{"id":2653,"date":"2018-06-19T20:39:24","date_gmt":"2018-06-19T20:39:24","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/24-nucleophilic-substitution-sn2-sn1\/"},"modified":"2020-06-23T17:08:49","modified_gmt":"2020-06-23T17:08:49","slug":"8-3-factors-affecting-rate-of-nucleophilic-substitution-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/8-3-factors-affecting-rate-of-nucleophilic-substitution-reactions\/","title":{"raw":"8.3. Factors affecting rate of nucleophilic substitution reactions","rendered":"8.3. Factors affecting rate of nucleophilic substitution reactions"},"content":{"raw":"<section class=\"mt-content-container\">\r\n<div class=\"mt-section\">\r\n<h1>Designing a \"good\" nucleophilic substitution<\/h1>\r\nIf you want to do well in this class, there are several things you need to work hard at: Being attentive in class, studying the notes and this textbook (especially before exams), practicing problems, and completing the quizzes and homeworks.\u00a0 As long as you do all of these things then you're likely to pass (though I can't give guarantees!).\u00a0 So there are many different factors that can affect your grade.\u00a0 In the same way, the outcome of a reaction (such as nucleophilic substition) depends on many different things - reactants, solvent, etc.\u00a0 When we want to make a chemical in a lab or on a chemical plant, we need to design the reaction so that it works well, and gives a good yield of the product in a reasonable time.\u00a0 The reactants and conditions we use will depend on what we're trying to do.\u00a0 In this section, we examine what factors will help an S<sub>N<\/sub>2 or S<sub>N<\/sub>1 reaction be successful.\r\n<h1>Factors affecting the S<sub>N<\/sub>2 reaction<\/h1>\r\n<p class=\"mt-align-justify\">As we saw in the previous section, in the S<sub>N<\/sub>2 reaction the rate of reaction depends on both the electrophile (usually an alkyl halide) and the nucleophile.<\/p>\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203831\/image009-2.png\" alt=\"image010.png\" width=\"393\" height=\"84\" \/>\r\n\r\nIn practice, the rates of S<sub>N<\/sub>2 reactions vary enormously, and for a practicable procedure we need to make sure that the reaction will happen at a reasonable rate.\u00a0 So what makes for a good S<sub>N<\/sub>2 reaction?\u00a0 We need to consider what makes a suitable nucleophile, and what makes a suitable electrophile.\r\n<h2>Nucleophile strength<\/h2>\r\nIn <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/6-5-lewis-acids-bases\/\">section 6.5<\/a>, we learnt what makes a nucleophile strong (reactive) or weak (unreactive).\u00a0 Anything which removes electron-density from the nucleophilic atom will make it less nucleophilic.\u00a0 We summarized the main points from 6.5 as follows:\r\n<ul>\r\n \t<li>Charge - negatively charged =&gt; stronger nucleophile<\/li>\r\n \t<li>Within a row - more electronegative atom =&gt; weaker nucleophile<\/li>\r\n \t<li>Within a column, size of atom.\u00a0 Polar protic solvent, bigger atom is better; polar aprotic solvent, smaller atom is better.<\/li>\r\n \t<li>Resonance - if the nucleophilic lone pair can be delocalized by resonance, it will make it less nucleophilic<\/li>\r\n \t<li>Steric hindrance - a hindered nucleophilic atom will tend to be less reactive, particularly when attacking a crowded electrophile.<\/li>\r\n<\/ul>\r\nRegarding the solvent, polar aprotic solvents such as DMSO, DMF, acetone or acetonitrile are popular choices for S<sub>N<\/sub>2 reactions, because rates are generally faster than with polar protic solvents (water, alcohols, etc.).\u00a0 This is because the nucleophile is almost \"naked\" in aprotic solvents, whereas in polar protic solvents it is surrounded by a cage of solvent molecules.\r\n\r\nIf we have a strong nucleophile, the S<sub>N<\/sub>2 reaction will happen faster; a weak nucleophile will react more slowly and may not even react.\u00a0 So in general we want a strong nucleophile.\r\n\r\n<\/div>\r\n<h2 class=\"mt-section\">The electrophile<\/h2>\r\n<h3 class=\"mt-section\">(a) Structure of the alkyl group<\/h3>\r\n<div class=\"mt-section\">\r\n\r\nIn the structure of the S<sub>N<\/sub>2 transition state, there are 90<sup>o<\/sup>\u00a0bond angles between the breaking bond to the leaving group and the three bonds which remain connected to the carbon as well as between the bond being made to the nucleophile and those same three bonds.\r\n<p class=\"mt-align-center\"><img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203918\/SN2ster1.gif\" alt=\"\" width=\"159\" height=\"142\" \/><\/p>\r\n<p class=\"mt-align-justify\">As long as the two of the groups attached to the carbon being attacked are small hydrogens, the repulsions which happen do not require much energy. If the groups attached to the carbon are larger, though, like methyl groups, the transition state energy increases, the activation energy increases, and the reaction becomes much slower.<\/p>\r\n<p class=\"mt-align-center\"><img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203920\/SN2ster2.gif\" alt=\"\" width=\"444\" height=\"136\" \/><\/p>\r\n<p class=\"mt-align-justify\">This means that the reactivity order for alkyl halides in S<sub>N<\/sub>2 reactions is:<\/p>\r\n<p class=\"mt-align-center\"><b>methyl &gt; primary &gt; secondary &gt; tertiary<\/b><\/p>\r\n<p class=\"mt-align-justify\">The practical outcome of this is that\u00a0S<sub>N<\/sub>2 reactions are generally reliable only when the alkyl halide is primary, though under the correct conditions secondary halides can react well also.<\/p>\r\n\r\n<\/div>\r\n<\/section>\r\n<h3 class=\"editable\">(b) Leaving group ability - what makes a good leaving group?<\/h3>\r\n<div class=\"mt-section\">\r\n\r\nIn our general discussion of nucleophilic substitution reactions, we have until now been designating the leaving group simply as \u201cX\". \u00a0As you\u00a0 may imagine, however, the nature of the leaving group is an important consideration: if the C-X bond does not break, the new bond between the nucleophile and electrophilic carbon cannot form, regardless of whether the substitution is S<sub>N<\/sub>1 or S<sub>N<\/sub>2.