Organic Chemistry Lecture Notes - Structure and Synthesis of Alkenes

Organic Chemistry, 9th Edition L. G. Wade, Jr. Chapter 7 Lecture Structure and...

Organic Chemistry, 9th Edition L. G. Wade, Jr. Chapter 7 Lecture Structure and Synthesis of Alkenes Chad Snyder, PhD Grace College © 2017 Pearson Education, Inc. © 2014 Pearson Education, Inc.  Introduction Alkenes are hydrocarbons with carbon–carbon double bonds. Alkenes are also called olefins, meaning “oil-forming gas.” The functional group of alkenes is the carbon–carbon double bond, which gives this group its reactivity. © 2017 Pearson Education, Inc.  Sigma Bonds of Ethylene © 2017 Pearson Education, Inc.  Orbital Description Sigma bonds around the double-bonded carbon are sp2 hybridized. Angles are approximately 120° and the molecular geometry is trigonal planar. Unhybridized p orbitals with one electron will overlap, forming the double bond (pi bond). © 2017 Pearson Education, Inc.  Bond Lengths and Angles sp2 hybrid orbitals have more s character than the sp3 hybrid orbitals. Pi overlap brings carbon atoms closer, shortening the C—C bond from 1.54 Å in alkanes down to 1.33 Å in alkenes. © 2017 Pearson Education, Inc.  Pi Bonding in Ethylene The pi bond in ethylene is formed by overlap of the unhybridized p orbitals of the sp2 hybrid carbon atoms. Each carbon has one unpaired electron in the p orbital. This overlap requires the two ends of the molecule to be coplanar. © 2017 Pearson Education, Inc.  Cis-Trans Interconversion Cis and trans isomers cannot be interconverted. No rotation around the carbon–carbon bond is possible without breaking the pi bond (264 kJ/mole). © 2017 Pearson Education, Inc.  Elements of Unsaturation Unsaturation is a structural element that decreases the number of hydrogens in the molecule by two. It is also called index of hydrogen deficiency. Double bonds and rings are elements of unsaturation. © 2017 Pearson Education, Inc.  Calculating Unsaturations To calculate the number of unsaturations, first find the number of hydrogens the carbons would have if the compounds were saturated. Then subtract the actual number of hydrogens and divide by 2. This calculation cannot distinguish between unsaturations from multiple bonds and those from rings. © 2017 Pearson Education, Inc.  Example: Calculate the Unsaturations for a Compound with Formula C5H8. First calculate the number of hydrogen atoms for a saturated compound with five carbons: (2 × C) + 2 (2 × 5) + 2 = 12 Now subtract from this number the actual number of hydrogen atoms in the formula and divide by 2: 12 – 8 = 4 = 2 unsaturations 2 2 The compound has two unsaturations. They can be two double bonds, two rings, or one double bond and one ring. © 2017 Pearson Education, Inc.  Elements of Unsaturation: Heteroatoms Halogens replace hydrogen atoms in hydrocarbons, so when calculating unsaturations, count halides as hydrogen atoms. Oxygen does not change the C:H ratio, so ignore oxygen in the formula. Nitrogen is trivalent, so it acts like half a carbon. Add the number of nitrogen atoms when calculating unsaturations. © 2017 Pearson Education, Inc.  Degrees of Unsaturation: The Nitrogen Atom © 2017 Pearson Education, Inc.  Example: Calculate the Unsaturations for a Compound with Formula C4H7Br. First calculate the number of hydrogens for a saturated compound with four carbons: (2 × C) + 2 + N (2 × 4) + 2 = 10 Now subtract from this number the actual number of hydrogens in the formula and divide by 2. Remember to count halides as hydrogens: 10 – 8 = 2 = 1 unsaturation 2 2 © 2017 Pearson Education, Inc.  Example: Calculate the Unsaturations for a Compound with Formula C6H7N. First calculate the number of hydrogens for a saturated compound with six carbons. Add the number of nitrogens: (2 × C) + 2 + N (2 × 6) + 2 + 1 = 15 Now subtract from this number the actual number of hydrogens in the formula and divide by 2: 15 – 7 = 8 = 4 unsaturations 2 2 © 2017 Pearson Education, Inc.  