Organic Chemistry Final Exam Review Notes
Terms in this set (404)
What is an S orbital?
An s orbital has a sphere of electron density and is lower in energy than the other orbitals of the same shell.
What is a p orbital?
A p orbital has a dumbbell shape and contains a node (no electron density) at the nucleus. It is higher in energy than an s orbital.
What is bonding?
Bonding is the joining of two atoms in a stable arrangement.
•Through bonding, atoms attain a complete outer shell of valence electrons (stable noble gas configuration).
•Atoms can form either ionic or covalent bonds to attain a complete outer shell (octet rule for second row elements).
• Ionic bonds result from the transfer of electrons from one
element to another.
• Covalent bonds result from the sharing of electrons between two nuclei.
What do we know about valence electrons?
Second row elements can have no more than eight electrons around them. For neutral molecules, this has two consequences:
• Atoms with one, two, three, or four valence electrons form one, two, three, or four bonds, respectively, in
neutral molecules (e.g., BF3, CH4).
•Atoms with five or more valence electrons form enough bonds to give an octet (e.g., NH3). This results in the following equation: Predicted number of bonds= 8- valence electrons
What are lone pairs?
When second-row elements form fewer than four bonds their octets consist of both bonding (shared) and nonbonding (unshared) electrons. Unshared electrons are also called lone pairs.
What is formal charge?
Formal charge is the charge assigned to individual atoms in a Lewis structure.
•Formal charge is calculated as follows:
•The number of electrons "owned" by an atom is determined by its number of bonds and lone pairs.
•An atom "owns" all of its unshared electrons and half of its shared electrons.
What is resonance?
Some molecules cannot be adequately represented by a single Lewis structure. For example:
•These structures are called resonance structures or resonance forms. A double-headed arrow is used to separate the two resonance structures.
•Resonance structures are two Lewis structures having the same placement of atoms but a different arrangement of electrons. Neither resonance structure is an accurate representation for (HCONH)¯. The true structure is a composite of both resonance forms and is called a resonance hybrid.
•The hybrid shows characteristics of both structures.
•Resonance allows certain electron pairs to be delocalized over two or more atoms, and this delocalization adds stability.
•A molecule with two or more resonance forms is said to be resonance stabilized.
What are the basic principles of resonance stability?
Resonance structures are not real. An individual resonance structure does not accurately represent the structure of a
molecule or ion. Only the hybrid does.
• Resonance structures are not in equilibrium with each other. There is no movement of electrons from one form to
• Resonance structures are not isomers. Two isomers differ in the arrangement of both atoms and electrons, whereas resonance structures differ only in the arrangement of
What are the rules of drawing resonance structures?
Rule : Two resonance structures differ in the position of multiple bonds and nonbonded electrons. The placement of atoms and single bonds always stays the same.
Rule : Two resonance structures must have the same number of unpaired electrons.
Rule : Resonance structures must be valid Lewis structures. Hydrogen must have two electrons and no second-row element can have more than eight electrons.
What is curved arrow notation?
Curved arrow notation is a convention that shows how electron position differs between two resonance forms.
•Curved arrow notation shows the movement of an electron pair. The tail of the arrow always begins at the electron pair, either in a bond or lone pair. The head points to where the electron pair
What is the occurrence of resonance?
Two different resonance structures can be drawn when a lone pair is located on an atom directly bonded to a double bond.
Multiple resonance structures can also be drawn when an atom bearing a (+) charge is bonded either to a double bond or an atom with a lone pair.
What is a resonance hybrid?
A resonance hybrid is a composite of all possible resonance structures. In the resonance hybrid, the electron pairs drawn in different locations in individual resonance forms are delocalized.
•When two resonance structures are different, the hybrid looks more like the "better" resonance structure.
•The "better" resonance structure is called the major contributor to the hybrid, and all others are minor contributors.
A "better" resonance structure is one that has more bonds and
How do you determine molecular shape?
Two variables define a molecule's structure: bond length
and bond angle. • Bond length decreases across a row of the periodic
table as the size of the atom decreases.
• Bond length increases down a column of the periodic table as the size of an atom increases.
•The number of groups surrounding a particular atom determines its geometry. A group is either an atom or a lone pair of electrons.
•The most stable arrangement keeps these groups as far away from each other as possible. This is exemplified by Valence Shell Electron Pair Repulsion (VSEPR) theory.
How do you draw structures for molecules?
•A solid line is used for a bond in the plane. •A wedge is used for a bond in front of the plane. •A dashed line is used for a bond behind the plane.
How do you draw condensed molecular structures?
•All atoms are drawn in, but the two-electron bond lines are generally omitted.
•Atoms are usually drawn next to the atoms to which they are bonded.
•Parentheses are used around similar groups bonded to the same atom.
•Lone pairs are omitted.
How do you draw a skeletal structure?
•Assume there is a carbon atom at the junction of any two lines or at the end of any line.
•Assume there are enough hydrogens around each carbon to make it tetravalent.
•Draw in all heteroatoms and the hydrogens directly bonded to them.
•A charge on a carbon atom takes the place of one hydrogen atom.
•The charge determines the number of lone pairs. Negatively charged carbon atoms have one lone pair and positively charged carbon atoms have none.
Orbitals and Bonding: Hydrogen
• When the 1s orbital of one H atom overlaps with the 1s
orbital of another H atom, a sigma () bond that concentrates electron density between the two nuclei is formed.
• This bond is cylindrically symmetrical because the electrons forming the bond are distributed symmetrically about an imaginary line connecting the two nuclei.
Orbitals and Bonding: Methane
•To account for the bonding patterns observed in more complex molecules, we must take a closer look at how the 2s and 2p
orbitals of atoms in the second row are utilized.
•In addition to its two core electrons, carbon has four valence electrons.
•In its ground state, carbon places two electrons in the 2s orbital and one each in 2p orbitals.
Note: The lowest energy arrangement of electrons for an atom is called its ground state.
•To solve this dilemma, chemists have proposed that atoms like carbon do not use pure s and pure p orbitals in forming bonds.
•Instead, atoms use a set of new orbitals called hybrid
•Hybridization is the combination of two or more atomic orbitals to form the same number of hybrid orbitals, each having the same shape and energy.
•In this description, carbon should form only two bonds because it has only two unpaired valence electrons.
•However, the resulting species, CH2, is very unstable and cannot be isolated under typical laboratory conditions.
•Note that in CH2, carbon would not have an octet of electrons.
Shape and Orientation of sp3
•The mixing of a spherical 2s orbital and three dumbbell shaped 2p orbitals together produces four hybrid orbitals, each having one large lobe and one small lobe.
•The four hybrid orbitals are oriented towards the corners of a tetrahedron, and form four equivalent bonds.
Bonding Using sp3 Hybrid Orbitals
•Each bond in CH4 is formed by overlap of an sp3 hybrid orbital of carbon with a 1s orbital of hydrogen.
•These four bonds point to the corners of a tetrahedron.
Each carbon is trigonal planar. Each carbon is sp2 hybridized.
Each carbon atom has two unhybridized 2p orbitals that are perpendicular to each other and to the sp hybrid orbitals.
•The side-by-side overlap of two 2p orbitals on one carbon with two 2p orbitals on the other carbon creates the second and third bonds of the triple bond.
•All triple bonds are composed of one sigma and two pi bonds.
Bond Length and Bond Strength
•As the number of electrons between two nuclei increases, bonds become shorter and stronger.
•Triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds.
•The length and strength of C—H bonds vary depending on the hybridization of the carbon atom.
Electronegativity is a measure of an atom's attraction for electrons in a bond. it will increase going up the periodic table and also as you move right.
•Electronegativity values are used to indicate whether the electrons in a bond are equally shared or unequally shared between two atoms.
•When electrons are equally shared, the bond is nonpolar.
•A carbon—carbon bond is nonpolar.
•C—H bonds are considered to be nonpolar because the electronegativity difference between C and H is small (e.g. electronegativity difference <0.5).
•Whenever two different atoms having similar electronegativities are bonded together, the bond is nonpolar.
• Polarity of a bond can help us predict its reactivity.
•Bonding between atoms of different electronegativity values results in unequal sharing of electrons.
•Example: In the C—O bond, the electrons are pulled away from C (2.5) toward O (3.4), the element of higher electronegativity. The bond is polar, or polar covalent. The bond is said to have dipole; that is, separation of charge.
•The d+ means the indicated atom is electron deficient.
•The d- means the indicated atom is electron rich.
•The direction of polarity in a bond is indicated by an
arrow with the head of the arrow pointing towards the more electronegative element.
•The tail of the arrow is drawn at the less electronegative element.
A polar molecule has either one polar bond, or two or more bond dipoles that reinforce each other. An example is wate
A nonpolar molecule has either no polar bonds, or two or more bond dipoles that cancel. An example is carbon dioxide:
Review - Brønsted-Lowry Acids and Bases
A Brønsted-Lowry acid is a proton donor.
•It must have a proton.
•A Brønsted-Lowry base is a proton acceptor. •It must be able to form a bond to a proton.
•A hydrogen atom without its electron is a proton.
•H+ = proton
Acidic and Basic Sites in Morphine
Some molecules contain both hydrogen atoms and lone pairs and thus, can act either as acids or bases, depending on the particular reaction.
Reactions of Brønsted-Lowry Acids and Bases
•A Brønsted-Lowry acid base reaction results in the transfer of a proton from an acid to a base.
•The electron pair of the base B: forms a new bond to the proton of the acid forming the conjugate acid of the base.
•The acid H-A loses a proton, leaving the electron pair in the H-A bond on A. This forms the conjugate base of the acid.
Acid Strength and pKa
•Acid strength is the tendency of an acid to donate a proton.
•The more readily a compound donates a proton, the stronger an acid it is.
•Acidity is measured by an equilibrium constant.
•When a Brønsted-Lowry acid H-A is dissolved in water, an acid-base reaction occurs, and an equilibrium constant can be written for the reaction.
•Because the concentration of the solvent H2O is essentially constant, the equation can be rearranged and a new equilibrium constant, called the acidity constant, Ka, can be defined.
Ka and pKa
It is generally more convenient when describing acid strength to use "pKa"values than Ka values.
Outcome of Acid-Base
• The position of the equilibrium depends on the relative
strengths of the acids and bases.
• Equilibrium always favors formation of the weaker acid
Because the pKa of the starting acid (25) is lower than that of the conjugate acid (38), equilibrium favors the products.
Comparing the Acidity of Any Two Acids
Always draw the conjugate bases.
• Determine which conjugate base is more stable.
• The more stable the conjugate base, the more acidic
Steps in Solving Acid-Base Reaction Equilibria
Step  Identify the acid and base in the starting materials. Assume -NH2 is the base because it bears a net
negative charge. That makes HCCH the acid.
Step  Draw the products of proton transfer and identify
the conjugate acid and base in the products. Acetylene gives up its proton to -NH2.
Step  Compare the pKa values of the acid and the
conjugate acid. Equilibrium favors formation of the weaker acid with the higher pKa. The pKa of NH3 is higher; therefore products are favored.
Factors that Determine Acid Strength
• General rule:
•Anything that stabilizes a conjugate base A:¯ makes the starting acid H-A more acidic.
• Four factors affect the acidity of H-A. These are:
•Element effects •Inductive effects •Resonance effects •Hybridization effects
Element Effects—Trends in the
Across a row of the periodic table, the acidity of H-A increases as the electronegativity of A increases. Positive or negative charge is stabilized when it is spread over a larger volume.
•Down a column of the periodic table, size, and not electronegativity, determines acidity.
•The acidity of H-A increases as the size of A increases.
•An inductive effect is the pull of electron density through
bonds caused by electronegativity differences of atoms.
•More electronegative atoms stabilize regions of high electron density by an electron withdrawing inductive effect.
•The more electronegative the atom and the closer it is to the site of the negative charge, the greater the effect.
•The acidity of H-A increases with the presence of electron withdrawing groups in A.
•An inductive effect is the pull of electron density through bonds caused by electronegativity differ-ences between atoms.
•The reason for the increased acidity of 2,2,2-trifluoroethanol is that the three electronegative fluorine atoms stabilize the negatively charged conjugate base.
•Delocalization of charge through resonance influences acidity.
•Acetic acid is more acidic than ethanol, even though both molecules have the negative charge on the same element, O.
Comparison of Ethoxide and Acetate ions
•The conjugate base of ethanol has a localized charge.
•The conjugate base of acetic acid is resonance delocalized.
Hybridization Effects on acidity
The greater the percent s character, the greater the acidity, and therefore the lower the pKa value.
Stability of Conjugate Bases and the effect on the acidity
The higher the percent of s-character of the hybrid orbital, the more stable the conjugate base.
HOW TO Determine the Relative Acidity of Protons
Step  Identify the atoms bonded to hydrogen, and use periodic trends to assign relative acidity.
• The most common H-A bonds in organic compounds are C-H, N-H and O-H. Because acidity increases left to right across a row, the relative acidity of these bonds is C-H<N-H<O-H. Therefore, H atoms bonded to C atoms are usually less acidic
than H atoms bonded to any heteroatom.
Step  If the two H atoms in question are bonded to the same element, draw the conjugate bases and look for other points of difference. Ask
• Do electron-withdrawing groups stabilize the conjugate base? • Is the conjugate base resonance stabilized? • How is the conjugate base hybridized?
Commonly Used Acids in Organic Chemistry
• The familiar acids HCl and H2SO4 are often used in organic reactions.
• Various organic acids are also commonly used (e.g., acetic acid and p-toluenesulfonic acid (TsOH)).
Commonly Used Bases in Organic Chemistry
- sodium hydroxide
- sodium methoxide
- sodium exthodixde
- sodium terbutoxide
- sodium amide
- sodium hydride
Characteristics of Strong Organic Bases
•Strong bases have weak conjugate acids with high pKa values, usually > 12.
•Strong bases have a net negative charge, but not all negatively charged species are strong bases. For example, none of the halides F¯, Cl¯, Br¯, or I¯, is a strong base.
•Carbanions, negatively charged carbon atoms, are especially strong bases. A common example is butyllithium.
Other Common Bases in Organic Chemistry
•Amines (e.g., triethylamine and pyridine) are organic bases.
•They are basic due to having a lone pair on N.
