Chapter 10 - Alkenes PDF

1/19/2025 ® Because learning changes everything...

1/19/2025 ® Because learning changes everything. Organic Chemistry Seventh Edition Janice Gorzynski Smith University of Hawai’i Chapter 10 Lecture Outline Prepared by Andrea Leonard University of Louisiana at Lafayette © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. © McGraw Hill LLC Alkene Structure Alkene Hybridization Alkenes are also called olefins. Recall that the double bond consists of a bond and a bond. Alkenes contain a carbon-carbon double bond. Each carbon is hybridized and trigonal planar, with bond Terminal alkenes have the double bond at the end of the angles of approximately. carbon chain. Internal alkenes have at least one carbon atom bonded to each end of the double bond. Cycloalkenes contain a double bond in a ring. Access the text alternative for slide images. © McGraw Hill LLC 2 © McGraw Hill LLC 5 Bond Dissociation Energy Bond dissociation energies of the C—C bonds in ethane (a σ bond only) and ethylene (one σ and one π bond) can be used to estimate the strength of the π component of the double bond. The π bond is much weaker than the σ bond of a C—C double bond, making it much more easily broken. Therefore, alkenes undergo many reactions that alkanes do not. © McGraw Hill LLC © McGraw Hill LLC 6 1  1/19/2025 Degrees of Unsaturation for Cyclic Alkenes 1 Molecules Containing Heteroatoms 1 Cycloalkenes having fewer than eight carbon atoms have a cis Ignore O atoms in the molecule (this divalent atom geometry. is a linker and has no effect on degree of A trans-cycloalkene must have a carbon chain long enough to connect the ends of the double bond without introducing too much strain. unsaturation). trans-Cyclooctene is the smallest, isolable trans cycloalkene, but it is considerably less stable than cis-cyclooctene, making it one of the few Add number of halogens to number of H's (they alkenes having a higher energy trans isomer. are equivalent to H). Subtract 1 H for each N present (N’s two connections allows extra H). E.g., C6H10OCl3N is equivalent to C6H12. Access the text alternative for slide images. © McGraw Hill LLC 7 © McGraw Hill LLC 10 Degrees of Unsaturation for Cyclic Alkenes 2 Molecules Containing Heteroatoms 2 Table 10.1 Properties of the Carbon–Carbon Double Bond Access the text alternative for slide images. Access the text alternative for slide images. © McGraw Hill LLC 8 © McGraw Hill LLC 11 Calculating Degrees of Unsaturation Naming Alkenes and Alkenols An acyclic alkene has the general structural formula CnH2n. Alkenes are unsaturated hydrocarbons because they have fewer than the Compounds that contain both a double bond and a hydroxy maximum number of hydrogen atoms per carbon. group are named as alkenols and the chain (or ring) is Cycloalkanes also have the general formula CnH2n. numbered to give the OH group the lower number. Each π bond or ring removes two hydrogen atoms from a molecule, and this introduces one degree of unsaturation. The number of degrees of unsaturation for a given molecular formula can be calculated by comparing the actual number of H atoms in a compound to the maximum number of H atoms possible for the number of carbons present if the molecule were an acyclic alkane. This procedure gives the total number of rings and/or π bonds in a molecule. Access the text alternative for slide images. © McGraw Hill LLC 9 © McGraw Hill LLC 12 2  1/19/2025 Naming Polyenes and Cyclic How to Assign the Prefixes E and Z Alkenes to an Alkene 2 Compounds with two double bonds are named as dienes by changing Step : Assign E or Z based on the location of the the “-ane” ending of the parent alkane to the suffix “-adiene”. two higher-priority groups (1). Compounds with three double bonds are named as trienes, and so forth. The E isomer has the two higher-priority groups on the In naming cycloalkenes, the double bond is located between C1 and same opposite sides. C2, and the “1” is usually omitted in the name. The Z isomer has the two higher-priority groups on the The ring is numbered clockwise or counterclockwise to give the first same side. substituent the lower number. Figure 10.1 © McGraw Hill LLC 13 © McGraw Hill LLC 16 Common Names of Alkenes and Cis and Trans Isomers of Alkenes Alkene Substituents Alkenes having one alkyl group bonded to each carbon Some alkene or alkenyl substituents have common names. atom can be differentiated using the prefixes – cis and The simplest alkene, CH2 =CH2, named in the IUPAC trans. system as ethene, is often called ethylene. cis has the two alkyl groups on the same side of the double bond. trans has the two alkyl groups on opposite sides of the double bond. Figure 10.2 Access the text alternative for slide images. © McGraw Hill LLC 14 © McGraw Hill LLC 17 How to Assign the Prefixes E and Z Physical Properties of Alkenes to an Alkene 1 Step : Assign priorities to the two substituents Most alkenes exhibit only weak van der Waals interactions, so their physical properties are similar to alkanes of comparable molecular weight. on