IUPAC Nomenclature Find the longest continuous carbon chain that includes the double-bonded carbons. Ending -ane changes to -ene. Number the chain so that the double bond has the lowest possible number. In a ring, the double bond is assumed to be between carbon 1 and carbon 2. © 2017 Pearson Education, Inc.  IUPAC and New IUPAC © 2017 Pearson Education, Inc.  Ring Nomenclature In a ring, the double bond is assumed to be between carbon 1 and carbon 2. © 2017 Pearson Education, Inc.  Multiple Double Bonds Give the double bonds the lowest numbers possible. Use di-, tri-, or tetra- before the ending -ene to specify how many double bonds are present. © 2017 Pearson Education, Inc.  Alkenes As Substitutents © 2017 Pearson Education, Inc.  Cis-Trans Isomers Also called geometric isomerism If there are similar groups on same side of the double bond, alkene is cis. If there are similar groups on opposite sides of the double bond, alkene is trans. Not all alkenes show cis-trans isomerism. © 2017 Pearson Education, Inc.  Cyclic Compounds Trans cycloalkenes are not stable unless the ring has at least eight carbons. All cycloalkenes are assumed to be cis unless otherwise specifically named trans. © 2017 Pearson Education, Inc.  Biochemistry Application © 2017 Pearson Education, Inc.  E-Z Nomenclature Use the Cahn–Ingold–Prelog rules to assign priorities to groups attached to each carbon in the double bond. If high-priority groups are on the same side, the name is Z (for zusammen). If high-priority groups are on opposite sides, the name is E (for entgegen). © 2017 Pearson Education, Inc.  Example Assign priority to the substituents according to their atomic number (1 is highest priority). If the highest priority groups are on opposite sides, the isomer is E. If the highest priority groups are on the same side, the isomer is Z. © 2017 Pearson Education, Inc.  Cyclic Stereoisomers Double bonds outside the ring can show stereoisomerism. © 2017 Pearson Education, Inc.  Stereochemistry in Dienes If there is more than one double bond in the molecule, the stereochemistry of all the double bonds should be specified. © 2017 Pearson Education, Inc.  Commercial Uses of Ethylene © 2017 Pearson Education, Inc.  Commercial Uses of Propylene © 2017 Pearson Education, Inc.  Addition Polymers © 2017 Pearson Education, Inc.  Physical Properties of Alkenes Low boiling points, increasing with mass Branched alkenes have lower boiling points. Less dense than water Slightly polar – Pi bond is polarizable, so instantaneous dipole–dipole interactions occur. – Alkyl groups are electron-donating toward the pi bond, so they may have a small dipole moment. © 2017 Pearson Education, Inc.  Polarity and Dipole Moments of Alkenes Cis alkenes have a greater dipole moment than trans alkenes, so they will be slightly polar. The boiling point of cis alkenes will be higher than the trans alkenes. © 2017 Pearson Education, Inc. © 2017 Pearson Education, Inc.  Heat of Hydrogenation Combustion of an alkene and hydrogenation of an alkene can provide valuable data as to the stability of the double bond. The more substituted the double bond, the lower its heat of hydrogenation. © 2017 Pearson Education, Inc.  Relative Heats of Hydrogenation More substituted double bonds are usually more stable. © 2017 Pearson Education, Inc.  Relative Stabilities © 2017 Pearson Education, Inc.  HINT Heats of hydrogenation are usually exothermic. A larger amount of heat given off implies a less stable alkene, because the less stable alkene starts from a higher potential energy. © 2017 Pearson Education, Inc. Chapter 7 36  Substituent Effects The isomer with the more substituted double bond has a larger angular separation between the bulky alkyl groups. © 2013 Pearson Education, Inc. Chapter 7 37  Disubstituted Isomers Stability: cis < geminal < trans isomer The less stable isomer has a higher exothermic heat of hydrogenation. CH3 CH3 cis-2-butene C C -120 kJ H H iso-butene (CH3)2C=CH2 -117 kJ H CH3 trans-2-butene C C -116 kJ CH3 H © 2013 Pearson Education, Inc. Chapter 7 38  Cycloalkenes A ring makes a major difference only if there is ring strain, either because of a small ring or because of a trans double bond. Rings that are five-membered or larger can easily accommodate double bonds, and these cycloalkenes react much like straight- chain alkenes. © 2017 Pearson Education, Inc.  