MECH MONSTER 6 Flashcards
free-radical halogenation
Rarely an effective method for the synthesis of alkyl halides. It usually produces mixtures of products because there are different kinds of hydrogen atoms that can be abstracted. Also, more than one halogen atom may react, giving multiple substitutions. The chlorination of propane can give a mixture of products.
free-radical halogenation
Free-radical bromination is highly selective in the synthesis of alkyl halides, and it gives good yields of products that have one type of hydrogen atom that is more reactive than the others. Isobutane has only one tertiary hydrogen atom, and this atom is preferentially abstracted to give a tertiary free radical.
free-radical halogenation
All the hydrogen atoms in cyclohexane are equivalent, so a medium yield of chlorocyclohexane results. Formation of dichlorides and trichlorides is possible, but these side reactions are controlled by using only a small amount of chlorine and an excess of cyclohexane.
allylic bromination
Either end of the resonance-stabilized allylic radical can react with bromine. In one of the products, the bromine atom appears in the same position where the hydrogen atom was abstracted. The other product results from reaction at the carbon atom that bears the radical in the second resonance form of the allylic radical. This second compound is said to be the product of an allylic shift.
allylic position
a carbon atom next to a carbon–carbon double bond
initiation step of allylic bromination
Initiation Step
Bromine absorbs light, causing formation of radicals.
first propagation step of allylic bromination
First Propagation Step
A bromine radical abstracts an allylic hydrogen.
second propagation step of allylic bromination
Second Propagation Step
Either radical carbon can react with bromine.
For efficient allylic bromination, a large concentration of bromine must be avoided because bromine can also add to the double bond. N-Bromosuccinimide (NBS) is often used as the bromine source in free-radical brominations because it combines with the HBr side product to regenerate a constant low concentration of bromine. No additional bromine is needed because most samples of NBS contain traces of Br2 to initiate the reaction.
allylic bromination
allylic bromination
free-radical halogenation (synthetically useful only in certain cases)
free-radical halogenation (synthetically useful only in certain cases)
allylic bromination
nucleophilic substitution
a nucleophile (Nuc-) replaces a leaving group (X- ) from a carbon atom, using its lone pair of electrons to form a new bond to the carbon atom.
dehydrohalogenation elimination
Both the halide ion and another substituent are lost. A new π bond is formed. The reagent (B-) reacts as a base, abstracting a proton from the alkyl halide. Most nucleophiles are also basic and can engage in either substitution or elimination, depending on the alkyl halide and the reaction conditions.
nucleophilic substitution
a nucleophile (-OCH3) replaces a leaving group (Br-) from a carbon atom, using its lone pair of electrons to form a new bond to the carbon atom.
elimination
when OH is protonated, H2O is the leaving group
elimination
both Br atoms are lost, iodide ion is a nucleophile that reacts at Br.
SN2 (second-order nucleophilic substitution)
Hydroxide ion is a strong nucleophile (donor of an electron pair) because the oxygen atom has unshared pairs of electrons and a negative charge. Iodomethane is called the substrate, meaning the compound that is attacked by the reagent. The carbon atom of iodomethane is electrophilic because it is bonded to an electronegative iodine atom. Electron density is drawn away from carbon by the halogen atom, giving the carbon atom a partial positive charge. The negative charge of hydroxide ion is attracted to this partial positive charge.
SN2 (second-order nucleophilic substitution)
Hydroxide ion attacks the back side of the electrophilic carbon atom, donating a pair of electrons to form a new bond. This one-step mechanism is supported by kinetic information. The rate is found to double when the concentration of either reactant is doubled. The reaction is therefore first order in each of the reactants and second order overall. The rate equation has the following form:
rate = kr[CH3I][-OH]
explain the reaction-energy diagram for the SN2 reaction of methyl iodide with hydroxide
The reaction-energy diagram for the SN2 reaction of methyl iodide with hydroxide shows only one energy maximum: the transition state. There are no intermediates.
The electrostatic potential maps of the reactants, transition state, and products show that the negatively charged nucleophile (red) attacks the electrophilic (blue) region of the substrate. In the transition state, the negative charge (red) is delocalized over the nucleophile and the leaving group. The negative charge leaves with the leaving group.
SN2 reaction
Takes place in a single (concerted) step. A strong nucleophile attacks the electrophilic carbon, forcing the group to leave. The order of reactivity for substrates is CH3X > 1° > 2°. (3° alkyl halides cannot react by this mechanism.)
which is a stronger nucleophile, a base or its congugate acid?
base
which is a stronger nucleophile
R-O-
R-OH
R-O-
SN2 halogen exchange reaction
Alkyl fluorides are difficult to synthesize directly, and they are often made by treating alkyl chlorides or bromides with KF under conditions that use a crown ether to dissolve the fluoride salt in an aprotic solvent, which enhances the normally weak nucleophilicity of the fluoride ion.
SN2 halogen exchange reaction
Iodide is a good nucleophile, and many alkyl chlorides react with sodium iodide to give alkyl iodides.
