{"id":12594,"date":"2026-05-20T12:58:29","date_gmt":"2026-05-20T12:58:29","guid":{"rendered":"https:\/\/www.vedprep.com\/exams\/?p=12594"},"modified":"2026-05-20T13:00:03","modified_gmt":"2026-05-20T13:00:03","slug":"oxidation-reactions","status":"publish","type":"post","link":"https:\/\/www.vedprep.com\/exams\/iit-jam\/oxidation-reactions\/","title":{"rendered":"Oxidation reactions (KMnO4, OsO4, Ozonolysis) For IIT JAM"},"content":{"rendered":"<p><strong>Oxidation reactions<\/strong> involving KMnO4, OsO4, and ozonolysis are crucial for IIT JAM, involving the conversion of alkenes and alkynes to carbonyl compounds, with KMnO4 and OsO4 being used for cleavage reactions and Ozonolysis for functional group introduction.<\/p>\n<h2><strong>Syllabus &#8211; Organic Chemistry for IIT JAM<\/strong><\/h2>\n<p>These reactions are crucial in organic chemistry and are covered in standard textbooks such as NCERT Organic Chemistry and Arihant Organic Chemistry. These books provide an in-depth understanding of the mechanisms and applications of these reactions to cover <strong>Oxidation reactions<\/strong> under the<a href=\"https:\/\/jam2026.iitb.ac.in\/files\/syllabus_CY.pdf\" rel=\"nofollow noopener\" target=\"_blank\"> <strong>IIT JAM syllabus<\/strong><\/a>.<\/p>\n<p>Key topics to focus on include:<\/p>\n<ul>\n<li>Oxidation of alcohols to aldehydes, ketones, and carboxylic acids using KMnO4 and other oxidizing agents<\/li>\n<li>Synthetic applications of OsO4 in the oxidation of alkenes<\/li>\n<li>Ozonolysis of alkenes and alkynes<\/li>\n<\/ul>\n<p>Mastering these concepts is essential for success in IIT JAM and other competitive exams.<\/p>\n<h2><strong>Oxidation Reactions (KMnO4, OsO4, Ozonolysis) For IIT JAM: An Overview<\/strong><\/h2>\n<p data-path-to-node=\"1\"><strong>Oxidation reactions<\/strong> are a massive part of organic chemistry, and if you are prepping for IIT JAM, CSIR NET, or GATE, you already know they show up everywhere. At its core, oxidation just means a molecule is losing electrons, changing its oxidation state. Today, we are breaking down three heavy hitters you absolutely need to master: <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>, <span class=\"math-inline\" data-math=\"\\text{OsO}_4\" data-index-in-node=\"349\">O<sub>s<\/sub>O<sub>4<\/sub><\/span>, and ozonolysis.<\/p>\n<p data-path-to-node=\"2\">Let&#8217;s start with <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>\u00a0(potassium permanganate). This is a fierce, strong oxidizing agent. It does not just tinker with a molecule; it can completely chop alkenes and alkynes down the middle to form carboxylic acids or ketones, depending on what groups are attached to the starting carbons. Because it reacts so visibly, chemists love using it to spot if a molecule has any hidden double or triple bonds.<\/p>\n<p data-path-to-node=\"3\">Then we have <span class=\"math-inline\" data-math=\"\\text{OsO}_4\" data-index-in-node=\"349\">O<sub>s<\/sub>O<sub>4<\/sub><\/span> (osmium tetroxide). Think of this as a precision surgeon compared to <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>&#8216;s sledgehammer. It targets alkenes to create neighborly diols (two <span class=\"math-inline\" data-math=\"-\\text{OH}\" data-index-in-node=\"176\">$-\\text{OH}$<\/span> groups right next to each other) without tearing the whole carbon skeleton apart. Here at <a href=\"https:\/\/www.vedprep.com\/online-courses\"><strong>VedPrep<\/strong><\/a>, we often see students pairing these two reagents together in their study notes because contrasting their reactivity makes remembering them so much easier.<\/p>\n<p data-path-to-node=\"4\">Finally, there is ozonolysis. This process uses ozone (<span class=\"math-inline\" data-math=\"\\text{O}_3\" data-index-in-node=\"55\">O<sub>3<\/sub><\/span>) to unzip a double bond and form a temporary, fragile structure called an ozonide. Depending on how you clean up the reaction afterward, that ozonide splits into two separate carbonyl compounds. It is hands-down one of the best ways to introduce aldehydes and ketones into a molecule.<\/p>\n<h2><strong>Worked Example: Oxidation of Alkene with KMnO4<\/strong><\/h2>\n<p data-path-to-node=\"7\">The oxidation of alkenes using <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span> is an organic chemistry classic, and it is a favorite for testing unsaturation in the lab. When you throw a harsh <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>\u00a0solution at an alkene under the right conditions, it breaks that double bond wide open.