Properties and Synthesis Alkenes & Alkynes I Properties and Synthesis

Alkenes & Alkynes Chapter 7 Physical Properties of Alkenes & Alkynes E-Z System Vicinal (Vic) & Geminal (Gem) Relative Stabilities of Alkenes & Cycloalkenes Zaitsev’s Rule & Hofmann Rule Dehydration of Alcohols Terminal Alkynes Syn & Anti Additions Hydrogenation of Alkynes Index of Hydrogen Deficiency

Physical Properties Factors Intermolecular Forces (IMF) Molecular Weight (MW) Symmetry

Boiling Point Trends-General Ethene -104 0C 2C Propene -47 0C 3C 1-Butene -6.3 0C 4C 1-Pentene 30 0C 5C 1-Hexene 63 0C 6C 1-Heptene 94 0C 7C As Molecular Weight Increases Boiling Point (BP) increases.

Boiling Point Trends-Branching Branching decreases boiling point due to decrease surface area which translates into fewer surface electrons participating in building intermolecular forces (IMF).

Boiling Point Trends-Branching 1-Pentene 30 0C 2-Methyl-1-butene 31 0C Basically no significant difference in BP.

Sharing the Burden (Concept Review) The more the molecule can spread out the burden (Charge, Pi electrons, free radical, etc.) the more stable the molecule will be.                                                                                                              3° 2° 1° Methyl most stable next-to-most stable next-to-least stable least stable

Sharing the Burden Ability to spread out the burden of the double bond translates into greater stability. Pi electrons have greater flow with more Beta-Carbons.

Boiling Point Trends-Branching Branching decreases boiling point due to decrease surface area which translates into fewer surface electrons participating in building intermolecular forces. Branching of Alkenes leads to greater flow of Pi electrons which can participate in building intermolecular forces. The two factors together cancel each other out resulting in no significant change in BP

Boiling Point Trends-Cis/Trans Analyze the handout (Chart with Physical Properties of Alkenes) Dipole Moments

Boiling Point Trends-Cis/Trans Cis has a higher BP than Trans due to the stronger dipole of the Cis.

BP Trends-Cis/Trans vs. Terminal Analyze the handout (Chart with Physical Properties of Alkenes) H CH3CH2 Dipole Moments 1-Butene

BP Trends-Cis/Trans & Terminal Cis & Trans have higher BP than terminal alkenes due to increase flow of Pi electrons resulting in stronger IMF. Cis has a higher BP than Trans due to the stronger dipole of the Cis.

Alkane vs. Alkene Boiling Point -88.6 0C -104 0C -42.1 0C -47 0C -0.5 0C -6.3 0C 36.1 0C 30 0C 68.7 0C 63 0C 98.4 0C 94 0C Decreased Decreased Decreased Decreased Decreased Decreased

Alkane vs. Alkene Boiling Point Adding a double bond decreases the BP due to decreased sigma electrons participating in building IMF. Why do Cis and Trans Butene have higher BP’s than Butane?

Melting Point Trends-General Ethene -169 0C 2C Propene -185 0C 3C 1-Butene -185 0C 4C 1-Pentene -165 0C 5C 1-Hexene -140 0C 6C 1-Heptene -119 0C 7C As Molecular Weight Increases Melting Point (MP) increases, except where symmetry plays a role.

Melting Point Trends-Branching No general trend If branching increases symmetry, melting point (MP) increases. If branching decreases symmetry, melting point decreases

Melting Point Trends-Cis/Trans Analyze the handout (Chart with Physical Properties of Alkenes) Planes of Symmetry One Plane of Symmetry Two Planes of Symmetry

Melting Point Trends-Cis/Trans Trans has a higher MP than Cis due to the greater symmetry. Trans has two planes of symmetry, while Cis only has one plane of symmetry.

Melting Point Trends-Cis/Trans Analyze the handout (Chart with Physical Properties of Alkenes) H CH3CH2 Planes of Symmetry 1-Butene One Plane of Symmetry Two Planes of Symmetry No

Melting Point Trends-Cis/Trans Cis & Trans have higher MP than terminal alkenes due to greater symmetry Trans has a higher MP than Cis due to the greater symmetry. Trans has two planes of symmetry, while Cis only has one plane of symmetry.

