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"; Organic Chemistry Exam 3 Chapter 6 & 7 Questions - Chem Homework Help
Organic Chemistry Exam 3 Chapter 6 & 7 Questions

Organic Chemistry,

9th Edition L. G. Wade, Jr.

Chapter 7 Lecture

Structure and Synthesis of Alkenes; Elimination

Introduction

  • Alkenes are hydrocarbons with carbon– carbon double bonds.
  • Alkenes are also called olefins, meaning “oilforming gas.”
  • The functional group of alkenes is the carbon– carbon double bond, which gives this group its reactivity.

Orbital Description

  • Sigma bonds around the double-bonded carbon are sp2 hybridized.
  • Angles are approximately 120º and the molecular geometry is trigonal planar.
  • Unhybridized p orbitals with one electron will overlap, forming the double bond (pi bond).

word image 1783

Bond Lengths and Angles

word image 1784

  • 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.

Pi Bonding in Ethylene

word image 1785

  • 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.

Cis-Trans Interconversion

word image 1786

  • Cis and trans isomers cannot be interconverted.
  • No rotation around the carbon–carbon bond is possible without breaking the pi bond (264 kJ/mole).

Elements of Unsaturation

word image 1787

  • Unsaturation: A structural element that decreases the number of hydrogens in the molecule by two.
  • Also called index of hydrogen deficiency.
  • Double bonds and rings are elements of unsaturation.

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.

IUPAC and New IUPAC

word image 1788

Ring Nomenclature

2

1-methylcyclopentene

1

3-methylcyclopentene

In a ring, the double bond is assumed to be between carbon 1 and carbon 2.

C

H

3

2

3

CH

3

1

Multiple Double Bonds

word image 1790

  • Give the double bonds the lowest numbers possible.
  • Use di-, tri-, tetra– before the ending -ene to

specify how many double bonds are present.

Cis-Trans Isomers

word image 1792

  • Also called geometric isomerism.
  • Similar groups on the same side of the double bond of an alkene is called cis.
  • Similar groups on the opposite sides of the double bond of an alkene is called trans.
  • Not all alkenes show cis-trans isomerism.

Cyclic Compounds

word image 1794

  • 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.

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).

Example

  • Assign priority to the substituents according to their atomic number (1 is highest priority).

word image 1796

1

2

1

2

  • If the highest priority groups are on opposite sides, the isomer is E.
 

•

If the highest priority

E-1-bromo-1-chloropropene

 

groups are on the same

side, the isomer is Z.

Cyclic Stereoisomers

word image 1798

• Double bonds outside the ring can show stereoisomerism.

Stereochemistry in Dienes

word image 1799

• If there is more than one double bond in the molecule, the stereochemistry of all the double bonds should be specified.

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.

word image 1802

Relative Stabilities

word image 1804

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.

    

Substituent Effects

• The isomer with the more substituted double bond has a larger angular separation between the bulky alkyl groups.

word image 1810

Disubstituted Isomers

  • Stability: cis < geminal < trans isomer
  • The less stable isomer has a higher exothermic heat of hydrogenation.

cis-2-butene

CH3 CH3

C

C

H H

-120 kJ

iso-butene

(CH3)2C=CH2

-117 kJ

trans-2-butene

H CH3

C

C

CH3 H

-116 kJ

Cycloalkenes

word image 1812

  • 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.

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.

word image 1813

Stability of Cycloalkenes

word image 1814

  • The cis isomer is more stable than trans in small cycloalkenes.
  • Small rings have additional ring strain.
  • 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.

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.

word image 1815

Solved Problem

Which of the following alkenes are stable?

word image 1817

word image 1819

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.

Chapter 7 28

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 may have a small dipole moment.

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.

word image 1822

Alkene Synthesis Overview

  • E2 dehydrohalogenation (-HX)
  • E1 dehydrohalogenation (-HX)
  • Dehalogenation of vicinal dibromides (-X2)
  • Dehydration of alcohols (-H2O)

Dehydrohalogenation by the E2

Mechanism

word image 1824

  • A strong base abstracts H+ as a double bond forms and the X– leaves from the adjacent carbon atom.
  • Tertiary and hindered secondary alkyl halides give alkenes in good yields.
  • Tertiary halides are the best E2 substrates because they are prone to elimination and cannot undergo SN2 substitution.

