{"id":8573,"date":"2014-10-24T17:25:36","date_gmt":"2014-10-24T21:25:36","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=8573"},"modified":"2025-02-16T11:05:08","modified_gmt":"2025-02-16T17:05:08","slug":"the-williamson-ether-synthesis","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2014\/10\/24\/the-williamson-ether-synthesis\/","title":{"rendered":"The Williamson Ether Synthesis"},"content":{"rendered":"<p><strong>The Williamson Ether Synthesis<\/strong><\/p>\n<ul>\n<li class=\"p1\"><span class=\"s1\">In the Williamson Ether Synthesis, an\u00a0<b>alkyl halide<\/b>\u00a0<i>(or sulfonate, such as a tosylate or mesylate)<\/i>\u00a0undergoes nucleophilic substitution (S<sub>N<\/sub>2) by an\u00a0<b>alkoxide<\/b>\u00a0to give an\u00a0<b>ether<\/b>.<\/span><\/li>\n<li class=\"p1\">Being an S<sub>N<\/sub>2 reaction, best results are obtained with <strong>primary alkyl<\/strong>\u00a0<strong>halides <\/strong>or <strong>methyl<\/strong> halides. Tertiary alkyl halides give elimination instead of ethers. Secondary alkyl halides will give a mixture of elimination and substitution.<\/li>\n<li>The <strong>alkoxide<\/strong> RO<sup>&#8211;<\/sup>\u00a0 can be those of methyl, primary, secondary, or tertiary alcohols.<\/li>\n<li>The reaction is often run with a mixture of the alkoxide and its parent alcohol (e.g. NaOEt\/EtOH or CH<sub>3<\/sub>ONa\/CH<sub>3<\/sub>OH). Alternatively, a strong base may be added to the alcohol to give the alkoxide. Sodium hydride (NaH) or potassium hydride (KH) are popular choices.<\/li>\n<li>When an alkoxide and alkyl halide are present on the same molecule, an <strong>intramolecular<\/strong> reaction may result to give a new ring. This works best for 5- and 6-membered rings.<\/li>\n<li>When planning the synthesis of ethers using the Williamson, take care to select the best starting materials for an S<sub>N<\/sub>2 reaction. Avoid planning a Williamson involving a tertiary alkyl halide or a phenyl halide!<\/li>\n<\/ul>\n<p><img fetchpriority=\"high\" decoding=\"async\" class=\"alignnone wp-image-35192\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2023\/07\/0-summary-of-the-williamson-ether-synthesis-forming-an-ether-from-alkyl-halide-and-alkoxide.gif\" alt=\"summary of the williamson ether synthesis forming an ether from alkyl halide and alkoxide\" width=\"640\" height=\"599\" \/><\/a><\/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 Williamson Ether Synthesis<\/a><\/li>\n<li><a href=\"#two\">Mechanism of the Williamson Ether Synthesis is SN2<\/a><\/li>\n<li><a href=\"#three\">Primary and Methyl Alkyl Halides Work Best<\/a><\/li>\n<li><a href=\"#four\">Solvent Choice In The Williamson<\/a><\/li>\n<li><a href=\"#five\">Intramolecular Williamson Ether Syntheses<\/a><\/li>\n<li><a href=\"#six\">Planning Ether Synthesis via the Williamson<\/a><\/li>\n<li><a href=\"#seven\">Summary<\/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. The Williamson Ether Synthesis<\/h2>\n<p>One of the simplest and most versatile ways for making <strong>ethers<\/strong> is the S<sub>N<\/sub>2 reaction between an alkoxide (<span style=\"color: #993366;\"><em>RO<sup>&#8211;<\/sup>, the conjugate base of an alcohol<\/em><\/span>) and an alkyl halide.<\/p>\n<p>Although this is a very old reaction &#8211; the first report was in 1850! &#8211; it just hasn&#8217;t been surpassed. It works well for making a variety of ethers and is known as the <strong>Williamson Ether Synthesis<\/strong>.<\/p>\n<p><img decoding=\"async\" class=\"alignnone wp-image-35457\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2023\/09\/1-typical-example-of-williamson-ether-synthesis-sn2-reaction-between-alkoxide-and-alcohol.gif\" alt=\"typical example of williamson ether synthesis sn2 reaction between alkoxide and alcohol\" width=\"640\" height=\"367\" \/><\/a><\/p>\n<p>S<sub>N<\/sub>2 reactions between neutral alcohols and alkyl halides are generally quite <strong>slow<\/strong>. Since the <strong>conjugate base of any species is a better nucleophile<\/strong>, the reaction is sped up considerably (<span style=\"color: #ff0000;\">Note 1<\/span>) by employing an <strong>alkoxide<\/strong> instead of a neutral alcohol. [<span style=\"color: #993366;\"><em>See article: <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2012\/06\/18\/what-makes-a-good-nucleophile\/\">What Makes A Good Nucleophile<\/a>?