<\/p>\n<p data-path-to-node=\"8\">However, let&#8217;s clear up a massive error in the textbook snippet above.<\/p>\n<p data-path-to-node=\"8\"><b data-path-to-node=\"9,0\" data-index-in-node=\"3\">CRITICAL IIT JAM CORRECTION:<\/b><\/p>\n<p data-path-to-node=\"9,0\">The original text claims that reacting 2-methyl-2-butene with hot, acidic <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>\u00a0gives 2,3-butanedione (a 1,2-dicarbonyl). <b data-path-to-node=\"9,0\" data-index-in-node=\"162\">This is entirely incorrect.<\/b><\/p>\n<p data-path-to-node=\"9,0\">Acidic, hot <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4 <\/sub><\/span>causes <b data-path-to-node=\"9,0\" data-index-in-node=\"223\">oxidative cleavage<\/b>. It splits the double bond completely. Let&#8217;s look at what actually happens.<\/p>\n<p data-path-to-node=\"10\"><strong>Question<\/strong><\/p>\n<p data-path-to-node=\"11\">What is the actual product of the reaction of 2-methyl-2-butene with <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"69\">$\\text{KMnO}_4$<\/span> in a hot, acidic medium?<\/p>\n<p data-path-to-node=\"12\"><strong>Step-by-Step Solution<\/strong><\/p>\n<ol start=\"1\" data-path-to-node=\"13\">\n<li>\n<p data-path-to-node=\"13,0,0\">Let&#8217;s look at our starting structure, 2-methyl-2-butene: <span class=\"math-inline\" data-math=\"\\text{(CH}_3)_2\\text{C=CH-CH}_3\" data-index-in-node=\"57\">(CH<sub>3<\/sub>)<sub>2<\/sub>C=CH-CH<sub>3<\/sub><\/span>.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"13,1,0\">Draw a line right through the double bond. Imagine cutting it in half.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"13,2,0\">On the left side, you have a carbon attached to two methyl groups: <strong><span class=\"math-inline\" data-math=\"\\text{(CH}_3)_2\\text{C=CH-CH}_3\" data-index-in-node=\"57\">(<\/span><\/strong><span class=\"math-inline\" data-math=\"\\text{(CH}_3)_2\\text{C=CH-CH}_3\" data-index-in-node=\"57\">CH<\/span><span class=\"math-inline\" data-math=\"\\text{(CH}_3)_2\\text{C=CH-CH}_3\" data-index-in-node=\"57\"><sub>3<\/sub>)<sub>2<\/sub>C<\/span>.\u00a0When cleaved, this fully substituted carbon oxidizes straight to a ketone: <b data-path-to-node=\"13,2,0\" data-index-in-node=\"168\">acetone<\/b> (propan-2-one).<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"13,3,0\">On the right side, you have a carbon with one hydrogen and one methyl group: <span class=\"math-inline\" data-math=\"\\text{(CH}_3)_2\\text{C=CH-CH}_3\" data-index-in-node=\"57\">=CH-CH<sub>3<\/sub><\/span>. Under these harsh, acidic conditions, this part initially forms an aldehyde, which is immediately oxidized further into a carboxylic acid: <b data-path-to-node=\"13,3,0\" data-index-in-node=\"233\">ethanoic acid<\/b> (acetic acid).<\/p>\n<\/li>\n<\/ol>\n<p data-path-to-node=\"14\">So, you do not get a dicarbonyl compound at all. You get a mixture of a ketone and a carboxylic acid.<\/p>\n<table data-path-to-node=\"15\">\n<thead>\n<tr>\n<td><strong>Reactant<\/strong><\/td>\n<td><strong>Reagent<\/strong><\/td>\n<td><strong>Actual Cleavage Products<\/strong><\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><span data-path-to-node=\"15,1,0,0\"><b data-path-to-node=\"15,1,0,0\" data-index-in-node=\"0\">2-methyl-2-butene<\/b><\/span><\/td>\n<td><span data-path-to-node=\"15,1,1,0\"><span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"0\">$\\text{KMnO}_4$<\/span> (Hot, Acidic)<\/span><\/td>\n<td><span data-path-to-node=\"15,1,2,0\"><b data-path-to-node=\"15,1,2,0\" data-index-in-node=\"0\">Acetone + Ethanoic Acid<\/b><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><strong>Common Misconception: Ozonolysis vs. KMnO4 Oxidation<\/strong><\/h2>\n<p data-path-to-node=\"18\">It is incredibly easy to mix up ozonolysis and <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>\u00a0oxidation because both reactions look like they are doing the exact same thing at first glance: attacking a double bond. But the final destination of your reaction depends heavily on your choice of reagent and workup.