each end of the C=C bond by using the priority Alkenes have low melting points and boiling points. rules for R,S nomenclature. Melting and boiling points increase as the number of carbons increases Divide the double bond in half and assign the numbers 1 due to increased surface area. and 2 to indicate the relative priority of the two groups on Alkenes are soluble in organic solvents and insoluble in water. each end. The C—C single bond between an alkyl group and one of the double bond carbons of an alkene is slightly polar because the sp3 hybridized alkyl carbon donates electron density to the sp2 hybridized alkenyl carbon. © McGraw Hill LLC 15 © McGraw Hill LLC 18 3  1/19/2025 Cis/Trans Differ in Physical Fatty Acids 1 Properties A consequence of the alkene dipole is that cis and trans isomeric Triacylglycerols are hydrolyzed to glycerol and three fatty alkenes often have somewhat different physical properties. acids of general structure RCOOH. cis-2-Butene has a higher boiling point (4°C) than trans-2-butene (1°C). In the cis isomer, the two Csp3 −Csp2 bond dipoles reinforce each other, yielding a small net molecular dipole. In the trans isomer, the two bond dipoles cancel. Saturated fatty acids have no double bonds in their long hydrocarbon chains, and unsaturated fatty acids have one or more double bonds in their hydrocarbon chains. As the number of double bonds in the fatty acid increases, the melting point decreases. Access the text alternative for slide images. © McGraw Hill LLC 19 © McGraw Hill LLC 22 Useful Products Formed From Fatty Acids 2 Ethylene Figure 10.3 Table 10.2 The Effect of Double Bonds on the Melting Point of Fatty Acids Increasing the number of double bonds in the fatty acid side chains decreases the melting point of the triacylglycerol. © McGraw Hill LLC 20 © McGraw Hill LLC 23 Naturally Occurring Alkenes 3-D Structure of C18 Fatty Acids The larger the number of Z double bonds, the more kinks in the hydrocarbon chain. This causes poorer stacking and less van der Waals interactions, leading to lower melting points. Figure 10.4 © McGraw Hill LLC Left: I. Rozenbaum & F. Cirou/PhotoAlto; Right: Jill Braaten/McGraw Hill 21 © McGraw Hill LLC 24 4  1/19/2025 Regioselectivity and Stereoselectivity Triacylglycerols of Alkene Formation Fats and oils are both triacylglycerols, but with The most stable alkene (Zaitsev product) is usually different physical properties. formed as the major product. Fats have higher melting points—they are solids at room temperature. Oils have lower melting points—they are liquids at room temperature. The composition (saturated vs. unsaturated) of the three fatty acids in the triacylglycerol determines whether it is a fat or an oil. © McGraw Hill LLC 25 © McGraw Hill LLC 28 Properties of Fatty Acids Addition Reactions Fats are derived from fatty acids having few or no The characteristic reaction of alkenes in addition—the π double bonds. bond is broken and two new σ bonds are formed. Oils are derived from fatty acids having a larger number of double bonds. Saturated fats are typically obtained from animal sources, whereas unsaturated oils are common in vegetable sources. Alkenes are electron rich, with the electron density of the π bond concentrated above and below the plane of the An exception to this generalization is coconut oil, molecule. which is largely composed of saturated alkyl side Therefore, alkenes act as nucleophiles and react with electrophiles. chains. Simple alkenes do not react with nucleophiles or bases, reagents that are themselves electron rich. © McGraw Hill LLC 26 © McGraw Hill LLC 29 Electrostatic Potential Plot of Preparation of Alkenes Ethylene Alkenes can be prepared from alkyl halides, The red electron-rich region of the bond is located tosylates, and alcohols via elimination reactions. above and below the plane of the molecule. © McGraw Hill LLC 27 © McGraw Hill LLC 30 5  1/19/2025 How to Draw the Products of an Syn and Anti Addition to Alkenes Addition Reaction Because the carbon atoms of a double bond are both trigonal planar, Locate the C—C double bond. the elements of X and Y can be added to them from the same side or from opposite sides. Identify the σ bond of the reagent that breaks. Syn addition takes place when both X and Y are added from the same Break the π bond of the alkene and the σ bond of the side. reagent, and form two new σ bonds to the C atoms of the Anti addition takes place when X and Y are added from opposite sides. double bond. © McGraw Hill LLC 31 © McGraw Hill LLC 34 Heat of Formation for Electrophilic Addition Reactions of Cyclohexene Addition In each reaction, the pi bond is broken, and two new sigma Addition reactions are exothermic because the two σ bonds bonds are formed. formed in the product are stronger than the σ and π bonds broken in the reactants. Figure 10.5 For example, ΔH degrees for the addition of HBr to ethylene is −14 kcal/mol. Figure 10.6 Access the text alternative for slide images. © McGraw Hill LLC 32 © McGraw Hill LLC 35 Hydrohalogenation—Electrophilic Mechanism of