{"id":6877,"date":"2013-01-18T08:32:09","date_gmt":"2013-01-18T13:32:09","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=6877"},"modified":"2026-04-30T13:16:53","modified_gmt":"2026-04-30T18:16:53","slug":"wrapup-the-quick-n-dirty-guide-to-sn1sn2e1e2","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2013\/01\/18\/wrapup-the-quick-n-dirty-guide-to-sn1sn2e1e2\/","title":{"rendered":"Wrapup: The Key Factors For Determining SN1\/SN2\/E1\/E2"},"content":{"rendered":"<p><strong>S<sub>N<\/sub>1\/S<sub>N<\/sub>2\/E1\/E2 &#8211; Summarizing The Key Factors That Determine Whether A Reaction Will Be SN1, SN2, E1 or E2<\/strong><\/p>\n<p>In this article we walk through the thought process that will let you determine if a given reaction goes through a S<sub>N<\/sub>1, S<sub>N<\/sub>2, E1, or E2 pathway.<\/p>\n<p>It assumes that you know how to draw the product of a reaction if you are told what pathway is operating. If you aren&#8217;t sure how to draw the product of a reaction if you&#8217;re told it&#8217;s S<sub>N<\/sub>1\/S<sub>N<\/sub>2\/E1\/E2, then I suggest going back to the individual articles on those topics for a refresher. [<a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/07\/13\/the-sn1-mechanism\/\">SN1<\/a> <a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/07\/04\/the-sn2-mechanism\/\">SN2<\/a> <a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/09\/19\/the-e1-reaction\/\">E1<\/a> <a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/09\/27\/the-e2-mechanism\/\">E2<\/a>]<\/p>\n<p><em><span style=\"color: #800080;\">Note that this series covers alkyl halides (and its relatives) but not alcohols. For more on alcohols, skip ahead to the alcohol chapter<\/span> (<a href=\"https:\/\/www.masterorganicchemistry.com\/2015\/04\/16\/elimination-reactions-of-alcohols\/\">Elimination reactions of alcohols<\/a>) \u00a0\u00a0<\/em><\/p>\n<ul>\n<li>Look for a <strong>good leaving group<\/strong> on an <strong>alkyl<\/strong> (sp<sup>3<\/sup>-hybridized) carbon.<\/li>\n<li>Determine if the carbon bearing the good leaving group is <strong>primary, secondary, tertiary<\/strong> or <strong>methyl.<\/strong><\/li>\n<li>Identify the <strong>strongest nucleophile\/base<\/strong> that is present.<\/li>\n<li>Note the <strong>temperature<\/strong>, if given (heat tends to favor elimination)<\/li>\n<li>The identity of the solvent can also indicate S<sub>N<\/sub>2, particularly with weakly basic nucleophiles. With strongly basic nucleophiles (RO(-) and HO(-) ) on secondary alkyl halides, double check with your instructor which reaction pathway tends to operate (E2 or S<sub>N<\/sub>2)<\/li>\n<\/ul>\n<p>This article used to be called the &#8220;Quick N&#8217; Dirty Guide to S<sub>N<\/sub>1\/S<sub>N<\/sub>2\/E1\/E2&#8243;. It&#8217;s been revised to make it a little less quick and a little less dirty, but hopefully more comprehensive and useful. [<a href=\"#noteone\">Note 1<\/a>]<\/p>\n<p><img fetchpriority=\"high\" decoding=\"async\" class=\"alignnone wp-image-36362\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2024\/05\/0-summary-sn1-sn2-e1-e2-decision-key-factors.gif\" alt=\"summary - sn1 sn2 e1 e2 decision key factors\" width=\"640\" height=\"415\" \/><\/a><\/p>\n<p><span style=\"color: #800080;\"><em>\u00a0You need to be logged in a <a style=\"color: #800080;\" href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\">MOC Member<\/a> to see all the quizzes in this post. It&#8217;s affordable ($10\/month) and <strong>many<\/strong> people have found it useful in preparing for their exams.\u00a0 There are roughly 3500 quizzes and counting, as well as accompanying &#8220;cheat sheets&#8221; and also the Reaction Guide.