Bulky Bases for E2 Reactions

word image 1825

  • If the substrate is prone to substitution, a bulky base can minimize the amount of substitution.
  • Large alkyl groups on a bulky base hinder its approach to attack a carbon atom (substitution), yet it can easily abstract a proton (elimination).

Hofmann Product

Bulky bases, such as potassium tert-butoxide, abstract the least hindered H+,giving the less substituted alkene as the major product (Hofmann product).

word image 1828

E2 Reactions Are Stereospecific

word image 1829

  • Depending on the stereochemistry of the alkyl halide, the E2 elimination may produce only the cis or only the trans isomer.
  • The geometry of the product will depend on the anticoplanar relationship between the proton and the leaving group.

Stereochemistry of E2 Elimination

word image 1831

  • Most E2 reactions go through an anticoplanar transition state.
  • This geometry is most apparent if we view the reaction with the alkyl halide in a Newman projection.

Solved Problem

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.)

word image 1833

Chapter 7 38

E2 Reactions on Cyclohexanes

word image 1836

  • 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.

Solved Problem

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.

word image 1838

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.

word image 1839

Chapter 7 41

Debromination of Vicinal

Dibromides

word image 1841

  • Vicinal dibromides are converted to alkenes by reduction with iodide ion in acetone.
  • Not an important reaction, but the mechanism is similar to the E2 dehydrohalogenation reaction.

E1 Elimination Mechanism

  • Tertiary and secondary alkyl halides:

3º > 2º

  • Carbocation intermediate.
  • Rearrangements are possible.
  • Weak nucleophiles such as water or alcohols.
  • Usually have SN1 products, too, since the solvent can

attack the carbocation directly.

word image 1843

Dehydration of Alcohols

word image 1844

  • Use concentrated H2SO4 or H3PO4 to shift the equilibrium and increase the yield of the reaction.
  • E1 mechanism.
  • Rearrangements are common.
  • Reaction obeys Zaitsev’s rule.

Dehydration Mechanism: E1

Step 1: Protonation of the hydroxyl group (fast equilibrium).

word image 1846

Step 2: Ionization to a carbocation (slow; rate limiting).

word image 1849

Step 3: Deprotonation to give the alkene (fast).

word image 1851

Organic Chemistry,

9th Edition L. G. Wade, Jr.

Chapter 7 Lecture

Structure and Synthesis of Alkenes; Elimination

Introduction

  • Alkenes are hydrocarbons with carbon– carbon double bonds.
  • Alkenes are also called olefins, meaning “oilforming gas.”
  • The functional group of alkenes is the carbon– carbon double bond, which gives this group its reactivity.

Orbital Description

  • Sigma bonds around the double-bonded carbon are sp2 hybridized.
  • Angles are approximately 120º and the molecular geometry is trigonal planar.
  • Unhybridized p orbitals with one electron will overlap, forming the double bond (pi bond).

word image 1789

Bond Lengths and Angles

word image 1791

  • 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.

Pi Bonding in Ethylene

word image 1793

  • 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.

Cis-Trans Interconversion

word image 1795

  • Cis and trans isomers cannot be interconverted.
  • No rotation around the carbon–carbon bond is possible without breaking the pi bond (264 kJ/mole).

Elements of Unsaturation

word image 1797

  • Unsaturation: A structural element that decreases the number of hydrogens in the molecule by two.
  • Also called index of hydrogen deficiency.
  • Double bonds and rings are elements of unsaturation.

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.

IUPAC and New IUPAC

word image 1800

Ring Nomenclature

2

1-methylcyclopentene

1

3-methylcyclopentene

In a ring, the double bond is assumed to be between carbon 1 and carbon 2.

C

H

3

2

3

CH

3

1

Multiple Double Bonds

word image 1801

  • Give the double bonds the lowest numbers possible.
  • Use di-, tri-, tetra– before the ending -ene to

specify how many double bonds are present.

