{"id":966,"date":"2017-10-19T15:12:53","date_gmt":"2017-10-19T15:12:53","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&#038;p=966"},"modified":"2018-10-03T18:27:17","modified_gmt":"2018-10-03T18:27:17","slug":"calculating-degree-of-unsaturation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/calculating-degree-of-unsaturation\/","title":{"raw":"Calculating Degree of Unsaturation","rendered":"Calculating Degree of Unsaturation"},"content":{"raw":"<div class=\"elm-header\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Objectives<\/h3>\r\n<div id=\"elm-main-content\" class=\"elm-content-container\">\r\n<div>\r\n<div id=\"skills\">\r\n\r\nAfter completing this section, you should be able to\r\n<ol>\r\n \t<li>determine the degree of unsaturation of an organic compound, given its molecular formula, and hence determine the number of double bonds, triple bonds and rings present in the compound.<\/li>\r\n \t<li>draw all the possible isomers that correspond to a given molecular formula containing only carbon (up to a maximum of six atoms) and hydrogen.<\/li>\r\n \t<li>draw a specified number of isomers that correspond to a given molecular formula containing carbon, hydrogen, and possibly other elements, such as oxygen, nitrogen and the halogens.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"elm-main-content\" class=\"elm-content-container\">\r\n<div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key TERMS<\/h3>\r\n<div>\r\n\r\nMake certain that you can define, and use in context, the key terms below.\r\n<ul>\r\n \t<li>degree of unsaturation<\/li>\r\n \t<li>saturated<\/li>\r\n \t<li>unsaturated<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<p class=\"boxtitle\">There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (<a class=\"internal\" title=\"Physical Chemistry\/Spectroscopy\/Magnetic Resonance\/Integration in NMR\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Spectroscopy\/Magnetic_Resonance_Spectroscopies\/Nuclear_Magnetic_Resonance\/NMR%3A_Experimental\/NMR%3A_Interpretation\/Integration_in_NMR\" rel=\"internal\">NMR<\/a>) and infrared radiation (<a title=\"Physical_Chemistry\/Spectroscopy\/Vibrational_Spectroscopy\/Infrared_Spectroscopy\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Spectroscopy\/Vibrational_Spectroscopy\/Infrared_Spectroscopy\" rel=\"internal\">IR<\/a>) are the primary ways\u00a0of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of $$\\pi$$ bonds and\/or cyclic rings.<\/p>\r\n\r\n<\/div>\r\n<div id=\"section_1\">\r\n<h3 class=\"editable\">Saturated and Unsaturated Molecules<\/h3>\r\nIn the lab, <a title=\"Physical Chemistry\/Physical Properties of Matter\/Solutions\/SOLUBILITY\/Types of Saturation\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Equilibria\/Solubilty\/Types_of_Saturation\" rel=\"internal\">saturation<\/a> may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds\u00a0with no rings is considered saturated.\r\n<table style=\"margin: auto;border-spacing: 1px;width: 576px\" border=\"0\" cellpadding=\"1\">\r\n<tbody>\r\n<tr>\r\n<td>\u00a0 CH<sub>3<\/sub>CH<sub>2<\/sub>CH<sub>3<\/sub><\/td>\r\n<td>\u00a0 <img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1958\/chewiki_sat.bmp?revision=1&amp;size=bestfit&amp;width=150&amp;height=94#fixme\" alt=\"chewiki_sat.bmp\" width=\"150px\" height=\"94px\" \/><\/td>\r\n<td><img class=\"internal\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1959\/chewiki_sat2_(3).bmp?revision=1&amp;size=bestfit&amp;width=113&amp;height=128#fixme\" alt=\"chewiki_sat2 (3).bmp\" width=\"113px\" height=\"128px\" \/><\/td>\r\n<td>\u00a01-methyoxypentane<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nUnlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and\/or ring(s).