\u00a0<\/em><\/span><\/p>\n<p><strong>Table of Contents<\/strong><\/p>\n<ol>\n<li style=\"list-style-type: none;\">\n<ol>\n<li><a href=\"#one\">The SN1\/SN2\/E1\/E2 Decision<\/a><\/li>\n<li><a href=\"#two\">Step 1: Identify a Good Leaving Group<\/a><\/li>\n<li><a href=\"#three\">Step 2: The Leaving Group Should Be On An Alkyl (sp3-hybridized) Carbon<\/a><\/li>\n<li><a href=\"#four\">Step 3: Identify The Carbon As Primary, Secondary, Tertiary (or Methyl)<\/a><\/li>\n<li><a href=\"#five\">Step 4: Identify The Base\/Nucleophile<\/a><\/li>\n<li><a href=\"#six\">Step 5: The Role of Temperature<\/a><\/li>\n<li><a href=\"#seven\">Step 6: Solvent, And SN2\/E2 With Secondary Alkyl Halides And Strong Nucleophiles<\/a><\/li>\n<li><a href=\"#notes\">Notes<\/a><\/li>\n<li><a href=\"#quiz\">Quiz Yourself!<\/a><\/li>\n<li><a href=\"#references\">(Advanced) References and Further Reading<\/a><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<hr \/>\n<h2><a id=\"one\"><\/a>1. S<sub>N<\/sub>1\/SN2\/E1\/E2 : Wrapup<\/h2>\n<p><span style=\"color: #993366;\"><em>(Formerly, &#8220;The Quick N&#8217; Dirty Guide To S<sub>N<\/sub>1\/S<sub>N<\/sub>2\/E1\/E2&#8243;)<\/em><\/span><\/p>\n<p>Here we are at the end of our series on determining whether a reaction is S<sub>N<\/sub>1, S<sub>N<\/sub>2, E1, or E2. It&#8217;s time to recap all of the steps in the deductive reasoning required to determine which reaction pathway is operating.<\/p>\n<p>Note that this article assumes that you are able to draw the correct\u00a0 product of a reaction if you already know what the most likely reaction pathway is. \u00a0If you are already told ahead of time what mechanism is operating, then <strong>you don&#8217;t need this article.\u00a0<\/strong><\/p>\n<p><span style=\"color: #000000;\">However, if you&#8217;re at all unsure on how to draw the product of an S<sub>N<\/sub>2 or E2 reaction if you&#8217;ve been given the starting materials and reactants, then\u00a0 you should to go back to articles on the various reactions for a refresher before coming back here.\u00a0<\/span><\/p>\n<p>[<a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/07\/13\/the-sn1-mechanism\/\">SN1<\/a> <a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/07\/04\/the-sn2-mechanism\/\">SN2<\/a> <a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/09\/19\/the-e1-reaction\/\">E1<\/a> <a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/09\/27\/the-e2-mechanism\/\">E2<\/a>]<\/p>\n<p>This series also assumes you know a few key concepts such as, &#8220;<a href=\"https:\/\/www.masterorganicchemistry.com\/2011\/04\/12\/what-makes-a-good-leaving-group\/\">What Makes a Good Leaving Group?<\/a>&#8220;, <a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/06\/18\/what-makes-a-good-nucleophile\/\">What Makes a Good Nucleophile?<\/a>, <a href=\"https:\/\/www.masterorganicchemistry.com\/2011\/03\/11\/3-factors-that-stabilize-carbocations\/\">What Factors Influence Carbocation Stability<\/a>? .<\/p>\n<p>With that throat-clearing out of the way&#8230;<\/p>\n<p>Let&#8217;s say you&#8217;re given a reaction and are asked to draw the major product <em>without<\/em> being told what mechanism is operating.<\/p>\n<p>In order to draw the major product,\u00a0you&#8217;ll need to figure out what <strong>mechanism<\/strong> the reaction proceeds through &#8211; <strong>if the reaction proceeds at all\u00a0<\/strong>(<span style=\"color: #800080;\"><em>yes, &#8220;no reaction&#8221; is also a possibility!<\/em><\/span>).<\/p>\n<p>In this post, we&#8217;ll walk through the thought process necessary to arrive at a reasonable conclusion.<\/p>\n<p>Once you&#8217;ve determined what mechanism is operating, <strong>then<\/strong> you&#8217;ll need to apply the pattern of bonds formed\/broken for each reaction to give you the correct product.