•They are weaker bases since they are neutral, not negatively
Lewis Acids and Bases
•A Lewis acid is an electron pair acceptor. •A Lewis base is an electron pair donor. •Lewis bases are structurally the same as BrØnsted-Lowry
bases. Both have an available electron pair—a lone pair or an electron pair in a pi bond.
•A BrØnsted-Lowry base always donates this electron pair to a proton, but a Lewis base donates this electron pair to anything that is electron deficient.
•Any species that is electron deficient and capable of accepting an electron pair is also a Lewis acid.
•All BrØnsted-Lowry acids are also Lewis acids, but the reverse is not necessarily true.
•Common examples of Lewis acids (which are not BrØnsted-Lowry acids) contain elements in group 3A of the periodic table that can accept an electron pair because they do not have filled valence shells of electrons.
Lewis Acid-Base Reactions
•In a Lewis acid-base reaction, a Lewis base donates an electron pair to a Lewis acid.
•This is illustrated in the reaction of BF3 with H2O. H2O donates an electron pair to BF3 to form a new bond.
Electrophiles and Nucleophiles
•Lewis acid-base reactions illustrate a general pattern in organic chemistry.
•Electron-rich species react with electron-poor species.
•A Lewis acid is also called an electrophile.
•When a Lewis base reacts with an electrophile other than a proton, the Lewis base is also called a nucleophile.
•In this example, BF3 is the electrophile and H2O is the nucleophile.
Lewis Acid-Base Reactions that Form One New Covalent Bond
•Note that in each reaction below, the electron pair is not removed from the Lewis base.
•Instead, it is donated to an atom of the Lewis acid and one new covalent bond is formed.
Drawing Products of Lewis Acid-Base Reactions
• In other Lewis acid-base reactions, one bond is
formed and one bond is broken.
• To draw the products of these reactions, keep in mind
the following steps: •Always identify the Lewis acid and base first. •Draw a curved arrow from the electron pair of the
base to the electron-deficient atom of the acid.
•Count electron pairs and break a bond when needed to keep the correct number of valence electrons.
•The reaction between cyclohexene and HCl can be treated as a Lewis acid-base interaction.
•HCl acts as the Lewis acid, and cyclohexene, having a bond, is the Lewis base.
Drawing the Product of the Reaction of HCl with Cyclohexene
•The electron pair in the bond of the Lewis base forms a new bond to the proton of the Lewis acid, generating a carbocation.
•The H-Cl bond must break, giving its two electrons to Cl, forming Cl¯.
•Because two electron pairs are involved, two curved arrows are needed.
• Most organic molecules contain a carbon backbone consisting of C-C and C-H bonds to which functional groups are attached.
• A functional group is an atom or a group of atoms with characteristic chemical and physical properties.
• Structural features of a functional group include: • Heteroatoms—atoms other than carbon or hydrogen. • Bonds most commonly occur in C-C and C-O double
• Functional groups are important!
• Functional groups distinguish one organic molecule from another.
• They determine a molecule's: • geometry • physical properties • reactivity
Reactivity of Functional Groups
•Heteroatoms and bonds confer reactivity on a particular molecule.
• Heteroatoms have lone pairs and create electron-deficient sites on carbon.
• A pi bond makes a molecule a base and a nucleophile,
and is easily broken in chemical reactions.
• Ethane belongs to a large group of organic compounds: Hydrocarbons
• Hydrocarbons are compounds made up of only the elements carbon and hydrogen. They may be aliphatic or aromatic.
• Aromatic hydrocarbons are so named because many of the earliest known aromatic compounds had strong, characteristic odors.
• The simplest aromatic hydrocarbon is benzene.
• The six-membered ring and three pi bonds of benzene comprise a single functional group, found in most aromatic compounds.
Functional Groups with Carbon-Heteroatom (C-Z) sigma bonds
• Several types of functional groups contain C-Z sigma bonds.
• The electronegative heteroatom Z creates a polar bond, making carbon electron deficient.
Functional Groups with C=O Group
• This group is called a "carbonyl group".
• The polar C-O bond makes the carbonyl carbon an electrophile, while the lone pairs on O allow it to react as a nucleophile and base.
• The carbonyl group also contains a pi bond that is more easily broken than a C-O sigma bond.
Importance of Functional Groups
A functional group determines all of the following properties of a molecule:
• bonding and shape • type and strength of intermolecular forces • physical properties • nomenclature • chemical reactivity
•Intermolecular forces are interactions that exist between molecules.
•Functional groups determine the type and strength of these interactions.
•Ionic and Covalent compounds have very different intermolecular interactions.
•Ionic compounds contain oppositely charged particles held together by extremely strong electrostatic interactions.
•These ionic interactions are much stronger than the intermolecular forces present between covalent molecules.
Intermolecular Forces in Covalent Molecules
• Covalent compounds are composed of discrete molecules.
• The nature of the forces between molecules depends on the functional group present.
• There are three different types of interactions, shown below in order of increasing strength:
van der Waals forces dipole-dipole interactions hydrogen bonding
van der Waals Forces
• van der Waals forces are also known as London forces.
• They are weak interactions caused by momentary changes in electron density in a molecule.
• They are the only attractive forces present in nonpolar compounds.
van der Waals Forces and Surface Area
•All compounds exhibit van der Waals forces.
•The larger the surface area of a molecule, the larger the attractive force between two molecules, and the stronger the intermolecular forces.
van der Waals Forces and Polarizability
•Polarizability is a measure of how the electron cloud around an atom responds to changes in its electronic environment. Larger atoms, like iodine, which have more loosely held valence electrons, are more polarizable than smaller atoms like fluorine, which have more tightly held electrons.
• Dipole-dipole interactions are the attractive forces between the permanent dipoles of two polar molecules.
• The dipoles in adjacent molecules (e.g., acetone below) align so that the partial positive and partial negative charges are in close proximity.
• These attractive forces caused by permanent dipoles are much stronger than weak van der Waals forces.
•Hydrogen bonding typically occurs when a hydrogen atom bonded to O, N, or F, is electrostatically attracted to a lone pair of electrons on an O, N, or F atom in another molecule.
•Hydrogen bonding is the strongest of the three types of intermolecular forces.
• The boiling point of a compound is the temperature at which liquid molecules are converted into gas.
• In boiling, energy is needed to overcome the attractive forces in the more ordered liquid state.
• The stronger the intermolecular forces, the higher the boiling point.
Boiling Point and Intermolecular Forces
• The relative strength of the intermolecular forces increases from pentane to butanal to 1-butanol.
• The boiling points of these compounds increase in the same order.
Other Factors Affecting Boiling Points
• For compounds with similar functional groups:
•The larger the surface area, the higher the boiling point.
•The more polarizable the atoms, the higher the boiling
• The melting point is the temperature at which a solid is converted to its liquid phase.
• In melting, energy is needed to overcome the attractive forces in the more ordered crystalline solid.
• The stronger the intermolecular forces, the higher the melting point.
For two compounds with similar functional groups, the more symmetrical the compound, the higher the melting point.
Melting Point Trends
• For covalent molecules of approximately the same molecular weight, the melting point depends upon the identity of the functional group.
• The stronger the intermolecular attraction, the higher the melting points (the same is true for boiling points).
Effect of Symmetry on Melting Points
• For compounds having the same functional group and similar molecular weights, the more compact and symmetrical the shape, the higher the melting point.
• A compact symmetrical molecule like neopentane packs well into a crystalline lattice whereas isopentane, which has a CH3 group dangling from a four-carbon chain, does not.
• Neopentane has a much higher melting point than isopentane
• Solubility is the extent to which a compound, called a solute, dissolves in a liquid, called a solvent.
• The energy needed to break up the interactions between the molecules or ions of the solute comes from new interactions between the solute and the solvent.
• "Like dissolves like."
• Compounds dissolve in solvents having similar
kinds of intermolecular forces.
• Polar compounds dissolve in polar solvents like water or alcohols capable of hydrogen bonding with the solute.
• Nonpolar or weakly polar compounds dissolve in: • nonpolar solvents (e.g., carbon tetrachloride [CCl4]
and hexane [CH3(CH2)4CH3]).
• weakly polar solvents (e.g., diethyl ether [CH3CH2OCH2CH3]).
Solubility of Ionic Compounds
• Most ionic compounds are soluble in water, but insoluble in organic solvents.
• To dissolve an ionic compound, the strong ion-ion interactions must be replaced by many weaker ion-dipole interactions.
Solubility of Organic Molecules
• An organic compound is water soluble only if it contains one polar functional group capable of hydrogen bonding with the solvent for every five carbon atoms it contains.
• For example, compare the solubility of butane and acetone in H2O and CCl4.
Water Solubility of Organic Molecules
• The size of an organic molecule with a polar functional group determines its water solubility.
• A low molecular weight alcohol like ethanol is water soluble.
• Cholesterol, with 27 carbon atoms and only one OH group, has a carbon skeleton that is too large for the OH group to solubilize by hydrogen bonding.
• Therefore, cholesterol is insoluble in water
Hydrophobic and Hydrophilic
• The nonpolar part of a molecule that is not attracted to H2O is said to be hydrophobic.
• The polar part of a molecule that can hydrogen bond to H2O is said to be hydrophilic.
• In cholesterol, for example, the hydroxy group is hydrophilic, whereas the carbon skeleton is hydrophobic.
• Soap molecules have two distinct parts: • There is a hydrophilic portion composed of ions called
the polar head.
• There is a hydrophobic carbon chain of nonpolar C-C and C-H bonds, called the nonpolar tail.
Structure of the Cell Membrane
• Phospholipids contain an ionic or polar head, and two long nonpolar hydrocarbon tails.
• In an aqueous environment, phospholipids form a lipid bilayer, with the polar heads oriented toward the aqueous exterior and the nonpolar tails forming a hydrophobic interior.
Functional Groups and Reactivity
• All functional groups contain a heteroatom, a pi bond or both.
• These features create electrophilic sites and nucleophilic sites in a molecule.
• Electron-rich sites (nucleophiles) react with electron poor sites (electrophiles).
• An electronegative heteroatom like N, O, or X makes a carbon atom electrophilic as shown below.
Nucleophilic Sites in Molecules
• A lone pair on a heteroatom makes it basic and nucleophili
• pi bonds create nucleophilic sites and are more easily broken than sigma bonds.
Reaction of pi Bonds with Electrophiles
•An electron-rich carbon reacts with an electrophile, symbolized as E+.
•For example, alkenes contain an electron-rich double bond, and so they react with electrophiles E+.
Reaction of Nucleophiles with Electrophiles
Alkyl halides possess an electrophilic carbon atom, so they react with electron-rich nucleophiles.
Acyclic Alkanes and Cycloalkanes
• Alkanes are aliphatic hydrocarbons having only C-C and C-H bonds. They can be categorized as acyclic or cyclic.
• They are also called saturated hydrocarbons because they have the maximum number of hydrogen atoms per carbon.
• Acyclic alkanes have the molecular formula CnH2n+2 (where n = an integer) and contain only linear and branched chains of carbon atoms.
• Cycloalkanes contain carbons joined in one or more rings.
• Because their general formula is CnH2n, they have two fewer H atoms than an acyclic alkane with the same number of carbons.
Tetrahedral Geometry of Carbon
• All C atoms in an alkane are surrounded by four groups, making them sp3 hybridized and tetrahedral, with all bond angles of 109.5°.
• The 3-D representations and ball-and-stick models for these alkanes indicate the tetrahedral geometry around each C atom.
• In contrast, the Lewis structures are not meant to imply any 3-D arrangement.
• The three-carbon alkane CH3CH2CH3, called propane, has a molecular formula C3H8.
• In the 3-D drawing that each C atom has two bonds in the plane (solid lines), one bond in front (on a wedge) and one bond behind the plane (on a dashed line).
Equivalent Structures of Propane
• For propane and higher molecular weight alkanes, the carbon skeleton can be drawn in a variety of ways and still represent the same molecule.
• For example, the three carbons of propane can be drawn in a horizontal row or with a bend.
• In a Lewis structure, the bends in a carbon chain do not matter.
• There are two different ways to arrange four carbons, giving two compounds with molecular formula C4H10, named butane and isobutane.
• Butane and isobutane are constitutional isomers—two different compounds with the same molecular formula.
• Constitutional isomers (also called structural isomers) differ in the way the atoms are connected to each other.
Classification of Carbon Atoms
• Carbon atoms in alkanes and other organic compounds are classified by the number of other carbons directly bonded to them.
• A primary (1o) carbon is bonded to one other C atom. • A secondary (2o) carbon is bonded to two other C atoms. • A tertiary (3o) carbon is bonded to three other C atoms. • A quaternary (4o) carbon is bonded to four other C atoms.
Classification of Hydrogen Atoms
• Hydrogen atoms are classified depending on the type of carbon atom to which they are bonded.
• A primary (1°) hydrogen is on a C bonded to one other
• A secondary (2°) hydrogen is on a C bonded to two other C atoms.
• A tertiary (3°) hydrogen is on a C bonded to three other C atoms.
• Cycloalkanes have molecular formula CnH2n and contain carbon atoms arranged in a ring.
• Simple cycloalkanes are named by adding the prefix cyclo-
to the name of the acyclic alkane having the same number of carbons.
• Carbon substituents bonded to a long carbon chain are called alkyl groups.
• An alkyl group is formed by removing one H atom from an alkane.
• To name an alkyl group, change the -ane ending of the parent alkane to -yl.
• Thus, methane (CH4) becomes methyl (CH3-) and ethane (CH3CH3) becomes ethyl (CH3CH2-).
Naming Three Carbon Alkyl Groups
• Naming three- or four-carbon alkyl groups is more complicated because the parent hydrocarbons have more than one type of hydrogen atom.
• For example, propane has both 1° and 2° H atoms, and removal of each of these H atoms forms a different alkyl group with a different name, propyl or isopropyl.
Conformations of Acyclic Alkanes
Conformations are different arrangements of atoms that are interconverted by rotation about single bonds.
Eclipsed and Staggered Conformations
• Names are given to two different conformations.
• In the eclipsed conformation, the C-H bonds on one carbon are directly aligned with the C-H bonds on the adjacent carbon.
• In the staggered conformation, the C-H bonds on one carbon bisect the H-C-H bond angle on the adjacent carbon.
Conformations and Dihedral Angle
• Rotating the atoms on one carbon by 60° converts an eclipsed conformation into a staggered conformation, and vice versa.
• The angle that separates a bond on one atom from a bond on an adjacent atom is called a dihedral angle.