\u00a0 There are two main factors: The strength of the C-X bond, and the stability of the X group after it has left.\u00a0 It turns out that the two factors lead to the same prediction for halogen leaving group ability:\r\n<p style=\"text-align: center\"><big>I &gt; Br &gt; Cl &gt; F<\/big><\/p>\r\n\r\n<h4 class=\"mt-align-justify\">C-X bond strength<\/h4>\r\n<p class=\"mt-align-justify\">Since the bond between the carbon and the leaving group is being broken in the transition state, the weaker this bond is the lower the activation energy and the faster the reaction. This leads to the following reactivity order for alkyl halides<\/p>\r\n<p class=\"mt-align-center\" style=\"text-align: center\"><b><img class=\"alignnone wp-image-5102\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/23165635\/AlkylHalideReactivity.png\" alt=\"Alkyl halide reactivity is RI&gt;RBr&gt;RCl&gt;RF\" width=\"321\" height=\"30\" \/><\/b><\/p>\r\n<p class=\"mt-align-justify\">Practically, alkyl fluorides are not used for S<sub>N<\/sub>2 reactions because the C-F bond is too strong. Often alkyl iodides are reactive enough to be difficult to store, so the the common choices for reactions are alkyl chlorides and alkyl bromides.<\/p>\r\n\r\n<h4>Stability of the group after leaving<\/h4>\r\nWhen the C-X bond breaks in a nucleophilic substitution, the pair of electrons in the bond goes with the leaving group.\u00a0 In this way, the leaving group is analogous to the conjugate base in a Br\u00f8nsted-Lowry acid-base reaction. When we were evaluating the strength of acids in chapter 7, what we were really doing was evaluating the stability of the conjugate base that resulted from the proton transfer.\u00a0 All of the concepts that we used to evaluate the stability of conjugate bases we can use again to evaluate leaving groups.\u00a0 In other words, the trends in basicity are parallel to the trends in leaving group potential - the weaker the base, the better the leaving group. Just as with conjugate bases, the most important question regarding leaving groups is this:\u00a0 when a leaving group leaves and takes a pair of electrons with it, how well is the extra electron density stabilized?\r\n\r\nIn laboratory synthesis reactions, halides often act as leaving groups.\u00a0 Iodide, which is the <em>least<\/em> basic of the four main halides, is also the <em>best<\/em> leaving group \u2013 it is the most stable as a negative ion. Fluoride is the least effective leaving group among the halides, because fluoride anion is the most basic.\r\n\r\nWe already know that the use of polar, aprotic solvents increases the reactivity of nucleophiles in S<sub>N<\/sub>2 reactions, because these solvents do not 'cage' the nucleophile and keep it from attacking the electrophile.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Factors favoring S<sub>N<\/sub>2<\/h3>\r\nTo design an effective S<sub>N<\/sub>2 reaction using an alkyl halide, we need:\r\n<ul>\r\n \t<li>An unhindered alkyl halide (preferably methyl or primary, but secondary may be possible)<\/li>\r\n \t<li>A good leaving group (preferably I or Br)<\/li>\r\n \t<li>A strong nucleophile<\/li>\r\n \t<li>A suitable solvent - polar aprotic is most effective<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1>Factors affecting the S<sub>N<\/sub>1 reaction<\/h1>\r\n<\/div>\r\n<div class=\"mt-section\">\r\n\r\nAs we learnt in <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/8-2-physical-chemistry-for-sn2-and-sn1-reactions\/\">section 8.2<\/a>, the nucleophile has no effect on the rate of an S<sub>N<\/sub>1 reaction.\u00a0 This means that we only need to consider the electrophile, usually an alkyl halide.\u00a0 Another feature of the S<sub>N<\/sub>1 reaction is that it is often prone to side reactions, which is why it is less used in synthesis than the S<sub>N<\/sub>2 reaction.\r\n<h2 class=\"mt-section\">The electrophile<\/h2>\r\nThis topic was examined in general in section 6.5., and also considered above for S<sub>N<\/sub>2.\u00a0 But a electrophile that is good for S<sub>N<\/sub>2 is not necessarily good for S<sub>N1<\/sub>, for reasons that will become clear.\u00a0 We also have a new factor to consider - the stability of the carbocation that is formed as a result of the heterolysis step.\r\n<h3 class=\"mt-section\">(a) Structure of the alkyl group<\/h3>\r\n<\/div>\r\nIn the vast majority of the nucleophilic substitution reactions you will see in this and other organic chemistry texts, the electrophilic atom is a carbon which is bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom is an attractive target for something that is electron-rich, <em>i.e<\/em>. a nucleophile. However, we must also consider the effect of steric hindrance on electrophilicity. In addition, we must discuss how the nature of the electrophilic carbon, and more specifically the stability of a potential carbocationic intermediate, influences the S<sub>N<\/sub>1 reaction.\r\n<div class=\"mt-section\">\r\n<h3 class=\"editable\">Steric effects on electrophilicity<\/h3>\r\nIn an S<sub>N<\/sub>1 mechanism, the nucleophile attacks an sp<sup>2<\/sup>-hybridized carbocation intermediate, which has trigonal planar geometry with \u2018open\u2019 120 angles.\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204048\/image049-3.png\" alt=\"image050.png\" width=\"495\" height=\"231\" \/>\r\n\r\nWith this open geometry, the empty p orbital of the electrophilic carbocation is no longer significantly shielded from the approaching nucleophile by the bulky alkyl groups. A carbocation is a very potent electrophile, and the nucleophilic step occurs very rapidly compared to the first (ionization) step.\u00a0 This is in direct contrast to the S<sub>N<\/sub>2 reaction, where bulky alkyl groups hinder the reaction.\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<h3 class=\"editable\">Stability of carbocation intermediates<\/h3>\r\nWe know that the rate-limiting step of an S<sub>N<\/sub>1 reaction is the first step - formation of the this carbocation intermediate.