Cyclopropene Cyclopropene has bond angles of about 60°, compressing the bond angles of the carbon–carbon double bond to half their usual value of 120°. The double bond in cyclopropene is highly strained. © 2017 Pearson Education, Inc.  Stability of Cycloalkene Cis isomer is more stable than trans in small cycloalkenes. Small rings have additional ring strain. It must have at least eight carbons to form a stable trans double bond. For cyclodecene (and larger), the trans double bond is almost as stable as the cis. © 2017 Pearson Education, Inc.  Bredt’s Rule A bridged, bicyclic compound cannot have a double bond at a bridgehead position unless one of the rings contains at least eight carbon atoms. © 2017 Pearson Education, Inc.  Solved Problem 1 Which of the following alkenes are stable? Solution Compound (a) is stable. Although the double bond is at a bridgehead, it is not a bridged bicyclic system. The trans double bond is in a ten-membered ring. Compound (b) is a Bredt’s rule violation and is not stable. The largest ring contains six carbon atoms, and the trans double bond cannot be stable in this bridgehead position. Compound (c) (norbornene) is stable. The (cis) double bond is not at a bridgehead carbon. Compound (d) is stable. Although the double bond is at the bridgehead of a bridged bicyclic system, there is an eight-membered ring to accommodate the trans double bond. © 2017 Pearson Education, Inc.  Alkene Synthesis Overview E2 dehydrohalogenation (–HX) E1 dehydrohalogenation (–HX) Dehalogenation of vicinal dibromides (–X2) Dehydration of alcohols (–H2O) © 2017 Pearson Education, Inc.  Elimination Reactions Elimination reactions produce double bonds. Also called dehydrohalogenation (–HX) © 2017 Pearson Education, Inc.  The E1 Reaction Unimolecular elimination Two groups are lost: a hydrogen and the halide. Nucleophile acts as base. The E1 and SN1 reactions have the same conditions, so a mixture of products will be obtained. © 2017 Pearson Education, Inc.  E1 Mechanism Step 1: Ionization to form a carbocation Step 2: Solvent abstracts a proton to form an alkene. © 2017 Pearson Education, Inc.  A Closer Look © 2017 Pearson Education, Inc.  E1 Energy Diagram The E1 and the SN1 reactions have the same first step: Carbocation formation is the rate-determining step for both mechanisms. © 2017 Pearson Education, Inc.  Zaitsev’s Rule If more than one elimination product is possible, the most-substituted alkene is the major product (most stable). major product (trisubstituted) © 2017 Pearson Education, Inc.  Alkene Stability © 2017 Pearson Education, Inc.  The E2 Reaction Elimination, bimolecular Requires a strong base This is a concerted reaction: The proton is abstracted, the double bond forms, and the leaving group leaves, all in one step. © 2017 Pearson Education, Inc.  The E2 Reaction © 2017 Pearson Education, Inc.  The E2 Mechanism Order of reactivity for alkyl halides 3° > 2° > 1° A mixture may form, but the Zaitsev product predominates. © 2017 Pearson Education, Inc.  Bulky Bases in E2 Eliminations; Hofmann Orientation If the substrate in an E2 elimination is prone to substitution, we can minimize the amount of substitution by using a bulky base. Bulky bases can accomplish dehydrohalogenations that do not follow the Zaitsev rule. They form the Hofmann product. © 2017 Pearson Education, Inc.  Zaitsev and Hofmann Products © 2017 Pearson Education, Inc.  E2 Stereochemistry The halide and the proton to be abstracted must be anti-coplanar ( = 180º) to each other for the elimination to occur. The orbitals of the hydrogen atom and the halide must be aligned so they can begin to form a pi bond in the transition state. The anti-coplanar arrangement minimizes any steric hindrance between the base and the leaving group. © 2017 Pearson Education, Inc.  