• For ethane in the staggered conformation, the dihedral angle for the C-H bonds is 60°; for eclipsed ethane, it is 0°
HOW TO Draw a Newman Projection
• End-on representations for conformations are commonly drawn using a convention called a Newman projection.
HOW TO Draw a Newman Projection:
Step : Look directly down the C-C bond (end-on), and draw a circle with a dot in the center to represent the carbons of the C-C bond.
Step : Draw in the bonds.
• Draw the bonds on the front C as three lines meeting at the center of the circle.
• Draw the bonds on the back C as three lines coming out of the edge of the circle. Step : Add the atoms on each bond.
Conformations of Ethane
• The staggered and eclipsed conformations of ethane interconvert at room temperature.
• The staggered conformations are more stable (lower in energy) than the eclipsed conformations.
• Electron-electron repulsion between bonds in the eclipsed conformation increases its energy compared with the staggered conformation, where the bonding electrons are farther apart.
Torsional Energy of Ethane
• The difference in energy between staggered and eclipsed conformers is ~3 kcal/mol, with each eclipsed C-H bond contributing 1 kcal/mol.
• The energy difference between staggered and eclipsed conformers is called torsional energy.
• Torsional strain is an increase in energy caused by eclipsing interactions.
Newman Projections - Butane
• An energy minimum and maximum occur every 60°as the conformation changes from staggered to eclipsed.
• Conformations that are neither staggered nor eclipsed are intermediate in energy.
• Butane and higher molecular weight alkanes have several C-C bonds, all capable of rotation.
Anti and Gauche Conformations
• A staggered conformation with two larger groups 180°from each other is called anti.
• A staggered conformation with two larger groups 60°from each other is called gauche.
• The staggered conformations are lower in energy than the eclipsed conformations.
• The relative energies of the individual staggered conformations depend on their steric strain.
• Steric strain is an increase in energy resulting when non-bonded atoms are forced too close to one another.
• Gauche conformations are generally higher in energy than anti conformations because of steric strain.
Conformation and Energy of
• The energy difference between the lowest and highest energy conformations is called a barrier to rotation.
Zigzag Skeletal Structures
• Since the lowest energy conformation has all bonds staggered and all large groups anti, alkanes are often drawn in zigzag skeletal structures to indicate this.
Angle Strain in Cycloalkanes
• Besides torsional strain and steric strain, the conformations of cycloalkanes are also affected by angle strain.
• Angle strain is an increase in energy when bond angles deviate from the optimum tetrahedral angle of 109.5°.
• Cycloalkanes with more than three C atoms in the ring are not flat molecules. They are puckered to reduce strain.
Axial and Equatorial Positions
• Each C in cyclohexane has two different kinds of hydrogens: (1) Axial hydrogens are located above and below the ring
(along a perpendicular axis).
(2) Equatorial hydrogens are located in the plane of the
ring (around the equator).
Conformational Change -
• Cyclohexanes undergo a conformational change called "ring-
• As a result of a ring flip, the up carbons become down
carbons, and the down carbons become up carbons.
• Axial and equatorial H atoms are also interconverted during a ring-flip; axial H atoms become equatorial H atoms, and equatorial H atoms become axial H atoms.
Boat Conformation of Cyclohexane
• Cyclohexane also can exist in a boat conformation.
• The boat forms of cyclohexane are 7 kcal/mol less stable than the chair forms.
• The boat conformation is destabilized by torsional strain because the hydrogens on the four carbon atoms in the plane are eclipsed.
• Additionally, there is steric strain because two hydrogens at either end of the boat, the "flag pole"hydrogens, are forced close to each other.
Chair Conformations and Energy
The two chair conformations of cyclohexane are different, so they are not equally stable.
• Larger axial substituents create destabilizing (and thus unfavorable) 1,3-diaxial interactions.
• In methylcyclohexane, each unfavorable H,CH3 interaction destabilizes the conformation by 0.9 kcal/mol, so Conformation 2 is 1.8 kcal/mol less stable than Conformation 1.
Preference of Equatorial Position in Substituted Cyclohexanes
• The larger the substituent on the six-membered ring, the higher the percentage of the equatorial conformation at equilibrium.
• With a very large substituent like tert-butyl [(CH3)3C-], essentially none of the conformation containing an axial tert-butyl group is present at room temperature
• There are two different 1,2-dimethylcyclopentanes—one having two CH3 groups on the same side of the ring and one having them on opposite sides of the ring.
• A and B are stereoisomers.
• A disubstituted cyclohexane, such as 1,4-dimethylcyclo-hexane, also has cis and trans stereoisomers.
• Each of these stereoisomers has two possible chair conformations.
Cis and Trans Stereoisomers
• Stereoisomers are isomers that differ only in the way the atoms are oriented in space.
• The prefixes cis and trans are used to distinguish these isomers.
• The cis isomer has two groups on the same side of the ring.
• The trans isomer has two groups on opposite sides of the ring
Trans Disubstituted Cycloalkanes
• Conformations 1 and 2 are not equally stable.
• Because conformation 2 has both CH3 groups in the roomier equatorial position, it is lower in energy.
• A trans isomer has two substituents on opposite sides, one up and one down.
• Whether substituents are axial or equatorial depends on the relative location of the two substituents (on carbons 1,2-, 1,3-, or 1,4-).
Cis Disubstituted Cycloalkanes
• A cis isomer has two groups on the same side of the ring, either both up or both down.
• In this example, Conformations 1 and 2 have two CH3 groups drawn up.
• Both conformations have one CH3 group axial and one equatorial, making them equally stable.
Oxidation and Reduction Reactions
• Oxidation results in an increase in the number of C-Z bonds. • Oxidation results in a decrease in the number of C-H bonds. • Reduction results in a decrease in the number of C-Z bonds. • Reduction results in an increase in the number of C-H bonds.
Combustion of Alkanes
• Alkanes undergo combustion—that is, they burn in the presence of oxygen to form carbon dioxide and water.
• This is an example of an oxidation-reduction reaction. Every C-H and C-C bond in the starting material is converted to a C-O bond in the product.
• What kind of reactions are these?
• Stereochemistry refers to the three-dimensional structure of a molecule.
• As a consequence of stereochemistry, apparently minor differences in 3-D structure can result in vastly different properties.
• We can observe this by considering starch and cellulose, which are both composed of the same repeating unit.
Stereochemistry of Starch and Cellulose
• In cellulose, the O atom joins two rings using equatorial bonds.
• In starch, the O atom joins two rings using one equatorial and one axial bond.
• Due to these differences in stereochemistry, humans can metabolize starch for energy but we cannot digest cellulose.
Constitutional Isomers in general
• Isomers are different compounds with the same molecular formula.
• The two major classes of isomers are constitutional isomers and stereoisomers.
• Constitutional/structural isomers have:
• different IUPAC names • same or different functional groups • different physical properties • different chemical properties
•Differ only in the way the atoms are oriented in space.
•Have identical IUPAC names (except for a prefix like cis or trans).
•Always have the same functional group(s).
•Differ in configuration (its particular three-dimensional arrangement).
Nonsuperimposable (Chiral) Objects
• Although everything has a mirror image, mirror images may or may not be superimposable.
• Some molecules are like hands. Left and right hands are mirror images, but they are not identical, or superimposable.
• A molecule (or object) that is not superimposable on its mirror image is said to be chiral
Superimposable (Achiral) Objects
• Other molecules are like socks. • Two socks from a pair are mirror images that are
superimposable. • A sock and its mirror image are identical.
• A molecule or object that is superimposable on its mirror image is said to be achiral.
• We can now consider several molecules to determine whether or not they are chiral.
• The molecule labeled A and its mirror image labeled B are not superimposable.
• No matter how you rotate A and B, all the atoms never align. • Thus, CHBrClF is a chiral molecule, and A and B are different
• A and B are stereoisomers—specifically, they are enantiomers.
• A carbon atom with four different groups is a tetrahedral stereogenic center.
In general, a molecule with no stereogenic centers will not be chiral (exceptions to this will be considered in Chapter 17).
• With one stereogenic center, a molecule will always be chiral.
• With two or more stereogenic centers, a molecule may or may not be chiral.
• To locate a stereogenic center, examine each tetrahedral carbon atom in a molecule, and look at the four groups—not the four atoms—bonded to it.
• Always omit from consideration all C atoms that cannot be tetrahedral stereogenic centers. These include:
• CH2 and CH3 groups • Any sp or sp2 hybridized C
Planes of Symmetry
• A plane of symmetry is a mirror plane that cuts the molecule in half, so that one half of the molecule is a reflection of the other half.
• Achiral molecules usually contain a plane of symmetry but chiral molecules do not.
Summary of Chirality
• Everything has a mirror image.
• The fundamental question is whether the molecule and its mirror image are superimposable.
• If not, they are chiral and do not contain a plane of
• If they are superimposable, they are achiral and will contain a plane of symmetry.
• The terms stereogenic center and chiral molecule are related but distinct.
• A chiral molecule must have one or more stereogenic centers.
Multiple Stereogenic Centers
• Larger organic molecules can have two, three, or even hundreds of stereogenic centers.
• To draw both enantiomers of a chiral compound such as 2-butanol, use the typical convention for depicting a tetrahedron.
• To form the first enantiomer, arbitrarily place the four groups—H, OH, CH3 and CH2CH3—on any bond to the stereogenic center.
• Then draw the mirror image
Stereogenic Centers on Rings
• Stereogenic centers may also occur at carbon atoms that are part of a ring.
• To find stereogenic centers on ring carbons, always draw the rings as flat polygons, and look for tetrahedral carbons that are bonded to four different groups.
R or S Stereogenic Centers
• Since enantiomers are two different compounds, they need to be distinguished by name.
• This is done by adding the prefix R or S to the IUPAC name of the enantiomer.
• To designate enantiomers as R or S, priorities must be assigned to each group bonded to the stereogenic center, in order of decreasing atomic number.
• The atom of highest atomic number gets the highest priority (1).
Assigning Priority for R and S
• If two atoms on a stereogenic center are the same, assign priority based on the atomic number of the atoms bonded to these atoms.
• One atom of higher atomic number determines the higher priority.
Assigning Priority for R and S-
• If two isotopes are bonded to the stereogenic center, assign priorities in order of decreasing mass number.
• Thus, in comparing the three isotopes of hydrogen, the order of priorities is:
Assigning Priority for R and S-
• To assign a priority to an atom that is part of a multiple bond, treat a multiply bonded atom as an equivalent number of singly bonded atoms.
• For example, the C of a C=O is considered to be bonded to two O atoms.
Orienting the Lowest Priority Group in Back
• If the lowest priority group is not facing towards back, rotate the molecule 120o around a stationary bond axis.
Finding All Possible Stereocenters
For a molecule with n stereogenic centers, the maximum number of stereoisomers is 2n.
• After drawing the compound and the mirror image, place B directly on top of A; and rotate B 180°and place it on top of A to see if the atoms align.
• In this case, the atoms of A and B do not align, making A and B nonsuperimposable mirror images—i.e., enantiomers.
• A and B are two of the four possible stereoisomers of 2,3-dibromopentane.
• Switching the positions of H and Br (or any two groups) on one stereogenic center of either A or B forms a new stereoisomer (labeled C in this example), which is different from A and B.
• The mirror imageof C is labeled D. • C and D are enantiomers. • Stereoisomers that are not mirror images of one another are called
diastereomers. • A and C are diastereomers.
• Compound C contains a plane of symmetry, and is achiral.
• A meso compound is an achiral compound that contains tetrahedral stereogenic centers. C is a meso compound.
• Meso compounds generally contain a plane of symmetry so that they possess two identical halves.
• Because one stereoisomer of 2,3-dibromobutane is superimposable on its mirror image, there are only three stereoisomers, not four.
R and S Assignments in Compounds with Two or More Stereogenic Centers
• When a compound has more than one stereogenic center, R
and S configurations must be assigned to each of them.
• Identical compounds have the same R,S designations at every tetrahedral stereogenic center.
• Enantiomers have exactly opposite R,S designations.
• Diastereomers have the same R,S designation for at least one stereogenic center and the opposite for at least one of the other stereogenic centers.
• The chemical and physical properties of two enantiomers are identical except in their interaction with chiral substances.
• They have identical physical properties, except for how they interact with plane-polarized light.
• Plane-polarized (polarized) light is light that has an electric vector that oscillates in a single plane.
• Plane-polarized light arises from passing ordinary light through a polarizer.
• A polarimeter is an instrument that allows polarized light to travel through a sample tube containing an organic compound and permits the measurement of the degree to which an organic compound rotates plane-polarized light.
• With achiral compounds, the light that exits the sample tube remains unchanged.
• A compound that does not change the plane of polarized light is said to be optically inactive.
Rotation of Plane-Polarized Light
• With chiral compounds, the plane of the polarized light is rotated through an angle .
• The angle is measured in degrees (°), and is called the observed rotation.
• A compound that rotates polarized light is said to be optically active.
Optical Activity Summary
• The rotation of polarized light can be clockwise or anticlockwise.
• If the rotation is clockwise, the compound is called dextrorotatory. The rotation is labeled d or (+).
• If the rotation is counterclockwise, the compound is called levorotatory. The rotation is labeled l or (-).
• Two enantiomers rotate plane-polarized light to an equal extent but in opposite directions. (e.g., if enantiomer A rotates polarized light +5°, the same concentration of enantiomer B rotates it -5°).
• No relationship exists between R and S prefixes and the (+) and (-) designations that indicate optical rotation.
• An equal amount of two enantiomers is called a racemic mixture or a racemate.
• A racemic mixture is optically inactive.
• Because two enantiomers rotate plane-polarized light to an equal extent but in opposite directions, the rotations cancel, and no rotation is observed.
Physical Properties of Racemic Mixtures
• Enantiomers have the same physical properties except for their behavior with plane polarized light.
• The physical properties of a racemic mixture can vary from those of the individual enantiomers.
• Specific rotation is a standardized physical constant for the amount that a chiral compound rotates plane-polarized light.
• Specific rotation is denoted by the symbol  and defined using a specific sample tube length (l, in dm), concentration (c in g/mL), temperature (25°C) and wavelength (589 nm).
• Enantiomeric excess (optical purity) is a measurement of how much one enantiomer is present in excess of the racemic mixture.
• It is denoted by the symbol ee. ee = % of one enantiomer - % of the other enantiomer.