Electrophilic Addition Addition of HX The mechanism of electrophilic addition consists of two successive Lewis acid-base reactions. Step – the alkene is the Lewis base that donates an electron pair to H-Br, the Lewis acid. Step – Br⁻ is the Lewis base that donates an electron pair to the carbocation, the Lewis acid. Two bonds are broken in this reaction—the weak π bond of the alkene and the HX bond—and two new σ bonds are formed—one to H and one to X. Recall that the H—X bond is polarized, with a partial positive charge on H. Because the electrophilic H end of HX is attracted to the electron-rich double bond, these reactions are called electrophilic additions. Access the text alternative for slide images. Access the text alternative for slide images. © McGraw Hill LLC 33 © McGraw Hill LLC 36 6  1/19/2025 Stereochemistry of Electrophilic Markovnikov’s Rule Addition With an unsymmetrical alkene, HX can add to the double Recall that trigonal planar atoms react with reagents from bond to give two constitutional isomers, but only one is two directions with equal probability. actually formed: Achiral starting materials yield achiral products. Sometimes new stereogenic centers are formed from hydrohalogenation. Markovnikov’s rule states that in the addition of HX to an unsymmetrical alkene, the H atom adds to the less substituted carbon atom—that is, the carbon that has the greater number of H atoms to begin with. © McGraw Hill LLC 37 © McGraw Hill LLC 40 Carbocation Stability and Stereochemistry of Carbocation Markovnikov’s Rule Formation The basis of Markovnikov’s rule is the formation of a The mechanism of hydrohalogenation illustrates why two carbocation in the rate-determining step of the mechanism. enantiomers are formed. In the addition of HX to an unsymmetrical alkene, the H Initial addition of H⁺ occurs from either side of the planar atom is added to the less substituted carbon to form the double bond. more stable, more substituted carbocation. Both modes of addition generate the same achiral carbocation. Either representation of this carbocation can be used to draw the second step of the mechanism. © McGraw Hill LLC 38 © McGraw Hill LLC 41 Hammond Postulate and Stereochemistry of Nucleophilic Electrophilic Addition Attack According to the Hammond postulate, the blue path is faster because Nucleophilic attack of Cl⁻ on the trigonal planar carbocation formation of the carbocation is an endothermic process. also occurs from two different directions, forming two Thus, the transition state to form the more stable secondary products, A and B, having a new stereogenic center. carbocation is lower in energy. A and B are enantiomers. The Ea for formation of the more stable secondary carbocation is lower than the Ea for formation of the primary carbocation. Since attack from either direction occurs with equal The secondary carbocation is formed faster. probability, a racemic mixture of A and B is formed. Figure 10.7 © McGraw Hill LLC 39 © McGraw Hill LLC 42 7  1/19/2025 Hydrohalogenation—Summary Electrophilic Addition of Alcohols Alcohols add to alkenes, forming ethers by the same mechanism. Table 10.3 Summary: Electrophilic Addition of HX to Alkenes For example, addition of CH3 OH to 2-methylpropene, forms Observation tert-butyl methyl ether (MTBE), a high-octane fuel additive. Mechanism The mechanism involves two steps. The rate-determining step forms a carbocation. Rearrangements can occur. Regioselectivity Markovnikov’s rule is followed. In unsymmetrical alkenes, H bonds to the less substituted C to form the more stable carbocation. Stereochemistry Syn and anti addition occur. © McGraw Hill LLC 43 © McGraw Hill LLC 46 Hydration—Electrophilic Addition of Halogenation—Electrophilic Water Addition of Halogen Hydration is the addition of water to an alkene to form an Halogenation is the addition of X2 (X = Cl or Br) to an alcohol. alkene to form a vicinal dihalide. Access the text alternative for slide images. © McGraw Hill LLC 44 © McGraw Hill LLC 47 Mechanism of Hydration Halogenation Details Halogens add to π bonds because halogens are polarizable. The electron-rich double bond induces a dipole in an approaching halogen molecule, making one halogen atom electron deficient and the other electron rich (Xδ+—Xδ−). The electrophilic halogen atom is then attracted to the nucleophilic double bond, making addition possible. Two facts demonstrate that halogenation follows a different mechanism from that of hydrohalogenation or hydration. No rearrangements occur. Only anti addition of X2 is observed. These facts suggest that carbocations are not intermediates. Access the text alternative for slide images. © McGraw Hill LLC 45 © McGraw Hill LLC 48 8  1/19/2025 Stereochemistry of Halonium Ring Halogenation Mechanism Opening In the second step, nucleophilic attack of Cl⁻ must occur from the backside. Since the nucleophile attacks from below and the leaving group departs from above, the two Cl atoms in the product are oriented trans to each other. Backside attack occurs with equal probability at either carbon