SGCh24 - Chapter Twenty-Four ORGANIC CHEMISTRY • • •...

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Unformatted text preview: Chapter Twenty-Four ORGANIC CHEMISTRY • • • Aliphatic Hydrocarbons Aromatic Hydrocarbons Functional Groups ALIPHATIC HYDROCARBONS STUDY OBJECTIVES 1. 2. 3. 4. Name four different types of aliphatic hydrocarbons and draw a structural formula of a typical example of each. Name hydrocarbons of these four groups given a structural formula. Given a systematic name, draw a structural formula. Write equations for addition reactions to alkenes and alkynes. Apply Markovnikov's rule concerning the addition of unsymmetrical reagents to alkenes. Organic Chemistry. The heart of organic chemistry is the carbon atom. Carbon is a key ingredient of about six million chemical compounds primarily because of its ability to form long chains of self-linked atoms. The existence of structural and geometric isomers contributes strongly to the number of organic or carboncontaining compounds. The h ydrocarbons are an important class of organic compound that consist only of the elements carbon and hydrogen. Hydrocarbons are divided into two classes: aliphatic and aromatic. A romatic hydrocarbons contain one or more benzene rings. A liphatic hydrocarbons do not contain benzene rings. Four groups of aliphatic hydrocarbons are known. These are alkanes, alkenes, alkynes, and cycloalkanes. Alkanes. The general formula for an alkane is Cn H2n+2 , where n is the number of carbon atoms in the molecule, n = 1, 2, 3…. When n = 1, we have the simplest member of the alkane family methane (CH4 ). The alkanes make up a homologous series; a series of compounds differing in the number of carbon atoms. As n increases one at a time, we can generate the formulas of the entire series of alkanes. The names and formulas of the first ten straight-chain alkanes are given in Table 24.1. The first part of each name represents the number of C atoms in the molecule. The ending '-ane' is common to all alkanes. If you learn these names, they will prove to be very useful in naming other organic compounds. Table 24.1 Names of the First Ten Alkanes. _________________________ Formula Name _________________________ CH4 Methane C 2 H6 Ethane C 3 H8 Propane C 4 H10 Butane C 5 H12 Pentane C 6 H14 Hexane C 7 H16 Heptane C 8 H18 Octane C 9 H20 Nonane C 10 H22 Decane _________________________ 4 63 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 64 / Organic Chemistry Alkenes. Alkenes contain C C double bonds, and members of this homologous series have the general formula Cn H2n . Alkenes are named with the same root word as alkanes to indicate the number of carbon atoms, but all names end in "ene." C 2 H4 C 3 H6 C 4 H8 CH2 CH2 CH3 CH CH2 CH3 CH2 CH CH2 Alkynes. Alkynes contain C alkynes end with "yne." C 2 H2 C 3 H4 C 4 H6 ethene (ethylene) propene butene C triple bonds, and have the general formula Cn H2n–2 . The names of CH CH CH3 C CH CH3 CH2 C CH ethyne (acetylene) propyne butyne Cycloalkanes. There is also a type of alkane that has atoms bonded into ring configurations. These are the cycloalkanes. They have the same general formula as the alkenes, C n H2n+2 , but do not have double bonds. Neither are these aromatics. Two cycloalkanes are shown below. H2 C—CH 2 \/ CH2 cyclopropane H2 C—CH 2 | | H2 C—CH 2 cyclobutane Structural Isomers. Until now, we have considered only aliphatic hydrocarbons that have straight chains of carbon atoms. Branching of hydrocarbon chains is very common. Butane (C4 H10 ) can be straight chained, and branched chained. CH3 —CH2 —CH2 —CH3 n-butane and CH3 —CH—CH3 | CH3 2-methylpropane Note that both molecules have the same molecular formula, but they have different arrangements of atoms (different structures). These are actually two distinguishable compounds with their different structures producing slightly different chemical and physical