Cis-Trans Isomers

word image 1803

  • Also called geometric isomerism.
  • Similar groups on the same side of the double bond of an alkene is called cis.
  • Similar groups on the opposite sides of the double bond of an alkene is called trans.
  • Not all alkenes show cis-trans isomerism.

Cyclic Compounds

word image 1805

  • 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.

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).

Example

  • Assign priority to the substituents according to their atomic number (1 is highest priority).

word image 1806

1

2

1

2

  • If the highest priority groups are on opposite sides, the isomer is E.
 

•

If the highest priority

E-1-bromo-1-chloropropene

 

groups are on the same

side, the isomer is Z.

Cyclic Stereoisomers

word image 1807

• Double bonds outside the ring can show stereoisomerism.

Stereochemistry in Dienes

word image 1808

• If there is more than one double bond in the molecule, the stereochemistry of all the double bonds should be specified.

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.

word image 1809

Relative Stabilities

word image 1811

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.

     

Substituent Effects

• The isomer with the more substituted double bond has a larger angular separation between the bulky alkyl groups.

word image 1816

Disubstituted Isomers

  • Stability: cis < geminal < trans isomer
  • The less stable isomer has a higher exothermic heat of hydrogenation.

cis-2-butene

CH3 CH3

C

C

H H

-120 kJ

iso-butene

(CH3)2C=CH2

-117 kJ

trans-2-butene

H CH3

C

C

CH3 H

-116 kJ

Cycloalkenes

word image 1818

  • 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.

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.

word image 1820

Stability of Cycloalkenes

word image 1821

  • The cis isomer is more stable than trans in small cycloalkenes.
  • Small rings have additional ring strain.
  • 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.

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.

word image 1823

Solved Problem

Which of the following alkenes are stable?

word image 1826

word image 1827

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.

Chapter 7 28

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 may have a small dipole moment.

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.

word image 1830

Alkene Synthesis Overview

  • E2 dehydrohalogenation (-HX)
  • E1 dehydrohalogenation (-HX)
  • Dehalogenation of vicinal dibromides (-X2)
  • Dehydration of alcohols (-H2O)

Dehydrohalogenation by the E2

Mechanism

word image 1832

  • A strong base abstracts H+ as a double bond forms and the X– leaves from the adjacent carbon atom.
  • Tertiary and hindered secondary alkyl halides give alkenes in good yields.
  • Tertiary halides are the best E2 substrates because they are prone to elimination and cannot undergo SN2 substitution.

Bulky Bases for E2 Reactions

word image 1834

  • If the substrate is prone to substitution, a bulky base can minimize the amount of substitution.
  • Large alkyl groups on a bulky base hinder its approach to attack a carbon atom (substitution), yet it can easily abstract a proton (elimination).

Hofmann Product

Bulky bases, such as potassium tert-butoxide, abstract the least hindered H+,giving the less substituted alkene as the major product (Hofmann product).

word image 1835

E2 Reactions Are Stereospecific

word image 1837

  • Depending on the stereochemistry of the alkyl halide, the E2 elimination may produce only the cis or only the trans isomer.
  • The geometry of the product will depend on the anticoplanar relationship between the proton and the leaving group.

Stereochemistry of E2 Elimination

word image 1840

  • Most E2 reactions go through an anticoplanar transition state.
  • This geometry is most apparent if we view the reaction with the alkyl halide in a Newman projection.

Solved Problem

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.)

word image 1842

Chapter 7 38

E2 Reactions on Cyclohexanes

word image 1845

  • 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.

Solved Problem

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.

word image 1847

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.

word image 1848

Chapter 7 41

Debromination of Vicinal

Dibromides

word image 1850

  • Vicinal dibromides are converted to alkenes by reduction with iodide ion in acetone.
  • Not an important reaction, but the mechanism is similar to the E2 dehydrohalogenation reaction.