of the three-membered ring to yield a racemic mixture. Access the text alternative for slide images. © McGraw Hill LLC 49 © McGraw Hill LLC 52 Stereochemical Outcome of Stability of Cation Intermediates Halogenation Carbocations are unstable because they have only cis-2-Butene yields two enantiomers, whereas trans-2- six electrons that surround carbon. butene yields a single achiral meso compound. Figure 10.8 Bridged halonium ions are unstable because of ring strain. © McGraw Hill LLC 50 © McGraw Hill LLC 53 Stereochemistry of Halonium Halohydrin Formation Formation Chlorination of cyclopentene affords both enantiomers of Treatment of an alkene with a halogen X2 and H2O trans-1,2-dichlorocyclopentane, with no cis products. forms a halohydrin by addition of the groups of X Initial addition of the electrophile Cl⁺ from (Cl2) occurs from and OH to the double bond. either side of the planar double bond to form a bridged chloronium ion. Access the text alternative for slide images. © McGraw Hill LLC 51 © McGraw Hill LLC 54 9  1/19/2025 Anti Stereochemistry in Halohydrin Mechanism of Halohydrin Formation Formation 2 With unsymmetrical alkenes, the preferred product has the electrophile X⁺ bonded to the less substituted carbon, and the nucleophile (H2O) bonded to the more substituted carbon. Even though X⁻ is formed in step of the mechanism, its concentration is small compared to H2O (often the solvent), so H2O and not X⁻ is the nucleophile. Access the text alternative for slide images. © McGraw Hill LLC 55 © McGraw Hill LLC 58 Generating Bromine in Halohydrin Regiochemistry of Halohydrin Formation Formation Although the combination of Br2 and H2O effectively forms As in the acid catalyzed ring opening of epoxides, bromohydrins from alkenes, other reagents can also be nucleophilic attack occurs at the more substituted carbon used. end of the bridged halonium ion because that carbon is Bromohydrins are also formed with N-bromosuccinimide better able to accommodate the partial positive charge in the transition state. (NBS) in aqueous DMSO [(CH3 )2S=O]. In H2O, NBS decomposes to form Br2, which then goes on to form a bromohydrin by the same reaction mechanism. © McGraw Hill LLC 56 © McGraw Hill LLC 59 Anti Stereochemistry in Halohydrin Summary of Halohydrin Formation Formation 1 Because the bridged halonium ion is opened by backside attack of H2O, addition of X and OH Table 10.4 Summary: Conversion of Alkenes to Halohydrins occurs in an anti fashion and trans products are Observation formed. Mechanism The mechanism involves three steps. The rate-determining step forms a bridged halonium ion. No rearrangements can occur. Regioselectivity The electrophile X⁺ bonds to the less substituted carbon. Stereochemistry Anti addition occurs. © McGraw Hill LLC 57 © McGraw Hill LLC 60 10  1/19/2025 Halohydrin Use in Synthesis Borane and Hydroboration 1 Halohydrins have been used in the synthesis of many BH3 is a reactive gas that exists mostly as a dimer, naturally occurring compounds. diborane (B2H6). Key steps in the synthesis of estrone, a female sex Borane is a strong Lewis acid that reacts readily with Lewis hormone, are illustrated below. bases. Figure 10.9 For ease of handling in the laboratory, it is commonly used as a complex with tetrahydrofuran (THF). © McGraw Hill LLC 61 © McGraw Hill LLC 64 Hydroboration–Oxidation 1 Borane and Hydroboration 2 Hydroboration–oxidation is a two-step reaction sequence The first step in hydroboration–oxidation is the addition of that converts an alkene into an alcohol. the elements of H and BH2 to the π bond of the alkene, forming an intermediate alkylborane. Hydroboration is the addition of borane (BH3) to an alkene, forming an alkylborane. Oxidation converts the C—B bond of the alkylborane to a C—O bond. © McGraw Hill LLC 62 © McGraw Hill LLC 65 Hydroboration–Oxidation 2 Hydroboration Mechanism Hydroboration–oxidation results in the addition of H2O to an The proposed mechanism involves concerted addition of H alkene. and BH2 from the same side of the planar double bond: the π bond and H—BH2 bond are broken as two new σ bonds are formed. Because four atoms are involved, the transition state is said to be four-centered. © McGraw Hill LLC 63 © McGraw Hill LLC 66 11  1/19/2025 Reactivity of Borane During Electronic Factors Affecting Hydroboration Regiochemistry of Hydroboration Because the alkylborane formed by the reaction with one equivalent of Electronic factors are also used to explain this alkene still has two B—H bonds, it can react with two more equivalents of alkene to form a trialkylborane. regioselectivity. We often draw hydroboration as if addition stopped after one equivalent If bond making and bond breaking are not of alkene reacts with BH3. completely symmetrical, boron bears a δ− charge Instead, all three B—H bonds actually react with three equivalents of an alkene to form a trialkylborane. in the transition state and carbon bears a δ+ Figure 10.10 charge. Since alkyl groups stabilize a positive charge, the more stable transition state has the partial