<\/em><\/span>]<\/p>\n<p>A common way to do the Williamson is to simply use the alkoxide nucleophile with its parent alcohol as solvent<span style=\"color: #993366;\"><em> (indeed, that&#8217;s how it was done in 1850!)\u00a0<\/em><\/span><\/p>\n<p>For example, the classic way to make diethyl ether is to treat the ethyl halide (<span style=\"color: #993366;\"><em>the chloride, bromide, or iodide all work, but not the fluoride<\/em><\/span>) with the ethoxide ion in ethanol.<\/p>\n<p><img decoding=\"async\" class=\"alignnone wp-image-35458\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2023\/09\/2-reaction-between-alcohols-and-alkyl-halides-is-very-slow-compared-to-alkoxides-which-are-much-better-nucleophiles.gif\" alt=\"-reaction between alcohols and alkyl halides is very slow compared to alkoxides which are much better nucleophiles\" width=\"640\" height=\"353\" \/><\/a><\/p>\n<p>For our purposes the identity of the leaving group (iodide, bromide, chloride, tosylate (OTs), mesylate (OMs) ) does not really matter, although it is important to know their relative leaving group abilities (<span style=\"color: #993366;\"><em>See article<span style=\"color: #993366;\">: <a style=\"color: #993366;\" 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><span style=\"color: #993366;\"><em>Likewise the identity of the alkali metal salt (Li<sup>+<\/sup> , Na<sup>+<\/sup>, K<sup>+<\/sup>, etc.) does not matter much for our purposes and we will use these metal salts interchangeably.\u00a0<\/em><\/span><\/p>\n<p>Since it&#8217;s not always an option to use the alcohol as solvent, another option is to generate the alkoxide by using a strong base that will irreversibly deprotonate the alcohol.<\/p>\n<p>A popular choice is the <strong>hydride<\/strong> ion H<sup>&#8211;\u00a0<\/sup>, which is the conjugate base of hydrogen gas (pK<sub>a<\/sub> about 35) and is also a poor nucleophile to boot. Sodium hydride (NaH) or potassium hydride (KH) can be added to the starting alcohol to generate the alkoxide. The hydrogen gas byproduct then bubbles out of solution into the atmosphere.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-35459\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2023\/09\/3-williamson-ether-synthesis-use-of-base-nah-or-kh-in-deprotonating-alcohol-to-give-alkoxide.gif\" alt=\"-reaction between alcohols and alkyl halides is very slow compared to alkoxides which are much better nucleophiles\" width=\"640\" height=\"317\" \/><\/a><\/p>\n<h2><a id=\"two\"><\/a>2. The Williamson Ether Synthesis Proceeds Through an S<sub>N<\/sub>2 Mechanism<\/h2>\n<p>The Williamson ether synthesis is a <strong>substitution<\/strong> reaction, where a bond is formed and broken on the same carbon atom. In this substitution reaction, a new C-O bond is formed, and a bond is broken between the carbon and the leaving group (LG) which is typically a halide or sulfonate.<\/p>\n<p>It proceeds through an <strong>S<sub>N<\/sub>2<\/strong> mechanism (<span style=\"color: #993366;\"><em>nucleophilic substitution, bimolecular<\/em><\/span>) where the nucleophile approaches the carbon atom from the <strong>backside<\/strong> of the carbon-leaving group bond.\u00a0 (<a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/07\/04\/the-sn2-mechanism\/\"><em><span style=\"color: #993366;\">See article: The S<sub>N<\/sub>2 Mechanism<\/span><\/em><\/a>)<\/p>\n<p>A pair of electrons from the nucleophile are donated into the sigma* (antibonding) orbital of the C-leaving group bond.<\/p>\n<p>This requires that the nucleophile actually makes its way to the orbital on the backside of the carbon!\u00a0 For this reason the S<sub>N<\/sub>2 is fastest for <strong>methyl<\/strong> and <strong>primary<\/strong> alkyl halides, and <strong>does not occur on tertiary alkyl halides<\/strong> due to the fact that nucleophiles can&#8217;t make their way through the tangled thicket of alkyl groups on the backside. (<span style=\"color: #993366;\"><em>See article: <span style=\"color: #993366;\"><a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2011\/07\/18\/steric-hindrance-is-like-a-fat-goalie\/\">Steric Hindrance Is Like A Fat Goalie<\/a><\/span><\/em><\/span>) .<\/p>\n<p>Substitution reactions of alkoxides with secondary alkyl halides\u00a0<strong>can<\/strong> occur, but often occur with significant elimination through the E2 pathway.