E2 Stereochemistry © 2017 Pearson Education, Inc.  Stereochemistry of E2 Elimination Most E2 reactions go through an anti-coplanar transition state. This geometry is most apparent if we view the reaction with the alkyl halide in a Newman projection. © 2017 Pearson Education, Inc.  E2 Reactions on Cyclohexanes An anti-coplanar conformation (180°) can only be achieved when both the hydrogen and the halogen occupy axial positions. The chair must flip to the conformation with the axial halide in order for the elimination to take place. © 2017 Pearson Education, Inc.  Solved Problem 3 Explain why the following deuterated 1-bromo-2-methylcyclohexane undergoes dehydrohalogenation by the E2 mechanism, to give only the indicated product. Two other alkenes are not observed. Solution In an E2 elimination, the hydrogen atom and the leaving group must have a trans-diaxial relationship. In this compound, only one hydrogen atom—the deuterium—is trans to the bromine atom. When the bromine atom is axial, the adjacent deuterium is also axial, providing a trans-diaxial arrangement. © 2017 Pearson Education, Inc.  Substitution or Elimination? The strength of the nucleophile determines the order: Strong nucleophiles or bases promote bimolecular reactions. Primary halides usually undergo SN2. Tertiary halides are a mixture of SN1, E1, or E2. They cannot undergo SN2. High temperature favors elimination. Bulky bases favor elimination. © 2017 Pearson Education, Inc. © 2017 Pearson Education, Inc.  Secondary Alkyl Halides Secondary alkyl halides are more challenging: – Strong nucleophiles will promote SN2/E2. – Weak nucleophiles promote SN1/E1. Strong nucleophiles with limited basicity favor SN2. Bromide and iodide are good examples of these. © 2017 Pearson Education, Inc.  Solved Problem 4 Predict the mechanisms and products of the following reaction. Solution © 2017 Pearson Education, Inc.  Solved Problem 5 Predict the mechanisms and products of the following reaction. Solution © 2017 Pearson Education, Inc.  E1 and SN1 Mechanisms © 2017 Pearson Education, Inc.  Dehydration of Alcohols Use concentrated H2SO4 or H3PO4 and remove low-boiling alkene as it forms to shift the equilibrium and increase the yield of the reaction. E1 mechanism Rearrangements are common. The reaction obeys Zaitsev’s rule. © 2017 Pearson Education, Inc.  Dehydration Mechanism: E1 Step 1: Protonation of the hydroxyl group (fast equilibrium) Step 2: Ionization to a carbocation (slow; rate limiting) © 2017 Pearson Education, Inc.  Dehydration Mechanism: Step 3 Step 3: Deprotonation to give the alkene (fast) © 2017 Pearson Education, Inc.  Solved Problem 6 Propose a mechanism for the sulfuric acid-catalyzed dehydration of t-butyl alcohol Solution The first step is protonation of the hydroxyl group, which converts it to a good leaving group. The second step is ionization of the protonated alcohol to give a carbocation. Abstraction of a proton completes the mechanism. © 2017 Pearson Education, Inc.  Catalytic Cracking of Alkanes A long-chain alkane is heated with a catalyst to produce an alkene and shorter alkane. Complex mixtures are produced. © 2017 Pearson Education, Inc.  Dehydrogenation of Alkanes Dehydrogenation is the removal of H2 from a molecule, forming an alkene (the reverse of hydrogenation). This reaction has an unfavorable enthalpy change but a favorable entropy change. © 2017 Pearson Education, Inc.  Solved Problem 2 Show that the dehalogenation of 2,3-dibromobutane by iodide ion is stereospecific by showing that the two diastereomers of the starting material give different diastereomers of the product. Solution Rotating meso-2,3-dibromobutane into a conformation where the bromine atoms are anti and coplanar, we find that the product will be trans-2-butene. A similar conformation of either enantiomer of the (±) diastereomer shows that the product will be cis-2-butene. (Hint: Your models will be helpful.) © 2017 Pearson Education, Inc.