E1 Elimination Mechanism

  • Tertiary and secondary alkyl halides:

3º > 2º

  • Carbocation intermediate.
  • Rearrangements are possible.
  • Weak nucleophiles such as water or alcohols.
  • Usually have SN1 products, too, since the solvent can

attack the carbocation directly.

word image 1852

Dehydration of Alcohols

word image 1853

  • Use concentrated H2SO4 or H3PO4 to shift the equilibrium and increase the yield of the reaction.
  • E1 mechanism.
  • Rearrangements are common.
  • Reaction obeys Zaitsev’s rule.

Dehydration Mechanism: E1

Step 1: Protonation of the hydroxyl group (fast equilibrium).

word image 1854

Step 2: Ionization to a carbocation (slow; rate limiting).

word image 1855

Step 3: Deprotonation to give the alkene (fast).

word image 1856

Organic Chemistry,

9th Edition L. G. Wade, Jr.

Chapter 6

Lecture

Alkyl Halides:

Nucleophilic Substitution

Classes of Alkyl Halides

H H

  • Alkyl halides: Halogen H C C is directly bonded to sp3

Br

carbon.

• Vinyl halides: Halogen

H H

alkyl halide

H

H

is bonded to sp2 carbon

C

C

Cl

of alkene. H vinyl halide

  • Aryl halides: Halogen is bonded to sp2 carbon on benzene ring.

I

aryl halide

Polarity and Reactivity

word image 1857

  • Halogens are more electronegative than C.
  • Carbon—halogen bond is polar, so carbon has partial

positive charge.

  • Carbon can be attacked by a nucleophile.
  • Halogen can leave with the electron pair.

IUPAC Nomenclature

  • Name as haloalkane (fluoro, chloro, bromo, iodo)
  • Choose the longest carbon chain, even if the halogen is not bonded to any of those carbons.
  • Use lowest possible numbers for position.

1 2

CH[1]CH2F CH3CHCH2CH3 CH3CH2CH2CHCH2CH2CH3

Cl

1 2 3 4 1 2 3 4 5 6 7

Examples

Br

CH

3

CH3CHCH2CH2CH2CHCH2CH2CH3

1 2 3 4 5 6 7 8 9

6-bromo-2-methylnonane

Br

F

H

H

1

3

cis-1-bromo-3-fluorocyclohexane

Systematic Common Names

  • An alkyl group is a substituent on a halide.
  • Useful only for small alkyl groups.

isobutyl bromide sec-butyl bromide

CH

3

CHCH

2

CH

3

Br

CH

3

CH

2

CH

CH

3

Br

CH

3

C

CH

3

Br

CH

3

tert-butyl bromide

Alkyl Halides Classification

  • Methyl halides: Halide is attached to a methyl group.
  • Primary alkyl halide: Carbon to which halogen is

bonded is attached to only one other carbon.

  • Secondary alkyl halide: Carbon to which halogen is

bonded is attached to two other carbons.

  • Tertiary alkyl halide: Carbon to which halogen is bonded is attached to three other carbons.

word image 1858

Types of Dihalides

word image 1859

  • Geminal dihalide: Two halogen atoms are bonded to the same carbon.
  • Vicinal dihalide: Two halogen atoms are bonded to adjacent carbons.

Dipole Moments

  • Electronegativities of the halides: F > Cl > Br > I
  • word image 1860 Bond lengths increase as the size of the halogen increases:

C—F < C—Cl < C—Br < C—I

  • Molecular dipoles depend on the geometry of the molecule.

Dipole Moments and Molecular Geometry

word image 1861

Notice how the four symmetrically oriented polar bonds of the carbon tetrahalides cancel to give a molecular dipole moment of zero.

Boiling Points

  • Greater intermolecular forces, higher b.p.
  • London forces greater for larger atoms.
  • Greater molar mass, higher b.p.
  • Spherical shape decreases b.p.

(CH3)3CBr CH3(CH2)3Br

73 °C 102 °C

Densities

  • Alkyl fluorides and alkyl chlorides (those with just one chlorine atom) are less dense than water (1.00 g/mL).
  • Alkyl chlorides with two or more chlorine atoms are denser than water.
  • All alkyl bromides and alkyl iodides are denser than water.