\r\n<table style=\"margin: auto;border-spacing: 1px;width: 600px\" border=\"0\" cellpadding=\"1\">\r\n<tbody>\r\n<tr>\r\n<td>\u00a0\u00a0 CH<sub>3<\/sub>CH=CHCH<sub>3<\/sub><\/td>\r\n<td>\u00a0 <img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1960\/chewiki_unsat1.bmp?revision=1&amp;size=bestfit&amp;width=75&amp;height=68#fixme\" alt=\"chewiki_unsat1.bmp\" width=\"75px\" height=\"68px\" \/><\/td>\r\n<td>\u00a0<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1961\/chewiki_unsat2.bmp?revision=1&amp;size=bestfit&amp;width=80&amp;height=62#fixme\" alt=\"chewiki_unsat2.bmp\" width=\"80px\" height=\"62px\" \/><\/td>\r\n<td>\u00a0<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1962\/chewiki_unsat3.bmp?revision=1&amp;size=bestfit&amp;width=110&amp;height=66#fixme\" alt=\"chewiki_unsat3.bmp\" width=\"110px\" height=\"66px\" \/><\/td>\r\n<td>3-chloro-5-octyne<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"section_2\">\r\n<h3 class=\"editable\">Calculating The Degree of Unsaturation (DoU)<\/h3>\r\nIf the molecular\u00a0formula is given, plug\u00a0in the numbers into this formula:\r\n\r\n\\[ DoU= \\dfrac{2C+2+N-X-H}{2} \\tag{7.2.1}\\]\r\n<ul>\r\n \t<li><em>C<\/em> is the number\u00a0of carbons<\/li>\r\n \t<li><em>N<\/em> is the number\u00a0of nitrogens<\/li>\r\n \t<li><em>X<\/em> is the number\u00a0of halogens (F, Cl, Br, I)<\/li>\r\n \t<li><em>H<\/em> is the number\u00a0of hydrogens<\/li>\r\n<\/ul>\r\nAs stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that\u00a0a saturated\u00a0molecule\u00a0has the maximum number of hydrogen atoms possible\u00a0to be\u00a0an acyclic alkane. Thus, the\u00a0number of hydrogens\u00a0can be represented by 2C+2, which is the general molecular representation of an <a class=\"internal\" title=\"Organic Chemistry\/Hydrocarbons\/Alkanes\" href=\"https:\/\/chem.libretexts.org\/Core\/Organic_Chemistry\/Hydrocarbons\/Alkanes\" rel=\"internal\">alkane<\/a>.\u00a0As an\u00a0example,\u00a0for\u00a0the molecular\u00a0formula\u00a0C<sub>3<\/sub>H<sub>4<\/sub> the number of actual hydrogens needed\u00a0for the compound to be saturated is 8 <em>[2C+2=(2x3)+2=8]<\/em>. The compound needs\u00a04 more hydrogens in order to be fully saturated <em>(expected number of hydrogens-observed number of hydrogens=8-4=4)<\/em>. Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as\u00a0saturated.\u00a0\u00a0Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C<sub>2<\/sub>H<sub>5<\/sub>Cl, there is one less hydrogen compared to ethane, C<sub>2<\/sub>H<sub>6<\/sub>.\r\n\r\nFor a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C<sub>3<\/sub>H<sub>9<\/sub>N compared to C<sub>3<\/sub>H<sub>8<\/sub>. Oxygen and sulfur\u00a0are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens\u00a0in ethanol,\u00a0C<sub>2<\/sub>H<sub>5<\/sub>OH, matches the number of hydrogens in ethane, C<sub>2<\/sub>H<sub>6<\/sub>.\r\n\r\nThe following chart\u00a0 illustrates the\u00a0possible combinations of the number of\u00a0double bond(s), triple bond(s), and\/or ring(s) for a\u00a0given degree of unsaturation. Each row corresponds to a different combination.\r\n<ul>\r\n \t<li>One degree of unsaturation is equivalent to 1 ring or 1 double bond (1\u00a0$$ \\pi $$\u00a0bond).<\/li>\r\n \t<li>Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2\u00a0$$ \\pi $$\u00a0bonds).<\/li>\r\n<\/ul>\r\n<table style=\"margin: auto;border-spacing: 0px;width: 448px\" border=\"1\" cellpadding=\"0\">\r\n<tbody>\r\n<tr>\r\n<td>\r\n<div><strong>DoU<\/strong><\/div><\/td>\r\n<td colspan=\"3\">\r\n<div><strong>Possible combinations of rings\/ \u00a0bonds<\/strong><\/div><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>\r\n<div># of rings<\/div><\/td>\r\n<td>\r\n<div># of double bonds<\/div><\/td>\r\n<td>\r\n<div># of triple bonds<\/div><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<div>1<\/div><\/td>\r\n<td>\r\n<div>1<\/div><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<td>\r\n<div>1<\/div><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<div>2<\/div><\/td>\r\n<td>\r\n<div>2<\/div><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<td>\r\n<div>2<\/div><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<td>\r\n<div>1<\/div><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>\r\n<div>1<\/div><\/td>\r\n<td>\r\n<div>1<\/div><\/td>\r\n<td>\r\n<div>0<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRemember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and\/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond,\u00a0<em>or<\/em>\u00a03 double bonds.\r\n<div>\r\n<div id=\"example\">\r\n<div class=\"textbox examples\">\r\n<h3>ExampLE<\/h3>\r\n<div id=\"section_2\">\r\n<div>\r\n<div id=\"example\">\r\n<p class=\"boxtitle\">\u00a0Benzene<\/p>\r\nWhat is the Degree of Unsaturation for Benzene?