<\/p>\n<ul data-path-to-node=\"19\">\n<li>\n<p data-path-to-node=\"19,0,0\"><b data-path-to-node=\"19,0,0\" data-index-in-node=\"0\">Ozonolysis<\/b> gives you ultimate control. If you use a <b data-path-to-node=\"19,0,0\" data-index-in-node=\"52\">reductive workup<\/b> (like zinc dust in water or dimethyl sulfide), you slice the double bond and slap an oxygen on each piece. If a carbon had a hydrogen attached, it stays an aldehyde.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"19,1,0\"><b data-path-to-node=\"19,1,0\" data-index-in-node=\"0\">KMnO4 (Hot, Acidic)<\/b> has no off-switch. It slices that bond but aggressively oxidizes any resulting aldehydes all the way up to carboxylic acids.<\/p>\n<\/li>\n<\/ul>\n<p data-path-to-node=\"20\">Imagine you are building a custom molecule for a project. If you want an aldehyde, <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>\u00a0will ruin your day by over-oxidizing it. You need reductive ozonolysis instead. Keeping these distinct outcomes straight is what separates a top-rank scorer from the rest of the pack on exam day.<\/p>\n<h2><strong>Application: Oxidation Reactions in Synthesis of Pharmaceuticals<\/strong><\/h2>\n<p data-path-to-node=\"23\">These reactions are not just theoretical puzzles we play with to pass exams; they are the literal foundation of modern medicine. For instance, think about how companies manufacture everyday painkillers like aspirin or paracetamol. <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>\u00a0is frequently used in industrial tracks to transform simple aromatic side chains into the essential carboxyl groups (-<span class=\"math-inline\" data-math=\"-\\text{COOH}\" data-index-in-node=\"362\">COOH<\/span>) that give these medicines their biological kick.<\/p>\n<p data-path-to-node=\"24\">When it comes to building highly complex structures like steroids or multi-ringed cancer therapies, stereochemistry is everything. One wrong orientation of an <span class=\"math-inline\" data-math=\"-\\text{OH}\" data-index-in-node=\"159\">-OH<\/span>\u00a0group can make a drug totally useless or even dangerous.<\/p>\n<p data-path-to-node=\"25\">Because <span class=\"math-inline\" data-math=\"\\text{OsO}_4\" data-index-in-node=\"8\">O<sub>s<\/sub>O<sub>4<\/sub><\/span>\u00a0delivers two hydroxyl groups to the exact same face of a double bond (called <i data-path-to-node=\"25\" data-index-in-node=\"98\">syn<\/i>-hydroxylation), pharmaceutical chemists rely on it heavily to build identical chiral centers every single time.<\/p>\n<p data-path-to-node=\"26\">Ozonolysis plays an equally vital role in drug discovery. Medicinal chemists often use it as a clean way to peel open a ring or chop a long carbon chain down to a specific aldehyde intermediate, which can then be easily modified into the final active pharmaceutical ingredient.<\/p>\n<h2><strong>Exam Strategy: Tips for IIT JAM Aspirants<\/strong><\/h2>\n<p data-path-to-node=\"29\">If you want to crack organic chemistry in IIT JAM, memorizing reactions like a grocery list will not cut it. The exam loves to throw unfamiliar, scary-looking structures at you to see if you actually understand the underlying electronics.<\/p>\n<p data-path-to-node=\"30\">Our core philosophy at <a href=\"https:\/\/www.vedprep.com\/online-courses\/iit-jam\"><strong>VedPrep<\/strong> <\/a>is focusing heavily on reaction mechanisms. Once you realize that a reagent is just hunting for electrons or looking to relieve ring strain, you can predict the product of almost any molecule they throw at you.<\/p>\n<p data-path-to-node=\"31\">Here are three quick tips to focus your study sessions:<\/p>\n<ul data-path-to-node=\"32\">\n<li>\n<p data-path-to-node=\"32,0,0\"><b data-path-to-node=\"32,0,0\" data-index-in-node=\"0\">Track your conditions:<\/b> Always look at the temperature and pH next to <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>. Cold, alkaline conditions (Baeyer\u2019s Reagent) do something completely different (<i data-path-to-node=\"32,0,0\" data-index-in-node=\"164\">syn<\/i>-dihydroxylation) than hot, acidic conditions (oxidative cleavage).