{"id":521,"date":"2017-10-04T20:56:06","date_gmt":"2017-10-04T20:56:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&#038;p=521"},"modified":"2018-10-03T17:20:51","modified_gmt":"2018-10-03T17:20:51","slug":"physical-properties-of-alkanes","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/physical-properties-of-alkanes\/","title":{"raw":"Physical Properties of Alkanes","rendered":"Physical Properties of Alkanes"},"content":{"raw":"<div class=\"elm-header\">\r\n<div class=\"elm-header-custom\">\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>arrange a number of given straight-chain alkanes in order of increasing or decreasing boiling point or melting point.<\/li>\r\n \t<li>arrange a series of isomeric alkanes in order of increasing or decreasing boiling point.<\/li>\r\n \t<li>explain the difference in boiling points between a given number of alkanes.<\/li>\r\n<\/ol>\r\n<\/div>\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 Term<\/h3>\r\nMake certain that you can define, and use in context, the key term below.\r\n<ul>\r\n \t<li>van der Waals force<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nAlkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless.\r\n<div id=\"section_1\">\r\n<h3 class=\"editable\">Boiling Points<\/h3>\r\nThe boiling points shown are for the \"straight chain\" isomers\u00a0of which there\u00a0is more than one. The first four <a title=\"Alkanes\" href=\"https:\/\/chem.libretexts.org\/Core\/Organic_Chemistry\/Hydrocarbons\/Alkanes\" rel=\"internal\">alkanes<\/a> are gases at room temperature, and solids do not\u00a0begin to appear until about $$C_{17}H_{36}$$, but this is imprecise\u00a0because different isomers typically have different melting and boiling points. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers!\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04205551\/bptsalkanes.gif\" alt=\"\" width=\"363px\" height=\"261px\" \/>\r\n\r\nCycloalkanes have boiling points that are approximately 20 K higher than the corresponding straight chain alkane.\r\n\r\nThere is not\u00a0a significant\u00a0<a title=\"Physical Chemistry\/Physical Properties of Matter\/Atomic and Molecular Properties\/Electronegativity\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Physical_Properties_of_Matter\/Atomic_and_Molecular_Properties\/Electronegativity\" rel=\"internal\">electronegativity <\/a>difference between carbon and hydrogen,\u00a0thus, there is\u00a0not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be <a title=\"Physical Chemistry\/Quantum Mechanics\/Atomic Theory\/Intermolecular Forces\/Van Der Waals Interactions\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Physical_Properties_of_Matter\/Atomic_and_Molecular_Properties\/Intermolecular_Forces\/Specific_Interactions\/Van_Der_Waals_Interactions\" rel=\"internal\">Van der Waals<\/a> dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size.\r\n\r\nWhere you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules.\r\n<div>\r\n<div id=\"example\">\r\n<div class=\"textbox examples\">\r\n<h3>Examples<\/h3>\r\n<div id=\"example\">\r\n<p class=\"boxtitle\">Example 1: Boiling Points of Alkanes<\/p>\r\nFor example, the boiling points of the three isomers of $$C_5H_{12}$$ are:\r\n<ul>\r\n \t<li>pentane: 309.2 K<\/li>\r\n \t<li>2-methylbutane: 301.0 K<\/li>\r\n \t<li>2,2-dimethylpropane: 282.6 K<\/li>\r\n<\/ul>\r\nThe slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less \"wriggly\"!\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<div>\r\n<div id=\"exercise\">\r\n\r\nFor each of the following pairs of compounds, select the substance which you expect to have the higher boiling point:\r\n<ol start=\"1\">\r\n \t<li>octane and nonane.<\/li>\r\n \t<li>octane and 2,2,3,3-tetramethylbutane.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_2\">\r\n<h3 class=\"editable\">Solubility<\/h3>\r\n<\/div>\r\n<\/div>\r\nAlkanes are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds.\r\n<div id=\"section_3\">\r\n<h4 class=\"editable\">Solubility in Water<\/h4>\r\nWhen a molecular substance dissolves in water, the following must occur:\r\n<ul>\r\n \t<li>break the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces.<\/li>\r\n \t<li>break the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds.<\/li>\r\n<\/ul>\r\nBreaking either of these attractions\u00a0requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is\u00a0relatively negligible; this is not\u00a0true of the hydrogen bonds in water.\r\n\r\nAs something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to\u00a0compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces \u00a0do not release a sufficient amount of energy to compensate for\u00a0the energy required\u00a0to break the hydrogen bonds in water. The alkane does not dissolve.