properties. Molecules that have the same molecular formula, but a different structure are called s tructural isomers . Straight chain hydrocarbons are called normal and use the symbol n. The straight chain form of C4 H10 has the name n-butane. Naming branched chain hydrocarbons is covered in the next section. In the alkane series, as the number of C atoms increases, the number of possible structural isomers increases dramatically. For example, C4 H10 has 2 isomers, C6 H14 has 5, and C10 H22 has 75 isomers. Nomenclature. The rules for naming hydrocarbons according to the IUPAC system are briefly summarized as follows: 1. 2. Back The systematic name of an alkane is based on the number of carbon atoms in the longest carbon chain. For alkanes, the longest carbon chain is given a name corresponding to the alkane with the same number of C atoms. The parent name of a compound with 5 carbon atoms in the longest carbon chain is pentane. Groups attached to the main chain are called s ubstituent groups . Substituent groups that contain only hydrogen and carbon atoms are called a lkyl groups . When an H atom is removed from an alkane the fragment is called an alkyl group. This group can be attached to the longest chain. Alkyl groups are named by dropping the ending -ane and adding yl to the alkane name. Therefore CH3 – is a methyl group, C2 H5 – is an ethyl group, and C3 H7 is a propyl group. Table 24.2 in the text lists the names of six common alkyl groups. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 65 3. 4. 5. 6. The locations of alkyl groups that are attached to the main chain must also be included in the name. Number the carbon atoms in the longest chain. The numbering should start at the end of the chain such that the side groups will have the smallest numbers. Prefixes such as di, tri, and tetra are used when more than one substituent of the same kind is present. Combine the names of substituent groups, their locations, and the parent chain into the hydrocarbon name. Arrange the substituent groups alphabetically followed by the parent name of the main chain. Alkanes can have many different types of substituents besides alkyl groups. Halogenated hydrocarbons contain fluorine, chloroine, bromine and iodine atoms which are named fluoro, chloro, bromo, and iodo, respectively. NH2 –, NO 2 –, and CH2 CH– are called amino, nitro, and vinyl, respectively. When naming alkenes we indicate the positions of the carbon-carbon double bonds, and in alkynes the position of the triple bond is numbered. For alkenes and alkynes, the parent name is derived from the longest chain that contains the double or triple bond. Then parent name will end in ene for alkenes and yne for alkynes. Number the C atoms in the main chain by starting at the end nearer to the double or triple bond, and use the lower number of the carbon atom in the C C bond. These rules are applied in Examples 24.1 and 24.2. Addition Reactions. Alkenes, alkynes, and aromatics are called unsaturated hydrocarbons. This means that they can acquire more hydrogen atoms in an addition reaction called hydrogenation. CH2 CH2 + H2 → CH 3 —CH3 CH CH + 2H2 → CH 3 —CH3 Alkanes are called saturated hydrocarbons because they cannot acquire additional hydrogen atoms. The carbon atoms in a saturated hydrocarbon are already bonded to the maximum number of H-atoms. CH3 —CH3 + H2 → no reaction In an addition reaction a small molecule such as H2 is added to an unsaturated hydrocarbon. The addition reaction occurs at the C C double bond. One atom of the small molecule links to one of the carbon atoms of the double bond, while the other atom attaches to the other carbon atom. Other examples are: CH2 CH2 CH2 CH2 + C l 2 → CH 2 Cl—CH 2 Cl CH2 + H2 O → CH 3 —CH2 OH CH2 + H Cl → CH 3 —CH2 Cl Markovnikov's Rule. The two carbons of a double bond suffer different fates during addition reactions with unsymmetrical reagents such as