<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"32,1,0\"><b data-path-to-node=\"32,1,0\" data-index-in-node=\"0\">Know your workups:<\/b> For ozonolysis, always check if the second step says <span class=\"math-inline\" data-math=\"\\text{Zn\/H}_2\\text{O}\" data-index-in-node=\"72\">Zn\/H<sub>2<\/sub><\/span> (reductive) or <span class=\"math-inline\" data-math=\"\\text{H}_2\\text{O}_2\" data-index-in-node=\"109\">H<sub>2<\/sub>O<sub>2<\/sub><\/span>\u00a0(oxidative). It completely alters whether you write down an aldehyde or a carboxylic acid as your answer.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"32,2,0\"><b data-path-to-node=\"32,2,0\" data-index-in-node=\"0\">Work backward:<\/b> Practice retrosynthesis. Look at a target dicarbonyl or keto acid and try to figure out what alkene must have been sliced open to make it.<\/p>\n<\/li>\n<\/ul>\n<h2><strong>Oxidation reactions (KMnO4, OsO4, Ozonolysis) For IIT JAM: Key Concepts<\/strong><\/h2>\n<p data-path-to-node=\"35\">Let&#8217;s pull all of this together into a clean, simple mental summary to help you visualize how these three tools function in a synthetic chemist&#8217;s toolkit:<\/p>\n<ul data-path-to-node=\"36\">\n<li>\n<p data-path-to-node=\"36,0,0\"><b data-path-to-node=\"36,0,0\" data-index-in-node=\"0\"><span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"334\">KMnO<sub>4<\/sub><\/span>:<\/b> Think of it as the heavy-duty oxidizer. It can easily convert primary alcohols all the way to carboxylic acids and secondary alcohols to ketones and aggressively cleave double or triple bonds when heated up.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"36,1,0\"><b data-path-to-node=\"36,1,0\" data-index-in-node=\"0\"><span class=\"math-inline\" data-math=\"\\text{OsO}_4\" data-index-in-node=\"349\">O<sub>s<\/sub>O<sub>4<\/sub><\/span>:<\/b> Think of it as the high-precision catalyst for adding functional groups. It gracefully adds two <span class=\"math-inline\" data-math=\"-\\text{OH}\" data-index-in-node=\"110\">$-\\text{OH}$<\/span> groups across the same side of an alkene double bond to yield a <i data-path-to-node=\"36,1,0\" data-index-in-node=\"185\">syn<\/i>-vicinal diol.<\/p>\n<\/li>\n<li>\n<p data-path-to-node=\"36,2,0\"><b data-path-to-node=\"36,2,0\" data-index-in-node=\"0\">Ozonolysis:<\/b> Think of it as the ultimate structure-mapper and unzipper. It cleanly cleaves carbon-carbon double or triple bonds, letting you introduce pristine carbonyl groups exactly where the unsaturation used to be.<\/p>\n<\/li>\n<\/ul>\n<h2><strong>Oxidation reactions (KMnO4, OsO4, Ozonolysis) For IIT JAM: Key Concepts<\/strong><\/h2>\n<p><strong>Oxidation reactions<\/strong> are a crucial part of organic chemistry, and understanding them is essential for students preparing for exams like IIT JAM, CSIR NET, and GATE. A strong oxidizing agent is a chemical species that readily gains electrons to cause oxidation. Potassium permanganate (KMnO\u2084) is a well-known strong oxidizing agent used in various <strong>oxidation reactions<\/strong>.<\/p>\n<p>KMnO4 is commonly used to oxidize alkenes, alkynes, and alkanes. It can also oxidize primary and secondary alcohols to carboxylic acids and ketones, respectively. The reaction conditions and the products formed depend on the reaction medium, such as acidic or basic.<\/p>\n<p>Another important oxidation reaction is the use of osmium tetroxide (OsO\u2084) as a catalyst for cleavage reactions. OsO4 is highly toxic and expensive, but it is effective in cleaving alkenes to form vicinal diols (diols with adjacent hydroxyl groups). This reaction is often used in combination with other reagents to achieve specific transformations.<\/p>\n<p>Ozonolysisis a process that involves the cleavage of an alkene or alkyne with ozone (O3), resulting in the introduction of functional groups such as carbonyl compounds. This reaction is useful for determining the structure of unsaturated compounds and is commonly used in organic synthesis.<\/p>\n<p>The following points summarizes the key points of these <strong>oxidation reactions<\/strong>:<\/p>\n<ul>\n<li>KMnO4: strong oxidizing agent<\/li>\n<li>OsO4: catalyst for cleavage reactions<\/li>\n<li>Ozonolysis :introduction of functional groups<\/li>\n<\/ul>\n<section>\n<h2 data-path-to-node=\"0\"><strong>Final Thoughts<\/strong><\/h2>\n<p data-path-to-node=\"1\">Mastering these core oxidation pathways isn\u2019t just about memorizing structural shifts\u2014it\u2019s about learning to balance reactivity with precision. Whether you are deploying the brute force of KMnO4, the stereospecific touch of OsO4, or the clean cutting power of ozonolysis, your success on the IIT JAM relies entirely on your command over the reaction conditions and workup steps. The exam will always try to trick you with complex structures, but if you look past the intimidating carbon skeletons and focus on the fundamental electronic shifts, the correct mechanism will clear itself up every single time.