\r\n<h3><strong>SOLUTION<\/strong><\/h3>\r\n[reveal-answer q=\"713260\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"713260\"]\r\n\r\nThe molecular formula for benzene is C<sub>6<\/sub>H<sub>6<\/sub><sub>. <\/sub>Thus,\r\n\r\nDoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1960\/chewiki_unsat1.bmp?revision=1&amp;size=bestfit&amp;width=75&amp;height=68#fixme\" alt=\"chewiki_unsat1.bmp\" width=\"75px\" height=\"68px\" \/>\r\n\r\nHowever, when given the molecular formula C<sub>6<\/sub>H<sub>6<\/sub>, benzene is only one of many possible structures (<a title=\"Isomers\" href=\"https:\/\/chem.libretexts.org\/Core\/Inorganic_Chemistry\/Coordination_Chemistry\/Properties_of_Coordination_Compounds\/Isomers\" rel=\"internal\">isomers<\/a>). The following structures all have DoB of 4 and have the same molecular formula as benzene.\r\n\r\n<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1951\/4_DoB_2.bmp?revision=1&amp;size=bestfit&amp;width=114&amp;height=62#fixme\" alt=\"4 DoB_2.bmp\" width=\"114px\" height=\"62px\" \/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1952\/4_DoB_3.bmp?revision=1&amp;size=bestfit&amp;width=89&amp;height=65#fixme\" alt=\"4 DoB_3.bmp\" width=\"89px\" height=\"65px\" \/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<img class=\"internal\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1953\/4DoB_1.bmp?revision=1&amp;size=bestfit&amp;width=118&amp;height=78#fixme\" alt=\"4DoB_1.bmp\" width=\"118px\" height=\"78px\" \/>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_3\">\r\n<h3 class=\"editable\">References<\/h3>\r\n<ol>\r\n \t<li>Vollhardt, K. P.C. &amp; Shore, N. (2007). <em>Organic Chemistry <\/em>(5<sup>th<\/sup>Ed.).\u00a0\u00a0New York: W. H. Freeman. (473-474)<\/li>\r\n \t<li>Shore, N. (2007). <em>Study Guide and Solutions Manual for Organic Chemistry <\/em>(5<sup>th<\/sup> Ed.). New York: W.H. Freeman. (201)<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"section_4\">\r\n<div class=\"textbox examples\">\r\n<h3>Examples<\/h3>\r\n<div id=\"section_4\">\r\n<h3 class=\"editable\">Problems<\/h3>\r\n<ol>\r\n \t<li>\u00a0Are the following molecules saturated or unsaturated:\r\n<ol>\r\n \t<li>\u00a0 <img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1955\/chewiki_prob1a.bmp?revision=1&amp;size=bestfit&amp;width=108&amp;height=75#fixme\" alt=\"chewiki_prob1a.bmp\" width=\"108px\" height=\"75px\" \/>\u00a0\u00a0\u00a0\u00a0(b.)<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1957\/chewiki_prob2b.bmp?revision=1&amp;size=bestfit&amp;width=134&amp;height=65#fixme\" alt=\"chewiki_prob2b.bmp\" width=\"134px\" height=\"65px\" \/>\u00a0\u00a0\u00a0\u00a0\u00a0(c.) <img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1956\/chewiki_prob1c.bmp?revision=1&amp;size=bestfit&amp;width=119&amp;height=75#fixme\" alt=\"chewiki_prob1c.bmp\" width=\"119px\" height=\"75px\" \/>\u00a0\u00a0\u00a0\u00a0(d.) C<sub>10<\/sub>H<sub>6<\/sub>N<sub>4<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Using the molecules from 1., give the degrees of unsaturation for each.<\/li>\r\n \t<li>Calculate the degrees of unsaturation for the following molecular formulas:\r\n<ol>\r\n \t<li>(a.) C<sub>9<\/sub>H<sub>20<\/sub><sub> \u00a0\u00a0\u00a0\u00a0<\/sub>(b.) C<sub>7<\/sub>H<sub>8<\/sub><sub> \u00a0\u00a0\u00a0\u00a0<\/sub>(c.) C<sub>5<\/sub>H<sub>7<\/sub>Cl \u00a0\u00a0\u00a0\u00a0(d.) C<sub>9<\/sub>H<sub>9<\/sub>NO<sub>4<\/sub><sub>\u00a0<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Using the molecular formulas from 3, are the molecules unsaturated or saturated.<\/li>\r\n \t<li>Using the molecular formulas from 3, if the molecules are saturated, how many rings\/double bonds\/triple bonds are predicted?<\/li>\r\n \t<li>(d.) <strong>unsaturated<\/strong>5.(a.) <strong>0 <\/strong>(Remember-a saturated molecule only contains single bonds)\r\n\r\n(b.) <em>The molecule can contain any of these combinations<\/em> <strong>(i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds\u00a0<\/strong>\r\n\r\n(c.) (<strong>i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds\u00a0<\/strong>\r\n\r\n(d.) <strong>(i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)... <\/strong><em>(As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and\/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)<\/em><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"section_5\">\r\n<h3 class=\"editable\">Answers<\/h3>\r\n1.