<\/p>\n<h2><a id=\"two\"><\/a>2. Step 1: Identify A Good Leaving Group<\/h2>\n<p>The first step in determining whether a reaction proceeds through S<sub>N<\/sub>1\/S<sub>N<\/sub>2\/E1\/E2 is to identify a <strong>good leaving group<\/strong> in the <strong>substrate <\/strong>(e.g. alkyl halide) where the reaction could conceivably take place.<\/p>\n<p>Good leaving groups are <strong>weak bases<\/strong> (<span style=\"color: #800080;\"><em>See article &#8211; <span style=\"color: #800080;\"><a style=\"color: #800080;\" href=\"https:\/\/www.masterorganicchemistry.com\/2011\/04\/12\/what-makes-a-good-leaving-group\/\">What Makes A Good Leaving Group?<\/a><\/span><\/em><\/span>).<\/p>\n<p>If you recall that &#8220;the stronger the acid, the weaker the conjugate base&#8221;, it stands to reason that the halide ions (Cl<sup>\u2013<\/sup> , Br<sup>\u2013<\/sup> , I<sup>\u2013<\/sup> <span style=\"color: #000000;\">but<\/span><span style=\"color: #ff0000;\"><strong> not <span style=\"color: #339966;\">F<\/span><\/strong><\/span>) and sulfonates (TsO<sup>\u2013<\/sup> and MsO<sup>\u2013<\/sup>\u00a0 ; <span style=\"color: #800080;\"><em>see <a style=\"color: #800080;\" href=\"https:\/\/www.masterorganicchemistry.com\/2015\/03\/10\/tosylates-and-mesylates\/\">Tosylates and Mesylates<\/a><\/em><\/span>) are great leaving groups.<\/p>\n<p>Carboxylates (RCO<sub>2<\/sub><sup>\u2013\u00a0<\/sup>) are the conjugate bases of carboxylic acids (<span style=\"color: #993366;\"><em>pK<sub>a<\/sub> of about 4<\/em><\/span>) and can also be decent leaving groups. Under acidic conditions (not covered in this series, but covered in the chapter on alcohols), water (H<sub>2<\/sub>O) and alcohols (ROH) can be good leaving groups.<\/p>\n<div class=\"wq-quiz-wrapper\" data-id=\"45648\"><style type=\"text\/css\" id=\"wq-flip-custom-css\">.wq-quiz-wrapper[data-id=\"45648\"] {\n--wq-question-width: 100%;\n--wq-question-color: #009cff;\n--wq-question-height: auto;\n--wq-font-color: #444;\n}\n\n\t\t\t.wq-quiz-wrapper[data-id=\"45648\"] {\n\t\t\t\t--wq-question-width: 600px;\n\t\t\t}\n\n\t\t\t@media screen and (max-width: 600px) {\n\t\t\t\t.wq-quiz-wrapper[data-id=\"45648\"] .wq_singleQuestionWrapper { width:100% !important; height:auto !important; }\n\t\t\t}\n\t\t<\/style><!-- wp quiz -->\n<div id=\"wp-quiz-45648\" class=\"wq_quizCtr single flip_quiz wq-quiz wq-quiz-45648 wq-quiz-flip wq-layout-single wq-skin-traditional wq-should-show-correct-answer\" data-quiz-id=\"45648\">\n<div class=\"wq-questions wq_questionsCtr\">\n\t<div class=\"wq-question wq_singleQuestionWrapper wq-question-9b0q2\" data-id=\"9b0q2\">\n\n\t\n\t<div class=\"item_top\">\n\t\t<div class=\"title_container\">\n\t\t\t<div class=\"wq_questionTextCtr\">\n\t\t\t\t<h4 class=\"wq-question-title\"><\/h4>\n\t\t\t<\/div>\n\t\t<\/div>\n\t<\/div>\n\n\t<div class=\"card \">\n\t\t<div class=\"front\" >\n\t\n\t\t\t\t\t<img decoding=\"async\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-images\/2676-Front.gif\" \/>\n\t\t\n\t\t\n\t\n\t\n\t\t\t<span class=\"top-desc\">Click to Flip<\/span>\n\t<\/div>\n\t\t<div class=\"back\" >\n\t\n\t\t\t\t\t<img decoding=\"async\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-images\/2676-Reverse.gif\" \/>\n\t\t\n\t\t\n\t\n\t<\/div>\n\t<\/div>\n\n\t\n<\/div>\n<\/div>\n<\/div>\n<!-- \/\/ wp quiz-->\n<\/div><!-- End .wq-quiz-wrapper -->\n<p>Some prominent examples of <strong>poor<\/strong> leaving groups include<\/p>\n<ul>\n<li><strong>Fluoride ion\u00a0<\/strong>&#8211; the C-F bond is extremely strong (about 130 kcal\/mol) and it is thermodynamically difficult to break the C-F bond.<\/li>\n<li>Hydroxy (HO) and alkoxy (RO) groups in the absence of acid<\/li>\n<li>Amines (NH<sub>2\u00a0<\/sub>, NHR, NR<sub>2<\/sub>) are too basic to be good leaving groups. However ammonium salts (-NR3<sup>+<\/sup>) can be good leaving groups.