•Consider the following example—If a mixture contains 75% of one enantiomer and 25% of the other, the enantiomeric excess is 75% - 25% = 50%.
•Thus, there is a 50% excess of one enantiomer over the racemic mixture.
• Generally, one does not know the % of each enantiomer but you can measure the specific rotation of the mixture and compare it to the specific rotation of a pure enantiomer.
• The enantiomeric excess can also be calculated if the specific rotation  of a mixture and the specific rotation  of a pure enantiomer are known.
ee = ([a] mixture/[a] pure enantiomer) x 100%.
Physical Properties of Stereoisomers
• Since enantiomers have identical physical properties, they cannot be separated by common physicaltechniques like distillation.
• Diastereomers and constitutional isomers have different physical properties, and therefore can be separated by common techniques
Chemical Properties of Enantiomers
•Two enantiomers have exactly the same chemical properties except for their reaction with chiral, non-racemic reagents.
•Many drugs are chiral and often must react with a chiral receptor or chiral enzyme to be effective.
•One enantiomer of a drug may effectively treat a disease whereas its mirror image may be ineffective or toxic.
Enantiomers and the Sense of Smell
•Research suggests that the odor of a particular molecule is determined more by its shape than by the presence of a particular functional group.
•Because enantiomers interact with chiral smell receptors, some enantiomers have different odors.
• Reactions are at the heart of Organic Chemistry.
• Virtually all chemical reactions are woven together by a few basic themes.
• Begin by looking for electron-rich or deficient sites at functional groups in the reacting molecules.
• These are often the location of bonds that might be easily broken.
• Learn about how reaction takes place (i.e., does it occur in one step or in a series of steps).
Writing Equations for Organic Reactions
• Equations for organic reactions are usually drawn with a single reaction arrow () between the starting material and product.
• The reagent, the chemical substance with which an organic compound reacts, is sometimes drawn on the left side of the equation with the other reactants.
• At other times, the reagent is drawn above the arrow.
• Although the solvent is often omitted from the equation, most organic reactions take place in liquid solvent.
• The solvent and temperature of the reaction may be added above or below the arrow.
Writing Equations for Sequential Reactions
• When two sequential reactions are carried out without drawing any intermediate compound, the steps are usually numbered above or below the reaction arrow.
• This convention signifies that the first step occurs before the second step, and the reagents are added in sequence, not at the same time
• A substitution is a reaction in which an atom or a group of atoms is replaced by another atom or group of atoms.
• In a general substitution, Y replaces Z on a carbon atom.
• Substitution reactions involve sigma bonds: one sigma bond breaks and another forms at the same carbon atom.
• While in some cases Z can be a hydrogen atom, the most common examples of substitution occur when Z is a heteroatom that is more electronegative than carbon.
• An elimination is a reaction in which elements of the starting material are "lost" and a pi bond is formed.
• In an elimination reaction, two groups X and Y are removed from a starting material.
• Two sigma bonds are broken, and a pi bond is formed between adjacent atoms.
• The most common examples of elimination occur when X = H and Y is a heteroatom more electronegative than carbon.
• An addition is a reaction in which elements are added to the starting material.
• In an addition reaction, new groups X and Y are added to the starting material. • A pi bond is broken and two sigma bonds are formed.
Relationship of Addition and Elimination Reactions
• Addition and elimination reactions are exactly opposite.
• A pi bond is formed in elimination reactions, whereas a pi
bond is broken in addition reactions. • Often these reactions are reversible.
• A reaction mechanism is a detailed description of how bonds are broken and formed as starting material is converted into product.
• A reaction can occur either in one step or a series of steps.
• Regardless of how many steps there are in a reaction, there are only two ways to break (cleave) a bond:
• the electrons in the bond can be divided equally
or unequally between the two atoms of the bond.
• Breaking a bond by equally dividing the electrons between the two atoms in the bond is called homolysis or homolytic cleavage.
• Homolysis generates two uncharged species with unpaired electrons.
• A reactive intermediate with a single unpaired electron is called a radical.
• Radicals are highly unstable because they contain an atom that does not have an octet of electrons.
• Breaking a bond by unequally dividing the electrons between the two atoms in the bond is called heterolysis or heterolytic cleavage.
• When two atoms have different electronegativities, the electrons end up on the more electronegative atom.
• A full-headed curved arrow shows the movement of an electron pair.
• Heterolysis generates a carbocation or a carbanion.
• A carbocation is an unstable intermediate containing a carbon surrounded by only six electrons.
• A carbanion is an unstable intermediate having a negative charge on carbon, which is not a very electronegative atom.
• Radicals and carbocations are electrophiles because they contain an electron-deficient carbon.
• Carbanions are nucleophiles because they contain a carbon with a lone pair.
• Bond formation occurs in two different ways. • Two radicals can each donate one electron to form a two-
• Two ions with unlike charges can come together, with the negatively charged ion donating both electrons to form the resulting two-electron bond.
• Bond formation always releases energy.
Bond Dissociation Energy
• Because bond breaking requires energy, bond dissociation energies are always positive numbers, and homolysis is always endothermic.
• Conversely, bond formation always releases energy, and thus is always exothermic.
Energy Associated with the H2 Bond
• As an example, the H-H bond requires +435 kJ/mol to cleave and releases -435 kJ/mol when formed.
• See Table 6.2 for a list of other bond energies.
Bond Dissociation Energy and Bond Strength
• Comparing bond dissociation energies is equivalent to comparing bond strength.
• The stronger the bond, the higher its bond dissociation energy.
• Bond dissociation energies decrease down a column of the periodic table.
• Generally, shorter bonds are stronger bonds
Enthalpy Change in Reactions
• Bond dissociation energies are used to calculate the enthalpy
change (H°) in a reaction in which several bonds are broken and formed
Determining H° for a Reaction
• Consider bonds being broken and formed.
Limitations on Bond Dissociation Energies
• Bond dissociation energies present overall energy changes only; they reveal nothing about the reaction mechanism or how fast a reaction proceeds.
• Bond dissociation energies are determined for reactions in the gas phase, whereas most organic reactions occur in a liquid solvent where solvation energy contributes to the overall enthalpy of a reaction.
• Though imperfect, using bond dissociation energies to
calculate H° gives a useful approximation of the energy changes that occur in a reaction.
Kinetics and Thermodynamics
• For a reaction to be practical, the equilibrium must favor products and the reaction rate must be fast enough to form them in a reasonable time.
• These two conditions depend on thermodynamics and kinetics respectively.
• Thermodynamics describes how the energies of reactants and products compare, and what the relative amounts of
reactants and products are at equilibrium.
• Kinetics describes reaction rates (how fast a reaction occurs).
• The equilibrium constant, Keq, is a mathematical expression that relates the amount of starting material and product at equilibrium.
The Equilibrium Constant
• The size of Keq expresses whether the starting materials or products predominate once equilibrium is reached.
• When Keq > 1, equilibrium favors the products and the equilibrium lies to the right as the equation is written.
• When Keq < 1, equilibrium favors the starting materials and the equilibrium lies to the left as the equation is written.
• For a reaction to be useful, the equilibrium must favor the products, and Keq > 1.
• The position of the equilibrium is determined by the relative energies of the reactants and products.
The Equilibrium Constant and Free Energy
• G° is the overall energy difference between reactants and products.
Relationship Between Equilibrium Constant and Free Energy
•G° is related to the equilibrium constant Keq by the following equation:
•When Keq > 1, log Keq is positive, making G° negative, and energy is released.
•Equilibrium favors the products.
•When Keq < 1, log Keq is negative, making G° positive, and energy is absorbed.
•Equilibrium favors the reactants.
Energy Difference and Equilibrium
• Compounds that are lower in energy have increased stability.
• The equilibrium favors the products when they are more stable (lower in energy) than the starting materials of a reaction.
• Because G° depends on the logarithm of Keq, a small change
in energy corresponds to a large difference in the relative amount of starting material and product at equilibrium.
Conformations and Equilibrium
• Monosubstituted cyclohexanes exist as two different chair conformations that rapidly interconvert at room temperature.
• The conformation having the substituent in the roomier equatorial position is favored.
• Knowing the energy difference between two conformations permits the calculation of the amount of each at equilibrium.
• Entropy is a measure of the randomness of a system. • The more freedom of motion or the more disorder present,
the higher the entropy.
• Gas molecules move more freely than liquid molecules and are higher in entropy.
• Cyclic molecules have more restricted bond rotation than similar acyclic molecules and are lower in entropy.
Role of Entropy in Total Energy Change
• The total energy change is due to two factors: the change in bonding energy and the change in disorder.
• The change in bonding energy can be calculated from bond dissociation energies.
• In most reactions that are not carried out at high temperature, the entropy term (TS°) is small compared to the enthalpy term (H°), and therefore, it is usually neglected.
Reactions in Which S° Plays a Role
• Reactions resulting in increased entropy are favored.
• S° is (+) when the products are more disordered than the reactants and (-) when the products are more ordered.
• Entropy changes are important when:
• the number of molecules of starting material differs from the number of molecules of product in the balanced
• an acyclic molecule is cyclized to a cyclic one, or a cyclic molecule is converted to an acyclic one
•An energy diagram is a schematic representation of the energy changes that take place as reactants are converted to products.
•An energy diagram plots the energy on the y axis versus the progress of reaction, often labeled as the reaction coordinate, on the x axis.
•The energy difference between reactants and products is H°.
•As a chemical reaction proceeds from reactants to products it passes through an unstable energy maximum called the transition state.
•The energy difference between the transition state and the starting material is called the energy of activation, Ea.
• The larger the Ea, the greater the amount of energy that is needed to break bonds, and the slower the reaction rate.
• The structure of the transition state is somewhere between the structures of the starting material and product.
• Any bond that is partially formed or broken is drawn with
a dashed line.
• Any atom that gains or loses a charge contains a partial charge in the transition state.
• Transition states are drawn in brackets, with a superscript double dagger (‡).
Energy Diagrams and Two-Step Reactions
• An energy diagram must be drawn for each step.
• The two energy diagrams must then be combined to form an energy diagram for the overall two-step reaction.
• Each step has its own energy barrier, with a transition state at the energy maximum.
Kinetics and Energy Diagrams
• Kinetics is the study of reaction rates (how fast does the reaction occur).
• Ea is the energy barrier that must be exceeded for reactants to be converted to products.
Factors Affecting Reaction Rates
• The higher the concentration, the faster the rate. • The higher the temperature, the faster the rate.
• G°, H°, and Keq do not determine the rate of a reaction.
• These quantities indicate the direction of the equilibrium and the relative energy of reactants and
• A rate law or rate equation shows the relationship between the reaction rate and the concentration of the reactants.
• It is experimentally determined by measuring the decrease in concentrations of reactants or the appearance of products over time.
• The rate constant k and the energy of activation Ea are inversely related; a high Ea corresponds to a small k.
• A rate equation contains concentration terms for all reactants in a one-step mechanism.
• A rate equation contains concentration terms for only the reactants involved in the rate-determining step in a multistep reaction.
• The order of a rate equation equals the sum of the exponents of the concentration terms in the rate equation.
Rate equations for Two-Step Reactions
• A two-step reaction has a slow rate-determining step, and a fast step.
• In a multistep mechanism, the reaction can occur no faster than its rate-determining step.
• Only the concentration of the reactants in the rate-determining step appears in the rate equation.
• Some reactions do not proceed at a reasonable rate unless a catalyst is added.
• A catalyst is a substance that speeds up the rate of a reaction.
• It is recovered unchanged in a reaction, and it does not
appear in the product.
• A catalyst lowers the activation energy, thus increasing the rate of the catalyzed reaction.
• The energy of the reactants and products is the same in both the catalyzed and uncatalyzed reactions, the position of equilibium is unaffected.
•Enzymes are biochemical catalysts composed of amino acids held together in a very specific three-dimensional shape.
•An enzyme contains a region called its active site which binds an organic reactant, called a substrate.
•The resulting unit is called the enzyme-substrate complex.
•Once bound, the organic substrate undergoes a very specific reaction at an enhanced rate.
•The products are then released. •What effect does an enzyme have on Ea?
• Alkyl halides are organic molecules containing a halogen atom bonded to an sp3 hybridized carbon atom.
• Alkyl halides are classified as primary (1°), secondary (2°), or
tertiary (3°), depending on the number of carbons bonded to the carbon with the halogen atom.
• The halogen atom in halides is often denoted by the symbol "X".
Types of Alkyl Halides
• Other types of organic halides include vinyl halides, aryl halides, allylic halides, and benzylic halides.
• Vinyl halides have a halogen atom (X) bonded to a C-C double bond.
• Aryl halides have a halogen atom bonded to a benzene ring.
• Allylic halides have X bonded to the carbon atom adjacent to a C-C double bond.
• Benzylic halides have X bonded to the carbon atom adjacent to a benzene ring.
The Polar Carbon-Halogen Bond
• The electronegative halogen atom in alkyl halides creates a polar C-X bond, making the carbon atom electron deficient.
• Electrostatic potential maps of four simple alkyl halides illustrate this point.
• This electron deficient carbon is a key site in the reactivity of alkyl halides.
Nucleophiles in Substitution Reactions
• Nucleophiles are Lewis bases that can be negatively charged or neutral.
• Negatively charged nucleophiles like HO¯ and HS¯ are used as salts with Li+, Na+, or K+ counterions to balance the charge.
• Since the identity of the counterion is usually inconsequential, it is often omitted from the chemical equation.
• When a neutral nucleophile is used, the substitution product bears a positive charge.
Drawing Products of Nucleophilic Substitution Reactions
• The overall effect of any nucleophilic substitution is the replacement of the leaving group by the nucleophile.
• To draw any nucleophilic substitution product: • Find the sp3 hybridized carbon with the leaving group.
• Identify the nucleophile, the species with a lone pair or
• Substitute the nucleophile for the leaving group and assign charges (if necessary) to any atom that is involved in bond breaking or bond formation.
The Leaving Group
• In a nucleophilic substitution reaction of R-X, the C-X bond is heterolytically cleaved, and the leaving group departs with the electron pair in that bond, forming X:¯.
• The more stable the leaving group X:¯, the better able it is to accept an electron pair.
• For example, H2O is a better leaving group than HO¯ because H2O is a weaker base.
• The weaker the base, the better the leaving group
• Conjugate bases of weaker acids are poorer leaving groups.
Nucleophiles and Base
• Nucleophiles and bases are structurally similar: both have a lone pair or a bond.