\u00a0 The rate of this step \u2013 and therefore, the rate of the overall substitution reaction \u2013 depends on the activation energy for the process in which the bond between the carbon and the leaving group breaks and a carbocation forms.\u00a0 According to Hammond\u2019s postulate (<a title=\"Organic Chemistry\/Organic Chemistry With a Biological Emphasis\/Chapter 6: Introduction to organic reactivity and catalysis\/Section 6.2: Energy diagrams\" href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/5-5-hammonds-postulate\/\" rel=\"internal\">section 5.5<\/a>), the more stable the carbocation intermediate is, the faster this first bond-breaking step will occur. In other words, the likelihood of a nucleophilic substitution reaction proceeding by a dissociative (S<sub>N<\/sub>1) mechanism depends to a large degree on the stability of the carbocation intermediate that forms.\r\n\r\nWe previously considered carbocation stability in <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/6-5-lewis-acids-bases\/\">section 6.5.<\/a>, and we found that in general, <em>more substituted carbocations are more stable<\/em>:\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204052\/image053-1.png\" alt=\"image054.png\" width=\"357\" height=\"150\" \/>\r\n\r\nA positively charged species such as a carbocation is very electron-poor, and thus anything which donates electron density to the center of electron poverty will help to stabilize it. Conversely, a carbocation will be <em>destabilized<\/em> by an electron withdrawing group.\r\n\r\nStabilization of a carbocation can also occur through <strong>resonance<\/strong>, which allows the \"burden\" of the negative charge to be delocalized, or shared, onto more than one atom. Resonance effects as a rule are more powerful than inductive effects.\u00a0 For example, a cation next to a double bond will delocalize the charge via type IV resonance, so an allylic carbocation is more stable than a comparable one that cannot do resonance:\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204058\/image059-2.png\" alt=\"image060.png\" width=\"289\" height=\"80\" \/>\r\n\r\nFinally, <strong>vinylic<\/strong> carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any S<sub>N<\/sub>1 reaction.\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204104\/image065-1.png\" alt=\"image066.png\" width=\"145\" height=\"114\" \/>\r\n\r\nWhen considering the possibility that a nucleophilic substitution reaction proceeds <em>via<\/em> an S<sub>N<\/sub>1 pathway, it is critical to evaluate the stability of the hypothetical carbocation intermediate.\u00a0 If this intermediate is not sufficiently stable, an S<sub>N<\/sub>1 mechanism must be considered unlikely, and the reaction probably proceeds by an S<sub>N<\/sub>2 mechanism.\r\n<div>\r\n<div>\r\n<h2>Leaving group<\/h2>\r\nThe C-X bond breaks in the rate determining step of S<sub>N<\/sub>1, just as it does in S<sub>N<\/sub>2, and in fact the rules are the same for determining a \"good\" leaving group.\u00a0 Again these are determined by the C-X bond strength and the stability of X after it has left.\u00a0 This means that we see the same trends as we did for S<sub>N<\/sub>2, where the larger halogens make better leaving groups, i.e.,\r\n<p style=\"text-align: center\"><big>I &gt; Br &gt; Cl &gt; F<\/big><\/p>\r\n\r\n<h2 class=\"mt-section\">Side reactions in S<sub>N<\/sub>1<\/h2>\r\n<h4 class=\"mt-section\">(a) Elimination<\/h4>\r\nIn all of our discussion so far about nucleophilic substitutions, we have ignored another important possibility.\u00a0 In many cases, including the two examples above, substitution reactions compete with a type of reaction known as <strong>elimination<\/strong>.\u00a0 This will be covered in detail soon, in <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/8-5-elimination-reactions\/\">section 8.5.<\/a>\u00a0 Consider, for example, the two courses that a reaction could take when 2-bromo-2-methylpropane reacts with water:\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204015\/image109.png\" alt=\"image110.png\" width=\"647\" height=\"234\" \/>\r\n\r\nWe begin with formation of the carbocation intermediate.\u00a0 In pathway \u2018a\u2019, water acts as a nucleophile \u2013 this is, of course, the familiar S<sub>N<\/sub>1 reaction.\u00a0 However, a water molecule encountering the carbocation intermediate could alternatively act as a <em>base<\/em> rather than as a nucleophile, plucking a proton from one of the methyl carbons and causing the formation of a new carbon-carbon p bond.\u00a0 This alternative pathway is called an elimination reaction, and in fact with the conditions above, both the substitution and the elimination pathways will occur in competition with each other.\u00a0 Elimination can be minimized by keeping the reaction cold, but some of this side-reaction is often inevitable.\r\n<h4>(b) Carbocation rearrangements<\/h4>\r\nThese will be covered very soon, in <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/8-4-rearrangements\/\">section 8.4.<\/a>\u00a0 If the carbocation can easily rearrange to a more stable carbocation, then rearrangement products are likely to be important, and the reaction may lead to mixtures.\r\n<h2>Solvent<\/h2>\r\nThe rates of S<sub>N<\/sub>1 reactions are generally increased by the use of a highly polar solvent, including protic (hydrogen bonding) solvents such as water or ethanol.\u00a0 In essence, a protic solvent increases the reactivity of the leaving group in an S<sub>N<\/sub>1 reaction, by helping to stabilize the products of the first (ionization) step. In the S<sub>N<\/sub>1 mechanism, remember, the rate determining step does <em>not<\/em> involve the nucleophilic species, so any reduction of nucleophilicity does not matter.\u00a0 What matters is that the charged products of the first step - the carbocation intermediate and the anionic leaving group - are stabilized best by a highly polar, protic solvent.