HCl, HBr, and H—OH (water). For example, two different compounds might possibly be formed by reaction of 1-butene with HCl. CH3 CH2 CH CH2 + H Cl → CH 3 CH2 CHCl—CH 3 or CH 3 CH2 CH2 —CH2 Cl observed product not found The rule that predicts which of the two products is formed is called M arkovnikov's rule . This rule states: In the addition of unsymmetrical reagents to alkenes, the positive group of the reagent adds to the carbon that already has the most hydrogen atoms. In HCl, HBr, and H2 O the hydrogen atoms have a partially positive charge because they are bonded to more electronegative atoms. Thus, in the above reaction with CH3 CH2 CH CH2 , the H atom from HCl bonds to the terminal C atom. The Cl atom bonds to the CH group. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 66 / Organic Chemistry Geometric Isomers of Alkenes. C C double bonds are completely rigid. Consider the structures of cis- and trans-2-butene. CH3 H \ / CC / \ H CH 3 trans-2-butene CH3 CH3 \ / CC / \ H H cis-2-butene Neither structure can rotate around the C C bond to become the other. These two molecules have different physical properties and can be separated from one another. Therefore, these two structures represent two different compounds and are isomers. Many pairs of such isomers have been isolated, and they have no tendency to undergo interconversion. The rigidity of the double bond gives rise to a new kind of isomerism: geometric isomerism , which is sometimes called cis-trans isomerism. The relative positions of atoms in space is known as a configuration. The isomers are named cis or trans depending on their configuration. The cis configuration has like groups on the same side of the double bond. CH 3 C H 3 \ / –––C C––– / \ H H The trans configuation has like groups on the opposite sides of the double bond. Trans means across. CH 3 H \ / –––C C––– / \ H CH3 ______________________________________________________________________ EXAMPLE 24.1 Naming Hydrocarbons Name the molecule that has the following structure: H 3 C CH 2 —CH3 | | CH3 —CH2 —C—CH—CH 2 —CH3 4| 3 2 1 5 CH 2 | 6 CH 2 —CH3 7 •Method of Solution First identify the longest continuous carbon chain. Two equivalent chains containing seven carbon atoms are noticeable. Number the carbon atoms as shown. Two ethyl groups appear at carbons 3 and 4, and one methyl group appears at carbon 4. The parent name is heptane. Placing the names of the substituent groups (alphabetically ethyl is before methyl) and their position numbers in front of the parent alkane name we get: 3,4-diethyl-4-methylheptane. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 67 •Comment If the chain had been numbered in reverse order the name would be 4,5-diethyl-5-methylheptane. Since 3,4 is smaller than 4,5 the first name is the correct one. ______________________________________________________________________ EXAMPLE 24.2 Naming Hydrocarbons Name the following hydrocarbon: 6 5 4 CH3 —CH2 —CH—CH3 | C H CH—CH3 3 2 1 •Method of Solution Locate the longest chain that contains the double bond. This chain contains six C atoms so the root name will be hexene. Number the chain so that the double bond has the smallest number, as shown. Note the methyl group on carbon 4. The name is 4-methyl-2-hexene. •Comment The 2 indicates the position of the C C double bond, which connects carbons numbered 2 and 3. _______________________________________________________________________ EXAMPLE 24.3 Addition Reactions Give the structure of the product of the following reaction. CH3 CH C—CH 3 + HCl | CH 3 •Method of Solution This solution involves the addition of an unsymmetrical reagent to a double bond. According to Markovnikov's rule the positive portion of the reagent (in this case an H atom) adds to the carbon atom in the double bond that already has the most hydrogen atoms. The Cl atom adds to the other C atom. The product will be Cl | CH3 CH2 —C—CH 3 | CH3 ____________________________________________________________________ EXERCISES 1. 