• They differ in what they attack. • Bases attack protons. • Nucleophiles attack other electron-deficient atoms
Nucleophiles vs. Bases
• Although nucleophilicity and basicity are interrelated, they are fundamentally different.
• Basicity is a measure of how readily an atom donates its
electron pair to a proton.
• It is characterized by an equilibrium constant, Ka in an acid-base reaction, making it a thermodynamic property.
• Nucleophilicity is a measure of how readily an atom donates its electron pair to other atoms.
• It is characterized by a rate constant, k, making it a kinetic property.
Nucleophilicityparallels basicity in three instances:
1. For two nucleophiles with the same nucleophilic atom, the
stronger base is the stronger nucleophile.
• The relative nucleophilicity of HO¯ and CH3COO¯, is determined by comparing the pKa values of their conjugate
acids (H2O = 15.7, and CH3COOH= 4.8).
• HO¯ is a stronger base and stronger nucleophile than
2. A negatively charged nucleophile is always a stronger
nucleophile than its conjugateacid. • HO¯ is a stronger base and stronger nucleophile than H2O. 3. Right-to-left across a row of the periodic table, nucleophilicity
increases as basicity increases:
Steric Effects on Nucleophile Strength
• Nucleophilicity does not parallel basicity when steric hindrance becomes important.
• Steric hindrance is a decrease in reactivity resulting from the presence of bulky groups at the site of a reaction.
• Steric hindrance decreases nucleophilicity but not basicity. • Sterically hindered bases that are poor nucleophiles are
called nonnucleophilic bases.
Solvent Effects on Nucleophilicity
• Most organic reactions are performed in a liquid solvent capable of dissolving the reactants, at least to some extent.
• Since substitution reactions involve polar starting materials, polar solvents are used to dissolve them.
• There are two type of polar solvents: protic and aprotic. • Nucleophilicity can be affected by the nature of the solvent.
Solvation by Polar Protic Solvents
• Polar protic solvents solvate both cations and anions well.
• If the salt NaBr is used as a source of the nucleophile Br¯ in H2O:
• The Na+ cations are solvated by ion-dipole interactions with
• The Br¯ anions are solvated by strong hydrogen bonding interactions.
Nucleophilicity in Polar Protic Solvents
• Smaller, more electronegative anions are solvated more strongly, effectively shielding them from reaction.
• In polar protic solvents, nucleophilicity increases down a column of the periodic table as the size of the anion increases.
• This is the opposite of basicity.
Polar Aprotic Solvents
• Polar aprotic solvents also exhibit dipole-dipole interactions, but they have no O-H or N-H bonds.
• They are incapable of hydrogen bonding
Nucleophilicity in Polar Aprotic Solvents
• Polar aprotic solvents solvate cations by ion-dipole interactions.
• Anions are not well solvated because the solvent cannot hydrogen bond to them.
• These anions are said to be "naked" and therefore, more
Nucleophilicity vs. Basicity in Polar Aprotic Solvents
• In polar aprotic solvents, nucleophilicity parallels basicity, and the stronger base is the stronger nucleophile.
• Because basicity decreases as size increases down a column, nucleophilicity decreases as well.
Bond Breaking and Making in Nucleophilic Substitution Mechanisms
• But what is the order of bond making and bond breaking?
• In theory, there are three possibilities. • Bond making and breaking occur at the same time. • Bond breaking occurs first. • Bond making occurs first
Nucleophilic Substitution Mechanisms - Concerted
1. Bond making and bond breaking occur at the same time.
• The mechanism is comprised of one step.
• In such a bimolecular reaction, the rate depends upon the concentration of both reactants.
• The rate equation is second order.
Nucleophilic Substitution Mechanisms - Bond Breaking First
2. Bond breaking occurs before bond making.
• The mechanism has two steps and a carbocation is formed as an intermediate.
• The first step is rate-determining. • The rate depends on the concentration of RX only. • The rate equation is first order.
Nucleophilic Substitution Mechanisms- Bond Making First
3. Bond making occurs before bond breaking.
• This mechanism has an inherent problem.
• The intermediate generated in the first step has 10 electrons around carbon, violating the octet rule.
• Because two other mechanistic possibilities do not violate a fundamental rule, this last possibility can be disregarded.
Kinetics and Mechanisms
Kinetic data show that the rate of reaction 1 depends on the concentration of both reactants, which suggests a bimolecular reaction with a one-step mechanism.
• This is an example of an SN2 (substitution nucleophilic bimolecular) mechanism.
Consider reaction 2 below:
• Kinetic data show that the rate of reaction 2 depends on the concentration of only the alkyl halide.
• This suggests a two-step mechanism in which the rate-determining step involves the alkyl halide only.
• This is an example of an SN1 (substitution nucleophilic unimolecular) mechanism.
SN2 Reaction Mechanism
• The mechanism of an SN2 reaction would be drawn as follows.
• Curved arrow notation is used to show the flow of electrons.
Transition States of SN2 Reactions
• The transition state always has partial bonds to the nucleophile and the leaving group.
• There can also be partial charges on the nucleophile and/or leaving group.
Inversion in SN2 Reactions
• The bond to the nucleophile in the product is always on the opposite side relative to the bond of the leaving group in the starting material.
• This is called backside attack.
• This results in inversion of configuration at a stereogenic center.
Substrate Reactivity in SN2
• As the number of R groups on the carbon with the leaving group increases, the rate of an SN2 reaction decreases.
• Methyl and 1° alkyl halides undergo SN2 reactions with ease. • 2° Alkyl halides react more slowly. • 3° Alkyl halides do not undergo SN2 reactions due to steric
• Bulky R groups near the reaction site make nucleophilic attack from the backside more difficult, slowing the reaction rate.
Energy Diagrams for SN2 Reactions
• The higher the Ea, the slower the reaction rate.
• Thus, any factor that increases Ea decreases the reaction rate.
SN1 Reaction Mechanism
• The mechanism of an SN1 reaction would be drawn as follows: Note the curved arrow formalism that is used to show the flow of electrons.
• Key features of the SN1 mechanism are that it has two steps, and carbocations are formed as reactive intermediates.
Stereochemistry of SN1 Reactions
To understand the stereochemistry of the SN1 reaction, we must examine the geometry of the carbocation intermediate.
Racemization in SN1 Reactions
• Loss of the leaving group in Step  generates a planar carbocation that is achiral.
• In Step , attack of the nucleophile can occur on either side to afford two products which are a pair of enantiomers.
• Because there is no preference for nucleophilic attack from either direction, an equal amount of the two enantiomers is formed—a racemic mixture.
• This process is called racemization.
Substrate Reactivity in SN1 Reactions
• The rate of an SN1 reaction is affected by the type of alkyl halide involved.
• The effect of the type of alkyl halide on SN1 reaction rates can be explained by considering carbocation stability.
• Carbocations are classified as primary (1°), secondary (2°), or tertiary (3°), based on the number of R groups bonded to the charged carbon atom.
• As the number of R groups increases, carbocation stability increases.
Inductive Effects and Carbocation Stability
• The order of carbocation stability can be rationalized through inductive effects and hyperconjugation.
• Inductive effects occur by the pull of electron density through bonds caused by electronegativity differences between atoms.
• Alkyl groups are electron donor groups that stabilize a positive charge because they contain several bonds, each containing electron density.
• As a result, alkyl groups are more polarizable than a hydrogen atom, and better able to donate electron density.
• In general, the more alkyl groups attached to a carbon with a positive charge, the more stable the cation will be.
Hyperconjugation and Carbocation Stability
• The order of carbocation stability is also a consequence of hyperconjugation.
• Hyperconjugation is the spreading out of charge by the overlap of an empty p orbital with an adjacent bond.
• This overlap delocalizes the positive charge on the carbocation over a larger volume, thus stabilizing it.
• For example: (CH3)2CH+ can be stabilized by hyperconjugation, but CH3
+ cannot .
The Hammond Postulate
• The Hammond postulate relates reaction rate to stability.
• It provides a qualitative estimate of the energy of a transition state.
• The Hammond postulate states that the transition state of a reaction resembles the structure of the species (reactant or product) to which it is closer in energy.
Endothermic Reaction Transition States
• In an endothermic reaction, the transition state resembles the products more than the reactants, so anything that stabilizes the product stabilizes the transition state also.
• Thus, lowering the energy of the transition state decreases Ea, which increases the reaction rate.
• If there are two possible products of different stability in an endothermic reaction, the transition state leading to the more stable product is lower in energy, so this reaction should occur faster.
Exothermic Reaction Transition States
• In the case of an exothermic reaction, the transition state resembles the reactants more than the products.
• Thus, lowering the energy of the products has little or no effect on the energy of the transition state.
• Since Ea is unaffected, the reaction rate is unaffected.
• The conclusion is that in an exothermic reaction, the more stable product may or may not form faster, since Ea is similar for both products.
Application of the Hammond Postulate to the SN1 Reaction
• In the SN1 reaction, the rate determining step is the formation of the carbocation, an endothermic process.
• According to the Hammond postulate, the stability of the carbocation determines the rate of its formation.
Predicting the Mechanism of Nucleophilic Substitutions Reactions
Predicting the Mechanism of Nucleophilic Substitutions Reactions
Nature of the Alkyl Halide
• The most important factor is the identity of the alkyl halide.
Effect of the Nucleophile
Strong nucleophiles (which usually bear a negative charge) present in high concentrations favor SN2 reactions.
• Weak nucleophiles, such as H2O and ROH favor SN1 reactions by decreasing the rate of any competing SN2 reaction.
• Consider what happens when the 2° alkyl halide A, which can react by either mechanism, is treated with either the strong nucleophile HO¯ or the weak nucleophile H2O.
Predicting SN1 or an SN2 -
• The strong nucleophile favors an SN2 mechanism
• The weak nucleophile favors an SN1 mechanism.
Effect of Leaving Groups
• A better leaving group increases the rate of both SN1 and SN2 reactions
Effect of Solvent on mechanisms
• Polar protic solvents like H2O and ROH favor SN1 reactions because the ionic intermediates (both cations and anions) are stabilized by solvation.
• Polar aprotic solvents favor SN2 reactions because nucleophiles are not well solvated, and therefore, are more nucleophilic.
Vinyl and Aryl Halides
• SN1 or SN2 reactions occur on sp3 hybridized carbons.
• Vinyl and aryl halides, which have a halogen attached to a sp2 hybridized carbon, do not undergo SN1 or SN2 reactions.
• Heterolysis of the C-X bond would form a highly unstable vinyl or aryl cation.
• Organic synthesis is the systematic preparation of a compound from a readily available starting material by one or many steps.
• Organic synthesis has produced many useful compounds (e.g., pharmaceuticals, pesticides, and polymers used in everyday life).
• Chemists may rely on synthesis to prepare useful substances such as a natural product produced by organisms, but in only minute amounts (e.g., Taxol used in cancer treatment).
• Nucleophilic substitution reactions, especially SN2, are used to introduce a wide variety of functional groups into a molecule, depending on the nucleophile.
Thinking Backwards in Organic Synthesis
• To carry out the synthesis of a particular compound, we must think backwards, and ask ourselves the question:
"What starting material and reagents are needed to make it?"
• If a nucleophilic substitution is being used, determine what alkyl halide and what nucleophile can be used to form a specific product.
Approaches Used in Organic Synthesis
• To determine the two components needed for synthesis, remember that the carbon atoms come from the organic starting material, in this case, a 1° alkyl halide.
• The functional group comes from the nucleophile, HO¯ in this case.
• With these two components, we can "fill in the boxes" to complete the synthesis.
General Features of Elimination
• Elimination reactions involve the loss of elements from the starting material to form a new pi bond in the product.
Elimination of HX
• In both example reactions a base removes the elements of an acid, HX, from the organic starting material.
• Removal of the elements HX is called dehydrohalogenation.
• Dehydrohalogenation is an example of Beta elimination.
• The curved arrow formalism shown below illustrates how four bonds are broken or formed in the process.
Common Bases for Dehydrohalogenation
• The most common bases used in elimination reactions are negatively charged oxygen compounds, such as HO¯ and its alkyl derivatives, RO¯, called alkoxides.
Drawing Products of Dehydrohalogenation
• Find the alpha carbon.
• Identify all beta carbons with H atoms.
• Remove the elements of H and X from the
• The double bond of an alkene consists of a sigma bond and a pi bond.
• Alkenes are classified according to the number of carbon atoms bonded to the carbons of the double bond.
Restricted Rotation About Double Bonds
• Recall that even though there is free rotation around single bonds, rotation about double bonds is restricted.
Stereoisomers of Alkenes
• Because of restricted rotation, two stereoisomers of 2-butene are possible.
• cis-2-Butene and trans-2-butene are diastereomers (i.e., non-mirror image stereoisomers).
• Whenever the two groups on each end of a carbon-carbon double bond are different from each other, cis-trans isomers are possible.
Stability of Trans Alkenes
• In general, trans alkenes are more stable than cis alkenes because the groups bonded to the double bond carbons are further apart, reducing steric interactions.
Stability in Alkenes
• The stability of an alkene increases as the number of R groups bonded to the double bond carbons increases.
• The higher the percent s-character, the more readily an atom accepts electron density.
• Thus, sp2 carbons are more able to accept electron density and sp3 carbons are more able to donate electron density.
• Increasing the number of electron donating groups on a carbon atom able to accept electron density makes the alkene more stable.
• The 2-butenes (disubstituted) are more stable than 1-butene (monosubstituted).
• trans-2-Butene is more stable than cis-2-butene (less crowding).
• There are two mechanisms of elimination—E2 and E1, just as there are two mechanisms of substitution, SN2 and SN1.
• The E2 mechanism is called bimolecular elimination. • The E1 mechanism is called unimolecular elimination. • The E2 and E1 mechanisms differ in the timing of bond
cleavage and bond formation, analogous to the SN2 and SN1 mechanisms.
• E2 and SN2 reactions have some features in common, as do E1 and SN1 reactions.
• The most common mechanism for dehydrohalogenation is the E2 mechanism.
• It exhibits second-order kinetics, and both the alkyl halide and the base appear in the rate equation.
rate = k[(CH3)3CBr][¯OH]
• The reaction is concerted—all bonds are broken and formed in a single step.
• There are close parallels between E2 and SN2 mechanisms in how the identity of the base, the leaving group, and the solvent affect the rate.
