《有机化学》课程教学课件（McMurry’s Organic Chemistry, 6th edition）Chapter 06 Alkenes - Structure and Reactivity

Alkenes:Structure and Reactivity Based on McMurry's Organic Chemistry,6th edition

Alkenes: Structure and Reactivity Based on McMurry’s Organic Chemistry, 6th edition

Alkene Are hydrocarbons with C=C They have fewer hydrogens than an alkane with the same number of carbons-CH2n for an alkene versus CH2n+2 for an alkane Unsaturated hydrocarbons A ring or a C=C in a molecule corresponds to a loss of 2 hydrogens from the alkane.Knowing this relationship,its possible to work backward from the molecular formula to calculate a molecule's degree of unsaturation-the number of rings and/or multiple bonds in the molecule two double bonds one ring,one double bond

Alkene „ Are hydrocarbons with C=C „ They have fewer hydrogens than an alkane with the same number of carbons - CnH2n for an alkene versus CnH2n+2 for an alkane „ Unsaturated hydrocarbons „ A ring or a C=C in a molecule corresponds to a loss of 2 hydrogens from the alkane. Knowing this relationship, its possible to work backward from the molecular formula to calculate a molecule’s degree of unsaturation – the number of rings and/or multiple bonds in the molecule two double bonds one ring, one double bond

Degree of Unsaturation With Other Elements ■ Organohalogens (X:F,CI,Br,I) Halogen replaces hydrogen Replace 2 Br by 2 H BrCH2CH-CHCH2Br = HCH2CH-CHCH2H C.H6Br2 =“C4Hg” One unsaturation: one double bond Add e2004 Thomson-Brooks/Cole Oxygen atoms-if connected by single bonds These don't affect the total count of H's O removed from here H2C=CHCH-CHCH2OH H2C=CHCH=CHCH2-H CsHg0=“CsH3”Two unsaturations: two double bonds 2004 Thomson-Brooks/Cole

Degree of Unsaturation With Other Elements „ Organohalogens (X: F, Cl, Br, I) „Halogen replaces hydrogen „ Oxygen atoms - if connected by single bonds „These don't affect the total count of H's

If C-N Bonds Are Present Nitrogen has three bonds-subtract one H for equivalent degree of unsaturation in hydrocarbon H CH2 H H CH2 H H CH2 N一H H CH2 H Removed H CsHgN=“C5Hg”Two unsaturations:one ring and one double bond C2004 Thomson-Brooks/Cole

If C-N Bonds Are Present „ Nitrogen has three bonds - subtract one H for equivalent degree of unsaturation in hydrocarbon

Naming of Alkenes Step 1:Name the parent hydrocarbon-Find longest continuous carbon chain containing the C=C CHCH2 H CH:CH2 H C=C C=C CH2CH2CH2 H CHCH2CH2 H Named as a pentene NOT as a hexene,since the double bond is not contained in the six-carbon chain C2004 Thomson-Brooks/Cole Step 2:Number carbons in chain so that double bond carbons and substituents have lowest possible numbers CH3 CHCH,CH2CH-CHCH3 CH CHCH-CHCH2CH 20o4Tmson- 21 1 23 45 6

Naming of Alkenes „ Step 1: Name the parent hydrocarbon - Find longest continuous carbon chain containing the C=C „ Step 2: Number carbons in chain so that double bond carbons and substituents have lowest possible numbers

■ Step 3:Write the full name-Rings have"cyclo"prefix;if more than one C=C,indicate the position and use suffixes-diene, triene CH3 CH CH CHCH-CHCH CH CHCH-CHCH2CH 21 1 3 45 6 2-Hexene 2-Methyl-3-hexene CH:CH2 H 2C=C1 CH3 CH CH2CH2 H 543 HC-g-CH-CH2 3 2-Ethyl-1-pentene 2-Methyl-1,3-butadiene

„ Step 3: Write the full name - Rings have “cyclo” prefix; if more than one C=C, indicate the position and use suffixes –diene, - triene

Electronic Structure of Alkenes ■Carbon atoms in a p orbitals bond double bond are sp2- hybridized ■Three equivalent orbitals at120° separation in plane Fourth orbital is atomic orbitals T bond p orbital sp2carbon Carbon-carbon double bond Combination of electrons in two sp2 orbitals of two atoms forms o bond between them Additive interaction of p 90 rotatio orbitals creates aπ bonding orbital ■ Occupiedπorbital prevents rotation about o- bond Broken bond after rotation bond (p orbitals are parallel) (p orbitals are perpendicular)

Electronic Structure of Alkenes „ Carbon atoms in a double bond are sp 2 - hybridized „ Three equivalent orbitals at 120º separation in plane „ Fourth orbital is atomic p orbital „ Combination of electrons in two sp 2 orbitals of two atoms forms σ bond between them „ Additive interaction of p orbitals creates a π bonding orbital „ Occupied π orbital prevents rotation about σ - bond

Cis-Trans Isomerism in Alkenes The presence of a carbon- carbon double can create two possible structures CH3 CH3 cis isomer-two similar C=C groups on same side of H the double bond ■trans isomer similar cis-2-Butene groups on opposite sides Each carbon must have two different groups for these H CH3 isomers to occur C-C Since bond rotation cannot occur,this means that the CH3 H cis-isomer cannot trans-2-Butene spontaneously interconvert to the trans

Cis-Trans Isomerism in Alkenes „ The presence of a carbon￾carbon double can create two possible structures „ cis isomer - two similar groups on same side of the double bond „ trans isomer similar groups on opposite sides „ Each carbon must have two different groups for these isomers to occur „ Since bond rotation cannot occur, this means that the cis-isomer cannot spontaneously interconvert to the trans

Cis,Trans Isomers Require That End Groups Must Differ in Pairs Cis-trans isomerism is not limited to disubstituted alkenes. It can occur whenever both double-bond carbons are attached to two different groups. If one of the double-bond carbons is attached to two identical groups,then cis-trans isomerism is not possible These two compounds are identical; they are not cis-trans isomers. These two compounds are not identical; they are cis-trans isomers. B 2004 Thomson/Brooks Cole

Cis, Trans Isomers Require That End Groups Must Differ in Pairs Cis-trans isomerism is not limited to disubstituted alkenes. It can occur whenever both double-bond carbons are attached to two different groups. If one of the double-bond carbons is attached to two identical groups, then cis-trans isomerism is not possible

Sequence Rules:The E,Z Designation If the higher-priority groups on each carbon are on the same side of the double bond,the alkene is designated Z If the higher-priority groups on each carbon are on opposite sides of the double bond,the alkene is designated E Lower Higher E double bond (Higher-priority groups Higher are on opposite sides.) Lower Higher Higher Z double bond (Higher-priority groups are on the same side.) Lower Lower 2004 Thomeson-Bro

Sequence Rules: The E,Z Designation „ If the higher-priority groups on each carbon are on the same side of the double bond, the alkene is designated Z „ If the higher-priority groups on each carbon are on opposite sides of the double bond, the alkene is designated E