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Alkenes E The double bond consists of a  bond and a  bond  bond from head-on overlap of sp 2 orbitals  bond from side-on overlap of p orbitals  bond.

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Presentation on theme: "Alkenes E The double bond consists of a  bond and a  bond  bond from head-on overlap of sp 2 orbitals  bond from side-on overlap of p orbitals  bond."— Presentation transcript:

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2 Alkenes E The double bond consists of a  bond and a  bond  bond from head-on overlap of sp 2 orbitals  bond from side-on overlap of p orbitals  bond  bond Rotating a double bond requires breaking the  bond NO FREE ROTATION at room temperature Contain at least one C=C double bond General formula: C n H 2n (like cycloalkanes) Each carbon atom in a C=C double bond is sp 2 hybridized A  bond is stronger than a  bond.  bonds  bond

3 Alkene Nomenclature (Naming) 1.Parent chain = longest chain that includes the double bond(s) 2.The double bonds have priority and must have the lowest number(s) possible 3.The first C atom in the C=C bond indicates the double bond’s location (or number in naming) 4.Name, number, & alphabetize substituents as usual 5.Replace –ane ending with –ene ending –Two double bonds: -diene; three double bonds: -triene 6.Put double bond number in front of entire root name (i.e. 2-pentene indicates the double bond starts on carbon 2) 7.Cyclic alkenes: number the atoms in the ring starting with the double bond 2-hexene 3-butyl-2,4-hexadiene

4 Naming Practice 4-ethyl-3,5-dimethyl-2-heptene 2,4-hexadiene 3,4-dimethyl-1,3-pentadiene 4-isopropyl-3,5-dimethyl- 1,3,5-heptatriene

5 Cis-trans isomerism in alkenes Substituents will stay on the same or opposite sides of the double bond (no C=C bond rotation) X 2-butene (same side) 2-butene (opposite sides) For cis-trans isomerism, each C in the double bond must have 2 different substituents attached (i.e. a C and a H, etc.) Determining cis or trans: follow the parent chain through the double bond cis-3-methyl-2-heptenecis trans- cis-

6 Cis/Trans Naming Practice trans-4-ethyl-3,5-dimethyl- 2-heptene trans,trans-2,4-hexadiene 3,4-dimethyl-1,3-pentadiene trans,trans-4-isopropyl-3,5- dimethyl-1,3,5-heptatriene No Cis or Trans Both H’s Both CH 3 ’s No Cis or Trans Both H’s

7 Important Common Names Vinyl (branch) Propylene Allyl (branch) Ethylene R R CH 2 HC i.e. polypropylene = milk jugs i.e. polyvinyl chloride = PVC pipe i.e. diallyllysergamide = derivative of LSD i.e. polyethylene = plastic bags

8 Arranging many double bonds Cumulated C=C double bonds all in a row: C=C=C=C Conjugated Single and double bonds alternate: −C=C−C=C−C=C− Isolated >1 single bond between double bonds: −C=C−C−C=C−C−C=C− Lycopene 11 conjugated double bonds 2 isolated double bonds

9 Reactions of Alkenes Alkanes – substitution reactions –R-H + A-B  R-A + H-B –R = “residue”, a generic alkyl group Alkenes – addition reactions + A-B  Thermodynamics:  H rxn = bonds broken – bonds formed = (  bond +  bond) – (  bond +  bond) Exothermic reaction    bond is electron-rich

10 Alkene Addition Reactions + A-B  Reaction Hydrohalogenation (addition of H-X, X = halogen) Hydration (addition of H 2 O) Addition of halogens Hydrogenation (addition of H 2 ) A-B H-F, H-Cl, H-Br, H-I H-OH Br-Br, Cl-Cl, F-F H-H

11 Reactions at the Double Bond    bond is electron-rich Acts as a nucleophile Attracts electrophiles to form bonds Electrophile: “electron-lover” Electron deficient reagent (often + charged cations) seeks e - in reactions Nucleophile: “nucleus-lover” electron-rich reagent (often – charged anions) tries to donate e - to an electrophile in reactions, forming bonds

12 Markovnikov’s Rule Consider the reaction Vladimir Markovnikov Two products are possible Experimentally, only 2-chloropropane is formed + 1-propene 1-chloropropane2-chloropropane Markovnikov’s Rule: The alkene carbon with the most H atoms gets the H Hydrohalogenation (H-X), hydration (H 2 O) of alkenes Why? Look at the reaction mechanism to find out... C C

13 Carbocation Stability Relative carbocation stability: 3° > 2° >> 1° > methyl >>>> 2°3°1°methyl Most stableLeast stable Why? Electrons can “drift” into the empty p orbital from C-H bonds on neighboring C atoms to help stabilize Result: The final product comes from the most stable carbocation intermediate