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-35460\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2023\/09\/4-mechanism-of-the-williamson-ether-synthesis-is-sn2.gif\" alt=\"-reaction between alcohols and alkyl halides is very slow compared to alkoxides which are much better nucleophiles\" width=\"640\" height=\"438\" \/><\/a><\/p>\n<p>In the transition state of the S<sub>N<\/sub>2 there is a five-coordinate geometry about the carbon with partial bonds to the nucleophile and to the leaving group.<\/p>\n<p>As the carbon-nucleophile bond strengthens and the carbon-leaving group bond weakens, the geometry of the carbon becomes <strong>inverted<\/strong>, like the <a href=\"\" class=\"custom-tooltip\" data-image=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2023\/08\/Supp-1-George-Bush-Inverted-Umbrella.jpg\" data-link=\"\" data-title=\"\" data-text=\"\">proverbial umbrella in a strong wind<\/a>. This is generally not noticeable unless the alkyl halide carbon is a chiral center.<\/p>\n<h2><a id=\"three\"><\/a>3. Primary and Methyl Alkyl Halides Work Best<\/h2>\n<p>The Williamson works best for primary and methyl alkyl halides. Let&#8217;s look at some examples:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-35464\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2023\/08\/5-primary-alkyl-halides-are-best-substrates-for-the-williamson-ether-synthesis-since-it-is-an-sn2-reaction.gif\" alt=\"primary alkyl halides are best substrates for the williamson ether synthesis since it is an sn2 reaction\" width=\"640\" height=\"485\" \/><\/a><\/p>\n<p>Note that the third example the stereochemistry of the C-O bond is unaffected. That&#8217;s because it&#8217;s only the geometry of the\u00a0<strong>electrophile\u00a0<\/strong>(i.e. the alkyl halide) in the S<sub>N<\/sub>2 that becomes inverted.<\/p>\n<p>If the\u00a0<strong>alkyl halide\u00a0<\/strong>carbon is chiral, inversion can occur, as it does for this secondary alkyl halide below.<\/p>\n<p>Since alkoxides are strong bases, there will be significant competition between S<sub>N<\/sub>2 and E2 reactions with secondary alkyl halides. Alkenes and ethers are generally obtained as mixtures. [<span style=\"color: #ff0000;\">Note 2<\/span>].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-35465\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2023\/08\/6-secondary-alkyl-halides-in-the-williamson-ether-synthesis-give-a-mixture-of-elimination-and-substitution-products.gif\" alt=\"secondary alkyl halides in the williamson ether synthesis give a mixture of elimination and substitution products\" width=\"640\" height=\"302\" \/><\/a><\/p>\n<p>With tertiary alkyl halides, the Williamson ether reaction fails completely, and only alkenes are obtained.<\/p>\n<p>Since S<sub>N<\/sub>2 and E2 reactions generally do not occur with sp2 hybridized carbons, another case where Williamson reactions fail is with\u00a0<strong>aryl\u00a0<\/strong>and\u00a0<strong>alkenyl\u00a0<\/strong>halides.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-35466\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2023\/08\/7-attempts-to-use-tertiary-alkyl-halides-in-the-williamson-ether-synthesis-sn2-results-in-elimination-via-e2-mechanism.gif\" alt=\"attempts to use tertiary alkyl halides in the williamson ether synthesis sn2 results in elimination via e2 mechanism\" width=\"640\" height=\"388\" \/><\/a><\/p>\n<h2><a id=\"four\"><\/a>4. Choice of Solvent For Ether Formation<\/h2>\n<p>A common choice of solvent for the Williamson is to use the parent alcohol of the alkoxide, such as ethanol when using sodium ethoxide.<\/p>\n<p>It is generally a bad idea to use an alcoholic solvent that is not the conjugate acid of the alkoxide, as discussed in the quiz below:<\/p>\n<div class=\"wq-quiz-wrapper\" data-id=\"35467\"><style type=\"text\/css\" id=\"wq-flip-custom-css\">.wq-quiz-wrapper[data-id=\"35467\"] {\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=\"35467\"] {\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=\"35467\"] .wq_singleQuestionWrapper { width:100% !important; height:auto !important; }\n\t\t\t}\n\t\t<\/style><!-- wp quiz -->\n<div id=\"wp-quiz-35467\" class=\"wq_quizCtr single flip_quiz wq-quiz wq-quiz-35467 wq-quiz-flip wq-layout-single wq-skin-traditional wq-should-show-correct-answer\" data-quiz-id=\"35467\">\n<div class=\"wq-questions wq_questionsCtr\">\n\t<div class=\"wq-question wq_singleQuestionWrapper wq-question-prdvv\" data-id=\"prdvv\">\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\/2565-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\/2565-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>When using sodium hydride (NaH) or potassium hydride (KH) a common choice of solvent is ethers such as tetrahydrofuran (THF), diethyl ether, or polar aprotic solvents such as dimethylsulfoxide (DMSO).