Bases in E2 Mechanisms
• E2 reactions are generally run with strong, negatively charged bases like¯OH and ¯OR.
• The base appears in the rate equation, so the rate of the E2 reaction increases as the strength of the base increases.
• Two strong, sterically hindered nitrogen bases called DBN and DBU are also sometimes used.
Effects of Leaving Group and Solvent on E2 Reactions
•Because the bond to the leaving group is partially broken in the transition state, the better the leaving group the faster the E2 reaction.
•Polar aprotic solvents increase the rate of E2 reactions.
Effect of Alkyl Halide Structure on E2 Reactions
• The SN2 and E2 mechanisms differ in how the alkyl halide structure affects the reaction rate.
• As the number of R groups on the carbon with the leaving group increases, the rate of the E2 reaction increases.
Transition States in E2 Mechanisms
• The increase in E2 reaction rate with increasing alkyl substitution can be rationalized in terms of transition state stability.
• In the transition state, the double bond is partially formed.
• This increases the stability of the double bond with alkyl substituents stabilizing the transition state (i.e., lowers Ea), which increases the rate of the reaction.
Product Stability and Rate of E2 Reactions
• Increasing the number of R groups on the carbon with the leaving group forms more highly substituted, more stable alkenes in E2 reactions.
• In the reactions below, since the disubstituted alkene is more stable, the 3° alkyl halide reacts faster than the 1° alkyl halide.
E2 Reaction in Organic Synthesis
The synthesis of both quinine and estradiol involve an E2 elimination as a key step.
The Zaitsev (Saytzeff) Rule
• When alkyl halides have two or more different Beta
carbons, more than one alkene product can be formed.
• The Zaitsev rule predicts that the major product in Beta elimination has the more substituted double bond.
• This is the more stable alkene.
Regioselectivity of E2 Reactions
• A reaction is regioselective when it yields predominantly or exclusively one constitutional isomer when more than one is possible.
• Thus, the E2 reaction is regioselective.
Stereoselectivity of E2 Reactions
•A reaction is stereoselective when it forms predominantly or exclusively one stereoisomer when two or more are possible.
•The E2 reaction is stereoselective because the more stable stereoisomer is formed preferentially
• The dehydrohalogenation of (CH3)3CCI with H2O to form (CH3)2C=CH2 can be used to illustrate the second general mechanism of elimination, the E1 mechanism.
• An E1 reaction exhibits first-order kinetics:
rate = k[(CH3)3CCI].
• The E1 reaction proceeds via a two-step mechanism: the bond to the leaving group breaks first before the
bond is formed.
• The slow step is unimolecular, involving only the alkyl halide.
• In an E2 reaction, the leaving group comes off as the
proton is removed, and the reaction occurs in one step.
• However, in an E1, the leaving group comes off before the
proton is removed, and the reaction occurs in two steps.
• Because E1 reactions often occur with a competing SN1 reaction, E1 reactions of alkyl halides are much less useful than E2 reactions.
Effect of Alkyl Halide Structure on E1 Reactions
• The rate of an E1 reaction increases as the number of R groups on the carbon with the leaving group increases.
Effect of Base on the E1 Reaction
• The strength of the base usually determines whether a reaction follows the E1 or E2 mechanism.
• Strong bases like ¯OH and ¯OR favor E2 reactions.
• Weaker bases like H2O and ROH favor E1 reactions.
Regioselectivity of E1 Reactions
• Zaitsev's rule applies to E1 reactions also.
• E1 reactions are regioselective, favoring formation of the more substituted, more stable alkene
SN1 and E1 Reactions
• SN1 and E1 reactions have exactly the same first step—
formation of a carbocation.
• They differ in what happens to the carbocat
Stereochemistry of E2 Reactions
• The transition state of an E2 reaction consists of four atoms from an alkyl halide—one hydrogen atom, two carbon atoms, and the leaving group (X)—all aligned in a plane.
• There are two ways for the C-H and C-X bonds to be coplanar.
Two Possible Geometries for E2
•E2 elimination occurs most often in the anti periplanar geometry.
•This arrangement allows the molecule to react in the lower energy staggered conformation, and allows the incoming base and leaving group to be further away from each other.
Anti Periplanar Geometry
• The requirement of anti periplanar geometry in an E2 reaction has important consequences for compounds containing six-membered rings.
• Chlorocyclohexane exists as two chair conformations. • Conformation X is preferred since the bulkier Cl group
is in the equatorial position.
Trans Diaxial Geometry for E2
• For E2 elimination, the C-Cl bond must be anti periplanar to the C-H bond on a beta carbon, and this occurs only when the H and Cl atoms are both in the axial position.
• The requirement for trans diaxial geometry means that elimination must occur from the less stable conformer, B.
E2 Reactions of Cis and Trans Isomers
• Consider the E2 dehydrohalogenation of cis- and trans-1-chloro-2-methylcyclohexane.
• The cis isomer exists as two conformations, A and B, each of which has one group axial and one group equatorial.
• E2 reaction must occur from conformation B, which contains an axial Cl atom.
Regiochemistry of E2 Reactions on Cyclohexanes
• Because conformation B has two different axial Beta hydrogens, labeled Ha and Hb, E2 reaction occurs in two different directions to afford two alkenes.
• The major product contains the more stable trisubstituted double bond, as predicted by the Zaitsev rule.
Axial Leaving Groups for E2 Reactions
• The trans isomer of 1-chloro-2-methylcyclohexane exists as two conformers: C, having two equatorial substituents, and D, having two axial substituents.
• E2 reaction must occur from D, since it contains an axial Cl atom.
Anti Zaitsev Products for E2 Reactions
• Because conformer D has only one axial beta H, the E2 reaction occurs only in one direction to afford a single product.
• The most substituted, "Zaitsev" alkene is not the major
product in this case.
Comparison of E1 and E2 Mechanisms
• The strength of the base is the most important factor in determining the mechanism for elimination.
E2 Reactions and Alkyne Synthesis
• A single elimination reaction produces a pi bond of an alkene.
• Two consecutive elimination reactions produce two pi bonds of an alkyne.
• Two elimination reactions are needed to remove two moles of HX from a dihalide substrate.
• Two different starting materials can be used—a vicinal dihalide or a geminal dihalide.
Bases for Alkyne Synthesis
• Stronger bases are needed to synthesize alkynes by dehydrohalogenation than are needed to synthesize alkenes.
• The typical base used is ¯NH2 (amide), used as NaNH2.
• KOC(CH3)3 can also be used with DMSO as solvent.
• Stronger bases are needed to break the stronger sp2 hybridized C-H bonds in the second elimination reaction.
When is the Reaction SN1, SN2, E1, or E2?
• Good nucleophiles that are weak bases favor substitution over elimination.
• These include I¯, Br¯, HS¯, ¯CN, and CH3COO¯.
Bulky Bases Favor Elimination
• Bulky nonnucleophilic bases favor elimination over substitution.
• KOC(CH3)3, DBU, and DBN are too sterically hindered to attack tetravalent carbon.
• They are, however, able to remove a small proton, favoring elimination over substitution.
Alcohols—Structure and Bonding
• Alcohols contain a hydroxy group (OH) bonded to an sp3 hybridized carbon.
• They are classified according to the number of alkyl groups attached to carbon bearing the OH.
Enols, Phenols, and Ether
• Compounds having a hydroxy group on a sp2
hybridized carbon—enols and phenols—undergo different reactions than alcohols.
• Ethers have two alkyl groups bonded to an oxygen
• Epoxides are ethers having the oxygen atom in a three-membered ring.
• Epoxides are also called oxiranes.
• The C-O-C bond angle for an epoxide must be 60°, a considerable deviation from the tetrahedral bond angle of 109.5°.
• Thus, epoxides have angle strain, making them more reactive than other ethers.
Oxygen Hybridization and Geometry
• The oxygen atom in alcohols, ethers, and epoxides is sp3 hybridized.
• Alcohols and ethers have a bent shape like that in H2O.
• The bond angle around the O atom in an alcohol or ether is similar to the tetrahedral bond angle of 109.5°.
• Because the O atom is much more electronegative than carbon or hydrogen, the C-O and O-H bonds are both polar.
Diols and Triols
• Compounds with two hydroxy groups are called diols or glycols.
• Compounds with three hydroxy groups are called triols.
• Simple ethers are usually assigned common names. To do so:
• Name both alkyl groups bonded to the oxygen, arrange these names alphabetically, and add the word ether.
• For symmetrical ethers, name the alkyl group and add the
Naming Complex Ethers
• More complex ethers are named using the IUPAC system.
• One alkyl group is named as a hydrocarbon chain, and the other is named as part of a substituent bonded to that chain:
• Name the simpler alkyl group as an alkoxy substituent by changing the -yl ending of the alkyl group to -oxy.
• Name the remaining alkyl group as an alkane, with the alkoxy group as a substituent bonded to this chain.
• Epoxides can be named in three different ways—
epoxyalkanes, oxiranes, or alkene oxides.
• To name an epoxide as an epoxyalkane, first name the alkane chain or ring to which the O atom is attached,
and use the prefix "epoxy" to name the epoxide as a
• Use two numbers to designate the location of the atoms to which the O is bonded.
Naming Epoxides as Oxiranes
• Epoxides bonded to a chain of carbon atoms can also be named as derivatives of oxirane, the simplest epoxide having two carbons and one oxygen atom in a ring.
• The oxirane ring is numbered to put the O atom at position one, and the first substituent at position two.
• No number is used for a substituent in a monosubstituted oxirane.
Naming Epoxides as Alkene Oxides
• Epoxides are also named as alkene oxides, since they are often prepared by adding an O atom to an alkene. To name an epoxide in this way:
• Mentally replace the epoxide oxygen with a double
bond. • Name the alkene. • Add the word oxide.
Hydrogen Bonding in Alcohols
• Alcohols, ethers, and epoxides exhibit dipole-dipole interactions because they have a bent structure with two polar bonds.
• Alcohols are capable of intermolecular hydrogen bonding. Thus, alcohols are more polar than ethers and epoxides.
• Steric factors affect hydrogen bonding.
Use of Crown Ethers
• The ability of crown ethers to complex cations can be exploited in nucleophilic substitution reactions.
• As the complexed cation goes into solution it carries the anion with it to maintain neutrality.
• The relatively unsolvated anion is extremely nucleophilic.
Preparation of Alchols and Ethers
• Alcohols and ethers are both common products of nucleophilic substitution.
• The preparation of ethers by the method shown in the last two equations is called the Williamson ether synthesis.
Williamson Ether Synthesis
• In theory, unsymmetrical ethers can be synthesized in two different ways.
• In practice, one path is usually preferred.
• The path involving alkoxide attack on a less hindered alkyl halide is preferred.
Preparation of Alkoxides
• An alkoxide salt is needed to make an ether.
• Alkoxides can be prepared from alcohols by a Brønsted-Lowry acid-base reaction.
• For example, sodium ethoxide (NaOCH2CH3) is prepared by treating ethanol with NaH.
• NaH is an especially good base for forming alkoxide because the by-product of the reaction, H2, is a gas that just bubbles out of the reaction mixture.
Forming Epoxides from Halohydrins
• Organic compounds that contain both a hydroxy group and a halogen atom on adjacent carbons are called halohydrins.
• In halohydrins, an intramolecular version of the Williamson ether synthesis can occur to form epoxides.
OH as a Leaving Group
• Unlike alkyl halides in which the halogen atom serves as a good leaving group, the OH group in alcohols is a very poor leaving group.
• For an alcohol to undergo nucleophilic substitution, OH must be converted into a better leaving group.
Substitution and Elimination Reactions of Alcohols
• Treatment of alcohols with a strong acid protonates the O converting the bad leaving group ¯OH into H2O, a good leaving group.
• This makes it possible to perform substitution and elimination reactions on alcohols.
Reactions of Alcohols—
• Dehydration, like dehydrohalogenation, is a beta elimination reaction in which the elements of OH and H are removed from the alpha and beta carbon atoms respectively.
Dehydration Requires Strong
Dehydration is typically carried out using H2SO4 and other strong acids, or phosphorus oxychloride (POCl3) in the presence of an amine base.
• Typical acids used for alcohol dehydration are H2SO4 or p-toluenesulfonic acid (TsOH).
Dehydration and Alcohol Substitution
Products that are more substituted better stabilize the transitions state
Zaitsev's Rule regarding alcohols
• When an alcohol has two or three carbons, dehydration is regioselective and follows the Zaitsev rule.
• The more substituted alkene is the major product when a mixture of constitutional isomers is possible.
Useful E1 Dehydration
• The E1 dehydration of 2o and 3o alcohols with acid gives clean elimination products without any by-products formed from an SN1 reaction.
• Clean elimination takes place because the reaction mixture contains no good nucleophile to react with the intermediate carbocation, so no competing SN1 reaction occurs.
• This makes the E1 dehydration of alcohols much more synthetically useful than the E1 dehydrohalogenation of alkyl halides.
Dehydration by E1 Mechanism
• 2° and 3° alcohols react by an E1 mechanism, whereas 1° alcohols react by an E2 mechanism. Why?
Dehydration by E1 Mechanism
Because of the instability of primaruy carbocations
E2 Dehydration of 1o Alcohols
• Since 1° carbocations are highly unstable, their dehydration cannot occur by an E1 mechanism involving a carbocation intermediate.
• However, 1° alcohols undergo dehydration by way of an E2 mechanism.
Enthalpy of Dehydration
• Entropy favors product formation in dehydration since one molecule of reactant forms two molecules of product.
• Enthalpy favors reactants, since the two sigma bonds broken in the reactant are stronger than the sigma and pi
bonds formed in the products.
Dehydration Reaction Equilibrium
• According to Le Châtelier's principle, a system at equilibrium will react to counteract any disturbance to the equilibrium.
• One consequence of this is that removing a product from a reaction mixture as it is formed drives the equilibrium to the right, forming more product.
• Thus, the alkene, which usually has a lower boiling point than the starting alcohol, can be removed by distillation as it is formed, thus driving the equilibrium to the right to favor production of more product.
• Often, when carbocations are intermediates, a less stable carbocation will be converted into a more stable carbocation by a shift of a hydrogen or an alkyl group.
• This is called a rearrangement.
• Because the migrating group in a 1,2-shift moves with two bonding electrons, the carbon it leaves behind now has only three bonds (six electrons), giving it a net positive (+) charge.
