{"id":385,"date":"2017-10-04T15:42:49","date_gmt":"2017-10-04T15:42:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&p=385"},"modified":"2017-10-24T16:05:19","modified_gmt":"2017-10-24T16:05:19","slug":"factors-that-determine-acid-strength","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/factors-that-determine-acid-strength\/","title":{"raw":"Factors That Determine Acid Strength","rendered":"Factors That Determine Acid Strength"},"content":{"raw":"
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\u00a0Periodic trends<\/h2>\r\n<\/div>\r\n
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<\/div>\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 methanol, but the concepts apply equally to more complex biomolecules, such as the side chains of alanine, lysine, and serine.\r\n\r\n\"image048.png\"\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: 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 ethyl anion, the negative charge is borne by carbon, while in the methylamine anion and methoxide anion the charges are located on a nitrogen and an oxygen, respectively.\u00a0 Remember the periodic trend in electronegativity (section 2.3A): it also increases as we move from left to right along a row, meaning that oxygen is the most electronegative of the three, and carbon the least.\u00a0 The more electronegative an atom, the better it is able to bear a negative charge<\/em>.\u00a0 Thus, the methoxide anion is the most stable (lowest energy, least basic) of the three conjugate bases, and the ethyl anion is the least stable (highest energy, most basic).\r\n\r\nWe can use the same set of ideas to explain the difference in basicity between water and ammonia.\r\n\r\n\"image050.png\"\r\n\r\nBy looking at the pKa<\/sub>values for the appropriate conjugate acids, we know that ammonia is more basic than water. Oxygen, as the more electronegative element, holds more tightly to its lone pair\u00a0 than the nitrogen.\u00a0 The nitrogen lone pair, therefore, is more likely to break away and form a new bond to a proton - it is, in other words, more basic. Once again, a more reactive (stronger) conjugate base means a less reactive (weaker) conjugate 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 halides: basicity, like electronegativity,\u00a0 increases as we move up the column.\r\n\r\n\"image052.png\"\r\n\r\nConversely, acidity in the haloacids increases as we move down<\/em> the column.\r\n\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 least<\/em> stable, and the most basic!\u00a0\u00a0 It turns out that when moving vertically in the periodic table, the 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 iodine ion, the negative charge is spread out over a significantly\u00a0 larger volume:\r\n\r\n\"image054.png\"\r\n\r\nThis illustrates a fundamental concept in organic chemistry that is important enough to put in red:\r\n
\r\n\r\nElectrostatic charges, whether positive or negative, are more stable when they are \u2018spread out\u2019 than when they are confined to one atom<\/em><\/strong>.<\/em>\r\n\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, the concept is applied only to the influence of atomic radius on anion stability.\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 acetic acid.\u00a0 HI, with a pKa<\/sub> of about -9,\u00a0 is one the strongest acids known.\r\n\r\nMore importantly to the study of biological organic chemistry, this trend tells us that thiols are more acidic than alcohols.\u00a0 The pKa<\/sub> of the thiol\u00a0 group on the cysteine side chain, for example, is approximately 8.3, while the pKa<\/sub> for the hydroxl on the serine side chain is on the order of 17.\r\n\r\nTo reiterate: acid strength increases as we move to the right along a row of the periodic table, and as we move down a column.\r\n\r\n\"image056.png\"\r\n
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Example<\/h3>\r\nDraw the structure of the conjugate base that would form if the compound below were to react with 1 molar equivalent of sodium hydroxide:\r\n\r\n\"image058.png\"\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n<\/div>\r\n
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The resonance effect<\/h3>\r\n
<\/div>\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 the same element acting as the proton donor\/acceptor.\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 pKa<\/sub> values of ethanol and acetic acid are very different.\u00a0 What makes a carboxylic acid so much more acidic than an alcohol?\u00a0 As before, we begin by considering the conjugate bases.\r\n\r\n\"image060.png\"\r\n\r\nIn both species, the negative charge on the conjugate base is held by an oxygen, so periodic trends cannot be invoked. For acetic acid, however, there is a key difference: a resonance contributor can be drawn in which the negative charge is localized on the second oxygen of the group. The two resonance forms for the conjugate base are equal in energy, according to our \u2018rules of resonance\u2019 (section 2.2C<\/a>). What this means, you may recall, is that the negative charge on the acetate ion is not located on one oxygen or the other: rather it is shared between the two.\u00a0 Chemists use the term \u2018delocalization of charge\u2019 to describe this situation. In the ethoxide ion, by contrast, the negative charge is \u2018locked\u2019 on the single oxygen \u2013 it has nowhere else to go.\r\n\r\nNow is the time to think back to that statement from the previous section that was so important that it got printed in bold font in its own paragraph \u2013 in fact, it is so important that we\u2019ll just say it again: \"Electrostatic charges, whether positive or negative, are more stable when they are \u2018spread out\u2019 than when they are confined to one atom.\"\u00a0 Now, we are seeing this concept in another context, where a charge is being \u2018spread out\u2019 (in other words, delocalized) 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 pKa<\/sub> units between ethanol and acetic acid (and remember, pKa<\/sub> is a log expression, so we are talking about a difference of over 1012<\/sup> between the acidity constants for the two molecules).\u00a0 The acetate ion is that much more stable than the ethoxide ion, all due to the effects of resonance delocalization.\r\n\r\nThe resonance effect also nicely explains why a nitrogen atom is basic when it is in an amine, but 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 second resonance contributor in which the nitrogen lone pair is part of a p bond.\r\n\r\n\"image062.png\"\r\n\r\nWhile the electron 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 electron density \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 part of a delocalized p system, and is very ready to form a bond with any acidic proton that might be nearby.\r\n\r\nOften it requires some careful thought to predict the most acidic proton on a molecule.\u00a0 Ascorbic acid, also known as Vitamin C, has a pKa<\/sub> of 4.1.\r\n\r\n\"image064.png\"\r\n\r\nThere are four hydroxyl groups on this molecule \u2013 which one is the most acidic?\u00a0 If we consider all four possible conjugate bases, we find that there is only one for which we can delocalized the negative charge over two<\/em> oxygen atoms.\r\n
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Example<\/h3>\r\nRank the compounds below from most acidic to least acidic, and explain your reasoning.\r\n\r\n\"image066.png\"\r\n
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The inductive effect<\/h3>\r\n
<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nCompare the pKa<\/sub> values of acetic acid and its mono-, di-, and tri-chlorinated derivatives:\r\n\r\n\"image068.png\"\r\n\r\nThe presence of the chlorines 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. Rather, the explanation for this phenomenon involves something called the inductive effect<\/strong>. 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 is called an electron-withdrawing group<\/strong>. Notice that the pKa<\/sub>-lowering effect of each chlorine atom, while significant, is not as dramatic as the delocalizing resonance effect illustrated by the difference in pKa<\/sub> values between an alcohol and a carboxylic acid.\u00a0 In general, resonance effects are more powerful than inductive effects<\/em>.\r\n\r\nThe\u00a0 inductive electron-withdrawing effect of the chlorines takes place through covalent bonds, and its influence decreases\u00a0 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
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Example<\/h3>\r\nExampleRank the compounds below from most acidic to least acidic, and explain your reasoning.\r\n\r\n\"image070.png\"\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n\r\n

Contributors<\/h3>\r\n