<\/p>\n<h2><a id=\"five\"><\/a>5. Intramolecular Williamson Reactions<\/h2>\n<p>If an alcohol and alkyl halide are present on the same molecule, there is the potential for an\u00a0<strong>intramolecular\u00a0<\/strong>Williamson ether reaction to occur.<\/p>\n<p>This will result in a new ring.<\/p>\n<p>Ring formation is best for 5- and 6-membered rings. [<span style=\"color: #ff0000;\">Note 3<\/span> ]<\/p>\n<p>The mechanism for the intramolecular Williamson reaction is identical to that for a normal S<sub>N<\/sub>2; it just looks weird the first time you see it.<\/p>\n<p>See if you can draw the mechanism:<\/p>\n<div class=\"wq-quiz-wrapper\" data-id=\"35468\"><style type=\"text\/css\" id=\"wq-flip-custom-css\">.wq-quiz-wrapper[data-id=\"35468\"] {\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=\"35468\"] {\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=\"35468\"] .wq_singleQuestionWrapper { width:100% !important; height:auto !important; }\n\t\t\t}\n\t\t<\/style><!-- wp quiz -->\n<div id=\"wp-quiz-35468\" class=\"wq_quizCtr single flip_quiz wq-quiz wq-quiz-35468 wq-quiz-flip wq-layout-single wq-skin-traditional wq-should-show-correct-answer\" data-quiz-id=\"35468\">\n<div class=\"wq-questions wq_questionsCtr\">\n\t<div class=\"wq-question wq_singleQuestionWrapper wq-question-444v1\" data-id=\"444v1\">\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\/2566-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\/2566-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>It&#8217;s also valuable to be able to work backwards from a final product to propose plausible starting materials. See if you can draw a feasible starting material that will result in the Williamson product below:<\/p>\n<div class=\"wq-quiz-wrapper\" data-id=\"35469\"><style type=\"text\/css\" id=\"wq-flip-custom-css\">.wq-quiz-wrapper[data-id=\"35469\"] {\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=\"35469\"] {\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=\"35469\"] .wq_singleQuestionWrapper { width:100% !important; height:auto !important; }\n\t\t\t}\n\t\t<\/style><!-- wp quiz -->\n<div id=\"wp-quiz-35469\" class=\"wq_quizCtr single flip_quiz wq-quiz wq-quiz-35469 wq-quiz-flip wq-layout-single wq-skin-traditional wq-should-show-correct-answer\" data-quiz-id=\"35469\">\n<div class=\"wq-questions wq_questionsCtr\">\n\t<div class=\"wq-question wq_singleQuestionWrapper wq-question-xg2v7\" data-id=\"xg2v7\">\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\/2567-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\/2567-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<h2><a id=\"six\"><\/a>6. Planning Syntheses of Ethers via The Williamson<\/h2>\n<p>All right. Given all we have said about the Williamson, let&#8217;s see some examples of using this reaction to plan some syntheses of ethers.<\/p>\n<p>With any ether there are\u00a0<strong>two\u00a0<\/strong>potential sites where a new C-O bond could be formed, which gives you <strong>two<\/strong> alkyl halide + alkoxide combinations to choose from.<\/p>\n<p>Let&#8217;s start with a fairly easy example. See if you can come up with reasonable starting materials for the synthesis of <strong>propyl methyl ether<\/strong> via the Williamson reaction.<\/p>\n<div class=\"wq-quiz-wrapper\" data-id=\"35470\"><style type=\"text\/css\" id=\"wq-flip-custom-css\">.wq-quiz-wrapper[data-id=\"35470\"] {\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=\"35470\"] {\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=\"35470\"] .wq_singleQuestionWrapper { width:100% !important; height:auto !important; }\n\t\t\t}\n\t\t<\/style><!-- wp quiz -->\n<div id=\"wp-quiz-35470\" class=\"wq_quizCtr single flip_quiz wq-quiz wq-quiz-35470 wq-quiz-flip wq-layout-single wq-skin-traditional wq-should-show-correct-answer\" data-quiz-id=\"35470\">\n<div class=\"wq-questions wq_questionsCtr\">\n\t<div class=\"wq-question wq_singleQuestionWrapper wq-question-oywq9\" data-id=\"oywq9\">\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\/2568-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\/2568-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><span style=\"color: #800080;\"><em>Note that for our purposes it doesn&#8217;t matter whether you choose Cl, Br, I or another good leaving group for your alkyl halide\/sulfonate, so long that it isn&#8217;t F.