Organic Chem Reactions

Front 1

What is needed for the addition of hydrogen to alkenes and alkynes?

Back 1

A platinum or palladium catalyst. 'Lindlar catalysts' are used to stop the reaction of alkynes at the alkene stage, otherwise, hydrogen adds until all pi bonds are reacted

Front 2

What is a good method to form new carbon carbon bonds using alkynes?

Back 2

Using NaNH2, you can convert an alkyne to an acetylide ion. This ion can react with primary alkyl halides or methyl halides.

Front 3

Describe the reaction of conjugated dienes and an electrophilic reagent.

Back 3

1,2 and 1,4 addition can occur. The hydrogen (i.e. off of H-Br) first adds to the double bond adjacent to the most substituted sp2 carbon and a cabocation is formed. The nucleophile then can add to that carbocation (1,2-addition), or can add to the 4th carbon on the conjugated system based on resonace contributers.

Front 4

What is electrophilic aromatic substitution (EAS)?

Back 4

EAS is when an electrophile adds to an aromatic compound in place of a hydrogen.

Front 5

What catalyst is needed in order to halogenate benzene?

Back 5

A Lewis acid (accepts electron pairs) like FeBr3 or FeCl3

Front 6

What is needed in the nitration of benzene?

Back 6

Nitric acid (HNO3) and sulfuric acid are needed. This adds NO2 to the ring.

Front 7

What is needed in the sulfonation of benzene?

Back 7

Fuming sulfuric acid (a solution of SO3 in sulfuric acid)

Front 8

Explain a Friedel-Crafts Acylation.

Back 8

An acyl groub is added by starting with an acyl chloride and a lewis acid (like AlCl3). The lewis acid gets reduced to AlCl4- and the benzene's pi electrons can attack the newly formed carbocation.

Front 9

How does benzene undergo alkylation?

Back 9

Similarly to the Friedel-Crafts acylation. Benzene and an alkyl halide react with the presence of a lewis acid.

Front 10

How can benzene an an alkene react?

Back 10

In the presence of an acid, the alkene forms a carbocation which can add to the benzene ring.

Front 11

How do substituents on benzene affect the orientation of a reaction?

Back 11

Activating substituents (electron donating and halogens) direct new reactions to the ortho and para positions, while deactivation substituents (electron withdrawing groups) direct the new reaction to the meta position.

Front 12

How can existing substituents be modified? (NO2, acyl groups, alkyl substituents)

Back 12

-NO2 can be reduced by hydrogen gas and a Pt/Pd catalyst to NH2. The same goes for an acyl group. The carbonyl oxygen will be replaced with two hydrogens.

Also, H2CrO4 and heat can also oxidize alkyl groups bonded to benzene and turn them into carboxylic acid. This happens regardless of the length of the alkyl substituent.

Front 13

Explain the basics behind a substitution and elimination reaction. What kinds of molecules undergo these reactions?

Back 13

A nucleophile attached to an sp3 carbon will either be substituted with another atom or group (substitution) and in an elimination, the nucleophilic atom or group will be removed, along with a hydrogen and a double bond forms. The nucleophilic atom or group that gets substituted or replaced is called the leaving group.

Front 14

What is an Sn2 reaction? What specific groups are involved?

Back 14

A substitution reaction that has two molecules involved in the rate determining step. It is a single step bi(2)molecular substitution reaction.

A nucleophile and an alkyl halide are involved. The alkyl halide undergoes back-side attact by the nucl. at the same time the leaving group (halide) leaves.

Front 15

Explain the factors involved in proving the Sn2 mechanism.

Back 15

1. The rate of the reaction depends on the concentration of both species (alkyl halide and nucleophile)

2. As the hydrogens of methyl bromide are successively replaced with methyl groups, the rate of the reaction gets slower

3. When an alkyl halide is bonded to an asymmetric center, the product of the reaction leads to one stereoisomer, whos configuration is the opposite of the initial alkyl halide's.

Front 16

Why does adding methyl groups in place of hydrogen in methyl bromide slow the reaction rate in an Sn2 reaction?

Back 16

Because the incoming nucleophile attacks the back of the carbon, larger groups off the carbon increase steric hindrance.

Front 17

Which types of alkyl halides are most reactive in an Sn2?

Back 17

Methyl halide> primary halide> secondary....

Front 18

Explain how the leaving group affects an Sn2 reaction. Why?