\r\n\r\n[reveal-answer q=\"567506\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"567506\"](a.)\u00a0unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)\r\n\r\n(b.) unsaturated\r\n\r\n(c.) saturated\r\n\r\n(d.) unsaturated[\/hidden-answer]\r\n\r\n2. <em>If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and\/or rings. However, you can also\u00a0determine the molecular\u00a0formula\u00a0and solve\u00a0for the degrees of unsaturation by\u00a0using the\u00a0formula.<\/em>\r\n\r\n[reveal-answer q=\"546426\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"546426\"](a.)\u00a02\r\n\r\n(b.) 2 (one double bond and the double\u00a0bond from the\u00a0carbonyl)\r\n\r\n(c.) 0\r\n\r\n(d.) 10[\/hidden-answer]\r\n\r\n3. <em>Use the formula to solve<\/em>\r\n\r\n[reveal-answer q=\"389197\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"389197\"](a.) 0\r\n\r\n(b.) 4\r\n\r\n(c.) 2\r\n\r\n(d.) 6[\/hidden-answer]\r\n\r\n4.\r\n\r\n[reveal-answer q=\"669637\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"669637\"]\r\n\r\n(a.) saturated\r\n\r\n(b.) unsaturated\r\n\r\n(c.) unsaturated\u00a0[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_6\">\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<div id=\"s61712\">\r\n<div id=\"section_7\">\r\n<h3 id=\"Questions-61712\">Questions<\/h3>\r\n<strong>1.<\/strong>\r\n\r\nCalculate degrees of unsaturation (DoU) for the following, and propose a structure for each.\r\n\r\nA \u2013 C<sub>5<\/sub>H<sub>8<\/sub>\r\n\r\nB \u2013 C<sub>4<\/sub>H<sub>4<\/sub>\r\n\r\n<b>2.\u00a0<\/b>\r\n\r\nCalculate the degree of unsaturation (DoU) for the following molecules\r\n\r\nA \u2013 C<sub>5<\/sub>H<sub>5<\/sub>N\r\n\r\nB \u2013 C<sub>5<\/sub>H<sub>5<\/sub>NO<sub>2<\/sub>\r\n\r\nC \u2013 C<sub>5<\/sub>H<sub>5<\/sub>Br\r\n\r\n<strong>3.<\/strong>\r\n\r\nThe following molecule is caffeine (C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub>), determine the degrees of unsaturation (DoU).\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05135319\/7.22.png\" alt=\"\" width=\"238\" height=\"203\" \/>\r\n\r\n<\/div>\r\n<div id=\"section_8\">\r\n<h3 id=\"Solutions-61712\">Solutions<\/h3>\r\n<strong>1.\r\n[reveal-answer q=\"543265\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"543265\"]<\/strong><img class=\"internal default alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05135321\/7.2.png\" alt=\"\" width=\"320\" height=\"214\" \/>\r\n\r\n<strong>[\/hidden-answer]<\/strong>\r\n\r\n&nbsp;\r\n\r\n<strong>2.<\/strong>\r\n\r\n[reveal-answer q=\"262440\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"262440\"]A = 4, B = 4, C = 3[\/hidden-answer]\r\n\r\n<strong>3.<\/strong>\r\n\r\n[reveal-answer q=\"330183\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"330183\"]6 DoU[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<h3 class=\"editable\">Contributors<\/h3>\r\n<ul>\r\n \t<li><a class=\"external\" title=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" href=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" target=\"_blank\" rel=\"external nofollow noopener\">Dr. Dietmar Kennepohl<\/a> FCIC (Professor of Chemistry, <a class=\"external\" title=\"http:\/\/www.athabascau.ca\/\" href=\"http:\/\/www.athabascau.ca\/\" target=\"_blank\" rel=\"external nofollow noopener\">Athabasca University<\/a>)<\/li>\r\n \t<li>Prof. Steven Farmer (<a class=\"external\" title=\"http:\/\/www.sonoma.edu\" href=\"http:\/\/www.sonoma.edu\" target=\"_blank\" rel=\"external nofollow noopener\">Sonoma State University<\/a>)<\/li>\r\n \t<li>Kim Quach (UCD)<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"elm-header\">\n<div class=\"textbox learning-objectives\">\n<h3>Objectives<\/h3>\n<div id=\"elm-main-content\" class=\"elm-content-container\">\n<div>\n<div id=\"skills\">\n<p>After completing this section, you should be able to<\/p>\n<ol>\n<li>determine the degree of unsaturation of an organic compound, given its molecular formula, and hence determine the number of double bonds, triple bonds and rings present in the compound.<\/li>\n<li>draw all the possible isomers that correspond to a given molecular formula containing only carbon (up to a maximum of six atoms) and hydrogen.