Organic Chemistry Study Sheet

Organic Chemistry Study Sheet
Things to Know
* Substitution reactions convert one group to another
* Elimination reactions convert alkyl halides or alcohols into alkenes
* Addition reactions add two groups across a double bond
Things to Consider
* Addition Reactions
o What are the identities of the groups added across the double bond?
o What is the expected regioselectivity (Mark or anti-Mark)?
o What is the expected stereospecificity (syn or anti)?
Some General Terms
Markovnikov Addition- The substituent is placed in the vinylic position bearing more hydrogen atoms (more substituted)
Anti-Markovnikov addition- The substituent is placed in the vinylic position bearing less hydrogen atoms (less substituted)
Regioselectivity
Syn addition

-Addition of H&X across a double bond
* Hydrohalogenation
o Markovnikov is achieved by adding pure HX.
o Anti-Markovnikov is achieved by adding HX with any trace of ROORs.
o Mechanism
o Regioselectivity depends on the preference of the reaction to proceed through the more stable carbocation.
o In many cases, a stereocenter is formed, resulting in the formation of enantiomers in a racemic mixture.
o Carbocation rearrangements will generally occur if they can, via a methyl or hydride shift. Therefore, a mixture of products will form. For example, a secondary carbocation can be converted to a tertiary carbocation. A nucleophilic attack can occur on either carbocation, resulting in a mixture
-Addition of H&OH across a double bond
* Acid-Catalyzed Hydration
o Most simple alkenes will undergo Markovnikov addition. The -OH group is placed in the more substituted position.
o H3O+ acts as a reagent, representing the presence of water and an acid source such as H2SO4
o Rate depends on the starting alkene, but generally, the more alkyl substituents they are, the faster the rate.
o Mechanism
o Tertiary carbocations will react faster than secondary ones.
o Since this is an equilibrium reaction, it is sensitive to changes in compliance with Le Chatlier’s Principle. Acid-catalyzed reactions are sensitive to temperature and [H2O].

o Intermediate carbocations can be attacked from either side so a new chirality center is generated, resulting in a racemic mixture of enantiomers.
* Oxymercuration-Demercuration
o Similar to acid-catalyzed reaction, but does not undergo a carbocation arrangement. So this is good for achieving a substantial majority of the Markovnikov product.
o Mechanism
* Hydroboration-Oxidation
o Another way to add water, but the goal is to produce the anti-Markovnikov product.
o In the case, where 2 new chirality centers are formed (as with a cyclo group), syn addition occurs.
o Borane (BH3) is somewhat similar in structure to a carbocation, but it lacks the octet of electrons, making it more reactive. In this reaction, it will react with another borane to produce diborane. The resonance structure of diborane is one of the few cases which can break the single bond and the charge density is spread over several atoms.
o To stabilize BH3, solvents such as THF can be used. THF donates electron density into the empty p orbital of boron.
o In hydroboration, the p bond attacking of borane triggers a simultaneous hydride shift (C-BH2 and C-H form. This first step explains the regeioselectivity, why it favors the anti-Markovnikov product.
* Electronic considerations- this first step does not have to be perfectly simultaneous so when the p bond attacks, one of the vinylic positions develops a partial positive charge. This partial charge triggers a hydride shift. One of the vinylic carbons develops a partial positive charge when the alkene interacts with borane. Positive charges prefers to develop on the more substituted carbon so the BH3 group has to be placed at the less substituted carbon
* Steric considerations- BH2 is larger than H so the transition state will be lower in energy if the BH2 group is placed in the less sterically hindered position.
o Stereospecificity comes into play because the first step is concerted. This produces syn addition.
* Zero chirality centers- not relevant
* One chirality center- both enantiomers are obtained because syn addition occurs on either side of the alkene with equal likelihood
* Two chirality centers- to fulfill syn addition requirements, a pair of enantiomers forms
-Addition of H2 across a double bond
* Catalytic Hydrogenation
o Reduces an alkene to an alkane in the presence of a metal catalyst
o Stereospecificity
* Zero chirality centers- only one product is formed
* One chirality center- both enantiomers form
* Two chirality centers- pair of enantiomers with syn addition forms
* The metal catalyst can be Pt, Pd, or Ni, to name a few. These three are heterogeneous catalysts because they do not dissolve in the reaction medium. This is the “alien abduction” reaction. The hydrogen atoms are absorbed to the surface of the metal and an alkene will come and pick these hydrogen atoms up on the same side.
o Catalysts
* Pt, Pd, Ni
* Homogeneous catalysts are soluble in the reaction medium. The most commonly used one for hydrogenation is Wilkinson’s catalyst. Syn-addition is observed.