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Factors favoring S<sub>N<\/sub>1<\/h3>\r\nTo design an effective S<sub>N<\/sub>1 reaction using an alkyl halide, we need:\r\n<ul>\r\n \t<li>A highly substituted alkyl halide (preferably tertiary or resonance-stabilized, but secondary may be possible), ideally one which will not lead to rearrangement<\/li>\r\n \t<li>A good leaving group (preferably I or Br)<\/li>\r\n \t<li>A non-basic nucleophile (to reduce the elimination side reaction)<\/li>\r\n \t<li>A suitable solvent - polar protic is most effective<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<h1>Predicting S<sub>N<\/sub>1 vs. S<sub>N<\/sub>2 mechanisms<\/h1>\r\n<div>\r\n\r\nWhen considering whether a nucleophilic substitution is likely to occur via an S<sub>N<\/sub>1 or S<sub>N<\/sub>2 mechanism, we really need to consider three factors:\r\n\r\n1) <strong>The electrophile:<\/strong> when the leaving group is attached to a methyl group or a primary carbon, an S<sub>N<\/sub>2 mechanism is favored (here the electrophile is unhindered by surrounded groups, and any carbocation intermediate would be high-energy and thus unlikely). When the leaving group is attached to a tertiary, allylic, or benzylic carbon, a carbocation intermediate will be relatively stable and thus an S<sub>N<\/sub>1 mechanism is favored.\r\n\r\n2) <strong>The nucleophile<\/strong>: powerful nucleophiles, especially those with negative charges, favor the S<sub>N<\/sub>2 mechanism. Weaker nucleophiles such as water or alcohols favor the S<sub>N<\/sub>1 mechanism.\r\n\r\n3) <strong>The solvent<\/strong>:\u00a0 Polar aprotic solvents favor the S<sub>N<\/sub>2 mechanism by enhancing the reactivity of the nucleophile. Polar protic solvents favor the S<sub>N<\/sub>1 mechanism by stabilizing the carbocation intermediate. S<sub>N<\/sub>1 reactions are frequently solvolysis reactions.\r\n\r\nFor example, the reaction below has a tertiary alkyl bromide as the electrophile, a weak nucleophile, and a polar protic solvent (we\u2019ll assume that methanol is the solvent).\u00a0 Thus we\u2019d confidently predict an S<sub>N<\/sub>1 reaction mechanism.\u00a0 Because substitution occurs at a chiral carbon, we can also predict that the reaction will proceed with racemization.\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\/19204006\/image106-1.png\" alt=\"image106.png\" width=\"642\" height=\"102\" \/>\r\n\r\nIn the reaction below, on the other hand, the electrophile is a secondary alkyl bromide \u2013 with these, both S<sub>N<\/sub>1 and S<sub>N<\/sub>2 mechanisms are possible, depending on the nucleophile and the solvent.\u00a0 In this example, the nucleophile (a thiolate anion) is strong, and a polar protic solvent is used \u2013 so the S<sub>N<\/sub>2 mechanism is heavily favored.\u00a0 The reaction is expected to proceed with inversion of configuration.\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\/19204009\/image108.png\" alt=\"image108.png\" width=\"557\" height=\"97\" \/>\r\n<div class=\"textbox examples\">\r\n<h3>Exercise<\/h3>\r\nDetermine whether each substitution reaction shown below is likely to proceed by an S<sub>N<\/sub>1 or S<sub>N<\/sub>2 mechanism.\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\/19204012\/image108a.png\" alt=\"image108a.png\" width=\"355\" height=\"188\" \/>\r\n\r\n[reveal-answer q=\"778981\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"778981\"]\r\n\r\na) S<sub>N<\/sub>2 (primary electrophile, strong nucleophile, polar aprotic solvent)\r\n\r\nb) S<sub>N<\/sub>1 (tertiary electrophile, weak nucleophile, protic solvent)\r\n\r\nc) S<sub>N<\/sub>2 (secondary electrophile, strong nucleophile, polar protic solvent)\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<section class=\"mt-content-container\">\n<div class=\"mt-section\">\n<h1>Designing a &#8220;good&#8221; nucleophilic substitution<\/h1>\n<p>If you want to do well in this class, there are several things you need to work hard at: Being attentive in class, studying the notes and this textbook (especially before exams), practicing problems, and completing the quizzes and homeworks.\u00a0 As long as you do all of these things then you&#8217;re likely to pass (though I can&#8217;t give guarantees!).\u00a0 So there are many different factors that can affect your grade.\u00a0 In the same way, the outcome of a reaction (such as nucleophilic substition) depends on many different things &#8211; reactants, solvent, etc.\u00a0 When we want to make a chemical in a lab or on a chemical plant, we need to design the reaction so that it works well, and gives a good yield of the product in a reasonable time.\u00a0 The reactants and conditions we use will depend on what we&#8217;re trying to do.\u00a0 In this section, we examine what factors will help an S<sub>N<\/sub>2 or S<sub>N<\/sub>1 reaction be successful.<\/p>\n<h1>Factors affecting the S<sub>N<\/sub>2 reaction<\/h1>\n<p class=\"mt-align-justify\">As we saw in the previous section, in the S<sub>N<\/sub>2 reaction the rate of reaction depends on both the electrophile (usually an alkyl halide) and the nucleophile.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203831\/image009-2.png\" alt=\"image010.png\" width=\"393\" height=\"84\" \/><\/p>\n<p>In practice, the rates of S<sub>N<\/sub>2 reactions vary enormously, and for a practicable procedure we need to make sure that the reaction will happen at a reasonable rate.\u00a0 So what makes for a good S<sub>N<\/sub>2 reaction?\u00a0 We need to consider what makes a suitable nucleophile, and what makes a suitable electrophile.<\/p>\n<h2>Nucleophile strength<\/h2>\n<p>In <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/6-5-lewis-acids-bases\/\">section 6.5<\/a>, we learnt what makes a nucleophile strong (reactive) or weak (unreactive).\u00a0 Anything which removes electron-density from the nucleophilic atom will make it less nucleophilic.\u00a0 We summarized the main points from 6.5 as follows:<\/p>\n<ul>\n<li>Charge &#8211; negatively charged =&gt; stronger nucleophile<\/li>\n<li>Within a row &#8211; more electronegative atom =&gt; weaker nucleophile<\/li>\n<li>Within a column, size of atom.\u00a0 Polar protic solvent, bigger atom is better; polar aprotic solvent, smaller atom is better.