<\/li>\n<li>draw a specified number of isomers that correspond to a given molecular formula containing carbon, hydrogen, and possibly other elements, such as oxygen, nitrogen and the halogens.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"elm-main-content\" class=\"elm-content-container\">\n<div>\n<div class=\"textbox key-takeaways\">\n<h3>Key TERMS<\/h3>\n<div>\n<p>Make certain that you can define, and use in context, the key terms below.<\/p>\n<ul>\n<li>degree of unsaturation<\/li>\n<li>saturated<\/li>\n<li>unsaturated<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p class=\"boxtitle\">There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (<a class=\"internal\" title=\"Physical Chemistry\/Spectroscopy\/Magnetic Resonance\/Integration in NMR\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Spectroscopy\/Magnetic_Resonance_Spectroscopies\/Nuclear_Magnetic_Resonance\/NMR%3A_Experimental\/NMR%3A_Interpretation\/Integration_in_NMR\" rel=\"internal\">NMR<\/a>) and infrared radiation (<a title=\"Physical_Chemistry\/Spectroscopy\/Vibrational_Spectroscopy\/Infrared_Spectroscopy\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Spectroscopy\/Vibrational_Spectroscopy\/Infrared_Spectroscopy\" rel=\"internal\">IR<\/a>) are the primary ways\u00a0of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of $$\\pi$$ bonds and\/or cyclic rings.<\/p>\n<\/div>\n<div id=\"section_1\">\n<h3 class=\"editable\">Saturated and Unsaturated Molecules<\/h3>\n<p>In the lab, <a title=\"Physical Chemistry\/Physical Properties of Matter\/Solutions\/SOLUBILITY\/Types of Saturation\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Equilibria\/Solubilty\/Types_of_Saturation\" rel=\"internal\">saturation<\/a> may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds\u00a0with no rings is considered saturated.<\/p>\n<table style=\"margin: auto;border-spacing: 1px;width: 576px\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td>\u00a0 CH<sub>3<\/sub>CH<sub>2<\/sub>CH<sub>3<\/sub><\/td>\n<td>\u00a0 <img decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1958\/chewiki_sat.bmp?revision=1&amp;size=bestfit&amp;width=150&amp;height=94#fixme\" alt=\"chewiki_sat.bmp\" width=\"150px\" height=\"94px\" \/><\/td>\n<td><img decoding=\"async\" class=\"internal\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1959\/chewiki_sat2_(3).bmp?revision=1&amp;size=bestfit&amp;width=113&amp;height=128#fixme\" alt=\"chewiki_sat2 (3).bmp\" width=\"113px\" height=\"128px\" \/><\/td>\n<td>\u00a01-methyoxypentane<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and\/or ring(s).<\/p>\n<table style=\"margin: auto;border-spacing: 1px;width: 600px\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td>\u00a0\u00a0 CH<sub>3<\/sub>CH=CHCH<sub>3<\/sub><\/td>\n<td>\u00a0 <img decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1960\/chewiki_unsat1.bmp?revision=1&amp;size=bestfit&amp;width=75&amp;height=68#fixme\" alt=\"chewiki_unsat1.bmp\" width=\"75px\" height=\"68px\" \/><\/td>\n<td>\u00a0<img decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1961\/chewiki_unsat2.bmp?revision=1&amp;size=bestfit&amp;width=80&amp;height=62#fixme\" alt=\"chewiki_unsat2.bmp\" width=\"80px\" height=\"62px\" \/><\/td>\n<td>\u00a0<img decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1962\/chewiki_unsat3.bmp?revision=1&amp;size=bestfit&amp;width=110&amp;height=66#fixme\" alt=\"chewiki_unsat3.bmp\" width=\"110px\" height=\"66px\" \/><\/td>\n<td>3-chloro-5-octyne<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"section_2\">\n<h3 class=\"editable\">Calculating The Degree of Unsaturation (DoU)<\/h3>\n<p>If the molecular\u00a0formula is given, plug\u00a0in the numbers into this formula:<\/p>\n<p>\\[ DoU= \\dfrac{2C+2+N-X-H}{2} \\tag{7.2.1}\\]<\/p>\n<ul>\n<li><em>C<\/em> is the number\u00a0of carbons<\/li>\n<li><em>N<\/em> is the number\u00a0of nitrogens<\/li>\n<li><em>X<\/em> is the number\u00a0of halogens (F, Cl, Br, I)<\/li>\n<li><em>H<\/em> is the number\u00a0of hydrogens<\/li>\n<\/ul>\n<p>As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that\u00a0a saturated\u00a0molecule\u00a0has the maximum number of hydrogen atoms possible\u00a0to be\u00a0an acyclic alkane. Thus, the\u00a0number of hydrogens\u00a0can be represented by 2C+2, which is the general molecular representation of an <a class=\"internal\" title=\"Organic Chemistry\/Hydrocarbons\/Alkanes\" href=\"https:\/\/chem.libretexts.org\/Core\/Organic_Chemistry\/Hydrocarbons\/Alkanes\" rel=\"internal\">alkane<\/a>.