Back 18

The weaker the basicity, the faster the rate. That is because weak bases are stable bases.

Front 19

What affects do the incoming nucleophile have on an Sn2 rate?

Back 19

The better the nucleophilicty of the incoming group, the faster the rate.

Front 20

Describe an Sn1 reaction.

Back 20

It is a substitution reaction where the rate determining step involves one molecule, which is an alkyl halide.

This reaction is a two step reaction involving the same species as an Sn1 (alkyl halide and nucleophile)

First, the leaving group leaves, then the nucleophile comes in and bonds to the newly formed carbocation.

Front 21

Explain factors in determining the E1 mechanism.

Back 21

1. The rate law shows that the rate is dependent on only the concentration of the alkyl halide.

2. When tert-butyl halides' methyl groups are successively replaced by hydrogens, the reaction rate increases. This is opposite from Sn2

3. When a halogen is bonded to an asymmetric center it forms two stereoisomer.

Front 22

Why do tertiary alkyl halides react faster than secondary, primary, and methyl halides in an Sn1 reaction.

Back 22

Because the extra electron density provided by the methyl groups helps stabilize the carbocation once the leaving group leaves.

Front 23

Explain the stereochemistry involved in an Sn1.

Back 23

Because the leaving group first leaves the alkyl halide, an sp2 carbon is formed and is no longer tetrahedral, but planar.

This allows the nucleophile to come in from either the top or bottom of the sp2 carbon's place, hence creating both isomer.

Front 24

How does the leaving group affect an Sn1 reaction?

Back 24

The weaker the base, the weaker its bonded to the carbon, and the easier it is to dissociate from that carbon.

This is the same as the Sn2 rule.

Front 25

How does the nucleophile affect the Sn1 reaction?

Back 25

Because the rate of the reaction is independent from the nucleophile, it does not have any affect on the reaction.

Front 26

Explain the E2 reaction.

Back 26

An E2 reaction is an elimination reaction where the rate determining step is bimolecular.

A base and an alkyl halide are involved. The base removes a proton from a beta-carbon (adjacent to the carbon bonded to the halogen) while the electron that held the hydrogen form a double bond with the alpha-carbon, which kicks out the halogen.

This reaction is one step.

Front 27

Explain the E1 reaction.

Back 27

An elimination reaction where the rate determining step is unimolecular. The rate depends only on the concentration of the alkyl halide.

The first step involves dissociation of the halide, creating a carbocation. In the second step, a base removes a beta-proton and those electrons from the broken bond form a double bond with the alpha-carbon.

Front 28

Give one factor in the reactivities of elimination reactions (E2 and E1)

Back 28

The leaving group that is the weakest base increases reactivity.

Front 29

Explain the regioselectivity of E2 reactions.

Back 29

Because an alkene if formed, the most stable alkene created in the product is favored. A simple rule tells that the more substituted alkene formed is most favored.

Thus, we can deduce that tertiary alkyl halides are more reactive than secondar, etc. because they will lead to the more substituted alkene.

This means that the base removes a proton from the most substituted beta-carbon.

Front 30

Explain why tertiary alkyl halides are more reactive than secondary, etc. in an E1 reaction.

Back 30

Because a carbocation is formed in E1, the most substituted gives the most electron density to the cation.

Front 31

Give an example of how one isomer over another is favored in E reactions.

Back 31

When 2-bromobutane reacts with a base, a double bond is formed between the 2 and 3 carbon, leaving a methyl group and a hydrogen on either side of the double bond.

The isomer that has the methyl groups on opposite sides of the bond is the major product because os steric reductions. This will be the E isomer.

Front 32

What is the difference between a base and a nucleophile?

Back 32

Bases remove protons (like in eliminations) and nucleophiles react with electron poor regions (like in substitutions).

Front 33

Which two factors determine if am alkyl halide will undergo Sn2/E2 or Sn1/E1?

Back 33

1. The concentration of the nucleophile/base

2. The reactivity of the nucleophile/base

Front 34

Why does increasing the concentration and/or reactivity of base increase the rate of Sn2/E2?

Back 34

Because the rate law for both includes that species in the law.

Similarly, because the nucl./base isn't in the rate determining step in Sn1/E1, it doesn't matter how strong the nucl./base is.

Front 35

Briefly describe the favorable conditions for Sn2/E2 and Sn1/E1 reactions.