* Wilkinson’s catalyst also forms a pair of enantiomers for 1 and 2 chirality centers.
* Asymmetric Catalytic Hydrogenation allows only one of the enantiomers for form and is possible with the use a chiral catalyst.
* Ru(BINAP)Cl2 is commonly used. BINAP serves as the ligand but does not have a chirality center. It is chiral though because the single bond does not freely rotate given steric hindrance.
-Addition of X2 (Br2 or Cl2) across an alkene
* Halogenation
o Occurs with Br and Cl because F is too violent and I produces low yields.
o Stereospecificity
* Two chirality centers- anti addition occurs. This is because the second step (Nucleophilic attack) is an SN2 reaction, requiring a back-side attack.
o Stereochemistry
* Starting cis will produce a pair of enantiomers
* Starting trans will produce a meso compound.
o Mechanism
-Addition of Br and OH across an alkene
* Halohydrin Formation
o Occurs in the presence of water so that the bromonium ion is captured by the water molecule.
o Since the bromonium intermediate is high in energy, it reacts with any nucleophile it encounters. The bromonium ion is captured by the water molecule before it can react with another bromide ion, but some dibromide will form.
o Mechanism
o Regiochemistry
* OH occupies the more substituted position. The partial positive charge on the ion begins on the bromine atom and ends up on the oxygen atom. In the transition state, the charge passes through the carbon, giving it a partial carbocationic character. Water will attack the more substituted carbon because it is more capable of stabilizing the partial positive. So transition state takes place at the more substituted position.
-Addition of OH and OH across an alkene
* Anti Dihydroxylation
o Alkene is converted to an epoxide, a three-membered, cyclic ether. This occurs via a rxn with a peroxy axid (RCO3H), commonly peroxyacetic acid or meta-Chloroperoxylbenzoic acid (MCPBA). They are strong oxidizing agents and essentially add an oxygen atom to the alkene. It is opened via acid-catalyzed or base-catalyzed conditions.
* Syn Dihydroxylation
o Alkene can be treated with osmium tetroxide (OsO4), adding in a concerted process so that the oxygen atoms attach to the alkene simultaneously. The cyclic osmate ester is treated with either sodium sulfite, Na2SO3 (aq), or sodium bisulfite (NaHSO3) to produce a diol. This yields high yields.
o Caveat is that OsO4 is expensive so we use it in catalytic amounts and add co-oxidants. These include N-methylmorpholine (NMO) or tert-butyl hydroperoxide.
o Mechanism(s):

-Cleaving of a C=C bond
* Oxidative Cleavage
o Ozone can react with an alkene to produce an initial, primary ozonide. It is then treated with a mild reducing agent such as dimethyl (DMS) or Zn/H2O.
o Example
Some General Terms
Geminal- both groups are attached to the same carbon
Vicinal- groups are attached to adjacent carbons
Terminal- triple bond as a terminating bond (end)
Internal- triple bond is an internal bond (somewhere in the middle)
Tautomers- constitutional isomers that rapidly interconvert via proton migration
-Preparation of Alkynes
* Using alkyl dihalides, treat with bases in two successive E2 reactions. First elimination can use many different bases, but second one requires a strong base. NaNH2 dissolved in NH3 should do the trick. This is commonly used for terminal alkynes. Three equivalents of the amide are needed, two for the two E2 reactions, one to deprotonate the alkyne and form the alkynide ion for protonation, as shown below.
* A terminal alkyne can be formed by treatment with water as a second step.
-Reduction of Alkynes
* Catalytic Hydrogenation
o Two equivalents of H2 with Pt will yield an alkane from an alkyne. A cis alkene will be too difficult to isolate because it is more reactive than the starting alkyne.
o Using a partially deactivated catalyst, such as a poisoned catalyst, can convert and isolate the cis alkene. Common poisoned catalysts include Lindlar’s catalyst and P-2 (Ni2B).
* Cis because the H atoms are added to the same face of the alkene via the alien abduction method.
o Mechanism(s)
1.
2.
* Dissolving Metal Reduction
o Treatment with reagents Na in NH3(l). These reagents requires low temperatures.
o Mechanism
o The first step forms a radical anion, which is a trans alkene because it is lower in energy. The cis alkene features steric hindrance between the radical and lone pair.