SN2 halogen exchange reaction
Alkyl fluorides are difficult to synthesize directly, and they are often made by treating alkyl chlorides or bromides with KF under conditions that use a crown ether to dissolve the fluoride salt in an aprotic solvent, which enhances the normally weak nucleophilicity of the fluoride ion.
SN2 halogen exchange reaction
Iodide is a good nucleophile, and many alkyl chlorides react with sodium iodide to give alkyl iodides.
describe the periodic trend in nucleophilicity
Nucleophilicity increases down the periodic table, following the increase in size and polarizability, and the decrease in electronegativity.
- stronger* **I- ** > **Br- ** > Cl- > F- weaker
- stronger* -SeH > -SH > -OH weaker
- stronger* (CH3CH2)3P > ** (CH3CH2)3N** weaker
rate the nucleophilicity
carboxylate ion
alkoxide ion
alcohol
Alkoxide ion is the strongest nucelophile (similar to hydroxide ion), a species with a negative charge is a stronger nucleophile than a similar neutral species. In particular, a base is a stronger nucleophile than its conjugate acid.
Carboxylate ion is a moderate nucleophile, as carboxylic acids easily dissociate into a carboxylate anion and a positively charged hydrogen ion (proton), much more readily than alcohols do (into an alkoxide ion and a proton), because the carboxylate ion is stabilized by resonance. The negative charge that is left after deprotonation of the carboxyl group is delocalized between the two electronegativeoxygen atoms in a resonance structure.
Alcohol (similar to water) is the weakest nucleophile of the group, as the conjugate acid of alkoxide.
contrast basicity and nucleophilicity
Basicity is defined by the equilibrium constant for abstracting a proton. Nucleophilicity is defined by the rate of attack on an electrophilic carbon atom. In both cases, the nucleophile (or base) forms a new bond. If the new bond is to a pro- ton, it has reacted as a base; if the new bond is to carbon, it has reacted as a nucleophile.
describe the periodic trend in polarisability as it relates to the SN2 reaction
Trends in size and polarisability reflect an atom’s ability to engage in partial bonding as it begins to attack an electrophilic carbon atom. As we go down a column in the periodic table, the atoms become larger, with more electrons at a greater distance from the nucleus. The electrons are more loosely held, and the atom is more polarisable: Its electrons can move more freely toward a positive charge, resulting in stronger bonding in the transition state. The increased mobility of its electrons enhances the atom’s ability to begin to form a bond at a relatively long distance.
contrast flouride ion with iodide ion in the SN2 reaction
Fluoride has tightly bound electrons that cannot begin to form a C—F bond until the atoms are close together. Iodide has more loosely bound outer electrons that begin bonding earlier in the reaction.
steric hindrance
when bulky groups interfere with a reaction by virtue of their size
protic solvent
A protic solvent is one that has acidic protons, usually in the form of O - H or N - H groups.
describe the effect of anion size on solvation
Small anions are solvated more strongly than large anions in a protic solvent because the solvent approaches a small anion more closely and forms stronger hydrogen bonds. When an anion reacts as a nucleophile, energy is required to “strip off” some of the solvent molecules, breaking some of the hydrogen bonds that stabilized the solvated anion. More energy is required to strip off solvent from a small, strongly solvated ion such as fluoride than from a large, diffuse, less strongly solvated ion like iodide.
aprotic solvent
solvents without O - H or N - H groups, most polar, ionic reagents are insoluble in simple aprotic solvents such as alkanes
describle solvent effects on nucleophilicity of anions
In contrast with protic solvents, aprotic solvents (solvents without O - H or N - H groups) enhance the nucleophilicity of anions. An anion is more reactive in an aprotic solvent because it is not so strongly solvated. There are no hydrogen bonds to be broken when solvent must make way for the nucleophile to approach an electrophilic carbon atom.
polar aprotic solvent
Polar aprotic solvents have strong dipole moments to enhance solubility, yet they have no O - H or N - H groups to form hydrogen bonds with anions. Examples of useful polar aprotic solvents are acetonitrile, dimethylformamide, and acetone.
Fluoride ion, normally a poor nucleophile in hydroxylic (protic) solvents, can be a good nucleophile in an aprotic solvent. Although KF is not very soluble in acetonitrile, 18-crown-6 solvates the potassium ions, and the poorly solvated (and therefore nucleophilic) fluoride ion follows.
examples of ions that are strong bases and poor leaving groups in the SN2 reaction
list any stereochemical effects of the SN2 reaction
Walden inversion
an asymmetric carbon atom will undergo inversion of configuration
solvolysis
A SN1 reaction that takes place with the solvent acting as the nucleophile, its rate does not depend on the concentration of the nucleophile (usually methanol). The rate depends only on the concentration of the substrate.
What is the first step in the SN1 mechanism
formation of carbocation (rate limiting)
What is the second step in the SN1 reaction
nucleophilic attack on the carbocation
What is the third step in the SN1 reaction
loss of proton to solvent (only if the nucleophile is an uncharged molecule like water or an alcohol, the positively charged product must lose a proton to give the final uncharged product)
which compounds do not undergo SN1 or SN2 reactions?