2. 3. Write the general formulas of alkanes, alkenes, and alkynes. Which of the classes of hydrocarbons in number 1 are unsaturated hydrocarbons? Give the IUPAC name for the compound with the following structure. CH 3 | CH3 —CH2 —CH2 —CHCH2 —CH3 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 68 / Organic Chemistry 4. Give the IUPAC name for the following compound. CH 3 | CH—CH—CH 3 CH3 —CH 5. Name the following compounds. a. CH 3 CHFCH 3 b. (CH 3 )3 CCH 2 CHClCH 3 6. Write the structure for 3,3-dimethyl pentane. 7. What is the product of the following addition reaction: Cl2 + CH3 CH 8. What is the product of the addition reaction HBr + CH3 CH CH2 ? 9. Namethe following: a. CH3 (CH2 )3 CH CHCH3 b . CH 3 (CH2 )3 C 10. Draw all the isomers of C 4 H8 . CH2 ? CH3 AROMATIC HYDROCARBONS STUDY OBJECTIVES 1. 2. Describe the structure of aromatic hydrocarbons. Determine the product of a substitution reaction. Benzene. Benzene is the parent compound of a class of hydrocarbons called aromatic hydrocarbons. The hydrogen to carbon ratio in the molecular formula of benzene C6 H6 suggests that it is highly unsaturated. In 1865 Kekule suggested that benzene had a cyclic structure with three double bonds. H H H H H H Benzene can be represented as two resonance structures. Alternatively, benzene can be represented in terms of delocalized molecular orbitals. Regardless of which symbol is used, the hydrogen atoms are usually not explicitly written. Rather we must remember that there is one attached to each carbon. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 69 Aromatic hydrocarbons contain benzene rings and do not have a simple general formula. Benzene is the simplest compound in this group. Nomenclature of Aromatic Hydrocarbons. The naming of monosubstitited benzenes is straightforward. Br NO2 bromobenzene nitrobenzene If more than one substituent is present, we must indicate the position of the two groups relative to each other. The prefixes ortho- (o), meta- (m), and para- (p) are used to indicate their relative positions. Cl Cl Cl Cl Cl Cl p-dichlorobenzene o-dichlorobenzenem-dichlorobenzene Reactions of Aromatic Compounds. Halogens react with benzene by a s ubstitution reaction rather than an addition reaction. In this reaction an atom or group of atoms replaces an H atom in the benzene ring. Chlorine atoms can be substituted for hydrogen atoms by reacting benzene and chlorine in the presence of an FeCl 3 catalyst. Cl + Cl2 FeCl 3 c atalys t + HCl EXERCISES 11. Name the following compounds. Br NO2 NO2 NO2 a. b. Br c. Br 12. How does benzene differ in reactivity from unsaturated aliphatic hydrocarbons? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 70 / Organic Chemistry FUNCTIONAL GROUPS STUDY OBJECTIVES 1. 2. Define the term functional group. Sketch the general structural formulas of alcohol, ether, ketone, aldehyde, carboxyl, ester, and amine groups. Write structural formulas for products of the oxidation of alcohols and aldehydes, for products of the reactions of carboxylic acids with alcohols, and for products of the reactions of esters with water and sodium hydroxide. Alcohols. Alkanes are quite inert toward most substances, the main exception being oxygen. The alkane portion of organic compounds is combustable. Certain groups of atoms within carbon compouinds always react in the same way regardless of the chain length of the alkane portion. The group of atoms that is largely responsible for the chemical behavior of a molecule is called a f unctional group . For instance, alcohols contain an alkane group and a hydroxyl group, OH. Methanol (CH3 OH), ethanol (C2 H5 OH), and n-propanol (C3 H7 OH) are the first three alcohols of a homologous series. In each of the three, chemical reactions occur at the OH group, rather than at the less reactive