• A 1,2-shift can convert a less stable carbocation into a more stable carbocation.
• Rearrangements are not unique to dehydration reactions.
• Rearrangements can occur whenever a carbocation is formed as a reactive intermediate.
• 2° carbocation A rearranges to the more stable 3° carbocation by a 1,2-hydride shift, whereas carbocation B does not rearrange because it is 3° to begin with.
Dehydration of Alcohols Using POCl3
• Some organic compounds decompose in the presence of strong acid, so other methods have been developed to convert alcohols to alkenes.
• A common method uses phosphorus oxychloride (POCl3) and pyridine (an amine base) in place of H2SO4 or TsOH.
• POCl3 serves much the same role as a strong acid does in acid-catalyzed dehydration.
• It converts a poor leaving group (¯OH) into a good leaving group.
• Dehydration then proceeds by an E2 mechanism.
Conversion of Alcohols to Alkyl Halides with HX
• Substitution reactions do not occur with alcohols unless
¯OH is converted into a good leaving group.
• The reaction of alcohols with HX (X = Cl, Br, I) is a general method to prepare 1°, 2°, and 3° alkyl halides.
Mechanism of Reaction of Alcohols with HX
• More substituted alcohols usually react more rapidly with HX:
• This order of reactivity can be rationalized by considering the reaction mechanisms involved.
• The mechanism depends on the structure of the R group.
Reactivity of Hydrogen Halides
• The reactivity of hydrogen halides increases with increasing acidity.
• Because Cl¯ is a poorer nucleophile than Br¯ or I¯, the reaction of 1o alcohols with HCl occurs only when an additional Lewis acid catalyst, usually ZnCl2, is added.
• Complexation of ZnCl2 with the O atom of the alcohol makes a very good leaving group that facilitates the SN2 reaction.
Conversion of Alcohols to Alkyl Halides with SOCl2 and PBr3
• Primary and 2° alcohols can be converted to alkyl halides using SOCl2 and PBr3.
• SOCl2 (thionyl chloride) converts alcohols into alkyl chlorides.
• PBr3 (phosphorus tribromide) converts alcohols into alkyl bromides.
• Both reagents convert ¯OH into a good leaving group in situ—that is, directly in the reaction mixture—as well as provide the nucleophile, either Cl¯ or Br¯, to displace the leaving group.
Tosylate as Leaving Group
• Alcohols can be converted into alkyl tosylates.
• An alkyl tosylate is composed of two parts: the alkyl group R, derived from an alcohol; and the tosylate (short for p-toluenesulfonate), which is a good leaving group.
• A tosyl group, CH3C6H4SO2¯, is abbreviated Ts, so an alkyl tosylate becomes ROTs.
Formation and Use of Tosylates
• Alcohols are converted to tosylates by treatment with p-toluenesulfonyl chloride (TsCl) in the presence of pyridine.
• This process converts a poor leaving group (¯OH) into a good one (¯OTs).
• Tosylate is a good leaving group because its conjugate acid, p-toluenesulfonic acid (CH3C6H4SO3H, TsOH) is a strong acid (pKa = -7).
Stereochemistry of Tosylate Formation
• (2S)-2-Butanol is converted to its tosylate with retention of configuration at the stereogenic center.
• The C-O bond of the alcohol is not broken when tosylate is formed. (retention of configuration)
Substitution and Elimination of Tosylates
• Because alkyl tosylates have good leaving groups, they undergo both nucleophilic substitution and
elimination, exactly as alkyl halides do.
• Generally, alkyl tosylates are treated with strong nucleophiles and bases, so the mechanism of substitution is SN2, and the mechanism of elimination is E2
SN2 Inversion When Replacing Tosylates
• Because substitution occurs via an SN2 mechanism, inversion of configuration results when the leaving group is bonded to a stereogenic center.
Reaction of Ethers with Strong Acid
• In order for ethers to undergo substitution or elimination reactions, their poor leaving group must first be converted into a good leaving group by reaction with strong acids such as HBr and HI.
• HBr and HI are strong acids that are also sources of good nucleophiles (Br¯ and I¯, respectively).
• When ethers react with HBr or HI, both C-O bonds are cleaved and two alkyl halides are formed as products.
Mechanism of Ether Cleavage
• The mechanism of ether cleavage is SN1 or SN2, depending on the identity of R.
• When 2° or 3° alkyl groups are bonded to the ether oxygen, the C-O bond is cleaved by an SN1 mechanism involving a carbocation.
• With methyl or 1° R groups, the C-O bond is cleaved by an SN2 mechanism.
Reactions of Epoxides
• Epoxides do not contain a good leaving group.
• Epoxides do contain a strained three-membered ring with two polar bonds.
• Nucleophilic attack opens the strained three-membered ring, making it a favorable process even with a poor leaving group.
Addition of Nucleophiles to Epoxides
• Nucleophilic addition to epoxides occurs readily with strong nucleophiles and with acids like HZ, where Z is a nucleophilic atom.
Mechanism of Epoxide Reactions
• Virtually all strong nucleophiles open an epoxide ring by a two-step reaction sequence:
• In step , the nucleophile attacks an electron-deficient carbon, by an SN2 mechanism, thus cleaving the C-O bond and relieving the strain of the three-membered ring.
• In step , the alkoxide is protonated with water to generate a neutral product with two functional groups on adjacent atoms.
• Common nucleophiles that open the epoxide ring include ¯OH, ¯OR, ¯CN, ¯SR, and NH3.
Acidic Epoxide Ring Opening
• Acids HZ that contain a nucleophile Z also open epoxide rings by a two-step sequence.
• HCl, HBr, and HI, as well as H2O and ROH in the presence of acid, all open an epoxide ring in this manner.
Regioselectivity of Epoxide Ring Opening
• Ring opening of an epoxide with either a strong nucleophile or an acid HZ is regioselective because one constitutional isomer is the major or exclusive product.
• The site selectivity of these two reactions is exactly opposite.
Alkene Structure from Ch. 10
• Alkenes are also called olefins. • Alkenes contain a carbon-carbon double bond. • Terminal alkenes have the double bond at the end of the
• Internal alkenes have at least one carbon atom bonded to each end of the double bond.
• Cycloalkenes contain a double bond in a ring.
• Recall that the double bond consists of a pi bond and a sigma bond.
• Each carbon is sp2 hybridized and trigonal planar, with bond angles of approximately 120°.
pi Bond Dissociation Energy
• Bond dissociation energies of the C-C bonds in ethane (a sigma bond only) and ethylene (one sigma and one pi bond) can be used to estimate the strength of the pi component of the double bond.
• The pi bond is much weaker than the sigma bond of a C-C double bond, making it much more easily broken.
• Therefore, alkenes undergo many reactions that alkanes do not.
• Cycloalkenes having fewer than eight carbon atoms must have a cis geometry.
• A trans-cycloalkene in a small ring cycloalkene (< 8 C's)
would introduce significant strain into the ring.
• trans-Cyclooctene is the smallest, isolable trans cycloalkene, but it is considerably less stable than cis-cyclooctene, making it one of the few alkenes having a higher energy trans isomer.
Calculating Degrees of Unsaturation
• An acyclic alkene has the general structural formula CnH2n.
• Alkenes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon.
• Cycloalkanes also have the general formula CnH2n.
• Each bond or ring removes two hydrogen atoms from a molecule, and this introduces one degree of unsaturation.
• The number of degrees of unsaturation for a given molecular formula can be calculated by comparing the actual number of H atoms in a compound to the maximum number of H atoms possible for the number of carbons present if the molecule were an acyclic alkane.
• This procedure gives the total number of rings and/or
bonds in a molecule
Degrees of Unsaturation for Molecules Containing Heteroatoms
• Ignore O atoms in the molecule (this divalent atom is a linker and has no effect on degree of unsaturation).
• Add number of halogens to number of H's (they are equivalent to H).
• Subtract 1 H for each N present (N's two
connections allows extra H). • Example: C6H10OCl3N is equivalent to C6H12. • Saturated formula: C6H14 • Degrees of unsaturation: C6H14 - C6H12 = H2 • 2 2 = 1 degree of unsaturation • Important: Degree of unsaturation is always a whole
• Compounds that contain both a double bond and a hydroxy group are named as alkenols.
• These compounds are named as alcohols because the OH has priority over the double bond.
• The chain (or ring) is numbered to give the OH group the lower number.
• Notice how the positions of both functional groups are specified.
Naming Polyenes and Cyclic Alkenes
• Compounds with two double bonds are named as dienes by
changing the "-ane" ending of the parent alkane to the suffix "-adiene".
• Compounds with three double bonds are named as trienes, and so forth.
• In naming cycloalkenes, the double bond is located between
C1 and C2, and the "1" is usually omitted in the name.
• The ring is numbered clockwise or counterclockwise to give the first substituent the lower number.
Physical Properties of Alkenes
• Most alkenes exhibit only weak van der Waals interactions, so their physical properties are similar to alkanes of comparable molecular weight.
• Alkenes have low melting points and boiling points.
• Melting and boiling points increase as the number of carbons increases because of increased surface area.
• Alkenes are soluble in organic solvents and insoluble in water.
• The C-C single bond between an alkyl group and one of the double bond carbons of an alkene is slightly polar because the sp3 hybridized alkyl carbon donates electron density to the sp2 hybridized alkenyl carbon
Cis/Trans Differ in Physical Properties
• A consequence of the alkene dipole is that cis and trans isomeric alkenes often have somewhat different physical properties.
• cis-2-Butene has a higher boiling point (4 °C) than trans-2-butene (1 °C).
• In the cis isomer, the two Csp3Csp2 bond dipoles reinforce each other, yielding a small net molecular dipole.
• In the trans isomer, the two bond dipoles cancel.
Preparation of Alkenes
• Alkenes can be prepared from alkyl halides, tosylates, and alcohols via elimination reactions.
Regioselectivity and Stereoselectivity of Alkene Formation
• The most stable alkene (Zaitsev product) is usually formed as the major product.
• The most stable stereoisomer (trans) is formed when possible.
Addition Reactions for alkenes
• The characteristic reaction of alkenes is addition—the bond is broken and two new bonds are formed.
• Alkenes are electron rich, with the electron density of the
bond concentrated above and below the plane of the molecule.
• Therefore, alkenes act as nucleophiles and react with
• Simple alkenes do not react with nucleophiles or bases, reagents that are themselves electron rich.
Syn and Anti Addition to Alkenes
• Because the carbon atoms of a double bond are both trigonal planar, the elements of X and Y can be added to them from the same side or from opposite sides.
• Syn addition takes place when both X and Y are added
from the same side.
• Anti addition takes place when X and Y are added from opposite sides.
Hydrohalogenation - Electrophilic Addition of HX
• Two bonds are broken in this reaction—the weak bond of the alkene and the HX bond—and two new bonds are formed—one to H and one to X.
• Recall that the H-X bond is polarized, with a partial positive charge on H.
• Because the electrophilic H end of HX is attracted to the electron-rich double bond, these reactions are called electrophilic additions.
Mechanism of Electrophilic Addition
• The mechanism of electrophilic addition consists of two successive Lewis acid-base reactions.
• Step  - the alkene is the Lewis base that donates an electron
pair to H-Br, the Lewis acid.
• Step  - Br¯ is the Lewis base that donates an electron pair to the carbocation, the Lewis acid.
Energy Diagram for Electrophilic Addition
• Each step has its own energy barrier with a transition state energy maximum.
• Since step  has a higher energy transition state, it is rate-determining.
• H° for step  is positive because more bonds are broken than formed, whereas H° for step  is negative because only bond making occurs.
Markovnikov's Rule - Predicting Regiochemistry of Additon
• With an unsymmetrical alkene, HX can add to the double bond to give two constitutional isomers, but only one is actually formed
• With an unsymmetrical alkene, HX can add to the double bond to give two constitutional isomers, but only one is actually formed:
• Markovnikov's rule states that in the addition of HX to an unsymmetrical alkene, the H atom adds to the less substituted carbon atom—that is, the carbon that has the greater number of H atoms to begin with.
Carbocation Stability and Markovnikov's Rule
• The basis of Markovnikov's rule is the formation of a carbocation in the rate-determining step of the mechanism.
• In the addition of HX to an unsymmetrical alkene, the H atom is added to the less substituted carbon to form the more stable, more substituted carbocation.
Stereochemistry of Electrophilic Addition
Recall that trigonal planar atoms react with reagents from two directions with equal probability.
• Achiral starting materials yield achiral products.
• Sometimes new stereogenic centers are formed from hydrohalogenation.
Stereochemistry of Carbocation Formation
• The mechanism of hydrohalogenation illustrates why two enantiomers are formed.
• Initial addition of H+ occurs from either side of the planar double bond.
• Both modes of addition generate the same achiral carbocation.
• Either representation of this carbocation can be used to draw the second step of the mechanism.
Stereochemistry of Nucleophilic Attack
• Nucleophilic attack of Cl¯ on the trigonal planar carbocation also occurs from two different directions, forming two products, A and B, having a new stereogenic center.
• A and B are enantiomers.
• Since attack from either direction occurs with equal probability, a racemic mixture of A and B is formed.
Stability of Cation vs. Bridged Halonium Ion Intermediates
• Carbocations are unstable because they have only six electrons that surround carbon.
• Halonium ions are unstable because of ring strain.
Stereochemistry of Halonium Formation
• Consider the chlorination of cyclopentene to afford both enantiomers of trans-1,2-dichlorocyclopentane, with no cis products. Why?
• Initial addition of the electrophile Cl+ from (Cl2) occurs from either side of the planar double bond to form a bridged chloronium ion.
Generating Bromine in Halohydrin Formation
• Although the combination of Br2 and H2O effectively forms bromohydrins from alkenes, other reagents can also be used.
• Bromohydrins are also formed with N-bromosuccinimide (NBS) in aqueous DMSO [(CH3)2S=O].
• In H2O, NBS decomposes to form Br2, which then goes on to form a bromohydrin by the same reaction mechanism.
Anti Stereochemistry in Halohydrin Formation
• Because the bridged halonium ion is opened by backside attack of H2O, addition of X and OH occurs in an anti fashion and trans products are formed.
• With unsymmetrical alkenes, the preferred product has the electrophile X+ bonded to the less substituted carbon, and the nucleophile (H2O) bonded to the more substituted carbon.