positive charge on the more substituted carbon. © McGraw Hill LLC 67 © McGraw Hill LLC 70 Other Sources of B-H for Example of Regiochemistry of Hydroboration Hydroboration Since only one B—H bond is needed for hydroboration, Figure 10.11 commercially available dialkylboranes having the general structure R2BH are sometimes used instead of BH3. A common example is 9-borabicyclo[3.3.1]nonane (9-BBN). © McGraw Hill LLC 68 © McGraw Hill LLC 71 Regiochemistry of Hydroboration Oxidation Following Hydroboration With unsymmetrical alkenes, the boron atom bonds to the Since alkylboranes react rapidly with water and less substituted carbon atom. spontaneously burn when exposed to air, they are oxidized, without isolation, with basic hydrogen peroxide (H2O2, ⁻OH). Oxidation replaces the C—B bond with a C—O bond, forming a new OH group with retention of configuration. This regioselectivity can be explained by considering steric factors. The larger boron atom bonds to the less sterically hindered, The overall result of this two-step sequence is syn addition more accessible carbon atom. of the elements of H and OH to a double bond in an “anti- Markovnikov” fashion. © McGraw Hill LLC 69 © McGraw Hill LLC 72 12  1/19/2025 Summary of Hydroboration— Use of Alkenes in Synthesis Oxidation Suppose we wish to synthesize 1,2-dibromocyclohexane from cyclohexanol. Table 10.5 Summary: Hydroboration-Oxidation of Alkenes Observation Mechanism The addition of H and BH2 occurs in one step. No rearrangements can occur. Regioselectivity The OH group bonds to the less substituted carbon atom. Stereochemistry Syn addition occurs. To solve this problem we must: OH replaces BH2 with retention of configuration. © McGraw Hill LLC 73 © McGraw Hill LLC 76 Example of Hydroboration-Oxidation Retrosynthetic Analysis in Synthesis Hydroboration-oxidation is one step toward making the Working backwards from the product to determine the antimalarial drug artemisinin. starting material from which it is made is called retrosynthetic analysis. Figure 10.12 © McGraw Hill LLC 74 © McGraw Hill LLC 77 Keeping Track of Reactions 1st Learn the basic type of reaction for a functional group. This provides overall organization to the reactions. 2nd Learn the specific reagents for each reaction. It helps to classify each reagent according the its properties. Is it an acid or base? Is it a nucleophile or electrophile? Is it an oxidizing or reducing agent? Finally, you MUST practice these reactions over and over Because learning changes everything. ® again by writing them. It is not enough just to look at them. www.mheducation.com © McGraw Hill LLC 75 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. 13  1/19/2025 Cyclic Alkenes 2 - Text Alternative Return to parent-slide containing images. The table summarizes the various properties of carbon-carbon double bond. It has two columns and three rows. The column header reads: property and result. The row header reads: restricted rotation, stereoisomerism and stability. Row-wise data as below: Row 1: Restricted rotation; The rotation around the C-C double bond is restricted. Rotation can occur only if the pi bond breaks and then re-forms, a process that is unfavourable (section 8.2B). Accessibility Content: Text Row 2: Stereoisomerism; Whenever the two groups on each end of a C double bond carbon are different from each other, two diastereomers are possible. Cis and trans-but-e-ene (drawn at the Alternatives for Images bottom of Table 10.1) are diastereomers (section 8.2B). Row 3: Trans alkenes are generally more stable than cis alkenes. The stability of an alkene increases as the number of R groups on the C double bond C increases (section 8.2C). The ball and structures of three compounds but-1-ene, cis-but-2-ene and trans-but-2-ene are given. The first compound has a 4-carbon chain. C1 is double bonded to C2. The second compound has a 4-carbon chain with a double bond between C2 and C3. C1 and C4 are on the same sides of carbon-carbon double bond. The third compound has a 4-carbon chain with a double bond between C2 and C3. C1 and C4 are on the opposite sides of carbon-carbon double bond. In all models, the carbon and hydrogen atoms are represented by gray and white spheres, respectively. An orange arrow from left to right is labeled increasing stability. Return to parent-slide containing images. © McGraw Hill LLC 79 © McGraw Hill LLC 82 Alkene Hybridization - Text Alternative Calculating Degrees of Unsaturation - Text Alternative Return to parent-slide containing images. Return to parent-slide containing images. The first structure of ethylene has a carbon atom double bonded to The first structure shows two propane rings fused together. The another carbon atom. The second structure has two gray spheres second structure shows a 4-carbon chain with a double bond between connected to each other. Each gray sphere is further connected to C1 and C2. C3 is double bonded to C4. The second structure shows two white spheres. Both gray spheres are s p 2 hybridized. The bond a 4-carbon chain with a triple bond