<\/li>\n<li>Resonance &#8211; if the nucleophilic lone pair can be delocalized by resonance, it will make it less nucleophilic<\/li>\n<li>Steric hindrance &#8211; a hindered nucleophilic atom will tend to be less reactive, particularly when attacking a crowded electrophile.<\/li>\n<\/ul>\n<p>Regarding the solvent, polar aprotic solvents such as DMSO, DMF, acetone or acetonitrile are popular choices for S<sub>N<\/sub>2 reactions, because rates are generally faster than with polar protic solvents (water, alcohols, etc.).\u00a0 This is because the nucleophile is almost &#8220;naked&#8221; in aprotic solvents, whereas in polar protic solvents it is surrounded by a cage of solvent molecules.<\/p>\n<p>If we have a strong nucleophile, the S<sub>N<\/sub>2 reaction will happen faster; a weak nucleophile will react more slowly and may not even react.\u00a0 So in general we want a strong nucleophile.<\/p>\n<\/div>\n<h2 class=\"mt-section\">The electrophile<\/h2>\n<h3 class=\"mt-section\">(a) Structure of the alkyl group<\/h3>\n<div class=\"mt-section\">\n<p>In the structure of the S<sub>N<\/sub>2 transition state, there are 90<sup>o<\/sup>\u00a0bond angles between the breaking bond to the leaving group and the three bonds which remain connected to the carbon as well as between the bond being made to the nucleophile and those same three bonds.<\/p>\n<p class=\"mt-align-center\"><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203918\/SN2ster1.gif\" alt=\"\" width=\"159\" height=\"142\" \/><\/p>\n<p class=\"mt-align-justify\">As long as the two of the groups attached to the carbon being attacked are small hydrogens, the repulsions which happen do not require much energy. If the groups attached to the carbon are larger, though, like methyl groups, the transition state energy increases, the activation energy increases, and the reaction becomes much slower.<\/p>\n<p class=\"mt-align-center\"><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19203920\/SN2ster2.gif\" alt=\"\" width=\"444\" height=\"136\" \/><\/p>\n<p class=\"mt-align-justify\">This means that the reactivity order for alkyl halides in S<sub>N<\/sub>2 reactions is:<\/p>\n<p class=\"mt-align-center\"><b>methyl &gt; primary &gt; secondary &gt; tertiary<\/b><\/p>\n<p class=\"mt-align-justify\">The practical outcome of this is that\u00a0S<sub>N<\/sub>2 reactions are generally reliable only when the alkyl halide is primary, though under the correct conditions secondary halides can react well also.<\/p>\n<\/div>\n<\/section>\n<h3 class=\"editable\">(b) Leaving group ability &#8211; what makes a good leaving group?<\/h3>\n<div class=\"mt-section\">\n<p>In our general discussion of nucleophilic substitution reactions, we have until now been designating the leaving group simply as \u201cX&#8221;. \u00a0As you\u00a0 may imagine, however, the nature of the leaving group is an important consideration: if the C-X bond does not break, the new bond between the nucleophile and electrophilic carbon cannot form, regardless of whether the substitution is S<sub>N<\/sub>1 or S<sub>N<\/sub>2.\u00a0 There are two main factors: The strength of the C-X bond, and the stability of the X group after it has left.\u00a0 It turns out that the two factors lead to the same prediction for halogen leaving group ability:<\/p>\n<p style=\"text-align: center\"><span style=\"font-size: larger;\">I &gt; Br &gt; Cl &gt; F<\/span><\/p>\n<h4 class=\"mt-align-justify\">C-X bond strength<\/h4>\n<p class=\"mt-align-justify\">Since the bond between the carbon and the leaving group is being broken in the transition state, the weaker this bond is the lower the activation energy and the faster the reaction. This leads to the following reactivity order for alkyl halides<\/p>\n<p class=\"mt-align-center\" style=\"text-align: center\"><b><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-5102\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/23165635\/AlkylHalideReactivity.png\" alt=\"Alkyl halide reactivity is RI&gt;RBr&gt;RCl&gt;RF\" width=\"321\" height=\"30\" \/><\/b><\/p>\n<p class=\"mt-align-justify\">Practically, alkyl fluorides are not used for S<sub>N<\/sub>2 reactions because the C-F bond is too strong. Often alkyl iodides are reactive enough to be difficult to store, so the the common choices for reactions are alkyl chlorides and alkyl bromides.<\/p>\n<h4>Stability of the group after leaving<\/h4>\n<p>When the C-X bond breaks in a nucleophilic substitution, the pair of electrons in the bond goes with the leaving group.\u00a0 In this way, the leaving group is analogous to the conjugate base in a Br\u00f8nsted-Lowry acid-base reaction. When we were evaluating the strength of acids in chapter 7, what we were really doing was evaluating the stability of the conjugate base that resulted from the proton transfer.\u00a0 All of the concepts that we used to evaluate the stability of conjugate bases we can use again to evaluate leaving groups.\u00a0 In other words, the trends in basicity are parallel to the trends in leaving group potential &#8211; the weaker the base, the better the leaving group. Just as with conjugate bases, the most important question regarding leaving groups is this:\u00a0 when a leaving group leaves and takes a pair of electrons with it, how well is the extra electron density stabilized?<\/p>\n<p>In laboratory synthesis reactions, halides often act as leaving groups.\u00a0 Iodide, which is the <em>least<\/em> basic of the four main halides, is also the <em>best<\/em> leaving group \u2013 it is the most stable as a negative ion. Fluoride is the least effective leaving group among the halides, because fluoride anion is the most basic.<\/p>\n<p>We already know that the use of polar, aprotic solvents increases the reactivity of nucleophiles in S<sub>N<\/sub>2 reactions, because these solvents do not &#8216;cage&#8217; the nucleophile and keep it from attacking the electrophile.