Back 35

Highly reactive bases and a good nucl. in high concentrations favor Sn2

Weak bases and poor nucleophiles in low concentration favor Sn1/E1.

Front 36

What can determine if an Sn2 or E2 reaction happens.

Back 36

In Sn2/E2 conditions, substitution of the alkyl halide affects the mechanism.

For primary alkyl halides, the major product is an Sn2 reaction.

A secondary produced both, but with E2 favoring.

A tertiary alkyl halide will always produce the elimination product.

Front 37

What can determine if an Sn1 or E1 reaction happens.

Back 37

Because both form a carbocation, alkyl halides have the same order of reactivity in Sn1/E1.

Front 38

Describe what is necessary in the substitution of alcohols.

Back 38

Since the -OH group is a poor leaving group, it can only be susbstituted by being made into a better leaving group (less nucleophilic). This can be done with an acid that protonates the -OH group to -OH2+.

Furthermore, only weakly basic nucleophiles can be used, since strong bases would be protonated in solution, and would no longer be nucleophilic.

Front 39

Describe the elimination of alcohols (dehydration).

Back 39

This reactions requires heat and an acid catalyst.

Front 40

What happens in the oxidation of alcohols?

Back 40

It increases the number of C-O bonds (decreasing the # of C-H bonds)

Front 41

What are some typical catalysts in the oxidation of alcohols and when are they used.

Back 41

Chromic acid (H2CrO4) is used to catalyze this.

pyridinium chlorochromate (PCC) is used to quench the oxidation of a primary alcohol at the aldehyde stage instead of furhtering it to a carboxylic acid.

Front 42

What types of alcohols can be oxidized?

Back 42

Only primary and secondary alcohols can be since tertiary alcohols don't have a C-H bond.

Front 43

Why don't amines undergo sub. or elim.?

Back 43

Because the leaving group, -NH2, is extremely basic. Even if protonated, it does not undergo the reactions. Compare the pKas of the leaving groups of halides, alcohols and amines.

HF-3.2
H2O-15.7
NH3-36

Front 44

Why are substitution reaction of ethers similar to alcohols?

Back 44

Because of the similarity of their leaving groups. R-O for ehters and O-H for alcohols.

Front 45

Explain the reactivity of carboxyl compounds.

Back 45

These reactions are categorized as nucleophilic acyl substitutions.

Because of the more electronegative oxygen, the carbon is somewhat electron deficient and therefore an electrophile.

A nucleophile can attack the carbon and cause a tetrahedral intermediate, with a negative charge on the oxygen.

If the carbon was bonded to a less basic atom or group, that group is expelled, with the new nucleophile attached to the carbon.

Front 46

What rule is there about the reaction of carboxyl compounds?

Back 46

Carboxyl compounds only react to form less reactive compounds because a stronger base (ie worse leaving group) comes in.

Front 47

Describe three reactions that esters can undergo. Give a trend.

Back 47

Hydrolysis-acid catalyzed rxn. where water adds to carboxyl carbon to form carboxylic acid.

Transesterification- acid catalyzed where an alcohol reacts to form a new ester.

aminolysis- amines react to form an amide.

The trend is, in each, the incoming nucleophile adds to the carboxyl carbon, and loses one hydrogen.

Front 48

How can carboxylix acids be transformed into more reactive species.

Back 48

Carboxylic acids are not very reactive to acyl substitution because the OH is a bad leaving group. One way to get around this is to react a carboxylic acid with thionyl chloride (SOCl2) to make acyl chlorides.

Then, since acyl chlorides are reactive, we can make carboxylic acid derivatives by simply adding the appropriate nucleophile (such as an ester or amine)

Front 49

What two carbonyl compounds do not undergo nucleophilic acyl substitution? What kind of reactions do these compounds undergo? Why?

Back 49

Aldehydes and ketones. They undergo nucleophilic addition because they don't have leaving groups to leave like other carbonyl compounds.

Front 50

What are Grignard Reagents used for? How are they prepared?

Back 50

These reagents are used to create a nucleophilic carbon that can add to a carbonyl carbon. The reagent is prepared by mixing an alkyl halide with magnesium powder in diethylether. The resulting reagent is R-MgBr

Front 51

What happens when a Grignard reagent is in the presence of an acid, even when its in low concentration and a very weak acid, such as water, an alcohol, and an amine?