-Treatment of Alkynes with HX
* Hydrohalogenation
o Installs the halogen via Markovnikov addition, i.e. in the more substituted position.
o Mechanism
o A vinylic carbocation is formed. The reaction is expected proceed through the more stable, secondary vinylic carbocation, but it can form through similar or prmary. HX addition to alkynes is slightly slower than addition to alkenes. The vinylic carbocation exhibits partial carbocationic character. The transition state will therefore be lower in energy when the partial charge forms on the more substituted position.
o Two successive addition reactions (xs HX) will yield a geminal dihalide.
* Radical addition
o Treatment with HX and ROOR. In terminal alkynes, the Br occupies the terminal position and a mixture of E and Z isomers form.
o Mechanism
-Addition of OH group in alkynes
* Acid-catalyzed hydration
o Formation of ketone occurs because the installation of OH results in the unstable enol.
o Slower than its alkene counterpart because a high-energy vinylic carbocation intermediate is formed. To rectify this, mercuric sulfate (HgSO4) can catalyze the reaction.
o Mechanism
o The p bond of the enol is protonate to generate a resonance-stabilized intermediate, which is then deprotonated to generate a ketone. The p bond position will move.
o Enols and ketones are tautomers. Interconversion between the two is known as keto-enol tautomerization, which is an equilibrium process. This means specific [enol] and [ketone] are established. Generally, ketone is favored and once the equilibrium is reached, these [ ] can be measured. It can be catalyzed with trace amounts of acid or base.
o An enol produced by hydration of alkene hydration will tautomerize to form a ketone.
o Acid-catalyzed hydration of asymmetric internal alkynes yield a mixture of ketones
o This is often used for a terminal alkyne, yielding a methyl ketone.
* Hydroboration-Oxidation of Alkynes
o Anti-Markovnikov addition of OH
o Produces an enol with undergoes tautomerization to form an aldehyde.
o Mechanism
o In acid-catalyzed conditions, enol is protonated, then deprotonated. In base-catalyzed conditions, enol is deprotated to form a resonance-stailized anion called enolate ion and is protonated to generate the aldehyde.
o This is a concerted process. Since alkynes have 2 p bonds, to prevent a second addition of BH3, dialkyl boranes can be used. These include disiamylborane and 9-BBN.
o Example
-Addition of X2
* Halogenation
1. Since alkynes have 2 p bonds, it can undergo a reaction to form a tetrahalide using two equivalents of the halogen as shown below.
2. If one equivalent is used, it will produce a dihalide via anti addition. The E product will be the major product.
-Formation of Carboxlic Acids
* Ozonolysis
1. Treatment with 1)O3, 2)H2O to carboxlic acids.
2. Terminal alkynes are converted to carbon dioxides on their terminal side.

-Addition of alkyl groups
* Alkylation
1. Treatment of a terminal alkyne with a strong base such as NaNH2 will produce an alkynide ion, which functions as a nucleophile. When treated with an alkyl halide, it undergoes an SN2 reaction to install methyl or primary alkyl halides. IF TREATED WITH SECONDARY OR TERTIARY, IT WILL NOT WORK. ALKYNIDE ION WILL FUCTION AS A BASE AND ELIMINATION PRODUCTS ARE OBTAINED.
2. Say a reactant has two terminal protons. In two separate alkylations, both sides of the terminal alkyne are alkylated. The NaNH2 and RX are added SEPARATELY. The NaNH2 is allowed to react completely before the RX is put in. It enables the installation of two different alkyl groups.
Things to know
* Unlike sp2 hybridized, trigonal planar carbocations, carbanions are sp3 hybridized and trigonal pyramidal. This is shown below
* Radicals exhibit a trigonal planar of rapidly inverting shallow pyramid structure. This affects the stereochemistry
* In terms of stability, tertiary radicals are most stable, working their way down to methyl radicals. The more stable, the lower the BDEs.