<\/li>\n<li>Hydride (H-) and alkyl<\/li>\n<li>Finally, cyano (CN) and thio (SR) are also generally poor leaving groups for S<sub>N<\/sub>1\/S<sub>N<\/sub>2\/E1\/E2 reactions.<\/li>\n<\/ul>\n<h2><a id=\"three\"><\/a>3. The Leaving Group Must Be On An sp<sup>3<\/sup> Hybridized Carbon<\/h2>\n<p>While a good leaving group is a necessary condition for S<sub>N<\/sub>1\/S<sub>N<\/sub>2\/E1\/E2 to take place, it&#8217;s not a sufficient condition.<\/p>\n<p>The leaving group also has to be bonded to a carbon that is capable of undergoing backside attack (S<sub>N<\/sub>2), capable of being a stable carbocation (S<sub>N<\/sub>1\/E1) or reasonably effective at stabilizing partial positive charge (S<sub>N<\/sub>2\/E2).<\/p>\n<p>For this reason <strong>only<\/strong>\u00a0<strong>alkyl\u00a0carbons<\/strong> &#8211;<em> i.e<\/em>. <strong>sp<sup>3 <\/sup>hybridized, tetrahedral carbon<\/strong> &#8211; can undergo S<sub>N<\/sub>1\/S<sub>N<\/sub>2\/E1\/E2, with one limited exception we&#8217;ll cover in the chapter on alkynes and sticks out like a sore thumb (<span style=\"color: #993366;\"><em>see <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2013\/06\/11\/alkynes-via-elimination-reactions\/\">Alkynes via Elimination Reactions<\/a><\/em><\/span>).<\/p>\n<p>[<em><span style=\"color: #800080;\">See article &#8211;<\/span> <a href=\"https:\/\/www.masterorganicchemistry.com\/2023\/01\/18\/where-will-substitution-elimination-reactions-occur\/\">Identifying Where SN1\/SN2\/E1\/E2 Reactions Happen<\/a> <\/em>]<\/p>\n\n<p class=\"p1\"><img decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/2677-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/a><\/p>\n<p><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC Member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><br \/>\nOnce you&#8217;ve identified a good leaving group on an sp<sup>3<\/sup>-hybridized carbon, you&#8217;re ready for the next step.<\/p>\n<h2><a id=\"four\"><\/a>4. Identify the Carbon As Primary, Secondary, Tertiary or Methyl<\/h2>\n<p>If you&#8217;ve identified a good leaving group on an alkyl (sp<sup>3<\/sup>-hybridized carbon) then the next step is to classify that carbon as primary, secondary, tertiary or methyl. [<span style=\"color: #800080;\"><em>See <span style=\"color: #800080;\"><a style=\"color: #800080;\" href=\"https:\/\/www.masterorganicchemistry.com\/2010\/06\/16\/1-2-3-4\/\">Primary, Secondary, Tertiary In Organic Chemistry<\/a><\/span><\/em><\/span>].<\/p>\n<p>The number of carbons directly attached to the carbon bearing the leaving group has a profound impact on its reactivity since it will affect its steric hindrance (less steric hindrance for primary, more steric hindrance for tertiary) and ability to stabilize positive charge as a carbocation (less stability for primary, more stability for tertiary).<\/p>\n<p>See article: [<a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/11\/21\/deciding-sn1sn2e1e2-1-the-substrate\/\">Deciding SN1\/SN2\/E1\/E2: The Substrate<\/a>]<\/p>\n\n<p class=\"p1\"><img decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/2678-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/a><\/p>\n<p><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a\u00a0 MOC member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><\/p>\n<p>At this stage, identifying the carbon bearing the leaving group\u00a0 as primary, methyl or tertiary carbons will be helpful at ruling certain reactions in or out.<\/p>\n<ul>\n<li>If it&#8217;s\u00a0<strong>methyl<\/strong>, it&#8217;s S<sub>N<\/sub>2<\/li>\n<li>If it&#8217;s <strong>primary<\/strong>, it&#8217;s S<sub>N<\/sub>2 <em>unless\u00a0<\/em>a strong, bulky base is added. [<a href=\"#notetwo\">Note 2<\/a>]<\/li>\n<li>If it&#8217;s <strong>secondary<\/strong>, you can&#8217;t rule anything out yet.