<\/p>\n<p data-path-to-node=\"1\">To learn more in detail from our faculty, watch our YouTube video:<\/p>\n<p class=\"responsive-video-wrap clr\"><iframe title=\"Reagents and Name Reaction in Organic Chemistry | CSIR NET | GATE | IIT JAM | DU | BHU |Chem Academy\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/1mZUlluWaoQ?list=PLdZcCa6mtW233hnUC42MCJjOFuX4_LTWv\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/p>\n<h2><strong>Frequently Asked Questions<\/strong><\/h2>\n<\/section>\n<style>#sp-ea-17678 .spcollapsing { height: 0; overflow: hidden; transition-property: height;transition-duration: 300ms;}#sp-ea-17678.sp-easy-accordion>.sp-ea-single {margin-bottom: 10px; border: 1px solid #e2e2e2; }#sp-ea-17678.sp-easy-accordion>.sp-ea-single>.ea-header a {color: #444;}#sp-ea-17678.sp-easy-accordion>.sp-ea-single>.sp-collapse>.ea-body {background: #fff; color: #444;}#sp-ea-17678.sp-easy-accordion>.sp-ea-single {background: #eee;}#sp-ea-17678.sp-easy-accordion>.sp-ea-single>.ea-header a .ea-expand-icon { float: left; color: #444;font-size: 16px;}<\/style><div id=\"sp_easy_accordion-1779281109\">\n<div id=\"sp-ea-17678\" class=\"sp-ea-one sp-easy-accordion\" data-ea-active=\"ea-click\" data-ea-mode=\"vertical\" data-preloader=\"\" data-scroll-active-item=\"\" data-offset-to-scroll=\"0\">\n\n<!-- Start accordion card div. -->\n<div class=\"ea-card ea-expand sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176780\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176780\" aria-controls=\"collapse176780\" href=\"#\"  aria-expanded=\"true\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-minus\"><\/i> What is the fundamental difference between oxidation with cold, alkaline KMnO4 and hot, acidic KMnO4?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse collapsed show\" id=\"collapse176780\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176780\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Cold, alkaline <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"15\">KMnO<sub>4<\/sub><\/span>\u00a0(known as <b data-path-to-node=\"4\" data-index-in-node=\"39\">Baeyer\u2019s Reagent<\/b>) is a mild oxidizer that performs <i data-path-to-node=\"4\" data-index-in-node=\"90\">syn<\/i>-dihydroxylation, adding two <span class=\"math-inline\" data-math=\"-\\text{OH}\" data-index-in-node=\"122\">-OH<\/span>\u00a0groups across the double bond without breaking the carbon skeleton. Hot, acidic <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"15\">KMnO<sub>4<\/sub><\/span>\u00a0is an aggressive oxidizer that causes complete <b data-path-to-node=\"4\" data-index-in-node=\"274\">oxidative cleavage<\/b>, snapping the <span class=\"math-inline\" data-math=\"\\text{C=C}\" data-index-in-node=\"307\">C=C<\/span> or <span class=\"math-inline\" data-math=\"\\text{C}\\equiv\\text{C}\" data-index-in-node=\"321\">C\u2261C<\/span>\u00a0bond entirely to yield ketones, carboxylic acids, or <span class=\"math-inline\" data-math=\"\\text{CO}_2\" data-index-in-node=\"397\">CO<sub>2<\/sub><\/span>.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176781\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176781\" aria-controls=\"collapse176781\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Why does OsO4 result in a syn-diol rather than an anti-diol?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse176781\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176781\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>The reaction goes through a concerted, single-step mechanism where the osmium tetroxide molecule adds to the alkene. Because the two oxygen atoms are tied to the same osmium metal center, they are forced to attack the \u03c0-bond from the <b data-path-to-node=\"6\" data-index-in-node=\"236\">same face<\/b> simultaneously. This forms a cyclic osmate ester intermediate, which delivers the <i data-path-to-node=\"6\" data-index-in-node=\"328\">syn<\/i> (same side) stereochemistry upon hydrolysis.