14 Mechanism of Hydrohalogenation Step 1: Alkene  electrons attack H +, forming a carbocation Step 2: Cl - nucleophile forms bond with carbocation or+ carbocation The carbocation formed in step 1 determines the final product

15 Hydrohalogenation Predict the product of the following hydrohalogenation reaction + HBr OR Markovnikov Product Anti-Markovnikov Product Remember: Markovnikov’s rule says that the H (from HBr) will bond to the alkene C with the most H’s

16 Remember: Markovnikov’s rule says that the H (from H 2 O) will bond to the alkene C with the most H’s Hydration Predict the product of the following hydration reaction + H 2 O OR Markovnikov Product Anti-Markovnikov Product

17 Mechanism of Hydration (R + H 2 O) Step 1:  bond attacks electrophilic H + catalyst forming carbocation Step 2: Nucleophilic H 2 O attacks carbocation, forming oxonium ion Step 3: Oxonium ion loses a proton, regenerating H + catalyst Most stable carbocation + Markovnikov product oxonium ion

18 Mechanism of Halogen Addition (R + X 2 ) Step 1: Br 2 is polarized by  electrons, Br + attaches to alkene Step 2:Br + blocks access to one face of the alkene Br - adds to the other face (anti addition) bromonium ion intermediate + trans isomer is the only product

19 Halogenation + Br 2 The Br’s will add to opposite sides of a RING (anti addition) The Br’s will be forced into a trans conformation ALWAYS Trans isomer

20 Mechanism of Hydrogenation (R + H 2 ) Occurs in the presence of a metal catalyst (like Pt) Pt H−H + Pt H H Step 1: H 2 adsorbs to catalyst surface Step 2: Both H atoms add to same face of alkene (syn addition) Pt H H cis isomer is the only product

21 Hydrogenation Occurs in the presence of a metal catalyst (like Pt) + H 2 /Pt Both H’s will add to the same side of a RING (syn addition) If branches are present, they will be forced into a cis conformation cis isomer

22 Hydrogenation of alkenes vegetable oils + H 2 /Pt unsaturated saturated Contains double bonds Contains NO double bonds

23 Hydrohalogenation Determine any reactant(s) that could yield the given product of the following hydrohalogenation reaction + HBr trans-3-methyl-2-hexene 3-methyl-1-hexene + HBr Major product using M’s rule Only possible reactant for this product

24 Hydrohalogenation Determine any reactant(s) that could yield the given product of the following hydrohalogenation reaction + HBr trans-3-methyl-2-hexene trans-3-methyl-3-hexene All three reactants could give this product 2-ethyl-1-pentene

25 Halogenation Determine any reactant(s) that could yield the given product of the following halogenation reaction trans-3,5-dimethyl-2-heptene Only one possible reactant in this case + Br 2

26 Alkene Reaction Summary Hydrohalogenation (+ HX) Hydration (+ H 2 O) Halogenation (+ X 2 ) Hydrogenation (H 2 /Pt) Markovnikov’s rule Cis/trans with rings

27 Polymerization of Alkenes Polymer: a large molecule made by linking together small repeat units called monomers Polymerization mechanism: radical chain reaction MonomerPolymer ethene (ethylene) polyethylene propene (propylene) polypropylene

28 E Alkynes Contain at least one C  C triple bond with sp-hybridized C atoms Triple bond: one  bond (sp orbitals), two  bonds (p orbitals) Naming: triple bond indicated by –yne ending Reactivity: same addition reactions as alkenes Use 2 equivalents of addition reagent (i.e. + 2HCl) Use Markovnikov’s rule in the same manner ethyne (acetylene) 4-methyl-1-pentyne

29 Alkyne Naming Practice 3,4-dimethyl-1-pentyne 2-methyl-3-hexyne 3-methyl-1-pentyne 2,5-dimethyl-3-hexyne

30 Hydrohalogenation of Alkynes Predict the product of the following hydrohalogenation reaction + 2HBr ? Break the reaction into two steps, adding 1 HBr each time to the multiple bond + HBr Final Product 1-pentyne Markovnikov’s rule still applies…

31 Hydration of Alkynes Predict the product of the following hydration reaction + 2H 2 O ? Break the reaction into two steps, adding H 2 O each time to the multiple bond + H 2 O Final Product 3,3-dimethyl-1-butyne Markovnikov’s rule still applies…

32 Halogenation of Alkynes Predict the product of the following halogenation reaction + 2Br 2 ? Break the reaction into two steps, adding Br 2 each time to the multiple bond + Br 2 Final Product 3,3-dimethyl-1-butyne

33 Hydrogenation of Alkynes Predict the product of the following hydrogenation reaction + 2H 2 ? This reaction will simply turn the alkyne to an alkene, and then to an alkane + H 2 Final Product 3,3-dimethyl-1-butyne

34 Alkyne Reaction Practice HCl → + 2H 2 O → + 2Br 2 → + 2H 2 → + 2HCl → AND


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