\u00a0<\/em><\/span><\/p>\n<p>With propyl methyl ether, there are actually two good choices for building the ether. You can either use propyl halide (primary) with methoxide, or methyl halide (methyl) with propoxide. Both of these S<sub>N<\/sub>2 reactions should work well.<\/p>\n<p><span style=\"color: #000000;\">Since elimination is absolutely impossible on methyl, I&#8217;d give a slight preference to using the propoxide\/CH<sub>3<\/sub>Br combination, but either one would work.\u00a0<\/span><\/p>\n<p>Ready for the next one? See if you can come up with a plan for the ether below:<\/p>\n<div class=\"wq-quiz-wrapper\" data-id=\"35471\"><style type=\"text\/css\" id=\"wq-flip-custom-css\">.wq-quiz-wrapper[data-id=\"35471\"] {\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=\"35471\"] {\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=\"35471\"] .wq_singleQuestionWrapper { width:100% !important; height:auto !important; }\n\t\t\t}\n\t\t<\/style><!-- wp quiz -->\n<div id=\"wp-quiz-35471\" class=\"wq_quizCtr single flip_quiz wq-quiz wq-quiz-35471 wq-quiz-flip wq-layout-single wq-skin-traditional wq-should-show-correct-answer\" data-quiz-id=\"35471\">\n<div class=\"wq-questions wq_questionsCtr\">\n\t<div class=\"wq-question wq_singleQuestionWrapper wq-question-qsuuh\" data-id=\"qsuuh\">\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\/2569-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\/2569-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>Working backwards gives us two sets of reactants. The one that will work\u00a0<strong> best<\/strong> will be the alkoxide of the secondary alcohol with a\u00a0<strong>methyl\u00a0<\/strong>halide, since S<sub>N<\/sub>2 reactions are fastest on methyl groups and elimination reactions (E2) are impossible and will not compete.<\/p>\n<p>A worse choice would be a secondary alkyl halide with methoxide. Note that since this is an S<sub>N<\/sub>2\u00a0 &#8211;\u00a0 which proceeds with inversion of configuration &#8211; and the secondary alkyl halide <strong>chiral<\/strong>, the starting alkyl halide will have the\u00a0<strong>opposite\u00a0<\/strong>configuration to that of the starting material.<\/p>\n<p>Let&#8217;s try <em>t<\/em>-butyl ethyl ether next.<\/p>\n<div class=\"wq-quiz-wrapper\" data-id=\"35472\"><style type=\"text\/css\" id=\"wq-flip-custom-css\">.wq-quiz-wrapper[data-id=\"35472\"] {\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=\"35472\"] {\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=\"35472\"] .wq_singleQuestionWrapper { width:100% !important; height:auto !important; }\n\t\t\t}\n\t\t<\/style><!-- wp quiz -->\n<div id=\"wp-quiz-35472\" class=\"wq_quizCtr single flip_quiz wq-quiz wq-quiz-35472 wq-quiz-flip wq-layout-single wq-skin-traditional wq-should-show-correct-answer\" data-quiz-id=\"35472\">\n<div class=\"wq-questions wq_questionsCtr\">\n\t<div class=\"wq-question wq_singleQuestionWrapper wq-question-15zm1\" data-id=\"15zm1\">\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\/2570-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\/2570-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>Our two choices of reactants include<\/p>\n<ul>\n<li>A tertiary alkoxide with a primary alkyl halide<\/li>\n<li>A tertiary alkyl halide with a primary alkoxide<\/li>\n<\/ul>\n<p>Hopefully by now it should be clear that the first choice is best, since <strong>primary\u00a0<\/strong>alkyl halides are excellent substrates for S<sub>N<\/sub>2 reactions and <strong>tertiary<\/strong> alkyl halides are not.<\/p>\n<p>Even though the alkoxide is tertiary, the reaction should still work fairly well. Being bulky, there will be more elimination (E2) than normal, but this could be minimized by keeping the reaction temperature relatively low. [<span style=\"color: #993366;\"><em>See article: <a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2012\/10\/24\/bulky-bases-in-elimination-reactions\/\">Bulky Bases In Elimination Reactions<\/a><\/em><\/span>]<\/p>\n<p>Our final quiz asks how to synthesize\u00a0phenyl methyl ether.<\/p>\n<div class=\"wq-quiz-wrapper\" data-id=\"35473\"><style type=\"text\/css\" id=\"wq-flip-custom-css\">.wq-quiz-wrapper[data-id=\"35473\"] {\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=\"35473\"] {\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=\"35473\"] .wq_singleQuestionWrapper { width:100% !important; height:auto !important; }\n\t\t\t}\n\t\t<\/style><!