{"id":7831,"date":"2021-06-08T21:57:53","date_gmt":"2021-06-08T21:57:53","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/alkyl-halides-and-alcohols\/"},"modified":"2021-10-13T16:59:23","modified_gmt":"2021-10-13T16:59:23","slug":"alkyl-halides-and-alcohols","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/alkyl-halides-and-alcohols\/","title":{"raw":"Alkyl Halides and Alcohols","rendered":"Alkyl Halides and Alcohols"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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    \r\n \t
  1. Define functional group<\/em>.<\/li>\r\n \t
  2. Identify and name a simple alkyl halide.<\/li>\r\n \t
  3. Identify and name a simple alcohol.<\/li>\r\n \t
  4. Predict the product(s) of an elimination reaction of an alkyl halide or an alcohol.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nA functional group<\/strong> is any collection of atoms and\/or bonds with certain characteristic chemical reactions. We have already seen two functional groups: the C\u2013C double bond and the C\u2013C triple bond. They undergo certain characteristic chemical reactions \u2014 for example, the addition of a halogen across the multiple bond.\r\n\r\nThe presence of a halogen atom (F, Cl, Br, or I; X is used to represent any halogen atom) is one of the simplest functional groups. Organic compounds that contain a halogen atom are called alkyl halides<\/strong>. We have already seen some examples of alkyl halides when the addition of halogens across double and triple bonds was introduced in the section \"Branched Hydrocarbons\"<\/a>; the products of these reactions were alkyl halides.\r\n\r\nA simple alkyl halide can be named like an ionic salt, first by stating the name of the parent alkane as a substituent group (with the -yl<\/em> suffix) and then the name of the halogen as if it were the anion. So CH3<\/sub>Cl has the common name of methyl chloride, while CH3<\/sub>CH2<\/sub>Br is ethyl bromide and CH3<\/sub>CH2<\/sub>CH2<\/sub>I is propyl iodide. However, this system is not ideal for more complicated alkyl halides.\r\n\r\nThe systematic way of naming alkyl halides is to name the halogen as a substituent, just like an alkyl group, and use numbers to indicate the position of the halogen atom on the main chain. The name of the halogen as a substituent comes from the stem of the element\u2019s name plus the ending -o<\/em>, so the substituent names are fluoro<\/em>-, chloro<\/em>-, bromo<\/em>-, and iodo<\/em>-. If there is more than one of a certain halogen, we use numerical prefixes to indicate the number of each kind, just as with alkyl groups. For example, this molecule\u00a0is 2-bromobutane.\r\n

    [latex]\\chemfig{-[:30](-[:90]Br)-[:-30]-[:30]}[\/latex]<\/p>\r\nAnd this molecule is 2,3-dichloropentane:\r\n

    [latex]\\chemfig{-[:30](-[:90]Cl)-[:-30](-[:-90]Cl)-[:30]-[:-30]}[\/latex]<\/p>\r\nIf alkyl groups are present, the substituents are listed alphabetically. Numerical prefixes are ignored when determining the alphabetical ordering of substituent groups.\r\n

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    Example 16.6<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nName this molecule.\r\n

    [latex]\\chemfig{-[:30](-[:90]Cl)-[:-30](-[:-60]Cl)(-[:-120])-[:30]-[:90]}[\/latex]<\/p>\r\nSolution<\/em>\r\nThe longest carbon chain has five C atoms, so the molecule is a pentane. There are two chlorine substituents located on the second and third C atoms, with a one-carbon methyl group on the third C atom as well. The correct name for this molecule is 2,3-dichloro-3-methylpentane.\r\n\r\nTest Yourself<\/em>\r\nName this molecule.\r\n

    [latex]\\chemfig{Br-[:-60](-[:-120]Br)-(-[:-60])-[:60]Br}[\/latex]<\/p>\r\nAnswer<\/em>\r\n1,1,2-tribromopropane\r\n\r\n<\/div>\r\n<\/div>\r\nAnother simple functional group is the covalently bonded OH group. This is the alcohol<\/strong>\u00a0functional group. It is not the hydroxide ion;\u00a0in organic chemistry, rather than being present as a negatively charged species, it is a covalently bonded functional group.\r\n\r\nLike alkyl halides, alcohols have a common naming system and a more formal system. The common system is similar to that of alkyl halides: name the alkyl group attached to the OH group, ending with the suffix -yl<\/em>, and add the word alcohol<\/em> as a second word. So CH3<\/sub>OH is methyl alcohol, CH3<\/sub>CH2<\/sub>OH is ethyl alcohol, and CH3<\/sub>CH2<\/sub>CH2<\/sub>OH is propyl alcohol.\r\n\r\nAs with alkyl halides, though, this system is limited (although for smaller alcohols, it is very common in everyday usage). The formal system of naming uses the name of the hydrocarbon containing the OH group and having the correct number of C atoms, dropping the final -e<\/em> of the name and appending the suffix -ol<\/em>. Thus CH3<\/sub>OH is methanol and CH3<\/sub>CH2<\/sub>OH is ethanol. For larger alcohol molecules, we use a number to indicate the position of the OH group on the longest carbon chain, similar to the number needed for alkenes and alkynes. Again, the carbon chain is numbered to give the OH group the lowest number, no matter how large the other numbers are. So CH3<\/sub>CH2<\/sub>CH2<\/sub>OH is 1-propanol, while CH3<\/sub>CHOHCH3<\/sub> is 2-propanol.\r\n\r\n[caption id=\"attachment_902\" align=\"aligncenter\" width=\"173\"]\"Isopropyl Figure 16.4 \"Isopropyl Alcohol.\" What you find labelled \"isopropyl alcohol\" in a medicine cabinet is more formally called \"2-propanol.\"[\/caption]\r\n\r\nAnother acceptable way of naming an alcohol \u2014 especially a more complicated molecule \u2014 is to name the OH group as the hydroxy substituent and give it a numerical position like an alkyl group or a halogen atom. Thus 2-propanol would be called 2-hydroxypropane by this convention.\r\n