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Factors favoring S<sub>N<\/sub>2<\/h3>\n<p>To design an effective S<sub>N<\/sub>2 reaction using an alkyl halide, we need:<\/p>\n<ul>\n<li>An unhindered alkyl halide (preferably methyl or primary, but secondary may be possible)<\/li>\n<li>A good leaving group (preferably I or Br)<\/li>\n<li>A strong nucleophile<\/li>\n<li>A suitable solvent &#8211; polar aprotic is most effective<\/li>\n<\/ul>\n<\/div>\n<h1>Factors affecting the S<sub>N<\/sub>1 reaction<\/h1>\n<\/div>\n<div class=\"mt-section\">\n<p>As we learnt in <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/8-2-physical-chemistry-for-sn2-and-sn1-reactions\/\">section 8.2<\/a>, the nucleophile has no effect on the rate of an S<sub>N<\/sub>1 reaction.\u00a0 This means that we only need to consider the electrophile, usually an alkyl halide.\u00a0 Another feature of the S<sub>N<\/sub>1 reaction is that it is often prone to side reactions, which is why it is less used in synthesis than the S<sub>N<\/sub>2 reaction.<\/p>\n<h2 class=\"mt-section\">The electrophile<\/h2>\n<p>This topic was examined in general in section 6.5., and also considered above for S<sub>N<\/sub>2.\u00a0 But a electrophile that is good for S<sub>N<\/sub>2 is not necessarily good for S<sub>N1<\/sub>, for reasons that will become clear.\u00a0 We also have a new factor to consider &#8211; the stability of the carbocation that is formed as a result of the heterolysis step.<\/p>\n<h3 class=\"mt-section\">(a) Structure of the alkyl group<\/h3>\n<\/div>\n<p>In the vast majority of the nucleophilic substitution reactions you will see in this and other organic chemistry texts, the electrophilic atom is a carbon which is bonded to an electronegative atom, usually oxygen, nitrogen, sulfur, or a halogen. The concept of electrophilicity is relatively simple: an electron-poor atom is an attractive target for something that is electron-rich, <em>i.e<\/em>. a nucleophile. However, we must also consider the effect of steric hindrance on electrophilicity. In addition, we must discuss how the nature of the electrophilic carbon, and more specifically the stability of a potential carbocationic intermediate, influences the S<sub>N<\/sub>1 reaction.<\/p>\n<div class=\"mt-section\">\n<h3 class=\"editable\">Steric effects on electrophilicity<\/h3>\n<p>In an S<sub>N<\/sub>1 mechanism, the nucleophile attacks an sp<sup>2<\/sup>-hybridized carbocation intermediate, which has trigonal planar geometry with \u2018open\u2019 120 angles.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204048\/image049-3.png\" alt=\"image050.png\" width=\"495\" height=\"231\" \/><\/p>\n<p>With this open geometry, the empty p orbital of the electrophilic carbocation is no longer significantly shielded from the approaching nucleophile by the bulky alkyl groups. A carbocation is a very potent electrophile, and the nucleophilic step occurs very rapidly compared to the first (ionization) step.\u00a0 This is in direct contrast to the S<sub>N<\/sub>2 reaction, where bulky alkyl groups hinder the reaction.<\/p>\n<\/div>\n<div class=\"mt-section\">\n<h3 class=\"editable\">Stability of carbocation intermediates<\/h3>\n<p>We know that the rate-limiting step of an S<sub>N<\/sub>1 reaction is the first step &#8211; formation of the this carbocation intermediate.\u00a0 The rate of this step \u2013 and therefore, the rate of the overall substitution reaction \u2013 depends on the activation energy for the process in which the bond between the carbon and the leaving group breaks and a carbocation forms.\u00a0 According to Hammond\u2019s postulate (<a title=\"Organic Chemistry\/Organic Chemistry With a Biological Emphasis\/Chapter 6: Introduction to organic reactivity and catalysis\/Section 6.2: Energy diagrams\" href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/5-5-hammonds-postulate\/\" rel=\"internal\">section 5.5<\/a>), the more stable the carbocation intermediate is, the faster this first bond-breaking step will occur. In other words, the likelihood of a nucleophilic substitution reaction proceeding by a dissociative (S<sub>N<\/sub>1) mechanism depends to a large degree on the stability of the carbocation intermediate that forms.<\/p>\n<p>We previously considered carbocation stability in <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/6-5-lewis-acids-bases\/\">section 6.5.<\/a>, and we found that in general, <em>more substituted carbocations are more stable<\/em>:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204052\/image053-1.png\" alt=\"image054.png\" width=\"357\" height=\"150\" \/><\/p>\n<p>A positively charged species such as a carbocation is very electron-poor, and thus anything which donates electron density to the center of electron poverty will help to stabilize it. Conversely, a carbocation will be <em>destabilized<\/em> by an electron withdrawing group.<\/p>\n<p>Stabilization of a carbocation can also occur through <strong>resonance<\/strong>, which allows the &#8220;burden&#8221; of the negative charge to be delocalized, or shared, onto more than one atom. Resonance effects as a rule are more powerful than inductive effects.\u00a0 For example, a cation next to a double bond will delocalize the charge via type IV resonance, so an allylic carbocation is more stable than a comparable one that cannot do resonance:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204058\/image059-2.png\" alt=\"image060.png\" width=\"289\" height=\"80\" \/><\/p>\n<p>Finally, <strong>vinylic<\/strong> carbocations, in which the positive charge resides on a double-bonded carbon, are very unstable and thus unlikely to form as intermediates in any S<sub>N<\/sub>1 reaction.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204104\/image065-1.png\" alt=\"image066.png\" width=\"145\" height=\"114\" \/><\/p>\n<p>When considering the possibility that a nucleophilic substitution reaction proceeds <em>via<\/em> an S<sub>N<\/sub>1 pathway, it is critical to evaluate the stability of the hypothetical carbocation intermediate.\u00a0 If this intermediate is not sufficiently stable, an S<sub>N<\/sub>1 mechanism must be considered unlikely, and the reaction probably proceeds by an S<sub>N<\/sub>2 mechanism.<\/p>\n<div>\n<div>\n<h2>Leaving group<\/h2>\n<p>The C-X bond breaks in the rate determining step of S<sub>N<\/sub>1, just as it does in S<sub>N<\/sub>2, and in fact the rules are the same for determining a &#8220;good&#8221; leaving group.