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176782\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176782\" aria-controls=\"collapse176782\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> How do you determine if an ozonolysis reaction is reductive or oxidative?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse176782\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176782\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Look closely at the reagent listed in the second step (the workup). If you see a reducing agent like zinc dust in water (<span class=\"math-inline\" data-math=\"\\text{Zn\/H}_2\\text{O}\" data-index-in-node=\"121\">Zn\/H<sub>2<\/sub>O<\/span>) or dimethyl sulfide (<span class=\"math-inline\" data-math=\"\\text{CH}_3\\text{SCH}_3\" data-index-in-node=\"165\">CH<sub>3<\/sub>SCH<sub>3<\/sub><\/span>), it is a <b data-path-to-node=\"8\" data-index-in-node=\"199\">reductive workup<\/b>. If you see an oxidizing agent like hydrogen peroxide (<span class=\"math-inline\" data-math=\"\\text{H}_2\\text{O}_2\" data-index-in-node=\"271\">H<sub>2<\/sub>O<sub>2<\/sub><\/span>), it is an <b data-path-to-node=\"8\" data-index-in-node=\"303\">oxidative workup<\/b>.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176783\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176783\" aria-controls=\"collapse176783\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> If I perform reductive ozonolysis on 2-methyl-2-butene, what are my final products?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse176783\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176783\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Reductive ozonolysis cuts the double bond cleanly and replaces it with oxygens without further oxidation. Slicing <span class=\"math-inline\" data-math=\"\\text{(CH}_3)_2\\text{C=CH-CH}_3\" data-index-in-node=\"114\">(CH<sub>3<\/sub>)<sub>2<\/sub>C=CH-CH<sub>3<\/sub><\/span>\u00a0gives you a three-carbon ketone (<b data-path-to-node=\"16\" data-index-in-node=\"179\">acetone<\/b>) and a two-carbon aldehyde (<b data-path-to-node=\"16\" data-index-in-node=\"215\">acetaldehyde<\/b>).<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176784\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176784\" aria-controls=\"collapse176784\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> How can I synthesize a 1,2-diketone from an internal alkyne without breaking the carbon-carbon \u03c3-bond?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse176784\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176784\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>You should use dilute, neutral, or slightly alkaline <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"53\">KMnO<sub>4<\/sub><\/span>\u00a0at room temperature. Under these controlled, non-acidic conditions, the triple bond is oxidized into a 1,2-dicarbonyl compound instead of undergoing complete cleavage.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176785\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176785\" aria-controls=\"collapse176785\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What product is formed when cyclohexene undergoes reductive ozonolysis?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse176785\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176785\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Because cyclohexene is a cyclic alkene, unzipping the double bond does not split the molecule into two pieces. Instead, it opens up the ring into a single, straight-chain dialdehyde called <b data-path-to-node=\"22\" data-index-in-node=\"189\">hexane-1,6-dial<\/b> (adipaldehyde).<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176786\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176786\" aria-controls=\"collapse176786\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Does stereochemistry matter when performing ozonolysis on cis vs trans isomers of an alkene?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse176786\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176786\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>No. Because ozonolysis completely cleaves the double bond and converts the sp\u00b2 carbons into planar carbonyl groups (<span class=\"math-inline\" data-math=\"\\text{C=O}\" data-index-in-node=\"116\">C=O<\/span>), any structural information regarding whether the starting material was <i data-path-to-node=\"24\" data-index-in-node=\"200\">cis<\/i> or <i data-path-to-node=\"24\" data-index-in-node=\"207\">trans<\/i> is completely lost in the final products.