<\/li>\n<li>If it&#8217;s <strong>tertiary<\/strong>, rule out S<sub>N<\/sub>2 due to steric hindrance (S<sub>N<\/sub>1, E1, E2 all possible)<\/li>\n<\/ul>\n<h2><a id=\"five\"><\/a>5. Identify The Strongest Base \/ Nucleophile<\/h2>\n<p>The next step is to identify the\u00a0<strong>nucleophile<\/strong> \/\u00a0<strong>base<\/strong> that is participating in the reaction.<\/p>\n<p>For our purposes, focus on the\u00a0<strong>strongest\u00a0<\/strong>base\/nucleophile that is present, and ignore anything that is weaker.<\/p>\n<p>By &#8220;strongest&#8221;, I generally mean focus on <strong>negatively charged<\/strong> bases\/nucleophiles first, as they will be more likely to participate in these reactions than neutral bases\/nucleophiles. [<a href=\"#notethree\">Note 3]<\/a><\/p>\n<p>For example, if you see NaOEt\/EtOH, focus on NaOEt\u00a0 (which, ignoring sodium, we can just think of as CH<sub>3<\/sub>CH<sub>2<\/sub>O(-) for our purposes), since the conjugate base is always the better nucleophile.<\/p>\n<p><span style=\"color: #800080;\"><em>See article: <a style=\"color: #800080;\" href=\"https:\/\/www.masterorganicchemistry.com\/2012\/11\/30\/deciding-sn1sn2e1e2-2-the-nucleophilebase\/\">Deciding SN1\/SN2\/E1\/E2: The\u00a0 Nucleophile\/Base<\/a><\/em><\/span><\/p>\n\n<p class=\"p1\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/2679-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/a><\/p>\n<p><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC member<\/strong><\/a> to see the backside with answer.<\/p>\n<p><\/p>\n<p>The identity of the nucleophile\/base is useful for separating out S<sub>N<\/sub>2\/E2 from S<sub>N<\/sub>1\/E1 pathways.<\/p>\n<ul>\n<li>If the substrate is\u00a0<strong>primary<\/strong> the only exception to S<sub>N<\/sub>2 is if the nucleophile\/base is <strong>strongly basic<\/strong> <em>and<\/em> <strong>bulky<\/strong>. Two prominent examples are the\u00a0<em>t<\/em>-butoxide ion\u00a0\u00a0<em>t<\/em>BuO(-) and <a href=\"https:\/\/www.masterorganicchemistry.com\/2011\/08\/05\/reagent-friday-lithium-di-isopropyl-amide-lda\/\"> lithium diisopropyl amide<\/a> (LDA) . <span style=\"color: #993366;\"><em>(A third, more rarely seen example is the strong bulky base DBU)<\/em><\/span>.\u00a0 All of these reagents will perform E2 instead of S<sub>N<\/sub>2, giving the less substituted alkene (non-Zaitsev) product. (<span style=\"color: #993366;\"><em>See article &#8211; <span style=\"color: #800080;\"><a style=\"color: #800080;\" href=\"https:\/\/www.masterorganicchemistry.com\/2012\/10\/24\/bulky-bases-in-elimination-reactions\/\">Bulky Bases In Elimination Reactions<\/a><\/span><\/em><\/span>).<\/li>\n<li>If the substrate is\u00a0<strong>secondary<\/strong> and the nucleophile is strong but <strong>weakly basic<\/strong>, it&#8217;s S<sub>N<\/sub>2, particularly in a polar aprotic solvent like DMF, DMSO, acetone or acetonitrile. &#8220;Weakly basic&#8221; here means the conjugate acid has a pK<sub>a<\/sub> of 12 and below. So thiolates\u00a0 RS(-) are &#8220;weakly basic&#8221; but hydroxide [HO(-)] and alkoxide [RO(-)] are not.<\/li>\n<li>If the substrate is <strong>secondary<\/strong> and the nucleophile is <strong>strongly basic<\/strong>, it&#8217;s E2. This is definitely the case for acetylides (the conjugate bases of terminal acetylenes, pK<sub>a<\/sub> = 25) and amides (the conjugate bases of amines, pK<sub>a<\/sub> = 35-38).<span style=\"color: #800080;\"><em> For hydroxides (HO-) and alkoxides (RO-) confirm with your instructor.<\/em><\/span><\/li>\n<li>If the substrate is <strong>secondary<\/strong> and the nucleophile is <strong>weak<\/strong>, it&#8217;s possibly S<sub>N<\/sub>1\/E1. In particular look out for examples where carbocation rearrangements can occur, because these are commonly tested in this situation.