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    Example 16.7<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nName this molecule as an alcohol and as a substituted alkane.\r\n

    [latex]\\chemfig{-[:-30]-[:30](-[:60])(-[:120]Cl)-[:-30]-[:30]OH}[\/latex]<\/p>\r\nSolution<\/em>\r\nThe longest carbon chain containing the OH group has four C atoms, so the parent hydrocarbon is butane. Because the OH group is on the first C atom, it is 1-butanol. There is a methyl group on the second C atom, as well as a Cl atom, so the formal name for this alcohol is 2-chloro-2-methyl-1-butanol. If naming the alcohol group as a substituent, it would be 2-chloro-1-hydroxy-2-methylbutane.\r\n\r\nTest Yourself<\/em>\r\nName this molecule as an alcohol and as a substituted alkane.\r\n

    [latex]\\chemfig{Cl-[:30](-[:-90]Cl)(-[:150]Cl)-[:30]-[:-30]OH}[\/latex]<\/p>\r\nAnswer<\/em>\r\n2,2,2-trichloroethanol\r\n\r\n<\/div>\r\n<\/div>\r\nMost alkyl halides are insoluble in H2<\/sub>O. Smaller alcohols, however, are very soluble in H2<\/sub>O because these molecules can engage in hydrogen bonding with H2<\/sub>O molecules. For larger molecules, however, the polar OH group is overwhelmed by the nonpolar alkyl part of the molecule. While methanol is soluble in H2<\/sub>O in all proportions, only about 2.6 g of pentanol will dissolve in 100 g of H2<\/sub>O. Larger alcohols have an even lower solubility in H2<\/sub>O.\r\n\r\nOne reaction common to alcohols and alkyl halides is the\u00a0elimination reaction<\/strong>, which is the removal of the functional group (either X or OH) and an H atom from an adjacent carbon. The general reaction can be written this way:\r\n

    [latex]\\begin{array}{ccccc}\r\n\\chemfig{H-C(-[:90]H)(-[:-90]H)-C(-[:90]X)(-[:-90]H)-H}&\\xrightarrow{\\text{catalyst}}&\\chemfig{(-[:120]H)(-[:-120]H)=(-[:60]H)(-[:-60]H)}&+&\\ce{HX} \\\\\r\n\\end{array}[\/latex]<\/p>\r\nwhere X can represent either the X or the OH group. The biggest difference between elimination in alkyl halides and elimination in alcohols is the identity of the catalyst: for alkyl halides, the catalyst is a strong base; for alcohols, the catalyst is a strong acid. For compounds in which there are H atoms on more than one adjacent carbon, a mixture of products results.\r\n

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    Example 16.8<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nPredict the organic product(s) of this reaction.\r\n

    [latex]\\begin{array}{ccc}\r\n\\chemfig{H-C(-[:90]H)(-[:-90]H)-C(-[:90]\\ce{OH})(-[:-90]H)-C(-[:90]H)(-[:-90]H)-H}&\\xrightarrow{\\text{acid}}&?\r\n\\end{array}[\/latex]<\/p>\r\nSolution<\/em>\r\nUnder these conditions, an HOH (otherwise known as H2<\/sub>O) molecule is eliminated, and an alkene forms. It does not matter which adjacent carbon loses the H atom; in either case the product will be propene:\r\n

    [latex]\\chemfig{H-[:-30](-[:-90]H)=[:30](-[:90]H)-[:-30](-[:-90]H)(-[:30]H)-[:-30]H}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nPredict the organic product(s) of this reaction.\r\n

    [latex]\\begin{array}{ccc}\r\n\\chemfig{H-C(-[:90]H)(-[:-90]H)-C(-[:90]Cl)(-[:-90]H)-C(-[:90]H)(-[:-90]H)-C(-[:90]H)(-[:-90]H)-H}&\\xrightarrow{\\text{base}}&?\r\n\\end{array}[\/latex]<\/p>\r\nAnswer<\/em>\r\n1-butene and 2-butene\r\n\r\n<\/div>\r\n<\/div>\r\n

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    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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