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176787\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176787\" aria-controls=\"collapse176787\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> I have a molecule with both an alcohol and an alkene. Which reagent should I use if I want to oxidize the double bond but leave the alcohol alone?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse176787\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176787\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>You should opt for <b data-path-to-node=\"28\" data-index-in-node=\"19\">ozonolysis<\/b>. Ozone is highly selective for carbon-carbon multiple bonds because of its cycloaddition mechanism. <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"130\">KMnO<sub>4<\/sub><\/span>, on the other hand, is notorious for non-selectively oxidizing primary and secondary alcohols into carboxylic acids and ketones alongside attacking the alkene.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176788\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176788\" aria-controls=\"collapse176788\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> How can Baeyer's reagent be used as a qualitative laboratory test?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse176788\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176788\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Baeyer\u2019s reagent (<span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"18\">KMnO<sub>4<\/sub><\/span>) starts as a deep, intense purple solution. When it reacts with an unsaturated compound (alkene or alkyne), the purple permanganate ion (<span class=\"math-inline\" data-math=\"\\text{MnO}_4^-\" data-index-in-node=\"169\">MnO<sub>4<\/sub>-<\/span>) is consumed, and a sludge-like brown precipitate of manganese dioxide (<span class=\"math-inline\" data-math=\"\\text{MnO}_2\" data-index-in-node=\"256\">MnO<sub>2<\/sub><\/span>) forms. The disappearance of the purple color confirms unsaturation.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-176789\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse176789\" aria-controls=\"collapse176789\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Is it possible for an alkene to yield a ketone and a carboxylic acid under reductive ozonolysis conditions?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse176789\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-176789\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>No. A reductive workup (<span class=\"math-inline\" data-math=\"\\text{Zn\/H}_2\\text{O}\" data-index-in-node=\"24\">Zn\/H<sub>2<\/sub>O<\/span>) will never yield a carboxylic acid from a simple alkene carbon because it lacks the oxidizing power to turn an aldehyde into an acid. If you get a carboxylic acid, it means your workup was oxidative or you used hot <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"262\">KMnO<sub>4<\/sub><\/span>.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1767810\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1767810\" aria-controls=\"collapse1767810\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> How do I solve a retrosynthesis problem where the question gives me the cleavage products and asks for the original alkene?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1767810\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-1767810\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Line up the two carbonyl (<span class=\"math-inline\" data-math=\"\\text{C=O}\" data-index-in-node=\"26\">C=O<\/span>) groups facing each other. Mentally erase the two oxygen atoms, and draw a carbon-carbon double bond (<span class=\"math-inline\" data-math=\"\\text{C=C}\" data-index-in-node=\"139\">C=C<\/span>) directly connecting those two carbonyl carbons. Double-check your alkyl substituents to make sure your total carbon count matches.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1767811\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1767811\" aria-controls=\"collapse1767811\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> Why does the IIT JAM exam frequently ask about cyclic ozonolysis products?