Vinyl halides and aryl halides generally do not undergo SN1 or SN2 reactions. An SN1 reaction would require ionization to form a vinyl or aryl cation, either of which is less stable than most alkyl carbocations. An SN2 reaction would require back-side attack by the nucleophile, which is made impossible by the repulsion of the electrons in the double bond or aromatic ring.
what is the effect of a solvent’s dielectric constant on ionisation rate of tert-butyl chloride and what does that mean in terms of substitution reactions?
Ionization of an alkyl halide requires formation and separation of positive and negative charges, similar to what happens when sodium chloride dissolves in water. Therefore, SN1 reactions require highly polar solvents that strongly solvate ions. Note that ionisation occurs much faster in highly polar solvents such as water and alcohols. Although most alkyl halides are not soluble in water, they often dissolve in highly polar mixtures of acetone and alcohols with water.
list any stereochemical effects of the SN1 reaction
The SN1 reaction is not stereospecific. In the SN1 mechanism, the carbocation intermediate is sp2 hybridized and planar. A nucleophile can attack the carbocation from either face, because the carbocation is planar and achiral; and attack from both faces gives both enantiomers of the product. Such a process, giving both enantiomers of the product (whether or not the two enantiomers are produced in equal amounts), is called racemization. The product is either racemic or at least less optically pure than the starting material.
methyl shift
Primary halides and methyl halides rarely ionize to carbocations in solution. If a primary halide ionizes, it will likely ionize with rearrangement. The methyl shift occurs while bromide ion is leaving, so that only the more stable tertiary carbocation is formed.
SN1: a reluctant first-order substrate can be forced to ionize by adding some silver nitrate (one of the few soluble silver salts) to the reaction. Silver ion reacts with the halogen to form a silver halide (a highly exothermic reaction), generating the cation of the alkyl group.
first step of E1 mechanism
Unimolecular ionization to give a carbocation (rate-limiting)
second step of E1 mechanism
Deprotonation by a weak base (often the solvent) gives the alkene (fast).
The E1 reaction almost always competes with the SN1 reaction. Whenever a carbocation is formed, it can undergo either substitution or elimination, and mixtures of products often result.
E1 product:
Ionization of the alkyl halide gives a carbocation intermediate, which loses a proton to give the alkene. Ethanol serves as a base in the elimination.
SN1 product:
SN1 results from nucleophilic attack on the carbocation. Ethanol serves as a nucleophile in the substitution.
Provide the next step in the SN1 mechanism
Solvent (a weak nucleophile) attacks the rearranged (in a hydride shift, not shown) carbocation, deprotonation (not shown) gives the rearranged product.
Provide the next step in the E1 mechanism
The weakly basic solvent removes either adjacent proton on the rearranged (in a hydride shift, not shown) carbocation (fast).
What are the four ways a carbocation can react to become more stable?
- React with its own leaving group to return to the reactant: R+ + :X- -> R - X
- React with a nucleophile to form a substitution product (SN1): R+ + Nuc:- -> R - Nuc
- Lose a proton to form an elimination product (an alkene) (E1) (shown)
- Rearrange to a more stable carbocation, then react further.
The order of stability of carbocations is: resonance-stabilised, 3° > 2° > 1°.
Zaitsev’s rule
In elimination reactions, the most substituted alkene usually predominates.
Zaitsev orientation
E1 and E2 elimination reactions that produce the most substituted alkene
E2 reaction
In the general mechanism of the E2 reaction, a strong base abstracts a proton on a carbon atom adjacent to the one with the leaving group. As the base abstracts a proton, a double bond forms and the leaving group leaves. Like the SN2 reaction, the E2 is a concerted reaction in which bonds break and new bonds form at the same time, in a single step. (shown, methoxide reacts as a base rather than as a nucleophile).
E2 mechanism
The concerted E2 reaction takes place in a single step. A strong base abstracts a proton on a carbon next to the leaving group, and the leaving group leaves. The product is an alkene.
predict the major product and list the mechanism used
E2
predict the major product and list the mechanism used
E2
Name and describe the two conformations required for an E2 dehydrohalogenation
The E2 is a stereospecific reaction, because different stereoisomers of the starting material react to give different stereoisomers of the product. This stereospecificity results from the anti-coplanar transition state that is usually involved in the E2.
The E2 is a stereospecific reaction, because different stereoisomers of the starting material react to give different stereoisomers of the product. This stereospecificity results from the anti-coplanar transition state that is usually involved in the E2.
E1 vs E2: orientation of elimination
E1, E2: Usually Zaitsev orientation.
In most E1 and E2 eliminations with two or more possible products, the product with the most substituted double bond (the most stable product) predominates. This principle is called Zaitsev’s rule, and the most highly substituted product is called the Zaitsev product.
SN1
SN2
E1
E2