\u00a0As an\u00a0example,\u00a0for\u00a0the molecular\u00a0formula\u00a0C<sub>3<\/sub>H<sub>4<\/sub> the number of actual hydrogens needed\u00a0for the compound to be saturated is 8 <em>[2C+2=(2&#215;3)+2=8]<\/em>. The compound needs\u00a04 more hydrogens in order to be fully saturated <em>(expected number of hydrogens-observed number of hydrogens=8-4=4)<\/em>. Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as\u00a0saturated.\u00a0\u00a0Hence, the DoB formula divides by 2. The formula subtracts the number of X&#8217;s because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C<sub>2<\/sub>H<sub>5<\/sub>Cl, there is one less hydrogen compared to ethane, C<sub>2<\/sub>H<sub>6<\/sub>.<\/p>\n<p>For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C<sub>3<\/sub>H<sub>9<\/sub>N compared to C<sub>3<\/sub>H<sub>8<\/sub>. Oxygen and sulfur\u00a0are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens\u00a0in ethanol,\u00a0C<sub>2<\/sub>H<sub>5<\/sub>OH, matches the number of hydrogens in ethane, C<sub>2<\/sub>H<sub>6<\/sub>.<\/p>\n<p>The following chart\u00a0 illustrates the\u00a0possible combinations of the number of\u00a0double bond(s), triple bond(s), and\/or ring(s) for a\u00a0given degree of unsaturation. Each row corresponds to a different combination.<\/p>\n<ul>\n<li>One degree of unsaturation is equivalent to 1 ring or 1 double bond (1\u00a0$$ \\pi $$\u00a0bond).<\/li>\n<li>Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2\u00a0$$ \\pi $$\u00a0bonds).<\/li>\n<\/ul>\n<table style=\"margin: auto;border-spacing: 0px;width: 448px\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td>\n<div><strong>DoU<\/strong><\/div>\n<\/td>\n<td colspan=\"3\">\n<div><strong>Possible combinations of rings\/ \u00a0bonds<\/strong><\/div>\n<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>\n<div># of rings<\/div>\n<\/td>\n<td>\n<div># of double bonds<\/div>\n<\/td>\n<td>\n<div># of triple bonds<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<div>1<\/div>\n<\/td>\n<td>\n<div>1<\/div>\n<\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<td>\n<div>1<\/div>\n<\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<div>2<\/div>\n<\/td>\n<td>\n<div>2<\/div>\n<\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<td>\n<div>2<\/div>\n<\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<td>\n<div>1<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>\n<div>1<\/div>\n<\/td>\n<td>\n<div>1<\/div>\n<\/td>\n<td>\n<div>0<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and\/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond,\u00a0<em>or<\/em>\u00a03 double bonds.<\/p>\n<div>\n<div id=\"example\">\n<div class=\"textbox examples\">\n<h3>ExampLE<\/h3>\n<div id=\"section_2\">\n<div>\n<div id=\"example\">\n<p class=\"boxtitle\">\u00a0Benzene<\/p>\n<p>What is the Degree of Unsaturation for Benzene?<\/p>\n<h3><strong>SOLUTION<\/strong><\/h3>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q713260\">Show Answer<\/span><\/p>\n<div id=\"q713260\" class=\"hidden-answer\" style=\"display: none\">\n<p>The molecular formula for benzene is C<sub>6<\/sub>H<sub>6<\/sub><sub>. <\/sub>Thus,<\/p>\n<p>DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1960\/chewiki_unsat1.bmp?revision=1&amp;size=bestfit&amp;width=75&amp;height=68#fixme\" alt=\"chewiki_unsat1.bmp\" width=\"75px\" height=\"68px\" \/><\/p>\n<p>However, when given the molecular formula C<sub>6<\/sub>H<sub>6<\/sub>, benzene is only one of many possible structures (<a title=\"Isomers\" href=\"https:\/\/chem.libretexts.org\/Core\/Inorganic_Chemistry\/Coordination_Chemistry\/Properties_of_Coordination_Compounds\/Isomers\" rel=\"internal\">isomers<\/a>). The following structures all have DoB of 4 and have the same molecular formula as benzene.