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1767811\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-1767811\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Cyclic substrates are excellent for testing whether you actually understand connectivity. It is incredibly easy to miscount carbons when transforming a drawing of a ring into a straight-chain dicarbonyl product under exam pressure.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<!-- Start accordion card div. -->\n<div class=\"ea-card  sp-ea-single\">\n\t<!-- Start accordion header. -->\n\t<h3 class=\"ea-header\">\n\t\t<!-- Add anchor tag for header. -->\n\t\t<a class=\"collapsed\" id=\"ea-header-1767812\" role=\"button\" data-sptoggle=\"spcollapse\" data-sptarget=\"#collapse1767812\" aria-controls=\"collapse1767812\" href=\"#\"  aria-expanded=\"false\" tabindex=\"0\">\n\t\t<i aria-hidden=\"true\" role=\"presentation\" class=\"ea-expand-icon eap-icon-ea-expand-plus\"><\/i> What is a safe way to distinguish between an alkene and an alkane using these reactions?\t\t<\/a> <!-- Close anchor tag for header. -->\n\t<\/h3>\t<!-- Close header tag. -->\n\t<!-- Start collapsible content div. -->\n\t<div class=\"sp-collapse spcollapse \" id=\"collapse1767812\" data-parent=\"#sp-ea-17678\" role=\"region\" aria-labelledby=\"ea-header-1767812\">  <!-- Content div. -->\n\t\t<div class=\"ea-body\">\n\t\t<p>Treating both samples with either cold <span class=\"math-inline\" data-math=\"\\text{KMnO}_4\" data-index-in-node=\"39\">KMnO<sub>4<\/sub><\/span> or <span class=\"math-inline\" data-math=\"\\text{OsO}_4\" data-index-in-node=\"56\">OsO<sub>4<\/sub><\/span>\u00a0will work. The alkene will react, causing a distinct color change or forming a diol, while the alkane will remain completely inert because <span class=\"math-inline\" data-math=\"\\text{C-C}\" data-index-in-node=\"208\">C-C<\/span> and <span class=\"math-inline\" data-math=\"\\text{C-H}\" data-index-in-node=\"223\">C-H<\/span>\u00a0single <span class=\"math-inline\" data-math=\"\\sigma\" data-index-in-node=\"241\">\u03c3<\/span>-bonds are too stable to react under these conditions.<\/p>\n\t\t<\/div> <!-- Close content div. -->\n\t<\/div> <!-- Close collapse div. -->\n<\/div> <!-- Close card div. -->\n<\/div>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Oxidation reactions involving KMnO4, OsO4, and Ozonolysis are crucial for IIT JAM, involving the conversion of alkenes and alkynes to carbonyl compounds. These reactions are covered in Unit 11 and Unit 12 of the CSIR NET \/ NTA syllabus. The topic of oxidation reactions, including reactions with KMnO4, OsO4, and ozonolysis, falls under the Unit 11: Alcohols, Phenols and Ethers and Unit 12: Aldehydes, Ketones and Carboxylic Acids of the CSIR NET \/ NTA syllabus.<\/p>\n","protected":false},"author":12,"featured_media":12593,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":"","rank_math_seo_score":85},"categories":[23],"tags":[13825,2923,13820,13819,13824,13821,13822,13823,2922],"class_list":["post-12594","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-iit-jam","tag-and-ozonolysis","tag-competitive-exams","tag-oso4","tag-oxidation-reactions-kmno4","tag-oxidation-reactions-with-kmno4","tag-ozonolysis-for-iit-jam","tag-ozonolysis-for-iit-jam-notes","tag-ozonolysis-for-iit-jam-questions","tag-vedprep","entry","has-media"],"acf":[],"_links":{"self":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts\/12594","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/users\/12"}],"replies":[{"embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/comments?post=12594"}],"version-history":[{"count":6,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts\/12594\/revisions"}],"predecessor-version":[{"id":17687,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/posts\/12594\/revisions\/17687"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/media\/12593"}],"wp:attachment":[{"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/media?parent=12594"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/categories?post=12594"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vedprep.com\/exams\/wp-json\/wp\/v2\/tags?post=12594"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}