between C1 and C2. Text above angle between white-gray-white sphere is 120 degrees. The third the second and third structure reads, two pi bonds. The fourth structure has a carbon atom connected to another carbon. Each structure shows a 4-membered carbon ring with a double bond carbon is dash bonded to a hydrogen atom and wedge bonded to between C1 and C2. Text reads, one ring and one pi bond. another hydrogen. The p orbitals of both carbons overlap to form a pi bond. The single bond between both carbons is marked sigma bond. Return to parent-slide containing images. Return to parent-slide containing images. © McGraw Hill LLC 80 © McGraw Hill LLC 83 Degrees of Unsaturation for Molecules Containing Cyclic Alkenes 1 - Text Alternative Heteroatoms 2 - Text Alternative Return to parent-slide containing images. Return to parent-slide containing images. The first model has an 8-membered ring with a trans double bond The text reads: Example: Give the IUPAC name of the following alkene. The structure given below is a 6-carbon chain with a double bond between C2-C3, the second, third, and fifth carbon atoms are between C1 and C2. The second model has an 8-membered ring with each single bonded to a methyl group. The text below it reads: a cis double bond between C1 and C2. In both models, the carbon Step (1) Find the longest carbon chain and change the ane ending of the parent alkane to -ene. The and hydrogen atoms are represented by gray and white spheres, struccture given below is a 6-carbon chain has a double bond between C2-C3. C2, C3, and C5 are each single bonded to a methyl group. Text below the structure reads, 6 C's in the longest chain. respectively. Hexane converts to hexene. Step (2) Number the carbon chain to give the double bond the lower number and apply all other rules of nomenclature. a. Number the chain, and name using the first number assigned to the C double bond C. The structure of an alkene is depicted. The carbon atoms are numbered 1 through 6. Alkene has a 6-carbon chain with a double bond between C2-C3. C2, C3, and C5 are each single bonded to a methyl group. Text reads, Number the chain to put the carbon double bond carbon at C2, not C4. Hex-2-ene. b. Name and number the substituents. The structure has a 6-carbon chain with a double bond between C2 and C3. Three methyl groups are at C2, C3, and C5. The answer is 2,3,5-trimethylhex-2-ene. Return to parent-slide containing images. Return to parent-slide containing images. © McGraw Hill LLC 81 © McGraw Hill LLC 84 14  1/19/2025 Common Names of Alkenes and Alkene Substituents - Text Heat of Formation for Electrophilic Addition - Text Alternative Alternative Return to parent-slide containing images. Return to parent-slide containing images. (Top Image) The methylene group has a carbon atom double bonded In the reaction, the reactant has a carbon atom double bonded to another to another carbon atom. The carbon on right has a wavy line. Vinyl carbon atom. Each carbon is single bonded to two hydrogen atoms. The group has a carbon atom double bonded to another carbon atom. The reactant reacts with hydrogen bromide. The product has a 4-carbon chain. C1 carbon on right is bonded to a wavy chain. The allyl group has a 3- is single bonded to two hydrogen atoms and a bromine atom. C2 is single bonded to three hydrogen atoms. The calculations for delta H degrees are as carbon chain with a double bond between C1 and C2. C3 is bonded follows: Bonds broken. Delta H degrees for C H 2 double bonded to C H 2 to a wavy chain. pi bond is positive 267. Delta H degrees for H single bonded to B r is positive (Bottom Image) Methylenecyclohexane has a cyclohexane ring. C1 is 368. Total is positive 635 kilojoules per mole. Text reads, Energy needed to double bonded to a methylene group. 1-vinylcyclopentene has a break bonds. Bonds formed. Delta H degrees for B r C H 2 C H 2 single bonded to H is negative 410. Delta H degrees for C H 3 C H 2 single bonded cyclopentene ring. C1 is single bonded to a vinyl group. 2- to B r is negative 285. Total is negative 695 kilojoules per mole. Text reads, allylcycloheptanol has a 7-membered carbon ring. C1 is single Energy released in forming bonds. overall delta H degrees. Sum in step bonded to a hydroxyl group. C2 is single bonded to an allyl group. and sum in step (positive 635 kilojoules per mole plus negative 695 kilojoules per mole) equals negative 60 kilojoules per mole. Text reads, the reaction is exothermic. Return to parent-slide containing images. Return to parent-slide containing images. © McGraw Hill LLC 85 © McGraw Hill LLC 88 Fatty Acids 1 - Text Alternative Mechanism of Electrophilic Addition - Text Alternative Return to parent-slide containing images. Return to parent-slide containing images. Triacylglycerol has a 3-carbon chain. Each carbon is single bonded to In the first step, cis-but-2-ene reacts with hydrogen single bonded to bromine that an oxygen atom, which is single bonded to a carbonyl group. The first carries three lone pairs of electrons to form carbocation. Cis-but-2-ene has a 4- carbon