<\/li>\n<li>If the substrate is <strong>tertiary<\/strong> and the nucleophile is <strong>weak<\/strong>, you&#8217;re likely looking at S<sub>N<\/sub>1\/E1<\/li>\n<li>If the substrate is\u00a0<strong>tertiary\u00a0<\/strong>and the nucleophile is\u00a0<strong>strongly basic<\/strong>, then expect E2.<\/li>\n<\/ul>\n<h2><a id=\"six\"><\/a>6. Heat Usually Favors Elimination<\/h2>\n<p>Sometimes the temperature is noted. If <strong>heat<\/strong> (or \u0394) is indicated, that&#8217;s generally a clue that <strong>elimination <\/strong>will be taking place, since elimination reactions are promoted by heat.<\/p>\n<p>With <strong>tertiary <\/strong>alkyl halides, E1 is generally the major pathway over S<sub>N<\/sub>1 in the presence of heat.<\/p>\n<p>[See article: <a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/12\/19\/sn1-vs-e1-temperature-sn2-vs-e2\/\">Deciding SN1\/SN2\/E1\/E2: The Temperature<\/a>]<\/p>\n\n<p class=\"p1\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/2680-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/a><\/p>\n<p><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><\/p>\n<p>With <strong>secondary<\/strong> alkyl halides, be alert for one possible exception to this &#8211; S<sub>N<\/sub>1 reactions with rearrangements.<\/p>\n<p>Generally, leaving groups on secondary alkyl carbons don&#8217;t pop off very easily. Secondary carbocations are fairly unstable, after all!<\/p>\n<p>In some instances, you may see <strong>&#8220;heat&#8221;<\/strong> in the presence of a secondary alkyl halide, which will increase the rate of the leaving group leaving to form the secondary carbocation. This can then be followed by a hydride or alkyl shift which will generate a more stable tertiary carbocation. Again, you might want to confirm this with your instructor.<\/p>\n<h2><a id=\"seven\"><\/a>7. The Role of Solvent. Also, Secondary Alkyl Halides With Strongly Basic Nucleophiles<\/h2>\n<p>You may note that we haven&#8217;t really discussed solvent here.<\/p>\n<p>In S<sub>N<\/sub>1\/E1 reactions, the nucleophile often <strong>is<\/strong> the solvent. So if you see a reaction where there is only &#8220;H<sub>2<\/sub>O&#8221; or &#8220;EtOH&#8221; , then you already know that the reaction is being run in a polar protic solvent with a poor nucleophile.<\/p>\n<p>For secondary alkyl halides with <strong>weakly basic<\/strong> nucleophiles,\u00a0<strong>polar aprotic solvents<\/strong> such as DMF, DMSO, acetone, and acetonitrile are generally good indicators for\u00a0<strong>S<sub>N<\/sub>2 reactions<\/strong>, since these solvents enhance the nucleophilicity of many anions. (<span style=\"color: #800080;\"><em>See article<span style=\"color: #800080;\">: <a style=\"color: #800080;\" href=\"https:\/\/www.masterorganicchemistry.com\/2012\/04\/27\/polar-protic-polar-aprotic-nonpolar-all-about-solvents\/\">All About Solvents<\/a><\/span><\/em><\/span>)<\/p>\n<p>A common dilemma arises in how to treat secondary alkyl halides with\u00a0<strong>alkoxide<\/strong> and\u00a0<strong>hydroxide<\/strong> nucleophiles in the presence of a\u00a0<strong>polar aprotic solvent.\u00a0<\/strong><\/p>\n<p>Looking at the experimental data, the only conclusion to draw is that E2 is still favored over S<sub>N<\/sub>2 in this situation. Check with your instructor what they do. Experimental results say E2, but there&#8217;s a lot of inconsistency.<\/p>\n<p>See article: <a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/12\/04\/deciding-sn1sn2e1e2-the-solvent\/\">Deciding SN1\/SN2\/E1\/E2 &#8211; Secondary Alkyl Halides With Strong Bases<\/a>]<\/p>\n\n<p class=\"p1\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/2681-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/a><\/p>\n<p><a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>Become a MOC member<\/strong><\/a> to see the clickable quiz with answers on the back.