* Radicals can exhibit resonance. Namely, an unpaired electron is allylic to a p bond. Resonance-stabilized radials are more stable than tertiary radicals. A BDE argument is also used.

* Remember, allylic =/= vinylic, the latter does not have a resonance structure and is less stable than a primary radical.
* Reactions proceed via three steps: initiation, propagation, termination
* Arrow-pushing patterns are given by:
1. Homolytic cleavage- requires large input of energy in form of heat (?) or light (hv)

2. Addition to a p bond- radical adds to a p bond, destroying the p bond and forming a new radical

3. Hydrogen abstraction- radical abstracts a hydrogen, not proton transfer (ionic step)

4. Halogen abstraction- radical abstracts a halogen

5. Elimination- the position bearing an electron is in the a position. In an elimination, a double bond will form between a and ß. The single bond at the ß position is cleaved so double bond and radical forms.

6. Coupling- two radicals join together in a bond

* Radical initiators are compounds with weak bonds that undergo homolytic bond cleavage easily. Acyl peroxides are used because O-O bond is weak. The radical from this is resonance stabilized
* Radical inhibitors prevent a process from continuing or starting. They destroy radicals. An example is hydroquinone. Using this, hydrogen abstraction occurs and a resonance-stabilized radical which is less reactive than the original radical, occurs. This radical destroys another radial via hydrogen abstraction to form benzoquinone.
-Replacement with a Halogen (Radicals involving alkanes)
* Chlorination
o Mechanism
o Chain reaction can occur in which xs Cl2 can be supplied to replace H with Cl.
o Bromination will not occur because the first propagation step, hydrogen abstraction, has a high activation energy.
o Occurs at the secondary position more readily than primary because the rate-determining hydrogen abstraction step has a transition state lower in energy for a secondary radical. A mixture of products does form however with secondary and primary positions incurring the halogen.

o Bromination is more selective than chlorination. The amount of radical character in bromination is larger and will be more sensitive to the substrate. The transition state is closer in energy to reactants so the bond between H and the carbon is only beginning to break and the inverse is true for bromination. Tertiary positions are favored in F, Cl, and Br.
o Stereochemistry
* A new chirality center can form if the reactant originally had none, producing a racemic mixture
* The process can also occur at an existing chirality center, producing a racemic mixture.
-Installation of Br in an allylic position (Radicals involving alkenes)
* Allylic Bromination
o addition of 1)Br2 and 2)hv. The problem with this is that Br2 can create competition between allylic bromination and ionic addition. So [Br2] needs to be kept at a minimum; this is achieved by using NBS. The N-Br bond is cleaved to produce a bromine radical. It undergots hydrogen abstraction to form a resonance-stabilized radical and HBr. Then the HBr reacts with NBS ionically to form Br2. The second propagation step involve halogen abstraction to form the allylic bromine and a bromine radical.

-Oxidation processes
* Autooxidation
o Oxidation of organic compounds in the presence of organic compounds
o Example
o Mechanism

o This results in the formation of hydroperoxide, which are unstable and decompose violently.
o Initiated by light.
o To prevent food from going rancid quickly, preservatives such as BHT and BHA. They function as radical inhibitors because they react with radicals to form resonance-stabilized radicals. They are called antioxidants because one molecule of either can prevent autooxidation.
o Vitamins E and C are natural antioxidants. E is hydrophobic b/c it has a long carbon chain. C is hydrophilic because it is composed of mostly OH groups.
-Radical Addition of Hr (Anti-Markovnikov)
* Mechanism

o The propagation steps are responsible for the stereochemistry. The first propagation is a tertiary carbon radical, which is more stable and is favored. The intermediate is a radical. The alkene will react with a bromine first, unlike in an ionic mechanism, where it will react with a proton first.
* The ionic produces Markovnikov
* The radical produces anti-Markovnikov
o HBr is thermodynamically favored since HI fails first propagation step, HCl fails second.
-Radical Polymerization
* Chain branching will occur
* In a substituted ethylene the following will occur