-- wp quiz -->\n<div id=\"wp-quiz-35473\" class=\"wq_quizCtr single flip_quiz wq-quiz wq-quiz-35473 wq-quiz-flip wq-layout-single wq-skin-traditional wq-should-show-correct-answer\" data-quiz-id=\"35473\">\n<div class=\"wq-questions wq_questionsCtr\">\n\t<div class=\"wq-question wq_singleQuestionWrapper wq-question-mbu3f\" data-id=\"mbu3f\">\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\/2571-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\/2571-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>Our choices are:<\/p>\n<ul>\n<li>A methyl halide with a phenyl alkoxide<\/li>\n<li>An aryl halide with methoxide<\/li>\n<\/ul>\n<p>Again, there is a very clear good choice here, as sp<sup>2<\/sup> hybridized carbons (aryl halides) do not undergo S<sub>N<\/sub>2 reactions <span style=\"color: #993366;\"><em>(just try doing a backside attack inside that phenyl ring!<\/em><\/span>). In constrast, the reaction between the phenoxide and CH<sub>3<\/sub>Br should work much better without any complication.<\/p>\n<h2><a id=\"seven\"><\/a>7. Summary<\/h2>\n<p>So what are the key takeaways here?<\/p>\n<p>If you&#8217;ve already learned the S<sub>N<\/sub>2 reaction, this should mainly be a refresher.<\/p>\n<ul>\n<li>S<sub>N<\/sub>2 reactions work well for methyl and primary alkyl halides, don&#8217;t work for tertiary, alkenyl or aryl halides, and are pretty borderline for secondary alkyl halides.<\/li>\n<li>Make sure you are familiar with drawing intramolecular examples of this reaction, because instructors tend to love throwing these types of mechanisms at you on exams.<\/li>\n<li>Get comfortable with planning the synthesis of ethers through Williamson reactions. There will generally be two\u00a0 alkoxide\/alkyl halide combinations to choose from. Choose the best SN2 reaction available.<\/li>\n<\/ul>\n<p>There are situations where the Williamson is\u00a0<strong>not<\/strong> the best choice for ether synthesis and we must resort to other methods. <span style=\"color: #993366;\"><em>For more, see <span style=\"color: #993366;\"><a style=\"color: #993366;\" href=\"https:\/\/www.masterorganicchemistry.com\/2014\/11\/07\/synthesis-of-ethers-2-back-to-the-future\/\">Ethers From Alkenes, Tertiary Alkyl Halides, and Oxymercuration<\/a><\/span>.\u00a0<\/em><\/span><\/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\/11\/07\/synthesis-of-ethers-2-back-to-the-future\/\" class=\"\"><span>Ethers From Alkenes, Tertiary Alkyl Halides and Alkoxymercuration<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2011\/07\/18\/steric-hindrance-is-like-a-fat-goalie\/\" class=\"\"><span>Steric Hindrance is Like a Fat Goalie<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/06\/18\/what-makes-a-good-nucleophile\/\" class=\"\"><span>What Makes A Good Nucleophile?<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2011\/04\/12\/what-makes-a-good-leaving-group\/\" class=\"\"><span>What makes a good leaving group?<\/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\/27\/the-e2-mechanism\/\" class=\"\"><span>The E2 Mechanism<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2014\/10\/06\/how-to-make-alcohols-more-reactive\/\" class=\"\"><span>Alcohols Can Act As Acids Or Bases (And Why It Matters)<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2012\/10\/24\/bulky-bases-in-elimination-reactions\/\" class=\"\"><span>Bulky Bases in Elimination Reactions<\/span><\/a><\/li><li><a href=\"https:\/\/www.masterorganicchemistry.com\/2015\/06\/12\/intramolecular-reactions-of-alcohols-and-ethers\/\" class=\"\"><span>Intramolecular Reactions of Alcohols and Ethers<\/span><\/a><\/li><\/ul><\/div>\n<p><span style=\"color: #800080;\"><em>The contents of a separate article, &#8220;The Williamson Ether Synthesis &#8211; Planning&#8221; has been combined into this article &#8211; Sept 2023.\u00a0<\/em><\/span><\/p>\n<p><strong>Note 1 &#8211;<\/strong> [Background rate of SN2]<\/p>\n<p><strong>Note 2\u00a0<\/strong>&#8211; [Secondary alkyl halide]<\/p>\n<p><strong><a id=\"notethree\"><\/a>Note 3\u00a0<\/strong>&#8211; A Final Note About Rates of Ring Formation<\/p>\n<p>I should end with a cautionary note about reactions that lead to ring formation. One question that comes up a lot is, &#8220;<em>when do I know when a new ring will form<\/em>?