\r\n<div id=\"note\">\r\n<blockquote>\r\n<p class=\"boxtitle\">Note: This is a simplification\u00a0because entropic effects are important when things dissolve.<\/p>\r\n<\/blockquote>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_4\">\r\n<h4 class=\"editable\">Solubility in organic solvents<\/h4>\r\nIn most organic solvents, the primary forces of attraction between the solvent molecules are <a title=\"Van der Waals Forces\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Physical_Properties_of_Matter\/Atomic_and_Molecular_Properties\/Intermolecular_Forces\/Van_der_Waals_Forces\" rel=\"internal\">Van der Waals<\/a> - either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus,\u00a0there is no barrier to solubility.\r\n\r\n<\/div>\r\n<div id=\"section_5\">\r\n<div class=\"textbox exercises\">\r\n<h3>Exercise<\/h3>\r\n<h3 class=\"editable\">Question<\/h3>\r\n<ol>\r\n \t<li>For each of the following pairs of compounds, select the substance you expect to have the higher boiling point.\r\n<ol>\r\n \t<li>octane and nonane.<\/li>\r\n \t<li>octane and 2,2,3,3\u2011tetramethylbutane.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<div id=\"section_6\">\r\n<h4 class=\"editable\">Answers:<\/h4>\r\n<p class=\"editable\">[reveal-answer q=\"167240\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"167240\"]1.Nonane will have a higher boiling point than octane, because it has a longer carbon chain than octane.<\/p>\r\n<p class=\"editable\">2. Octane will have a higher boiling point than 2,2,3,3\u2011tetramethylbutane, because it branches less than 2,2,3,3\u2011tetramethylbutane, and therefore has a larger \u201csurface area\u201d and more van der Waals forces.<\/p>\r\n\r\n<blockquote>\r\n<p class=\"editable\"><strong>Note: <\/strong>The actual boiling points arenonane, 150.8\u00b0C\r\noctane, 125.7\u00b0C\r\n2,2,3,3\u2011tetramethylbutane, 106.5\u00b0C<\/p>\r\n<\/blockquote>\r\n<p class=\"editable\">[\/hidden-answer]<\/p>\r\n\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>Jim Clark (<a class=\"external\" title=\"http:\/\/www.chemguide.co.uk\" href=\"http:\/\/www.chemguide.co.uk\" target=\"_blank\" rel=\"external nofollow noopener\">Chemguide.co.uk<\/a>)<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"elm-header\">\n<div class=\"elm-header-custom\">\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>arrange a number of given straight-chain alkanes in order of increasing or decreasing boiling point or melting point.<\/li>\n<li>arrange a series of isomeric alkanes in order of increasing or decreasing boiling point.<\/li>\n<li>explain the difference in boiling points between a given number of alkanes.<\/li>\n<\/ol>\n<\/div>\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 Term<\/h3>\n<p>Make certain that you can define, and use in context, the key term below.<\/p>\n<ul>\n<li>van der Waals force<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless.<\/p>\n<div id=\"section_1\">\n<h3 class=\"editable\">Boiling Points<\/h3>\n<p>The boiling points shown are for the &#8220;straight chain&#8221; isomers\u00a0of which there\u00a0is more than one. The first four <a title=\"Alkanes\" href=\"https:\/\/chem.libretexts.org\/Core\/Organic_Chemistry\/Hydrocarbons\/Alkanes\" rel=\"internal\">alkanes<\/a> are gases at room temperature, and solids do not\u00a0begin to appear until about $$C_{17}H_{36}$$, but this is imprecise\u00a0because different isomers typically have different melting and boiling points. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers!<\/p>\n<p><img decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04205551\/bptsalkanes.gif\" alt=\"\" width=\"363px\" height=\"261px\" \/><\/p>\n<p>Cycloalkanes have boiling points that are approximately 20 K higher than the corresponding straight chain alkane.<\/p>\n<p>There is not\u00a0a significant\u00a0<a title=\"Physical Chemistry\/Physical Properties of Matter\/Atomic and Molecular Properties\/Electronegativity\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Physical_Properties_of_Matter\/Atomic_and_Molecular_Properties\/Electronegativity\" rel=\"internal\">electronegativity <\/a>difference between carbon and hydrogen,\u00a0thus, there is\u00a0not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be <a title=\"Physical Chemistry\/Quantum Mechanics\/Atomic Theory\/Intermolecular Forces\/Van Der Waals Interactions\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Physical_Properties_of_Matter\/Atomic_and_Molecular_Properties\/Intermolecular_Forces\/Specific_Interactions\/Van_Der_Waals_Interactions\" rel=\"internal\">Van der Waals<\/a> dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size.<\/p>\n<p>Where you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules.<\/p>\n<div>\n<div id=\"example\">\n<div class=\"textbox examples\">\n<h3>Examples<\/h3>\n<div id=\"example\">\n<p class=\"boxtitle\">Example 1: Boiling Points of Alkanes<\/p>\n<p>For example, the boiling points of the three isomers of $$C_5H_{12}$$ are:<\/p>\n<ul>\n<li>pentane: 309.2 K<\/li>\n<li>2-methylbutane: 301.0 K<\/li>\n<li>2,2-dimethylpropane: 282.6 K<\/li>\n<\/ul>\n<p>The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less &#8220;wriggly&#8221;!<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<div>\n<div id=\"exercise\">\n<p>For each of the following pairs of compounds, select the substance which you expect to have the higher boiling point:<\/p>\n<ol start=\"1\">\n<li>octane and nonane.<\/li>\n<li>octane and 2,2,3,3-tetramethylbutane.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_2\">\n<h3 class=\"editable\">Solubility<\/h3>\n<\/div>\n<\/div>\n<p>Alkanes are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds.<\/p>\n<div id=\"section_3\">\n<h4 class=\"editable\">Solubility in Water<\/h4>\n<p>When a molecular substance dissolves in water, the following must occur:<\/p>\n<ul>\n<li>break the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces.<\/li>\n<li>break the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds.<\/li>\n<\/ul>\n<p>Breaking either of these attractions\u00a0requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is\u00a0relatively negligible; this is not\u00a0true of the hydrogen bonds in water.<\/p>\n<p>As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to\u00a0compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces \u00a0do not release a sufficient amount of energy to compensate for\u00a0the energy required\u00a0to break the hydrogen bonds in water. The alkane does not dissolve.<\/p>\n<div id=\"note\">\n<blockquote>\n<p class=\"boxtitle\">Note: This is a simplification\u00a0because entropic effects are important when things dissolve.<\/p>\n<\/blockquote>\n<\/div>\n<\/div>\n<div id=\"section_4\">\n<h4 class=\"editable\">Solubility in organic solvents<\/h4>\n<p>In most organic solvents, the primary forces of attraction between the solvent molecules are <a title=\"Van der Waals Forces\" href=\"https:\/\/chem.libretexts.org\/Core\/Physical_and_Theoretical_Chemistry\/Physical_Properties_of_Matter\/Atomic_and_Molecular_Properties\/Intermolecular_Forces\/Van_der_Waals_Forces\" rel=\"internal\">Van der Waals<\/a> &#8211; either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus,\u00a0there is no barrier to solubility.<\/p>\n<\/div>\n<div id=\"section_5\">\n<div class=\"textbox exercises\">\n<h3>Exercise<\/h3>\n<h3 class=\"editable\">Question<\/h3>\n<ol>\n<li>For each of the following pairs of compounds, select the substance you expect to have the higher boiling point.\n<ol>\n<li>octane and nonane.<\/li>\n<li>octane and 2,2,3,3\u2011tetramethylbutane.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div id=\"section_6\">\n<h4 class=\"editable\">Answers:<\/h4>\n<p class=\"editable\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q167240\">Show Answer<\/span><\/p>\n<div id=\"q167240\" class=\"hidden-answer\" style=\"display: none\">1.Nonane will have a higher boiling point than octane, because it has a longer carbon chain than octane.<\/p>\n<p class=\"editable\">2. Octane will have a higher boiling point than 2,2,3,3\u2011tetramethylbutane, because it branches less than 2,2,3,3\u2011tetramethylbutane, and therefore has a larger \u201csurface area\u201d and more van der Waals forces.<\/p>\n<blockquote>\n<p class=\"editable\"><strong>Note: <\/strong>The actual boiling points arenonane, 150.8\u00b0C<br \/>\noctane, 125.7\u00b0C<br \/>\n2,2,3,3\u2011tetramethylbutane, 106.5\u00b0C<\/p>\n<\/blockquote>\n<p class=\"editable\"><\/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>Jim Clark (<a class=\"external\" title=\"http:\/\/www.chemguide.co.uk\" href=\"http:\/\/www.chemguide.co.uk\" target=\"_blank\" rel=\"external nofollow noopener\">Chemguide.co.uk<\/a>)<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":311,"menu_order":7,"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-521","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/521","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\/311"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/521\/revisions"}],"predecessor-version":[{"id":2245,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/521\/revisions\/2245"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/521\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/media?parent=521"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=521"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/contributor?post=521"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/license?post=521"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}