alkane group. The hydroxyl group is the functional group in each, and the three alcohols react toward other reagents in the same way. Reactions of alcohols discussed in the textbook are oxidation by oxidizing agents, displacement of hydrogen by alkali metals, and esterification. An oxidizing agent removes two H atoms, one from the hydroxyl group and one from the adjacent carbon atom, forming an aldehyde (shown below). Permanganate ion is an oxidizing agent. – MnO4 CH3 CH2 CH2 –OH → CH3 CH2 CHO Quite often, in organic reactions, we do not write completely balanced equations. Here the emphasis is on the chemical change of the alcohol functional group. Alkali metals displace H2 (g) from alcohols. Potassium is a reducing agent. 2CH3 CH2 OH + 2K → 2CH 3 CH2 OK + H 2 Esterification will be discussed under carboxylic acids. Ethers. The functional group in ethers is C—O—C or, more generally, R—O—R, where R stands for any alkyl group such as those listed in Table 24.2 of the textbook. Ethers and alcohols are structural isomers. As illustrated in the following diagram, both dimethyl ether and ethanol have the same chemical formula. CH3 OCH3 dimethyl ether Tb –25°C ← C 2 H6 O → CH3 CH2 OH ethanol Tb 78°C The boiling points given show that ethers are much more volatile than the corresponding alcohols. This is because of the absence of hydrogen bonding in ethers. Alcohols contain the polar hydroxyl group, Oδ – –Hδ +, which can participate in hydrogen bonding to neighboring alcohol molecules. For many years diethyl ether, CH3 CH2 OCH2 CH3 , was used as an anesthetic. Aldehydes and Ketones. The funtional group in these compounds is the carbonyl group, C aldehydes the carbonyl group is at the end of the alkane chain. In ketones it is not a terminal group. O || CH3 CH2 CH2 CH or CH3 CH2 CH2 CHO butanal (an aldehyde) Back Forward Main Menu TOC O. In O || CH2 CH2 CCH 3 or CH3 CH2 COCH3 2-butanone (a ketone) Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 71 Aldehydes are easily oxidized to carboxylic acids. O O – | | MnO4 || CH3 CH2 CH → C H 3 CH2 C—OH Ketones generally are less reactive than aldehydes. O – || MnO4 CH3 CH2 CCH 3 → no reaction Carboxylic Acids. The carboxyl group is the functional group in organic acids. O || —C—OH abbreviated as —COOH This group is weakly acidic. The ionizable hydrogen atom accounts for the acidic properties of carboxylic acids. CH3 COOH(aq) CH 3 COO– (aq) + H+(aq) A homologous series of organic acids can be generated starting with formic acid. HCOOH CH3 COOH CH 3 CH2 COOH CH3 CH2 CH2 COOH methanoic acid (formic acid) ethanoic acid (acetic acid) propanoic acid butanoic acid Fatty acids are carboxylic acids that contain more than four carbon atoms. Two fatty acids are: CH3 (CH2 )14 COOH CH3 (CH2 )16 COOH palmitic acid stearic acid Esters. The ester functional group is O || —C—OR or —COOR where R stands for any alkyl group. This functional group resembles the carboxylic acid group except that the R group replaces the H atom. Carboxylic acids react with alcohols to form the compounds called esters. The reaction that changes an acid to an ester is called an e sterification reaction. O O || || CH3 C— OH + H —OCH3 → CH3 C—OCH3 + H 2 O acetic acid methanol methyl acetate Esterification is an example of a condensation reaction. A "condensation" is a reaction in which parts of two molecules become joined by formation of a new covalent bond to make a new, larger molecule. Esterification reactions are reversible, and at equilibrium a mixture is formed that contains all four substances. Saponification is the alkaline hydrolysis of an ester. In this reaction the base reacts directly with an ester to split it into a salt of an organic acid and an alcohol. C 2 H5 COOCH3 + NaOH → C 2 H5 COONa + CH 3 OH Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 72 / Organic Chemistry Amines. Organic bases contain the —NH2 functional group, which is