\u00a0 Again these are determined by the C-X bond strength and the stability of X after it has left.\u00a0 This means that we see the same trends as we did for S<sub>N<\/sub>2, where the larger halogens make better leaving groups, i.e.,<\/p>\n<p style=\"text-align: center\"><span style=\"font-size: larger;\">I &gt; Br &gt; Cl &gt; F<\/span><\/p>\n<h2 class=\"mt-section\">Side reactions in S<sub>N<\/sub>1<\/h2>\n<h4 class=\"mt-section\">(a) Elimination<\/h4>\n<p>In all of our discussion so far about nucleophilic substitutions, we have ignored another important possibility.\u00a0 In many cases, including the two examples above, substitution reactions compete with a type of reaction known as <strong>elimination<\/strong>.\u00a0 This will be covered in detail soon, in <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/8-5-elimination-reactions\/\">section 8.5.<\/a>\u00a0 Consider, for example, the two courses that a reaction could take when 2-bromo-2-methylpropane reacts with water:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/19204015\/image109.png\" alt=\"image110.png\" width=\"647\" height=\"234\" \/><\/p>\n<p>We begin with formation of the carbocation intermediate.\u00a0 In pathway \u2018a\u2019, water acts as a nucleophile \u2013 this is, of course, the familiar S<sub>N<\/sub>1 reaction.\u00a0 However, a water molecule encountering the carbocation intermediate could alternatively act as a <em>base<\/em> rather than as a nucleophile, plucking a proton from one of the methyl carbons and causing the formation of a new carbon-carbon p bond.\u00a0 This alternative pathway is called an elimination reaction, and in fact with the conditions above, both the substitution and the elimination pathways will occur in competition with each other.\u00a0 Elimination can be minimized by keeping the reaction cold, but some of this side-reaction is often inevitable.<\/p>\n<h4>(b) Carbocation rearrangements<\/h4>\n<p>These will be covered very soon, in <a href=\"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/8-4-rearrangements\/\">section 8.4.<\/a>\u00a0 If the carbocation can easily rearrange to a more stable carbocation, then rearrangement products are likely to be important, and the reaction may lead to mixtures.<\/p>\n<h2>Solvent<\/h2>\n<p>The rates of S<sub>N<\/sub>1 reactions are generally increased by the use of a highly polar solvent, including protic (hydrogen bonding) solvents such as water or ethanol.\u00a0 In essence, a protic solvent increases the reactivity of the leaving group in an S<sub>N<\/sub>1 reaction, by helping to stabilize the products of the first (ionization) step. In the S<sub>N<\/sub>1 mechanism, remember, the rate determining step does <em>not<\/em> involve the nucleophilic species, so any reduction of nucleophilicity does not matter.\u00a0 What matters is that the charged products of the first step &#8211; the carbocation intermediate and the anionic leaving group &#8211; are stabilized best by a highly polar, protic solvent.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Factors favoring S<sub>N<\/sub>1<\/h3>\n<p>To design an effective S<sub>N<\/sub>1 reaction using an alkyl halide, we need:<\/p>\n<ul>\n<li>A highly substituted alkyl halide (preferably tertiary or resonance-stabilized, but secondary may be possible), ideally one which will not lead to rearrangement<\/li>\n<li>A good leaving group (preferably I or Br)<\/li>\n<li>A non-basic nucleophile (to reduce the elimination side reaction)<\/li>\n<li>A suitable solvent &#8211; polar protic is most effective<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h1>Predicting S<sub>N<\/sub>1 vs. S<sub>N<\/sub>2 mechanisms<\/h1>\n<div>\n<p>When considering whether a nucleophilic substitution is likely to occur via an S<sub>N<\/sub>1 or S<sub>N<\/sub>2 mechanism, we really need to consider three factors:<\/p>\n<p>1) <strong>The electrophile:<\/strong> when the leaving group is attached to a methyl group or a primary carbon, an S<sub>N<\/sub>2 mechanism is favored (here the electrophile is unhindered by surrounded groups, and any carbocation intermediate would be high-energy and thus unlikely). When the leaving group is attached to a tertiary, allylic, or benzylic carbon, a carbocation intermediate will be relatively stable and thus an S<sub>N<\/sub>1 mechanism is favored.<\/p>\n<p>2) <strong>The nucleophile<\/strong>: powerful nucleophiles, especially those with negative charges, favor the S<sub>N<\/sub>2 mechanism. Weaker nucleophiles such as water or alcohols favor the S<sub>N<\/sub>1 mechanism.<\/p>\n<p>3) <strong>The solvent<\/strong>:\u00a0 Polar aprotic solvents favor the S<sub>N<\/sub>2 mechanism by enhancing the reactivity of the nucleophile. Polar protic solvents favor the S<sub>N<\/sub>1 mechanism by stabilizing the carbocation intermediate. S<sub>N<\/sub>1 reactions are frequently solvolysis reactions.<\/p>\n<p>For example, the reaction below has a tertiary alkyl bromide as the electrophile, a weak nucleophile, and a polar protic solvent (we\u2019ll assume that methanol is the solvent).\u00a0 Thus we\u2019d confidently predict an S<sub>N<\/sub>1 reaction mechanism.\u00a0 Because substitution occurs at a chiral carbon, we can also predict that the reaction will proceed with racemization.<\/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\/19204006\/image106-1.png\" alt=\"image106.png\" width=\"642\" height=\"102\" \/><\/p>\n<p>In the reaction below, on the other hand, the electrophile is a secondary alkyl bromide \u2013 with these, both S<sub>N<\/sub>1 and S<sub>N<\/sub>2 mechanisms are possible, depending on the nucleophile and the solvent.\u00a0 In this example, the nucleophile (a thiolate anion) is strong, and a polar protic solvent is used \u2013 so the S<sub>N<\/sub>2 mechanism is heavily favored.\u00a0 The reaction is expected to proceed with inversion of configuration.<\/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\/19204009\/image108.png\" alt=\"image108.png\" width=\"557\" height=\"97\" \/><\/p>\n<div class=\"textbox examples\">\n<h3>Exercise<\/h3>\n<p>Determine whether each substitution reaction shown below is likely to proceed by an S<sub>N<\/sub>1 or S<sub>N<\/sub>2 mechanism.