&#8221; [Shortcut spoiler: <strong>yes to 5 and 6 (and 3)<\/strong>, generally &#8220;no&#8221; to rings 7 and above]<\/p>\n<p>Great question! This is one of those issues that makes organic chemistry &#8220;hard&#8221; for the beginner, but &#8220;deep&#8221; and &#8220;interesting&#8221; for the lifelong practitioner because there are several key factors that often work in opposite directions.<\/p>\n<p>First of all: <strong>Not all rings form at the same rate<\/strong>.\u00a0That is, the <strong>rate<\/strong> at which a ring will form is,\u00a0 to some extent, dependent on the length of the chain.<\/p>\n<p>How does this fit in with what we already know about substitution reactions?<\/p>\n<p>Remember that the rate of a substitution reaction is proportional to the concentration of nucleophile and the concentration of electrophile. But what happens when the nucleophile and electrophile are on the same molecule? For this we use a concept called &#8220;effective concentration&#8221; which is to say that the <strong>reaction rate will be related to how much time the nucleophile spends in the vicinity of the electrophile.<\/strong> There isn&#8217;t space to go into this in detail in this post, but let&#8217;s use this velcro straps on this shoe as an overly simplistic example.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-15255\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/F1-intramolecular-reactions-sort-of-like-velcro-straps-the-straps-have-to-be-long-enough-to-reach-the-electrophile-.jpg\" alt=\"intramolecular reactions sort of like velcro straps the straps have to be long enough to reach the electrophile\" width=\"325\" height=\"276\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>If the velcro straps are<strong> too short<\/strong>, then the nucleophile (strap) can&#8217;t reach the electrophile (on the shoe). If the velcro straps are <strong>too long<\/strong> (imagine if they were each a foot long, for instance!) , then the shoes will be annoying to put on because of the decreasing likelihood that the nucleophile will be in the vicinity of the electrophile (an example of low &#8220;effective concentration&#8221;). The rate of formation for very large rings will approach the rate of intermolecular reactions.<\/p>\n<p>Of course molecules are more complicated than belts (or velcro straps) because of the ideal\u00a0109\u00b0 angles of tetrahedral carbons. That creates some additional complexity, notably the issue of\u00a0<strong>ring strain.\u00a0<\/strong><\/p>\n<p>For example you are probably aware by this point that 3 and 4 membered rings are quite strained, whereas rings of size 5, 6, and 7 are relatively unstrained. [<em>See article &#8211; <a href=\"https:\/\/www.masterorganicchemistry.com\/2014\/04\/03\/cycloalkanes-ring-strain-in-cyclopropane-and-cyclobutane\/\">Ring Strain in Cyclopropane and Cyclobutane<\/a><\/em>]<\/p>\n<p>If you&#8217;re just starting out you&#8217;re likely unaware that rings of size 8-11 are strained for a very interesting reason (transannular strain) and then rings of size 12 and above are generally unstrained.<\/p>\n<p>For a student in an introductory course, a good rule of thumb is: <strong>Formation of 5 and 6 membered rings is fast.<\/strong>\u00a0<strong>Formation of rings of size 7 and above is slow.\u00a0<\/strong>As for the smaller\u00a0ring sizes, we&#8217;ve seen examples where 3 membered rings form (<a href=\"https:\/\/www.masterorganicchemistry.com\/2015\/01\/26\/epoxides-the-outlier-of-the-ether-family\/\">from halohydrins<\/a>). Seen less often, but also fast is the formation of 4 membered rings.<\/p>\n<p>This is a vague generalization. &#8220;Give me numbers!&#8221; you might be saying. My copy of March says the following. Note that this is for a different reaction than the Williamson ether synthesis [formation of cyclic esters through SN2 of carboxylates with alkyl halides], but the trend should hold.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-15253\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/2019\/12\/F2-good-rule-of-thumb-for-ring-closures-releative-rate-of-ring-formation-is-5-fastest-then-56-then-4-then-7-then-formation-of-8-membered-rings-is-very-slow-baldwins-rules.gif\" alt=\"good rule of thumb for ring closures releative rate of ring formation is 5 fastest then 56 then 4 then 7 then formation of 8 membered rings is very slow baldwins rules\" width=\"630\" height=\"239\" \/><\/p>\n<p>For a more quantitative approach, I suggest you look into this paper on <a href=\"https:\/\/en.wikipedia.org\/wiki\/Baldwin%27s_rules\">ring closure kinetics.