called an amino group. Molecules containing an amino group are called amines. The amines may be considered as derivatives of ammonia (NH3 ). Amines have the general formula R3 N, where R may be H, an alkyl group, or an aromatic group. Molecules of the type RNH 2 are called primary amines. Those with R2 NH are secondary amines, and R3 N is a tertiary amine. H | CH3 —N—H methylamine a primary amine CH 3 | CH3 —N—H dimethylamine a secondary amine CH3 | CH3 —N—CH3 trimethylamine a tertiary amine ____________________________________________________________________ EXAMPLE 24.4 Identifying Functional Groups Indicate the functional groups in the following formulas: a. C 5 H11 OH b. CH3 CHO c. C 3 H7 OCH3 d. CH3 COC 2 H5 e. CH3 COOCH3 •Method of Solution Learning to recognize functional groups requires memorization of their structural formulas. Table 24.5 (textbook) shows a number of the important functional groups. a. C 5 H11 OH contains a hydroxyl group. It is an alcohol. b. c. d. e. O || CH3 CHO is a way to represent CH3 C—H on a single line of type. C O is a carbonyl group. CH3 CHO is an aldehyde. C 3 H7 OCH3 contains a C—O—C group and is an ether. CH3 COC 2 H5 is a way to represent a ketone on a single line. C O is a carbonyl group. CH3 COOCH3 is a condensed structural formula for: O || CH3 —C—O—CH3 CH3 COOCH3 is an ester functional group. ________________________________________________________________________ EXAMPLE 24.5 Predicting Reaction Products Predict the product or products of the following reactions: 2– a. CH3 CH(OH)CH3 + Cr2 O7 → – b. CH3 CHO + MnO4 → c. HCOOH + CH 3 OH → d. CH3 CH2 COOCH3 + NaOH → •Method of Solution a. This is the reaction of a secondary alcohol with an oxidizing agent. The product is a ketone: O || CH3 —C—CH 3 . Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 73 b. c. This is the reaction of an aldehyde with an oxidizing agent. The product is a carboxylic acid: CH3 COOH. The reaction of an alcohol with an acid yields an ester and water. O || HCOCH3 + H2 O d. This is a saponificaton of an ester to form the sodium salt of a carboxylic acid and an alcohol. O || CH3 CH2 CONa + CH 3 OH ________________________________________________________________________ EXERCISES 13. 14. 15. 16. 17. 18. 19. Draw structures for the following functional groups: aldehyde, amine, carboxylic acid, ester. Give two chemical properties of alcohols. Oxidation of what type of compound yields a ketone? What elements are present in an amine group? Which of the following functional groups exhibit hydrogen bonding? carboxylic acids, alcohols, ethers. What are the reactants in a saponificaton reaction that yields C6 H13 COONa and C2 H5 OH? Identify the functional group or type of molecule in the following. a. CH 3 OH b. CH3 CHO c. CH 3 COCH3 d. CH3 COOCH3 e. CH 3 CH2 CH3 _______________________________________________________________________________ PRACTICE TEST 1. Give the systematic name for each of the following structural formulas. a. CH3 (CH2 )7 CH3 b. CH 3 | CH3 —CH2 —CH—CH CH 2 c. d. CH3 —CH2 —C e. Back CH3 —CH2 —CH—CH2 —OH | CH 2 —CH3 CH 2 —CH3 | CH3 —CH2 —CH—CH— CH—CH 3 | | CH2 —CH2 CH3 | CH3 Forward Main Menu C—CH 2 —CH—CH3 | C 3 H7 TOC Study Guide TOC Textbook Website MHHE Website 4 74 / Organic Chemistry f. CH 3 | C H—CH 2 —CH3 | CH3 —CH2 —CH | CH 2 —CH2 —CH3 g. H \ C / CH3 CH 2 CH3 / C \ H h. i. 2. 3. 4. 