<\/p>\n<p><img decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1951\/4_DoB_2.bmp?revision=1&amp;size=bestfit&amp;width=114&amp;height=62#fixme\" alt=\"4 DoB_2.bmp\" width=\"114px\" height=\"62px\" \/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<img decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1952\/4_DoB_3.bmp?revision=1&amp;size=bestfit&amp;width=89&amp;height=65#fixme\" alt=\"4 DoB_3.bmp\" width=\"89px\" height=\"65px\" \/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<img decoding=\"async\" class=\"internal\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1953\/4DoB_1.bmp?revision=1&amp;size=bestfit&amp;width=118&amp;height=78#fixme\" alt=\"4DoB_1.bmp\" width=\"118px\" height=\"78px\" \/><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_3\">\n<h3 class=\"editable\">References<\/h3>\n<ol>\n<li>Vollhardt, K. P.C. &amp; Shore, N. (2007). <em>Organic Chemistry <\/em>(5<sup>th<\/sup>Ed.).\u00a0\u00a0New York: W. H. Freeman. (473-474)<\/li>\n<li>Shore, N. (2007). <em>Study Guide and Solutions Manual for Organic Chemistry <\/em>(5<sup>th<\/sup> Ed.). New York: W.H. Freeman. (201)<\/li>\n<\/ol>\n<\/div>\n<div id=\"section_4\">\n<div class=\"textbox examples\">\n<h3>Examples<\/h3>\n<div id=\"section_4\">\n<h3 class=\"editable\">Problems<\/h3>\n<ol>\n<li>\u00a0Are the following molecules saturated or unsaturated:\n<ol>\n<li>\u00a0 <img decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1955\/chewiki_prob1a.bmp?revision=1&amp;size=bestfit&amp;width=108&amp;height=75#fixme\" alt=\"chewiki_prob1a.bmp\" width=\"108px\" height=\"75px\" \/>\u00a0\u00a0\u00a0\u00a0(b.)<img decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1957\/chewiki_prob2b.bmp?revision=1&amp;size=bestfit&amp;width=134&amp;height=65#fixme\" alt=\"chewiki_prob2b.bmp\" width=\"134px\" height=\"65px\" \/>\u00a0\u00a0\u00a0\u00a0\u00a0(c.) <img decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/1956\/chewiki_prob1c.bmp?revision=1&amp;size=bestfit&amp;width=119&amp;height=75#fixme\" alt=\"chewiki_prob1c.bmp\" width=\"119px\" height=\"75px\" \/>\u00a0\u00a0\u00a0\u00a0(d.) C<sub>10<\/sub>H<sub>6<\/sub>N<sub>4<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Using the molecules from 1., give the degrees of unsaturation for each.<\/li>\n<li>Calculate the degrees of unsaturation for the following molecular formulas:\n<ol>\n<li>(a.) C<sub>9<\/sub>H<sub>20<\/sub><sub> \u00a0\u00a0\u00a0\u00a0<\/sub>(b.) C<sub>7<\/sub>H<sub>8<\/sub><sub> \u00a0\u00a0\u00a0\u00a0<\/sub>(c.) C<sub>5<\/sub>H<sub>7<\/sub>Cl \u00a0\u00a0\u00a0\u00a0(d.) C<sub>9<\/sub>H<sub>9<\/sub>NO<sub>4<\/sub><sub>\u00a0<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Using the molecular formulas from 3, are the molecules unsaturated or saturated.<\/li>\n<li>Using the molecular formulas from 3, if the molecules are saturated, how many rings\/double bonds\/triple bonds are predicted?<\/li>\n<li>(d.) <strong>unsaturated<\/strong>5.(a.) <strong>0 <\/strong>(Remember-a saturated molecule only contains single bonds)\n<p>(b.) <em>The molecule can contain any of these combinations<\/em> <strong>(i) 4 double bonds (ii) 4 rings (iii) 2 double bonds+2 rings (iv) 1 double bond+3 rings (v) 3 double bonds+1 ring (vi) 1 triple bond+2 rings (vii) 2 triple bonds (viii) 1 triple bond+1 double bond+1 ring (ix) 1 triple bond+2 double bonds\u00a0<\/strong><\/p>\n<p>(c.) (<strong>i) 1 triple bond (ii) 1 ring+1 double bond (iii) 2 rings (iv) 2 double bonds\u00a0<\/strong><\/p>\n<p>(d.) <strong>(i) 3 triple bonds (ii) 2 triple bonds+2 double bonds (iii) 2 triple bonds+1 double bond+1 ring (iv)&#8230; <\/strong><em>(As you can see, the degrees of unsaturation only gives the sum of double bonds, triple bonds and\/or ring. Thus, the formula may give numerous possible structures for a given molecular formula.)<\/em><\/li>\n<\/ol>\n<\/div>\n<div id=\"section_5\">\n<h3 class=\"editable\">Answers<\/h3>\n<p>1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567506\">Show Answer<\/span><\/p>\n<div id=\"q567506\" class=\"hidden-answer\" style=\"display: none\">(a.)\u00a0unsaturated (Even though rings only contain single bonds, rings are considered unsaturated.)<\/p>\n<p>(b.) unsaturated<\/p>\n<p>(c.) saturated<\/p>\n<p>(d.) unsaturated<\/p><\/div>\n<\/div>\n<p>2. <em>If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and\/or rings. However, you can also\u00a0determine the molecular\u00a0formula\u00a0and solve\u00a0for the degrees of unsaturation by\u00a0using the\u00a0formula.