<\/a><\/p>\n<hr \/>\n<h2><strong><a id=\"quiz\"><\/a>Quiz Yourself!<\/strong><\/h2>\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\/1368-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/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. <br \/>\n<\/p>\n<p class=\"p1\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/1375-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/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. <br \/>\n<\/p>\n<p class=\"p1\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/1678-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/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. <br \/>\n<\/p>\n<p class=\"p1\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/2572-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/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. <br \/>\n<\/p>\n<p class=\"p1\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/2573-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/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. <br \/>\n<\/p>\n<p class=\"p1\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-26714\" src=\"https:\/\/www.masterorganicchemistry.com\/wp-content\/uploads\/quiz-previews\/3054-Front-Image-Only.png\" alt=\"\" width=\"640\" height=\"616\" \/><\/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<hr \/>\n<h2><strong><a id=\"references\"><\/a>(Advanced) References and Further Reading<\/strong><\/h2>\n<ol>\n<li><strong>XLV. Theory of \u00e6therification.<br \/>\n<\/strong>Alexander Williamson (1850) , The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 37:251, 350-356,<br \/>\n<strong>DOI: <\/strong><a href=\"https:\/\/www.tandfonline.com\/doi\/abs\/10.1080\/14786445008646627\">080\/14786445008646627<\/a><br \/>\nThe original Williamson paper. When Williamson reported the reaction in 1850 he didn&#8217;t know what an S<sub>N<\/sub>2 was &#8211; scientists didn&#8217;t even know what electrons were, for that matter &#8211; which again goes to show that the science of organic chemistry developed through a lot of empirical observations first, and the theory developed later.<\/li>\n<li><strong>Equilenin 3-Benzyl Ether<br \/>\n<\/strong> M. Hoehn, Clifford R. Dorn, and Bernard A. Nelson<br \/>\n<em>The Journal of Organic Chemistry<\/em> <strong>1965<\/strong> <em>30<\/em> (1), 316-316<br \/>\n<strong>DOI: <\/strong><a href=\"https:\/\/pubs.acs.org\/doi\/abs\/10.1021\/jo01012a520\">DOI: 10.1021\/jo01012a520<\/a><br \/>\nOne of the reactions in this paper is a classic Williamson reaction \u2013 protection of the alcohol in dehydroestrone as a benzyl ether, using benzyl chloride.<\/li>\n<li><strong>Total Synthesis of (+)-7-Deoxypancratistatin: A Radical Cyclization Approach<br \/>\n<\/strong>Gary E. Keck, Stanton F. McHardy, and Jerry A. Murry<strong><br \/>\n<\/strong><em>Journal of the American Chemical Society<\/em> <strong>1995<\/strong> <em>117<\/em> (27), 7289-7290<br \/>\n<strong>DOI: <\/strong><a href=\"https:\/\/pubs.acs.org\/doi\/abs\/10.1021\/ja00132a047\">1021\/ja00132a047<\/a><br \/>\nIn modern organic synthesis, the Williamson reaction is used for the <em>protection<\/em> of reactive alcohols in a substrate. Common protecting groups include methoxymethyl (MOM) and 2-methoxyethoxymethyl (MEM). MOM protection is employed in this total synthesis by Prof. Keck and coworkers.<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The Williamson Ether Synthesis In the Williamson Ether Synthesis, an\u00a0alkyl halide\u00a0(or sulfonate, such as a tosylate or mesylate)\u00a0undergoes nucleophilic substitution (SN2) by an\u00a0alkoxide\u00a0to give an\u00a0ether. <\/p>\n","protected":false},"author":1,"featured_media":35192,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[1420],"tags":[1031,1029,842,848,471,473,850,1030,271,1028],"post_folder":[],"class_list":["post-8573","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-alcohols-epoxides-ethers","tag-alcohol","tag-alkoxide","tag-alkyl-halide","tag-backside-attack","tag-base","tag-e2","tag-ethers","tag-nah","tag-sn2","tag-wiliamson"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.7 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>The Williamson Ether Synthesis &#8211; Master Organic Chemistry<\/title>\n<meta name=\"description\" content=\"The Williamson Ether Synthesis is still the best method for making (most) ethers. 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