Reactions of Alkyl Halides

word image 1862 word image 1863

The SN2 Reaction

word image 1864

  • The halogen atom on the alkyl halide is replaced with the nucleophile (HO–).
  • Since the halogen is more electronegative than carbon, the C—I bond breaks heterolytically and the iodide ion leaves.

S 2 Mechanism

N

word image 1865

  • Bimolecular nucleophilic substitution (SN2).
  • Concerted reaction: New bond forming and old bond breaking at same time.
  • Reaction is second order overall.
  • Rate = k[alkyl halide][nucleophile].

SN2 Energy Diagram

word image 1866

  • The SN2 reaction is a one-step reaction.
  • Transition state is highest in energy.

Uses for SN2 Reactions

word image 1867

SN2: Nucleophilic Strength

  • Stronger nucleophiles react faster.
  • Strong bases are strong nucleophiles, but not all strong nucleophiles are basic.

word image 1868

Basicity Versus Nucleophilicity

word image 1869

  • Basicity is defined by the equilibrium constant for abstracting a proton.
  • Nucleophilicity is defined by the rate of attack on the electrophilic carbon atom

    

Steric hindrance (

bulkiness

)

hinders nucleophilicity

more than it hinders basicity.

Trends in Nucleophilicity

  • A negatively charged nucleophile is stronger than its neutral counterpart:

word image 1870

  • Nucleophilicity decreases from left to right:

word image 1871

  • Increases down periodic table as the size and polarizability increase:

word image 1872

Polarizability Effect

word image 1873

Larger atoms have a soft shell that can start to overlap the carbon atom from a farther distance.

Solvent Effects: Protic Solvents

word image 1874

  • Polar protic solvents have acidic hydrogens (O—H or

N—H) that can solvate the nucleophile, reducing their nucleophilicity.

  • Nucleophilicity in protic solvents increases as the size of the atom increases.

Solvent Effects: Aprotic Solvents

word image 703

  • Polar aprotic solvents do not have acidic protons and therefore cannot hydrogen bond.
  • Some aprotic solvents are acetonitrile, DMF, acetone, and DMSO.
  • SN2 reactions proceed faster in aprotic solvents.

Crown Ethers

  • Crown ethers

word image 1875 word image 1876

solvate the cation, so the nucleophilic strength of the anion increases.

  • Fluoride becomes a good nucleophile.

word image 1877

Leaving Group Ability

The best leaving groups are:

  • Electron-withdrawing helps to polarize the carbon atom.
  • Stable once they have left.
  • Polarizable, help to stabilize the transition state.

word image 1878

Structure of Substrate in SN2

Reactions

  • Relative rates for SN2:

CH3X > 1° > 2° >> 3°

  • Tertiary halides do not react via the SN2 mechanism,

due to steric hindrance.

word image 1879

Effect of Substituents on the Rates of SN2 Reactions

word image 1880

Stereochemistry of SN2

SN2 reactions will result in an inversion of configuration.

word image 1881

Back-Side Attack in the SN2 Reaction

word image 1882

The SN1 Reaction

  • The SN1 reaction is a unimolecular nucleophilic substitution.
  • It has a carbocation intermediate.
  • Rate = k[alkyl halide]
  • Racemization occurs.

SN1 Mechanism

Step 1: Formation of the carbocation.

word image 1883

Step 2: Attack of the nucleophile.

word image 1884

• If the nucleophile is an uncharged molecule like water or an alcohol, the positively charged product must lose a proton to give the final uncharged product.

1 Mechanism: Step 1

word image 1885

Formation of carbocation (rate-determining step)

1 Mechanism: Step 2

word image 1886

• The nucleophile attacks the carbocation, forming the product.

1 Mechanism: Step 3

word image 1887

• If the nucleophile was neutral, a third step (deprotonation) will be needed.

SN1 Energy Diagram

  • Forming the carbocation is an endothermic step.

word image 1888

  • Step 2 is fast with a low activation energy.

Rates of SN1 Reactions

  • Order of reactivity follows stability of carbocations (opposite to SN2).
  • 3° > 2° > 1° >> CH3X
  • More stable carbocation requires less energy to form.
  • A better leaving group will increase the rate of the reaction.