<\/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\/19204012\/image108a.png\" alt=\"image108a.png\" width=\"355\" height=\"188\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q778981\">Show Solution<\/span><\/p>\n<div id=\"q778981\" class=\"hidden-answer\" style=\"display: none\">\n<p>a) S<sub>N<\/sub>2 (primary electrophile, strong nucleophile, polar aprotic solvent)<\/p>\n<p>b) S<sub>N<\/sub>1 (tertiary electrophile, weak nucleophile, protic solvent)<\/p>\n<p>c) S<sub>N<\/sub>2 (secondary electrophile, strong nucleophile, polar protic solvent)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\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-2653\">\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>24: Nucleophilic Substitution, SN2, SN1. <strong>Authored by<\/strong>: Kirk McMichaelu00a0(Washington State University). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_-_A_%22Carbonyl_Early%22_Approach_(McMichael)\/24%3A_Nucleophilic_Substitution%2C_SN2%2C_SN1\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_-_A_%22Carbonyl_Early%22_Approach_(McMichael)\/24%3A_Nucleophilic_Substitution%2C_SN2%2C_SN1<\/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>8.3: More about nucleophiles. <strong>Authored by<\/strong>: Tim Soderberg, (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_08%3A_Nucleophilic_substitution_reactions_I\/8.3%3A_More_about_nucleophiles\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_08%3A_Nucleophilic_substitution_reactions_I\/8.3%3A_More_about_nucleophiles<\/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><li>8.5: Leaving groups. <strong>Authored by<\/strong>: Tim Soderberg (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_08%3A_Nucleophilic_substitution_reactions_I\/8.5%3A_Leaving_groups\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_08%3A_Nucleophilic_substitution_reactions_I\/8.5%3A_Leaving_groups<\/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>8.4: Electrophiles and carbocation stability. <strong>Authored by<\/strong>: Tim Soderberg (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_08%3A_Nucleophilic_substitution_reactions_I\/8.4%3A_Electrophiles_and_carbocation_stability\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_08%3A_Nucleophilic_substitution_reactions_I\/8.4%3A_Electrophiles_and_carbocation_stability<\/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>25: Elimination - E2 and E1. <strong>Authored by<\/strong>: Kirk McMichael (Washington State University). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_-_A_%22Carbonyl_Early%22_Approach_(McMichael)\/25%3A_Elimination_-_E2_and_E1\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_-_A_%22Carbonyl_Early%22_Approach_(McMichael)\/25%3A_Elimination_-_E2_and_E1<\/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>Organic Chemistry with a Biological Emphasis Volume I. <strong>Authored by<\/strong>: Timothy Soderberg. <strong>Provided by<\/strong>: University of Minnesota, Morris. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/digitalcommons.morris.umn.edu\/chem_facpubs\/1\">https:\/\/digitalcommons.morris.umn.edu\/chem_facpubs\/1<\/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":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"24: Nucleophilic Substitution, SN2, SN1\",\"author\":\"Kirk McMichaelu00a0(Washington State University)\",\"organization\":\"\",\"url\":\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_-_A_%22Carbonyl_Early%22_Approach_(McMichael)\/24%3A_Nucleophilic_Substitution%2C_SN2%2C_SN1\",\"project\":\"Chemistry LibreTexts\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"8.3: More about nucleophiles\",\"author\":\"Tim Soderberg, (University of Minnesota, Morris)\",\"organization\":\"\",\"url\":\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_08%3A_Nucleophilic_substitution_reactions_I\/8.3%3A_More_about_nucleophiles\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"8.5: Leaving groups\",\"author\":\"Tim Soderberg (University of Minnesota, Morris)\",\"organization\":\"\",\"url\":\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_08%3A_Nucleophilic_substitution_reactions_I\/8.5%3A_Leaving_groups\",\"project\":\"Chemistry LibreTexts\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"8.4: Electrophiles and carbocation stability\",\"author\":\"Tim Soderberg (University of Minnesota, Morris)\",\"organization\":\"\",\"url\":\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_08%3A_Nucleophilic_substitution_reactions_I\/8.4%3A_Electrophiles_and_carbocation_stability\",\"project\":\"Chemistry LibreTexts\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"25: Elimination - E2 and E1\",\"author\":\"Kirk McMichael (Washington State University)\",\"organization\":\"\",\"url\":\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_-_A_%22Carbonyl_Early%22_Approach_(McMichael)\/25%3A_Elimination_-_E2_and_E1\",\"project\":\"Chemistry LibreTexts\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Organic Chemistry with a Biological Emphasis Volume I\",\"author\":\"Timothy Soderberg\",\"organization\":\"University of Minnesota, Morris\",\"url\":\"https:\/\/digitalcommons.morris.umn.edu\/chem_facpubs\/1\",\"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-2653","chapter","type-chapter","status-publish","hentry"],"part":25,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2653","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":45,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2653\/revisions"}],"predecessor-version":[{"id":5104,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2653\/revisions\/5104"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/25"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/2653\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/media?parent=2653"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=2653"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/contributor?post=2653"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/license?post=2653"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}