5. CH3 —CH—CH2 —CH3 | OH CH3 —OH Draw structural formulas for: a. 3-ethyl-4-methyl-4-isopropylheptane b. 3-bromo-2,5-dimethyl-trans-3-hexene c. 3-methyl-3-hexanol d. trichloroacetic acid Draw structural formulas of all isomers of C 5 H10 . Include both structural and geometric isomers. List the functional groups by name that are shown in the following molecules. a. CH3 –CH CH–CH–CH 2 –NH2 b . O O O | || || || OH HO–C–CH2 –C–CH 2 –CH2 –CH Give the structure of the organic product of each of the following reactions: a. OH | CH 3 CH2 CHCH3 agent b. O || CH3 CH agent CH + Cl2 → c. CH3 C d. C 3 H8 + C l 2 + H2 light Pt catalys t e. f. CH3 CH2 CH CHCH 2 CH3 + H2 O g. 2– CH3 CHCH3 + Cr2 O7 dil → | OH h. Back O || 2– CH3 CH2 CCH 3 + Cr2 O7 Forward Main Menu → TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 75 + H2O i. j. O || CH3 COH + CH3 OH k. CH3 CH2 CH 6. a. H 2 SO4 dil H+ CHCH 3 + H2 Sketch structural formulas of the hydrolysis products of the following ester: CH 3 O | || CH3 —CH—C—CH 3 + H 2 O b. Pt H+ Sketch structures of the products of the following saponification reaction: CH3 (CH2 )12 COO(CH2 )4 CH3 + NaOH → 7. A certain reagent converts methanol into methanoic acid. In this reaction the alcohol reacts as: a. a Lewis base b. a Lewis acid c. a catalyst d. an oxidizing agent e. a reducing agent. 8. Define the following terms: a. an alkene b. homologous series c. an addition reaction d. functional group. 9. How many structural isomers are there of C4 H10 O? 10. Show how the following chemical change can be carried out in two steps. CH3 CH2 CH2 OH → CH3 CHCH3 | OH ANSWERS Exercises 1. 2. 3. 4. 5. 6. Back C n H2n+2 , C n H2n , and C n H2n–2 alkenes and alkynes 3-methylhexane 4-methyl-2-pentene a. 2-fluoropropane b. 2-chloro-4,4-dimethylpentane CH 3 | CH3 —CH2 —C—CH 2 —CH3 | CH 3 Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 76 / Organic Chemistry 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. CH3 CHCl—CH 2 Cl CH3 CHBr—CH3 a. 2-heptene b. 1-hexyne Six isomers. 1–Butene, cis- and trans-2-butene, 2-methyl propene and 2 cycloalkanes. a. o-dinitrobenzene b. p-dibromobenzene c. m-bromonitrobenzene Benzene is much less reactive than alkenes. Rather than addition reactions, benzene undergoes substitution reactions. O H O O || | || || —C—H —N—H—C—O—H —C—O—R Alcohols react with carboxylic acids to form esters. Also alcohols are easily oxidized. A secondary alcohol Nitrogen and hydrogen Carboxylic acids and alcohols An ester; C 6 H13 COOC2 H5 and NaOH. a. alcohol b. aldehyde c. ketone d. ester e. alkane Practice Test 1. n-nonane b. 3-methyl-1-pentene 6-methyl-3-nonyne e. 3,4-diethyl-2-methylheptane trans-2-pentene h. 2-butanol C 2 H5 CH3 | | CH3 —CH2 —CH—C—CH 2 —CH2 —CH3 | C H 3 —CH—CH3 b. 2. a. d. g. a. CH 3 || CH3 —CH—C | Br H C—CH—CH 3 | CH 3 c. 4. Back CH 3 | CH3 —CH2 —C—CH 2 —CH2 —CH3 | OH d. 3. c. 2-ethyl-1-butanol f. 4-ethyl-3-methylheptane i. methanol CCl 3 –COOH There are five alkenes: 1-pentene, 3-methyl-1-butene, 2-methyl-1-butene, and cis- and trans--2-pentene. There are three cycloalkanes: cyclopentane, methylcyclobutane, and dimethylcyclopropane. a. Carbon-carbon double bond, hydroxyl, amine b. Carboxyl, carbonyl (ketone), carbonyl (aldehyde) Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 77 5. a. b. c. d. O || CH3 CH2 CCH 3 CH3 COOH CH3 CCl 2 CHCl 2 C 3 H7 Cl, C 3 H6 Cl 2 , C 3 H5 Cl 3 ,etc. e. f. g. CH3 CH2 CH2 CHCH2 CH3 | OH O || CH3 CCH 3 h. no reaction i. OH j. k. 6. a. O || CH3 COCH3 CH3 CH2 CH2 CH2 CH3 CH 3 O | || CH3 —CH—C—OH + CH3 OH b. CH3 (CH2 )12 COONa + CH 3 (CH2 )4 OH 7. b. 8. a. b. c. A straight chain or branch chain hydrocarbon with one or more double bonds Organic compounds with the same functional group that differ only in the number of carbon atoms. A reaction in which a small molecule is added to an organic compound. The organic molecule must contain a C C bond or C O (carbonyl) group. d. A characteristic grouping of atoms that imparts certain chemical and physical properties when incorporate into organic molecules 9. Seven (four alcohols and three ethers) 10. CH3 CH2 CH2 OH CH3 CH CH3 CH conc CH 2 + H 2 O dil CH 2 + H2 O CH 3 CHCH3 | OH _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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This note was uploaded on 09/15/2009 for the course CHEM 102 taught by Professor Bastos during the Spring '08 term at Adelphi.

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