<\/em><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q546426\">Show Answer<\/span><\/p>\n<div id=\"q546426\" class=\"hidden-answer\" style=\"display: none\">(a.)\u00a02<\/p>\n<p>(b.) 2 (one double bond and the double\u00a0bond from the\u00a0carbonyl)<\/p>\n<p>(c.) 0<\/p>\n<p>(d.) 10<\/p><\/div>\n<\/div>\n<p>3. <em>Use the formula to solve<\/em><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q389197\">Show Answer<\/span><\/p>\n<div id=\"q389197\" class=\"hidden-answer\" style=\"display: none\">(a.) 0<\/p>\n<p>(b.) 4<\/p>\n<p>(c.) 2<\/p>\n<p>(d.) 6<\/p><\/div>\n<\/div>\n<p>4.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q669637\">Show Answer<\/span><\/p>\n<div id=\"q669637\" class=\"hidden-answer\" style=\"display: none\">\n<p>(a.) saturated<\/p>\n<p>(b.) unsaturated<\/p>\n<p>(c.) unsaturated\u00a0<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_6\">\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<div id=\"s61712\">\n<div id=\"section_7\">\n<h3 id=\"Questions-61712\">Questions<\/h3>\n<p><strong>1.<\/strong><\/p>\n<p>Calculate degrees of unsaturation (DoU) for the following, and propose a structure for each.<\/p>\n<p>A \u2013 C<sub>5<\/sub>H<sub>8<\/sub><\/p>\n<p>B \u2013 C<sub>4<\/sub>H<sub>4<\/sub><\/p>\n<p><b>2.\u00a0<\/b><\/p>\n<p>Calculate the degree of unsaturation (DoU) for the following molecules<\/p>\n<p>A \u2013 C<sub>5<\/sub>H<sub>5<\/sub>N<\/p>\n<p>B \u2013 C<sub>5<\/sub>H<sub>5<\/sub>NO<sub>2<\/sub><\/p>\n<p>C \u2013 C<sub>5<\/sub>H<sub>5<\/sub>Br<\/p>\n<p><strong>3.<\/strong><\/p>\n<p>The following molecule is caffeine (C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub>), determine the degrees of unsaturation (DoU).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05135319\/7.22.png\" alt=\"\" width=\"238\" height=\"203\" \/><\/p>\n<\/div>\n<div id=\"section_8\">\n<h3 id=\"Solutions-61712\">Solutions<\/h3>\n<p><strong>1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q543265\">Show Answer<\/span><\/p>\n<div id=\"q543265\" class=\"hidden-answer\" style=\"display: none\"><\/strong><img loading=\"lazy\" decoding=\"async\" class=\"internal default alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05135321\/7.2.png\" alt=\"\" width=\"320\" height=\"214\" \/><\/p>\n<p><strong><\/div>\n<\/div>\n<p><\/strong><\/p>\n<p>&nbsp;<\/p>\n<p><strong>2.<\/strong><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q262440\">Show Answer<\/span><\/p>\n<div id=\"q262440\" class=\"hidden-answer\" style=\"display: none\">A = 4, B = 4, C = 3<\/div>\n<\/div>\n<p><strong>3.<\/strong><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q330183\">Show Answer<\/span><\/p>\n<div id=\"q330183\" class=\"hidden-answer\" style=\"display: none\">6 DoU<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<h3 class=\"editable\">Contributors<\/h3>\n<ul>\n<li><a class=\"external\" title=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" href=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" target=\"_blank\" rel=\"external nofollow noopener\">Dr. Dietmar Kennepohl<\/a> FCIC (Professor of Chemistry, <a class=\"external\" title=\"http:\/\/www.athabascau.ca\/\" href=\"http:\/\/www.athabascau.ca\/\" target=\"_blank\" rel=\"external nofollow noopener\">Athabasca University<\/a>)<\/li>\n<li>Prof. Steven Farmer (<a class=\"external\" title=\"http:\/\/www.sonoma.edu\" href=\"http:\/\/www.sonoma.edu\" target=\"_blank\" rel=\"external nofollow noopener\">Sonoma State University<\/a>)<\/li>\n<li>Kim Quach (UCD)<\/li>\n<\/ul>\n<\/div>\n<\/div>\n","protected":false},"author":44985,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-966","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/966","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/users\/44985"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/966\/revisions"}],"predecessor-version":[{"id":2276,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/966\/revisions\/2276"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/23"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/966\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/media?parent=966"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=966"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/contributor?post=966"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/license?post=966"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}