*
Alcohols
* The carbon bearing the hydroxyl group is the a position.
* Because of hydrogen bonding interactions that occur for alcohols, the boiling pt is higher.
* Every alcohol has two regions.
o Hydrophobic where there are atoms such as carbon and hydrogen
o Hydrophilic where there is a hydroxyl group
* To evaluate the stability, deprotonate and assess the stability of the conjugate base. The conjugate base of an alcohol is an alkoxide ion, which exhibits a negative charge on the oxygen atom.

* There are two common ways to deprotonate alcohols
o A strong base such as NaH can be used. The hydride deprotonates the alcohol to generate hydrogen gas, which bubbles out of the solution.
o It is more practical to use Li, Na, or K as these metals will react with the alcohol and release H2 gas.
* Acidity
o Resonance- A resonance stabilized conjugate base means the acid is strong. A phenolate ion, resulting from the deprotonation of phenol, is more acidic than say a cyclohexane because it is resonance stabilized. Phenol therefore doesn’t need to be deprotonated by a strong base such as NaH, instead it can be deprotonated by hydroxide.
o Induction- Conjugate bases that are stabilized by electron-withdrawing effects are generally more acidic
o Solvation Effects- Less sterically hindered compounds are easily solvated and stabilized by a solvent so they would be more stable than sterically hindered compounds.
* Preparation of Alcohols
o Substitution reactions can convert primary and tertiary substrates into alcohols (via NaOH for SN2 and water for SN1), but secondary substrates aren’t effectively converted using substitution reactions.
o Acid-catalyzed hydration will produce a Markovnikov product, oxymercuration-demercuration will produce a Markovnikov product without a carbocation rearrangement (thereby limiting the amount of products formed), and hydroboration-oxidation will produce an anti-Markovnikov product.

* Alcohols can also be prepared via reduction.
* Oxidation states are a method of bookkeeping in Organic Chemistry, employing formal charge. Bonds are broken heterolytically so the more electronegative atom will theoretically take all the electrons while the less electronegative will take none. A formal charge calculation is then performed.
o Increase in oxidation number is an oxidation reaction
o Decrease in oxidation number is a reduction reaction.
* Reduction
o Ketones and aldehydes can be converted to alcohols using a reducing agent.
* Hydrogenation in the presence of a metal catalyst will work, but requires higher temperature and pressure.
* Sodium borohydride (NaBH4) works. It is a source of hydride and the solvent function as a source of a proton. The solvent can be ethanol, methanol or water. The first step involves the transfer of hydride to the carbonyl group.
* It functions as a delivery agent of the nucleophilic hydride. This cannot be achieved using sodium hydride because the hydride is not polarizable and will act as a base not a nucleophile.
* Mechanism
* Lithium aluminum hydride (LAH) works too. It is a delivery agent as well, but it is a much stronger reagent. It reacts violently with water and therefore, a protic solvent cannot react while LAH is still in the reaction flask. H3O+ can also act as a proton source.
* Mechanism
o NaBH4 and LAH are generally more favored because catalytic hydrogenation requires high temperature and pressure.

* Derivative of these two agents can also be used. R groups on the aluminum can be alkyl, cyano, or alkoxy to name a few. The choice of R groups can make the compound either electron donating or withdrawing to modify the reactivity.
* Because LAH is so reactive, it is less selective than NaBH4. LAH will react with a carboxylic acid or an ester to produce an alcohol, but NaBH4 will not.
* Mechanism for reduction of an ester

* In ester reduction, LAH delivers hydride to carbonyl group but LLG causes the carbonyl to reform. This reformed carbonyl can be attacked by hydride again. The LG is a methoxide ion, which is no good. It doesn’t function in E2 and SN2 reactions because the intermediate is high in energy and exhibits a negatively charged oxygen atom.
* Preparation of Diols
o Compounds with two hydroxyl groups. Diol is added to the end of the name and the hydroxyl groups are identified with numbers. Diols can be prepared via the previous reducing agents mentioned.
o Syn and anti dihydroxylation of an alkene will also work.