<\/p>\n<p><\/p>\n<h2><a id=\"eight\"><\/a>8. Summary<\/h2>\n<p>Once you have a reasonable idea of what reaction pathway is operating, makes sure you have adequate practice in actually\u00a0<em>drawing<\/em> the final product of these reactions. Hopefully this article has been useful in this regard, but if you need more practice, consult the original articles on substitution and elimination, or join the <a href=\"https:\/\/www.masterorganicchemistry.com\/moc-membership\/\"><strong>MOC Membership<\/strong><\/a> for a host of representative practice problems.<\/p>\n<hr \/>\n<h2><strong><a id=\"notes\"><\/a>Notes<\/strong><\/h2>\n<div class=\"related-articles\"><p><strong>Related Articles<\/strong><\/p><ul><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2014\/01\/10\/reactions-of-alkyl-halides\/\" class=\"\"><span>Alkyl Halide Reaction Map And Summary<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/organic-chemistry-practice-problems\/sn1-sn2-e1-e2-practice-problems\/\" class=\"\"><span>SN1 SN2 E1 E2 Practice Problems (MOC Membership)<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/07\/13\/the-sn1-mechanism\/\" class=\"\"><span>The SN1 Mechanism<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/07\/04\/the-sn2-mechanism\/\" class=\"\"><span>The SN2 Mechanism<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/09\/19\/the-e1-reaction\/\" class=\"\"><span>The E1 Reaction<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/09\/27\/the-e2-mechanism\/\" class=\"\"><span>The E2 Mechanism<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2023\/01\/18\/where-will-substitution-elimination-reactions-occur\/\" class=\"\"><span>Identifying Where Substitution and Elimination Reactions Happen<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/11\/21\/deciding-sn1sn2e1e2-1-the-substrate\/\" class=\"\"><span>Deciding SN1\/SN2\/E1\/E2 (1) \u2013 The Substrate<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/11\/30\/deciding-sn1sn2e1e2-2-the-nucleophilebase\/\" class=\"\"><span>Deciding SN1\/SN2\/E1\/E2 (2) \u2013 The Nucleophile\/Base<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/12\/19\/sn1-vs-e1-temperature-sn2-vs-e2\/\" class=\"\"><span>SN1 vs E1 and SN2 vs E2 : The Temperature<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/12\/04\/deciding-sn1sn2e1e2-the-solvent\/\" class=\"\"><span>Deciding SN1\/SN2\/E1\/E2  \u2013 The Solvent<\/span><\/a><\/li><\/ul><\/div>\n<p><strong><a id=\"noteone\"><\/a>Note 1.\u00a0<\/strong>From the original Quick N&#8217; Dirty Guide To S<sub>N<\/sub>1\/S<sub>N<\/sub>2\/E1\/E2:<\/p>\n<p><em>&#8220;Writing this post makes me feel like a nun giving out condoms. \u00a0I realize there will be many who are reading this an hour before their exam and are completely clueless on this subject. All I have to say is, God help you. And do more fricking practice problems so you don\u2019t put yourself in this situation next time.&#8221;<\/em><\/p>\n<p><strong><a id=\"notetwo\"><\/a>Note 2. <\/strong>\u00a0primary carbons that can form relatively stable carbocations (i.e. allylic\/benzylic) may proceed through the S<sub>N<\/sub>1\/E1 pathway if only poor nucleophiles are present.<\/p>\n<p><strong><a id=\"three\"><\/a>Note 3<\/strong>. Some competent <strong>neutral <\/strong>bases\/nucleophiles include:<\/p>\n<ul>\n<li><strong>Thiols:<\/strong> (R-SH, H<sub>2<\/sub>S) Can be considered to be good nucleophiles, will participate in SN2 (but not E2) reactions.<\/li>\n<li><strong>Amines<\/strong>: (R-NH<sub>2<\/sub>, R<sub>2<\/sub>NH, R<sub>3<\/sub>N); will act as nucleophiles (S<sub>N<\/sub>2) with primary alkyl halides; expect elimination (E2) with secondary and tertiary alkyl halides.