electron-withdrawing, the resonance effect is electron-donating).\u00a0 As a general rule a resonance effect is more powerful than an inductive effect &#8211; so overall, the methoxy group is acting as an electron donating group. A good rule of thumb to remember:<\/p>\n<div class=\"textbox key-takeaways\">\n<p style=\"text-align: center\"><strong>When resonance and induction compete, resonance usually wins!<\/strong><\/p>\n<\/div>\n<div>\n<div class=\"textbox exercises\">\n<h3>Exercise<\/h3>\n<p>Rank the three compounds below from lowest pKa to highest, and explain your reasoning.\u00a0 <em>Hint<\/em> &#8211; think about both resonance and inductive effects!<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203639\/figE7-4-5.png\" alt=\"\" width=\"339\" height=\"117\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q951666\">Show Solution<\/span><\/p>\n<div id=\"q951666\" class=\"hidden-answer\" style=\"display: none\">\n<p>Compound C has the lowest pK<sub>a<\/sub> (most acidic): the oxygen acts as an electron withdrawing group by induction.\u00a0 Compound A has the highest pK<sub>a<\/sub> (the oxygen is in a position to act as an electron donating group by resonance, thus destabilizing the negative charge of the conjugate base).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Further reading<\/h3>\n<ul>\n<li><a href=\"https:\/\/chem.libretexts.org\/LibreTexts\/Purdue\/Purdue_Chem_26100%3A_Organic_Chemistry_I_(Wenthold)\/Chapter_01%3A_Introduction_and_Review\/1.8_Acids_and_Bases\/Overview_of_Acids_and_Bases\">Overview of Acids and Bases<\/a><\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4634 alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/01151246\/static_qr_code_without_logo1-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/p>\n<ul>\n<li><a href=\"https:\/\/chem.libretexts.org\/Homework_Exercises\/Exercises%3A_Organic_Chemistry\/Organic%3A_Acid%2F%2FBase_Practice_Problems\">Base Practice problems<\/a><\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-thumbnail wp-image-4635 alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/01151706\/static_qr_code_without_logo2-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2569\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>1.22 How Substituents Affect the Strength of an Acid. <strong>Authored by<\/strong>: Tim Soderbergu00a0(University of Minnesota, Morris). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Map%3A_Organic_Chemistry_(Bruice)\/01._Electronic_Structure_and_Bonding_(Acids_and_Bases)\/1.22____How_Substituents_Affect_the_Strength_of_an_Acid\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Map%3A_Organic_Chemistry_(Bruice)\/01._Electronic_Structure_and_Bonding_(Acids_and_Bases)\/1.22____How_Substituents_Affect_the_Strength_of_an_Acid<\/a>. <strong>Project<\/strong>: Chemistry LibreTexts. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><li>7.4: Acid-base properties of phenols. <strong>Authored by<\/strong>: Tim Soderbergu00a0(University of Minnesota, Morris). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_07%3A_Organic_compounds_as_acids_and_bases\/7.4%3A_Acid-base_properties_of_phenols\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_07%3A_Organic_compounds_as_acids_and_bases\/7.4%3A_Acid-base_properties_of_phenols<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":311,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"1.22 How Substituents Affect the Strength of an Acid\",\"author\":\"Tim Soderbergu00a0(University of Minnesota, Morris)\",\"organization\":\"\",\"url\":\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Map%3A_Organic_Chemistry_(Bruice)\/01._Electronic_Structure_and_Bonding_(Acids_and_Bases)\/1.22____How_Substituents_Affect_the_Strength_of_an_Acid\",\"project\":\"Chemistry LibreTexts\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"7.4: Acid-base properties of phenols\",\"author\":\"Tim Soderbergu00a0(University of Minnesota, Morris)\",\"organization\":\"\",\"url\":\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_07%3A_Organic_compounds_as_acids_and_bases\/7.4%3A_Acid-base_properties_of_phenols\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2569","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2569","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2569\/revisions"}],"predecessor-version":[{"id":4799,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2569\/revisions\/4799"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/23"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2569\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/media?parent=2569"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=2569"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/contributor?post=2569"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/license?post=2569"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}