chain. C2 is double bonded to C3. A curved arrow from the double bond carbonyl carbon is single bonded to R. The second carbonyl carbon is points toward the hydrogen atom of hydrogen bromide. Another curved arrow from single bonded to R prime. The third carbonyl carbon is single bonded the single bond between hydrogen and bromine points toward the bromine atom. to R double prime. R groups have 11–19 C’s. Three ester groups are The first step is slow. The carbocation has a 4-carbon chain. C2 carries a positive labeled in red. charge. C3 is single bonded to a hydrogen atom. Text reads, new bond shown in red. In the second step, carbocation reacts with bromide ion to form a product. The bromide ion has four lone pairs of electrons. A curved arrow from one of the lone pairs of electrons on the bromide ion points toward the positive charge of carbocation. The product has a 4-carbon chain. C2 is single bonded to a bromine atom with three lone pairs of electrons. The text below the equation reads: 1. The pi bond of the alkene attacks the H of H B r to form a new C-H bond and a carbocation in the rate-determining in step. 2. Nucleophilic attack of B r negative on the carbocation forms the new C-Br bond. Return to parent-slide containing images. Return to parent-slide containing images. © McGraw Hill LLC 86 © McGraw Hill LLC 89 Hydrohalogenation—Electrophilic Addition of HX - Text Hydration—Electrophilic Addition of Water - Text Alternative Alternative Return to parent-slide containing images. Return to parent-slide containing images. The reactant that has a carbon atom double bonded to another In the first reaction, cyclohexene reacts with hydrogen single bonded carbon with two open single bonds at each carbon reacts with H to a hydroxyl group in the presence of sulfuric acid to form a product. single bonded to X. Text below the reactant reads, a pi bond is Text below cyclohexene reads, a pi bond is broken. Hydrogen of broken. The hydrogen of H single bonded to X carries a partial hydrogen single bonded to hydroxyl group carries a partial positive positive charge and X carries a partial negative charge. X equals C l, charge and hydroxyl group carries a partial negative charge. The B r, I. The product has a 4-carbon chain. C2 has an open single bond product has a cyclohexane ring. C1 and C2 are each single bonded to and is single bonded to X. C3 has an open single bond and is single a hydroxyl group and a hydrogen atom, respectively. In the second bonded to a hydrogen atom. reaction, propene reacts with hydrogen single bonded to a hydroxyl group in the presence of sulfuric acid to form 2-propanol. Hydrogen of hydrogen single bonded to hydroxyl group carries a partial positive charge and hydroxyl group carries a partial negative charge. 2- propanol has a 3-carbon chain. C2 is single bonded to a hydroxyl group. 2-propanol is present in rubbing alcohol and hand sanitizer. Return to parent-slide containing images. Return to parent-slide containing images. © McGraw Hill LLC 87 © McGraw Hill LLC 90 15  1/19/2025 Mechanism of Hydration - Text Alternative Mechanism of Halohydrin Formation - Text Alternative Return to parent-slide containing images. Return to parent-slide containing images. In the first step, cyclohexene reacts with H single bonded to O H 2 with a positive charge and a lone pair of In the first step, the reactant reacts with X single bonded to X where each X carries three lone pairs of electrons. The electrons on the oxygen atom to form carbocation and water that has two lone pairs of electrons on the reactant has a carbon atom double bonded to another carbon atom. Each carbon has two open single bonds. A oxygen atom. A curved arrow from the double bond points toward hydrogen atom of H single bonded to O curved arrow from the double bond of reactant points toward first X of X single bonded to X. Another curved arrow H 2. Another curved arrow from the single bond of H single bonded to O H 2 points toward the oxygen from one lone pair of first X points toward the double bond of reactant. Third curved arrow from the single bond atom. Carbocation has a cyclohexane ring. C1 carries a positive charge. C2 is single bonded to a hydrogen between both X points toward one of the lone pairs of second X. The first step is slow. A bridged halonium ion is formed that has a 3-membered ring made of two carbon atoms and an X with a positive charge and two lone pairs of atom. Text below carbocation reads, new bond shown in red. In the second step, carbocation reacts with electrons. C2 and C3 each have a wedge bond and a dash bond. The side product is a halide ion with four lone pairs water to form intermediate. A curved arrow from one of the lone pairs on oxygen atom of water points of electrons. In the second step, the bridged halonium ion reacts with water to form an intermediate. A curved arrow toward the positive charge. The intermediate has a cyclohexane ring. C1 is single bonded to the oxygen from one of the lone pairs of electrons on oxygen of water points toward C2 of bridged ion. Another curved