<\/li>\n<li><strong>Phosphines\u00a0<\/strong>(PPh<sub>3<\/sub>) : Good nucleophiles, weak bases. Expect S<sub>N<\/sub>2 reactions with primary and secondary alkyl halides.<\/li>\n<\/ul>\n<hr \/>\n<h2><strong><a id=\"quiz\"><\/a>Quiz Yourself!<\/strong><\/h2>\n<p>[Quizzes]<\/p>\n<hr \/>\n<h2><strong><a id=\"references\"><\/a>(Advanced) References and Further Reading<\/strong><\/h2>\n<ol>\n<li><strong>An Improved Decision Tree for Predicting a Major Product in Competing Reactions<\/strong><br \/>\nKate J. Graham<br \/>\n<em>Journal of Chemical Education<\/em> <strong>2014<\/strong> 91 (8), 1267-1268<br \/>\n<strong>DOI<\/strong>: <a href=\"https:\/\/pubs.acs.org\/doi\/10.1021\/ed400908g\">10.1021\/ed400908g\u00a0<\/a><br \/>\nFrom the abstract: &#8220;<em>When organic chemistry students encounter competing reactions, they are often overwhelmed by the task of evaluating multiple factors that affect the outcome of a reaction. The use of a decision tree is a useful tool to teach students to evaluate a complex situation and propose a likely outcome. Specifically, a decision tree can help students predict a major product in substitution and elimination reactions<\/em>.&#8221;<\/li>\n<li><strong>Comparing Nucleophilic Substitutions Versus Elimination Reactions In Comprehensive Introductory Organic Chemistry Textbooks<br \/>\n<\/strong>Donna J. Nelson<br \/>\n<em>Proceedings of the Oklahoma Academy of Sciences<strong>\u00a0<\/strong><\/em><strong>2022<\/strong>,\u00a0<em>102<\/em>, 105-114<br \/>\n<strong>DOI:\u00a0<\/strong>Direct link (<a href=\"https:\/\/ojs.library.okstate.edu\/osu\/index.php\/OAS\/article\/view\/9447\/8444\">PDF<\/a>)<br \/>\nA comparison of how 17 commonly used organic chemistry textbooks treat the key factors that differentiate S<sub>N<\/sub>1, S<sub>N<\/sub>2, E1, and E2 reactions (base strength, temperature, steric effects, nucleophilicity).<\/li>\n<\/ol>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-36161\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2024\/04\/F1-Donna-Nelson-textbook-study-impact.png\" alt=\"Donna Nelson textbook study sn1 sn2 e1 e2 impact\" width=\"640\" height=\"452\" srcset=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2024\/04\/F1-Donna-Nelson-textbook-study-impact.png 1094w, https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2024\/04\/F1-Donna-Nelson-textbook-study-impact-300x212.png 300w, https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2024\/04\/F1-Donna-Nelson-textbook-study-impact-1024x723.png 1024w, https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2024\/04\/F1-Donna-Nelson-textbook-study-impact-768x542.png 768w, https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2024\/04\/F1-Donna-Nelson-textbook-study-impact-760x536.png 760w\" sizes=\"(max-width: 640px) 100vw, 640px\" \/><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>SN1\/SN2\/E1\/E2 &#8211; Summarizing The Key Factors That Determine Whether A Reaction Will Be SN1, SN2, E1 or E2 In this article we walk through the <\/p>\n","protected":false},"author":1,"featured_media":36362,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[1417],"tags":[471,472,473,201,868,825,502,271,838,279],"post_folder":[],"class_list":["post-6877","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sn1sn2e1e2","tag-base","tag-e1","tag-e2","tag-elimination","tag-heat","tag-nucleophile","tag-sn1","tag-sn2","tag-solvent","tag-substitution"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.7 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Wrapup: The Key Factors For Determining SN1\/SN2\/E1\/E2<\/title>\n<meta name=\"description\" content=\"How do you tell if a reaction goes through SN1, SN2, E1 or E2? 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