* Preparation via Grignard Reagents
o A Grignard reagent is one formed from an alkyl halide and magnesium.

o The carbon in a Grignard reagent is more electronegative so it will withdraw electron density from magnesium via induction. A partial negative charge arises on the carbon atom. This can essentially be treated as an ionic bond between Mg and C. Grignard reagents can attack a wide range of electrophiles including the carbonyl group of ketones and aldehydes.
o Mechanism

o An R group is introduced with a Grignard reagent. Similar to LAH, water cannot be simultaneously present as the Grignard reagent is a strong base and will deprotonate water. Therefore, Grignard reagents should be kept away from moisture.
o Grignard reagents will react with ketones or aldehydes to produce alcohol.

o They will also react with esters to produce alcohols with two new R groups.
o Mechanism with Ester
o It won’t react with a carboxylic acid because it is not possible to form Grignard reactions in even mildly acidic protons, such as the proton of a hydroxyl group. It will attack itself to produce an alkoxide.
o Synthesis reactions will usually have more than one possibility. Example:

* Protection of Alcohols
o In the presence of a hydroxyl group, a Grignard reagent won’t form. However, this problem can be circumvented.
* Protect the hydroxyl group by converting it into a protecting group.
* Form the Grignard reagent and perform the reaction
* Deprotect and revert to a hydroxyl group.
* The protecting group can be a trimethylsilyl ether (OTMS), formed from trimethylsilyl chloride, TMSCl.
* This reaction is said to be SN2- like because the hydroxyl group functions as a nucleophile to attack Si and Cl ion is the LG. This may seem odd because the first step occurs at a tertiary substrate but an Si bond is longer than carbon atoms, allowing for a back-side attack.
* Once the desired Grignard reaction is performed, the trimethylsilyl group can be removed with H3O+ or fluoride ion (supplied by TBAF).

o Overall process
* Preparation of phenols
o Involves the formation and oxidation of cumene
* Substitution and Elimination Reactions with Alcohols
o SN1 will convert an alcohol into an alkyl halide

o SN2 will convert alcohols to different things
* Primary alcohols will react with HBr. The hydroxyl group is protonated to form a better leaving group. This reaction won’t work well for HCl, so ZnCl2 has to be used as a catalyst.

* An alcohol can be converted into a tosylate, which can undergo a nucleophilic attack. The configuration is inverted in the second step (SN2 process).

* Primary and secondary alcohols will react with SOCl2 or PBr3
o E1 and E2
* Elimination reactions occur. The more substituted alkene is favored. A strong base is used in E2 and no carbocation rearrangements occur.
* E1

* E2
* Oxidation
o Primary alcohols have two protons in the a position so it can be oxidized twice.
o Secondary alcohols have one so it can be oxidized once.
o Tertiary positions don’t have any so they won’t undergo oxidation.
o A common oxidizing agent is chromic acid, H2CrO4. The mechanism for oxidation with chromic acid is shown below.

o Primary alcohols oxidized with chromic acid produce carboxylic acid. The aldehyde intermediate is difficult to isolate.

o The aldehyde can be obtained using selective oxidizing reagents, one that reacts with the alcohol but not the aldehyde. PCC works for this. Methylene chloride (CH2Cl2) is typically used as the solvent.

o Secondary alcohols can be oxidized once to form a ketone, with either chromic acid or PCC. Sodium dichromate is less expensive, but PCC is more gentle and is preferred if other sensitive functional groups are present.
* Oxidation of Phenols
o Undergoes oxidation more readily than primary or secondary alcohols despite not having protons in the a position.
o The product is a benzoquinone, which can be converted to hydroquinones.
Suppose a Grignard reaction is performed with an aldehyde, followed by an oxidation to a ketone.
The following is a mechanism for converting aldehydes into ketones.
Retrosynthetic problem