Back 51

The acid protonates the very nucleophilic carbon to form an alkane.

Front 52

Explain the reaction of a Grignard reagent with carbonyl compounds like aldehydes and ketones?

Back 52

The Grignard reagent attacks the carbonyl carbon and an alkoxide ion is formed.

An acid then protonates the ion and forms an OH group.

Front 53

Explain the difference between the Class I and II carbonyl compounds.

Back 53

Class I have leaving groups that can be substituted out with another nucleophile. Groups like a halide, alcohol, ester, and amine.

Class II compounds, like aldehydes and ketones, cannot be substituted and only undergo nucleophilic addition.

Front 54

Explain how Grignar reagents react with class I carbonyl compounds. What rule is there when more than one equivalent of Grignard reagent is added to the carbonyl?

Back 54

The nucleophilic carbon from the grignard reagent attacks the carbonyl carbon and a tetrahedral intermediate is formed. The now negatively graged oxygen creates a pi bond again with the carbonyl carbon and the leaving group is expelled.

Once the leaving group has left, Another Grignard reagent can add and form another alkoxide ion. An acid can add to this and make it an alcohol.

Front 55

How do hydride ions react with class II carbonyl compounds? Name a typical source for the hydroxide ion.

Back 55

They hydride acts as a very good nucleophile and attacks the cabonyl carbon. A tetrahedral intermediate is formed and an acid is later added to add to the oxygen.

Sodium borohydride (NaBH4) is usually used to give an alkoxide ion.

Front 56

How do hydride ions react with Group I carbonyl compounds? What differences are there between group II compounds.

Back 56

Similarly to group II compounds, but that LiAlH4 is used as a source for hydride ions.

Also, the first hydride addition to a carboxylic acid, for example, creates a ketone, which then creates an alcohol.

Front 57

What kind of reactions do hydrides undergo?

Back 57

Reduction reactions.

Front 58

Describe the reactions of that aldehydes/ketones undergo with primary amines.

Back 58

Basically, the oxygen and nitrogen switch places, while the nitrogen brings along its substituents, the oxygen is eliminated as water. The double bond is between the carbon and nitrogen. Trace amounts of acid are needed.

When a primary amine (like methylamine and ethylamine) reacts with a ketone or aldehyde, the C-O double bond is replaced with a C-N double bond, and water is eliminated.

Trace amounts of acid are needed. THe amine attacks the carbonyl carbon, the negative oxygen picks up a hydrogen from the acid. Then the positive nitrogen dissociates a proton, which the oxygen can pick up to now become a good -H2O+ leaving group. The nitrogen's lone pair makes a pi bond with the previous carbonyl carbon and water is eliminated. Then the nitrogen looses its last hydrogen.

Front 59

Describe the reactions between secondary amines and aldehdes/ketones.

Back 59

Basically, the oxygen and nitrogen switch places. The nitrogen brings along is substituents and the oxygen is eliminated as water. A new double bond is formed between the alpha and beta carbon.

Front 60

Explain an aldol condensation.

Back 60

It is where two ketones or aldehydes combine to form one new product. A carbon carbon bond is formed.

This is base catalyzed, as an alpha carbon has a hydrogen removed and becomes a nucleophile. That nucleophile can attack the electrophilic carbonyl carbon on the other molecule.

A second step where acid is added is required.

Front 61

When will terminal alkynes add water?

Back 61

When an acid and a mercury ion are in solution.

Front 62

Explain how ozone can react with alkenes.

Back 62

Ozone will react with alkenes by taking the carbon carbon double bond, and splitting it to make two c-o double bonds.

Usually, dimethyl sulfer is used in this reaction in the second step.

Front 63

What is an imine? How is it formed?

Back 63

An imine is a compound with a carbon nitrogen double bond. It is formed when an aldehyde or ketone reacts with a primary amine and trace amounts of acid.

Front 64

What is an enamine? How are they made?

Back 64

They are compounds that have a carbon carbon double bond between the alpha and beta carbons realative to the nitrogen.

They are made when a ketone or aldehyde reacts with a secondary amine.

Front 65

What does NaNO2, and cold HCl do to a nitro group on benzene?

Back 65

The nitrogen on the ring forms a triple bond with the incoming nitrogen. The inner nitrogen (closest to benzene) has a positive charge. This makes a good leaving group.