Alkane vs. Alkene Melting Point Increased -183 0C -169 0C -188 0C -185 0C -138 0C -185 0C -130 0C -165 0C -95 0C -140 0C -910C -119 0C Non-Significant Decreased Decreased Decreased Decreased

Alkane vs. Alkene Melting Point Adding a double bond in a central location increases symmetry and thus increases MP. Adding a double bond in a non-central location decreases symmetry and thus decreases MP. How does Cis and Trans Butene MP compared to the MP of Butane? How do you explain this?

Saturated & Unsaturated Saturated – The carbon atoms of the molecule are complete filled with Hydrogen (No double or triple bonds) Unsaturated-Some carbon atoms have room for additional hydrogen (Double or triple bonds present) Polyunsaturated-Many carbon atoms have room for additional hydrogen (Many Double or triple bonds present)

Saturated & Unsaturated Saturated – CH3CH2CH2CH2CH2CH2CH2CH3 Unsaturated- CH3CH=CHCH2CH2CH2CH2CH3 CH3CH=CHCH2CH=CHCH2CH3 Polyunsaturated- CH2=CHCH=CHCH=CHCH=CH2 Fatty Acid- CH3CH2CH2CH2CH2CH2CH2COOH

Saturated & Unsaturated Fatty Acids Source: http://www.rpi.edu/dept/bcbp/molbiochem/MBWeb/mb2/part1/fatcatab.htm

Saturated & Unsaturated Fats Source: http://www.rpi.edu/dept/bcbp/molbiochem/MBWeb/mb2/part1/fatcatab.htm

Saturated & Unsaturated Fats

Saturated & Unsaturated Fatty Acids

Phospholipids Source: http://www.rpi.edu/dept/bcbp/molbiochem/MBWeb/mb1/part2/lipid.htm#animat2

Cell Membrane Fluidity

R-S System (Concept Review)

E-Z System E Entgegen (Opposite) Z Zusammen (Together)

E-Z System (Z) Together (E) Opposite

E-Z System Problem 7.1 Page 284 (A-D) Answers: (E)-1-bromo-1-chloro-1-pentene (E)-2-bromo-1-chloro-1-iodo-1-butene (Z)-3,5-dimethyl-2-hexene (Z)-1-chloro-1-iodo-2-methyl-1-butene

Vic vs. Gem Vic CH2BrCH2Br Gem CHBr2CH3 Vicinal Latin Vicinus meaning adjacent Used when groups are on adjacent Carbons Gem CHBr2CH3 Geminal Latin Geminus meaning twins Used when groups are on the same Carbon

Stabilities of Alkenes-General Ability to spread out the burden of the double bond translates into greater stability.

Stabilities of Alkenes-(Cis/Trans) Trans is more stable due to less crowding, less electron repulsion.

Stabilities of Alkenes-(Cis/Trans) Cis is more stable due to less ring strain, less electron repulsion. Ring strain electron repulsion is stronger than crowding electron repulsion.

Vision & Cis-Trans Stability

Stabilities of Alkenes-(Cis/Trans) Gem is more stable than Vic due to greater electron flow.

Zaitsev Rule Zaitsev, Saytzeff, Saytseff, Saytzev States that the major product of an elimination reaction will be the more stable alkene. * Major Product

Zaitsev Rule Zaitsev, Saytzeff, Saytseff, Saytzev States that the Major product of an elimination reaction will be the more stable alkene Ea driven reaction (Kinetic Control)

Ea driven reaction (Kinetic Control) Rate of formation of the product with the lower Ea is faster, thus it accumulates faster.