Regiochemistry of Halohydrin Formation
• As in the acid catalyzed ring opening of epoxides, nucleophilic attack occurs at the more substituted carbon end of the bridged halonium ion because that carbon is better able to accommodate the partial positive charge in the transition state.
Hydroboration-Oxidation of alkenes
• Hydroboration-oxidation is a two-step reaction sequence that converts an alkene into an alcohol.
• BH3 is a reactive gas that exists mostly as a dimer, diborane (B2H6).
• Borane is a strong Lewis acid that reacts readily with Lewis bases.
• For ease of handling in the laboratory, it is commonly used as a complex with tetrahydrofuran (THF).
• Because the alkylborane formed by the reaction with one equivalent of alkene still has two B-H bonds, it can react with two more equivalents of alkene to form a trialkylborane.
• We often draw hydroboration as if addition stopped after one equivalent of alkene reacts with BH3.
• Instead all three B-H bonds actually react with three equivalents of an alkene to form a trialkylborane
• Since only one B-H bond is needed for hydroboration, commercially available dialkylboranes having the general structure R2BH are sometimes used instead of BH3.
• A common example is 9-borabicyclo[3.3.1]nonane (9-BBN).
With unsymmetrical alkenes, the boron atom bonds to the less substituted carbon atom.
• This regioselectivity can be explained by considering steric factors.
• The larger boron atom bonds to the less sterically hindered, more accessible carbon atom.
• Since alkylboranes react rapidly with water and spontaneously burn when exposed to air, they are oxidized, without isolation, with basic hydrogen peroxide (H2O2, ¯OH).
• Oxidation replaces the C-B bond with a C-O bond, forming a new OH group with retention of configuration.
• The overall result of this two-step sequence is syn addition of the elements of H and OH to a double bond in an "anti-Markovnikov" fashion.
Electronic Factors Affecting Regiochemistry of Hydroboration
• Electronic factors are also used to explain this regioselectivity.
• If bond making and bond breaking are not completely symmetrical, boron bears a - charge in the transition state and carbon bears a + charge.
• Since alkyl groups stabilize a positive charge, the more stable transition state has the partial positive charge on the more substituted carbon.
Oxidation and Reduction of
•Oxidation results in an increase in the number of C − Z bonds (usually C − O bonds) or a decrease in
the number of C-H bonds.
•Reduction results in a decrease in the number of
C − Z bonds (usually C − O bonds) or an increase in the number of C − H bonds.
Oxidation and Reduction of Hydrocarbons
•Sometimes two carbon atoms are involved in a single oxidation or reduction reaction, and the net change in the
number of C − H or C − Z bonds at both atoms must be taken
•The conversion of an alkyne to an alkene, or an alkene to an alkane, are examples of reduction because each process adds
two new C − H bonds to the starting material.
Reductions of Organic Molecules
•There are three types of reductions differing in how H2 is added.
•The simplest reducing agent is H2. Reductions using H2 are carried out with a metal catalyst.
•A second way is to add two protons and two electrons to a substrate—that is, H2 = 2H+ + 2e−.
•Reductions of this sort use alkali metals as a source of electrons, and liquid ammonia as a source of protons.
•These are called dissolving metal reductions.
The third way to add H2 is to add hydride (H¯) and a proton (H+).
•The most common hydride reducing agents contain a hydrogen atom bonded to boron or aluminum.
•Simple examples include sodium borohydride (NaBH4) and lithium aluminum hydride (LiAlH4).
•NaBH4 and LiAlH4 deliver H¯ to the substrate, and then a proton is added from H2O or an alcohol.
•The addition of H2 occurs only in the presence of a metal catalyst, and thus it is called catalytic hydrogenation.
•The catalyst consists of a metal—usually Pd, Pt, or Ni, adsorbed onto a finely divided inert solid, such as charcoal.
•H2 adds in a syn fashion.
Rate of Catalytic Hydrogenation
•The mechanism explains two facts about hydrogenation:
•Rapid, sequential addition of H2 occurs from the side of the alkene complexed to the metal surface, resulting in syn addition.
•Less crowded double bonds complex more readily to the catalyst surface, resulting in faster reaction.
Hydrogenation Data & Degrees of Unsaturation
•Degrees of unsaturation include both rings and bonds.
•Because H2 will add to bonds but not C − C σ bonds of rings, one can use hydrogenation to distinguish the nature of unsaturation in a molecule.
•This is done by comparing the degrees of unsaturation before and after a molecule is treated with H2.
•The number of degrees of unsaturation lost = the number of bonds.
•The number of degrees of unsaturation remaining in the product = the number of rings in the original compound.
•For example, if a molecule with a formula C8H12 was converted to C8H14 upon hydrogenation, the original molecule contains one bond and two rings.
•Carbonyl groups (C=O) in a molecule can also undergo hydrogenation to form alcohols since they contain a bond.
Reduction of Alkynes Using Lindlar's Catalyst
•Reduction of an alkyne to a cis alkene is a stereoselective reaction, because only one stereoisomer is formed.
Dissolving Metal Reduction of Alkynes
•In a dissolving metal reduction (such as Na in NH3), the elements of H2 are added in an anti fashion to form a trans alkene.
Peroxide Oxidizing Agents
•There are two main categories of oxidizing agents: 1. Reagents that contain an oxygen-oxygen bond 2. Reagents that contain metal-oxygen bonds •Oxidizing agents containing an O-O bond include O2,
O3 (ozone), H2O2 (hydrogen peroxide), (CH3)3COOH (tert-butyl hydroperoxide), and peroxyacids.
•Peroxyacids (or peracids) have the general formula RCO3H.
Oxidizing Agents with Metal Oxygen Bonds
•The most common oxidizing agents with metal-oxygen bonds contain either chromium +6 or manganese +7.
•Common Cr6+ reagents include CrO3 and sodium or potassium dichromate (Na2Cr2O7 and K2Cr2O7).
•Pyridinium chlorochromate (PCC) is a more selective Cr6+ oxidant.
•The most common Mn7+ reagent is KMnO4 (potassium permanganate).
•Other oxidizing agents that contain metals include OsO4 (osmium tetroxide) and Ag2O [silver(I) oxide].
•Epoxidation is the addition of a single oxygen atom to an alkene to form an epoxide.
•Epoxidation is typically carried out with a peroxyacid.
•Dihydroxylation is the addition of two hydroxy groups to a double bond, forming a 1,2-diol or glycol.
•Depending on the reagent, the two new OH groups can be added to the opposite sides (anti addition) or the same side (syn addition) of the double bond.
Catalytic Syn Dihydroxylation
•Dihydroxylation can also be carried out by using a catalytic amount of OsO4, if the oxidant N-methylmorpholine N-oxide (NMO) is added.
•In the catalytic process, dihydroxylation of the double bond converts the Os8+ oxidant into an Os6+ product, which is then re-oxidized by NMO to Os8+.
Ozonolysis: Oxidative Cleavage of Alkenes
•Oxidative cleavage of an alkene breaks both the and bonds of the double bond to form two carbonyl compounds.
•Cleavage with ozone (O3) is called ozonolysis.
Oxidative Cleavage of Rings
•Ozonolysis of dienes or other polyenes results in oxidative cleavage of all C=C bonds.
•When oxidative cleavage involves a double bond that is part of a ring, the ring opens up affording a single chain with two carbonyls at the carbons where the double bonds were originally.
•Oxidative cleavage is a valuable tool in structure determination, helping to pinpoint the location of double bonds in complex alkene structures.
Oxidative Cleavage of Alkynes
•Alkynes undergo oxidative cleavage of the and both bonds of the triple bond.
•Internal alkynes are oxidized to carboxylic acids (RCOOH).
•Terminal alkynes afford a carboxylic acid and CO2 from the sp hybridized C-H bond.
Chromium Oxidizing Reagents
•Oxidation of alcohols to carbonyl compounds is typically carried out with Cr6+ oxidants, which are reduced to Cr3+ products.
•CrO3, Na2Cr2O7, and K2Cr2O7 are strong, nonselective oxidants used in aqueous acid (H2SO4 + H2O).
•PCC is soluble in CH2Cl2 (dichloromethane) and can be used without strong acid present, making it a more selective, milder oxidant.
Oxidation of 2° Alcohols
•Any of the Cr6+ oxidants effectively oxidize 2° alcohols to ketones.
Oxidation of 1° Alcohols
•Primary alcohols are oxidized to either aldehydes or carboxylic acids, depending on the reagent.
•Alkynes contain a carbon-carbon triple bond.
•An alkyne has the general molecular formula CnH2n−2, giving it four fewer hydrogens than the maximum possible for the number of carbons present.
•The triple bond introduces two degrees of unsaturation.
•Terminal alkynes have the triple bond at the end of the carbon chain so that a hydrogen atom is directly bonded to a carbon atom of the triple bond.
•Internal alkynes have a carbon atom bonded to each carbon atom of the triple bond.
•Recall that the triple bond consists of 2 bonds and 1 bond.
•Each carbon is sp hybridized with a linear geometry and bond angles of 180o.
•Like trans cycloalkenes, cycloalkynes with small rings are unstable.
•The carbon chain must be long enough to connect the two ends of the triple bond without introducing too much strain.
•Cyclooctyne is the smallest isolated cycloalkyne, though it decomposes upon standing at room temperature after a short time.
Physical Properties of Alkynes
•The physical properties of alkynes resemble those of hydrocarbons of similar shape and molecular weight.
•Alkynes have low melting points and boiling points.
•Melting point and boiling point increase as the number of carbons increases.
•Alkynes are soluble in organic solvents and insoluble in water.
•The simplest alkyne, H−CC−H, named in the IUPAC system as ethyne, is more often called acetylene, its common name.
•The two-carbon alkyl group derived from acetylene is called an ethynyl group.
•Acetylene (H−CC−H) is a colorless gas that burns in oxygen to form CO2 and H2O.
•The combustion of acetylene releases more energy per mole of product formed than any other hydrocarbons.
•When combined with oxygen, it burns with a very hot
flame and is used in welding.
Preparation of Alkynes
•Alkynes are prepared by elimination reactions.
•A strong base removes two equivalents of HX from a vicinal or geminal dihalide to yield an alkyne through two successive E2 elimination reactions.
Preparation of Alkynes from Alkenes
•Since vicinal dihalides are readily made from alkenes, one can convert an alkene to the corresponding alkyne in a two-step process involving:
•Halogenation of an alkene.
•Double dehydrohalogenation of the resulting vicinal dihalide.
General Addition Reactions of Alkynes
•Like alkenes, alkynes undergo addition reactions because they contain relatively weak bonds.
•Two sequential reactions can take place: addition of one equivalent of reagent forms an alkene, which can then add a second equivalent of reagent to yield a product having four new bonds.
Hydrohalogenation-Electrophilic Addition of HX
•Alkynes undergo hydrohalogenation, i.e., the addition of hydrogen halides, HX (X = Cl, Br, I).
•Two equivalents of HX are usually used: addition of one mole forms a vinyl halide, which then reacts with a second mole of HX to form a geminal dihalide.
Hydrohalogenation of Alkynes vs. Alkenes
•Electrophilic addition of HX to alkynes is slower than electrophilic addition of HX to alkenes, even though alkynes are more polarizable and have more loosely held electrons than alkenes.
•Markovnikov addition in step  places the H on the terminal carbon to form the more substituted carbocation A, rather than the less substituted carbocation B.
Halogen Stabilization of Carbocations
•Resonance stabilizes a molecule by delocalizing charge and electron density.
•Halogens stabilize an adjacent positive charge by resonance. •Carbocation A is stabilized by resonance.
Halogenation of Alkynes
•Halogens X2 (X = Cl or Br) add to alkynes just as they do to alkenes.
•Addition of one mole of X2 forms a trans dihalide, which can then react with a second mole of X2 to yield a tetrahalide.
Hydration of Alkynes
•In the presence of strong acid or Hg2+ catalyst, the elements of H2O add to the triple bond to form an enol initially.
•The enol is unstable and rearranges to a ketone.
•Tautomers are constitutional isomers that differ in the location of a double bond and a hydrogen atom.
•A and B are tautomers: A is the enol form and B is the keto form of the tautomer.
•An enol tautomer has an O−H group bonded to a C=C.
•A keto tautomer has a C=O and an additional C−H bond.
•Equilibrium favors the keto form largely because the C=O is much stronger than a C=C.
•Tautomerization, the process of converting one tautomer into another, is catalyzed by both acid and base.
Hydration of Internal vs. Terminal Alkynes
•Internal alkynes undergo hydration with concentrated acid to form ketones.
•Terminal alkynes require the presence of an additional Hg2+ catalyst (usually HgSO4) to yield methyl ketones by Markovnikov addition of water.
•Hydroboration−oxidation is a two-step reaction sequence that also converts an alkyne to a carbonyl compound.
•Addition of borane forms an organoborane. •Oxidation with basic H2O2 forms an enol. •Tautomerization of the enol forms a carbonyl compound. •The overall result is addition of H2O to a triple bond.
Hydroboration - Oxidation of Internal vs. Terminal Alkynes
•Hydroboration−oxidation of an internal alkyne forms a ketone, just as the acid-catalyzed hydration did.
•However, hydroboration−oxidation of a terminal alkyne forms an aldehyde.
•BH3 adds to the less substituted, terminal carbon resulting in anti-Markovnikov addition of water.
Formation of Acetylide Ions
•Sp hybridized C − H bonds are considerably more acidic than sp2 and sp3 hybridized C − H bonds.
•Therefore, terminal alkynes are readily deprotonated with strong base in a Brønsted-Lowry acid-base reaction.
•The resulting ion is called the acetylide ion.
Reactions of Acetylide Ions
•Acetylide anions are strong nucleophiles and react with unhindered alkyl halides to yield products of nucleophilic substitution.
•The mechanism of substitution is SN2, and thus the reaction is fastest with CH3X and 1o alkyl halides.
Elimination vs. Substitution with Acetylide Ions
•Steric hindrance around the leaving group causes 2° and 3° alkyl halides to preferentially undergo elimination by an E2 mechanism, as shown with 2-bromo-2-methylpropane.
•Thus, nucleophilic substitution with acetylide anions forms new carbon-carbon bonds in high yield only with unhindered CH3X and 1° alkyl halides.
•Conjugation occurs whenever p orbitals can overlap on three or more adjacent atoms.
•The four p orbitals on adjacent atoms make a 1,3-diene a conjugated system.
•The p orbital at the allylic position is in conjugation with the double bond.
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