Solvent Effect

• A polar protic solvent is best used because it can solvate both ions strongly through hydrogen bonding.

word image 1889

Stereochemistry of S 1

N

word image 1890

  • Carbocations are sp2 hybridized and trigonal planar. The lobes of the empty p orbital are on both sides of the trigonal plane.
  • Nucleophilic attack can occur from either side, producing mixtures of retention and inversion of configuration if the carbon is chiral.

Carbocation Stability

word image 1891

Carbocations are stabilized by inductive effect and by hyperconjugation.

Rearrangements

word image 1892

  • Carbocations can rearrange to form a more stable carbocation.
  • Move the smallest group on the adjacent carbon:
  • Hydride shift: H– on adjacent carbon moves.
  • Methyl shift: CH3– on adjacent carbon moves.

Hydride and Methyl Shifts

word image 1893

  • In this reaction, the methyl group on the adjacent carbon will move (along with both bonding electrons) to the primary carbon, displacing the bromide and forming a more stable tertiary carbocation.
  • The smallest groups on the adjacent carbon will move: If there is a hydrogen, it will give a hydride shift.

SN1 or SN2 Mechanism?

SN2

SN1

CH3X > 1º > 2º

3º > 2º

Strong nucleophile

Weak nucleophile (may also be used as solvent)

Polar aprotic solvent

Polar protic solvent

Rate = k[alkyl halide][Nuc]

Rate = k[alkyl halide ]

Inversion at chiral carbon

Racemization

No rearrangements

Rearranged products

Elimination Reactions

word image 1894

  • Elimination reactions produce double bonds.
  • Also called dehydrohalogenation (-HX).

The E1 Reaction

  • Unimolecular elimination.
  • Two groups 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.

E1 Mechanism

Step 1: Ionization to form a carbocation.

word image 1895

Step 2: Solvent abstracts a proton to form an alkene.

word image 1896

A Closer Look

word image 1897

E1 Energy Diagram

word image 1898

The E1 and the SN1 reactions have the same first step: Carbocation formation is the rate-determining step for both mechanisms.

Double Bond Substitution Patterns

word image 1899

tetrasubstituted trisubstituted disubstituted monosubstituted

  • In elimination reactions, the major product of the reaction is the more substituted double bond: Zaitsev’s rule.
  • The more substituted double bond is more stable.

Zaitsev’s Rule

• If more than one elimination product is possible, the most-substituted alkene is the major product (most stable).

major product

word image 1900

(trisubstituted)

E1 Competes with the SN1

word image 1901

  • Substitution results from nucleophilic attack on the carbocation.
  • Ethanol serves as a base in the elimination and as a nucleophile in the substitution.

A Carbocation Can…

  • React with a nucleophile to form a substitution product (SN1).
  • Lose a proton to form an elimination product (E1).
  • Rearrange to give a more stable carbocation, then react further.

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.

The E2 Mechanism

word image 1902

  • Order of reactivity for alkyl halides: 3°> 2 °> 1°
  • Mixture may form, but Zaitsev product predominates.

E2 Stereochemistry

  • The halide and the proton to be abstracted must be anti-coplanar (θ = 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.

word image 1903

E1 or E2 Mechanism?

E1

E2

3º > 2º

3º > 2º

Base strength (usually weak)

Strong base required

Good ionizing solvent

Solvent polarity not important

Rate = k[alkyl halide]

Rate = k[alkyl halide][base]

No required geometry

Coplanar leaving groups

(usually anti)

Rearranged products

No rearrangements

Substitution or Elimination?

  • Strength of the nucleophile determines 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.

Time to Practice!

  • SN2, SN1, E2, E1 Reaction Mechanism Examples:

https://youtu.be/pKJ0z7N6W5w https://youtu.be/MrzUZSETgFc

  • SN2, SN1, E2, E1 Practice Problem Examples:

https://youtu.be/lkcZaL3Nxps

  1. 2-chlorobutane 4-(2-fluoroethyl)heptane ↑

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