Ea driven reaction (Kinetic Control)

Hofmann Rule Opposite of Zaitsev Rule Applies when major product of an elimination reaction is the less stable alkene. Collision Driven Reaction

Hofmann Rule

Ea vs. Collision CH3CHBrCH(CH3)2 + NU:- E2 E2 SN2 CH3CH=C(CH3)2 CH2=CHCH(CH3)2

Ea vs. Collision SN2 vs. E2 E2 (Zaitsev) vs. E2 (Hofmann)

Dehydration of Alcohols Elimination Reaction Acid Catalyzed Products alkene or alkyne & water

Dehydration of 20 & 30 Alcohols E1 Mechanism Step 1: Protonation Add H+ to the hydroxyl group (Acid Catalysis) Makes a better leaving group (H2O vs. OH-)

Dehydration of 20 & 30 Alcohols Step 2: Dissociation Leaving group departs from substrate. Carbocation formed

Dehydration of 20 & 30 Alcohols Step 3: Nucleophilic Attack Nucleophile attacks a Beta-Carbon. Alkene or Alkyne formed

Acid-Catalyzed Dehydration of Secondary or Tertiary Alcohols Summary P

Dehydration of 10 Alcohols E2 Mechanism Step 1: Protonation Add H+ to the hydroxyl group (Acid Catalysis) Makes a better leaving group (H2O vs. OH-)

Dehydration of 10 Alcohols Step 2: Nucleophilic Attack Attacks Beta Hydrogen Leaving group departs from substrate. Alkene or Alkyne formed

Acid-Catalyzed Dehydration of Primary Alcohols Summary P. 303 Step 1 Step 2

Molecular Rearrangement

Molecular Rearrangement Molecular Rearrangement occurs to partially stabilize the carbocation

Terminal Alkynes Triple bond at the end of the carbon chain Moderate acidity Stronger acid than terminal alkenes and alkanes Rare in Nature

Terminal Alkynes- Creating a Nucleophile

Terminal Alkynes- Nucleophilic Attack

Terminal Alkynes-Advantages Can create any size carbon chain Can place the triple bond in any location Can then add functional groups via the triple bond.

Syn & Anti Additions

Hydrogenation of Alkynes Addition of a hydrogen molecule (H2) across the triple bond. Three reactions, leading to three different products. Platinum & H2 P-2 Catalyst (Nickel Boride) Lithium

Hydrogenation of Alkynes Platinum & H2 Hydrogen adds across both Pi bonds to produce an Alkane

Hydrogenation of Alkynes P-2 Catalyst Adds across only the first Pi bond Syn Addition Z or Cis alkene arrangement

Hydrogenation of Alkynes Lithium Adds across only the first Pi bond Anti Addition E or Trans alkene arrangement

Index of Hydrogen Deficiency Abbreviated IHD Counts missing pairs of hydrogen Each double bond has an IHD of 1 Each ring has an IHD of 1 Each triple bond has an IHD of 2 General formula for a straight chain alkane is RnH2n+2

Index of Hydrogen Deficiency

Index of Hydrogen Deficiency Hydrogenation remove double and triple bonds Hydrogenation will not effect ring structures

Index of Hydrogen Deficiency Hydrogenation remove double and triple bonds Hydrogenation will not effect ring structures

IHD Example Problems P. 185 problems 4.20, 4.21

IHD & Functional Groups Alkane CH3CH3 6 H Halogen CH3CH2Br 5 H Need to add one hydrogen for each halogen

IHD & Functional Groups Alkane CH3CH3 6 H Alcohol CH3CH2OH 6 H Ignore Oxygen

IHD & Functional Groups Alkane CH3CH3 6 H Aldehyde CH3CHO 4 H Contains carbonyl group (C=O) Thus contains one double bond Thus IHD = 1 Ignore Oxygen

IHD & Functional Groups Alkane CH3CH3 6 H Amine CH3CH2NH2 7 H Subtract one hydrogen for each nitrogen present.

IHD & Functional Groups For compounds containing halogen atoms, count the halogen atoms as hydrogen atoms. For compounds containing oxygen, ignore the oxygen. For compounds containing nitrogen, subtract one hydrogen for each nitrogen present.

Problems (p.329) 7.27 7.31 7.34 7.35 7.37 7.38 7.43 7.44 7.45 7.48 – 7.51