arrow atom of a water molecule. The oxygen carries a lone pair of electrons and a positive charge. In the second from the bond between C2 and X points toward one of the lone pairs of X. The intermediate has a 2-carbon chain. step, the intermediate leads to the formation of cyclohexanol and a hydronium ion with a positive charge C1 and C2 each have two open bonds. C1 is single bonded to X with three lone pairs of electrons. C2 is single and a lone pair of electrons on the oxygen atom. A curved arrow from one of the lone pairs of electrons on bonded to the oxygen atom of water. The oxygen atom carries a positive charge and a lone pair of electrons. In the oxygen of water points toward one of the hydrogen bonded to oxygen atom of intermediate. Another curved third step, a curved arrow from one of lone pairs of electrons of water points toward one of the hydrogen atom single arrow from the single bond between oxygen and hydrogen points toward the oxygen. In the third step, bonded to oxygen at C2. Another curved arrow from the single bond between oxygen and one hydrogen points cyclohexanol has a cyclohexane ring. C1 is single bonded to a hydroxyl group with two lone pairs of toward the oxygen atom. The product has a 2-carbon chain. C1 and C2 each have two open single bonds. C1 is electrons on the oxygen atom. The text below the equation reads: single bonded to a hydroxyl group with two lone pairs of electrons on the oxygen atom. C2 is single bonded to X with three lone pairs of electrons. The text below the equation reads: 1. The pi bond of the alkene attacks the H of H 3 O positive to form a new C-H bond and a carbocation in the rate-determining step. 1. Four bonds are broken or formed to generate an unstable bridged halonium ion that contains a three-membered ring. The electron pair in the pi bond and a lone pair on a halogen are used to form two new C-X bonds, and the X-X 2. Nucleophilic attack of H 2 O on the carbocation forms the new C-O bond. bond is cleaved. 3. Removal of a proton with H 2 O forms a neutral alcohol. Because the acid used in step (1) is 2. Nucleophilic attack of H 2 O ring opens the bridged halonium ion and forms a new C-O bond. regenerated in step (3), the reaction is acid-catalyzed. 3. Loss of a proton forms the halohydrin. Return to parent-slide containing images. Return to parent-slide containing images. © McGraw Hill LLC 91 © McGraw Hill LLC 94 Halogenation Mechanism - Text Alternative Return to parent-slide containing images. In the first step, the reactant reacts with X single bonded to another X. Each X carries three lone pairs of electrons. The reactant has a carbon atom double bonded to another carbon atom. Each carbon has two open bonds. A curved arrow from the double bond points toward first X of X single bonded to X. Another curved arrow from one lone pair of first X points toward the double bond. Third curved arrow from the single bond between both X points toward the second X. The first step is slow. A bridged halonium ion is formed that has a 3- membered ring made of an X and two carbon atoms. X carries a positive charge and two lone pairs of electrons. C2 and C3 each have a dash bond and a wedge bond. In the second step, a curved arrow from one of the lone pairs of electrons on X negative points toward C2 of halonium ion. Another curved arrow from the single bond between C2 and X points toward the X atom. The product has a 4-carbon chain. C2 and C3 each have an open single bond. C2 and C3 are each single bonded to X with three lone pairs of electrons. The text below the equation reads: 1. Four bonds are broken or formed to generate an unstable bridged halonium ion that contains a three-membered ring. The electron pair in the pi bond and a lone pair on a halogen are used to form two new C-X bonds, and the X-X bond is cleaved. 2. Nucleophilic attack of X negative ring opens the bridged halonium ion and forms a new C- X bond. Return to parent-slide containing images. © McGraw Hill LLC 92 Stereochemistry of Halonium Formation - Text Alternative Return to parent-slide containing images. Cyclopentene has a 5-membered ring with a double bond between C1 and C2. Each chlorine atom of chlorine molecule carries three lone pairs of electrons. A curved arrow from the double bond points toward one of the lone pairs of first chlorine atom of chlorine molecule. Second curved arrow from the other lone pair of chlorine atom points toward the double bond of cyclopentene. Third curved arrow from the single bond of chlorine molecule points toward the second chlorine atom. The first chloronium ion has a cyclopentane ring. C1 and C2 are wedge bonded to a chlorine atom with a positive charge and two lone pairs of electrons. Text reads, from above. The second chloronium ion has a cyclopentane ring. C1 and C2 are dash bonded to a chlorine atom with a positive charge and two lone pairs of electrons. Text reads, from below. The side product is a chloride ion with a negative charge and four lone pairs of electrons. Return to parent-slide containing images. © McGraw Hill LLC 93 16