Mechanism Pointers

Question 1

Addition of organometallic reagents

Answer 1

In these reactions you either want to add an R group (R-MgBr) or you want to add a H (Metal Hydrides) from which both, mechanisms start from the M-R/H bond which initiate the attack on the carbonyl group.

Question 2

The Wittig Reaction

Answer 2

1) Mechanism begins from negative carbon and carbonyl oxygen attacks Phosphorus immediately.
2) FORMS THE BOX
3) Pair of electrons go into making P=O bond as it is a strong bond therefore a big driving force to make it.
4) Pair of electrons also go to forming the actual alkene

STEPS 3 and 4 occur simultaneously in the same step.

Question 3

Formation of Diols

Answer 3

Forming diols can be done with Acid or Base.

1) Acid: Protonate then add water then remove 1 Hydrogen.
2) Base: Mechanism begins with Water to then form OH- (using a base against water), then reaction proceeds as Nucleophilic attack, when oxygen is minus take hydrogen from the base that took it from the water in the beginning step

Question 4

Formation of Hemi-Acetals

Answer 4

Alcohol comes in after protonating the carbonyl oxygen

Question 5

EXAM: Formation of Acetals

Answer 5

Hemi-Acetal mechanism but instead of stopping, you do proton transfer –> remove H2O –> add second alcohol –> stabilise by X- taking extra H on newly attached alcohol.

Question 6

EXAM: Formation of Acetals (Equilibrium)

Answer 6

To favour formation of acetals, use excess alcohol or remove water to push equilibrium.

Question 7

Formation of Imines

Answer 7

This reaction consists of a primary amine and a ketone/aldehyde

Question 8

Formation of Enamines

Answer 8

This reaction consists of a secondary amine and a ketone/aldehyde

Question 9

Amine Alkylation

Answer 9

Instead of iminium ion going forward to form an imine, it gets reduced by H3BCN- Present in -OAc and Na+, this formes an amine with other products:

H2BCN
NaOAc

Question 10

The Strecker Reaction

Answer 10

The Strecker Reaction refers to the formation of Amino Nitrite.

The mechanism begins with 2 Ions: NH4+, Cl-, and H2O

After this Iminium ion to be formed and then attacked by CN- that is introduced to then form Amino Nitrite

Question 11

Formation of enol

Answer 11

This reaction goes by acid catalysis or base.

1) Acid: oxygen gets protonated then X- attacks acidic hydrogen on alpha carbon
2) Base: Attacks acidic hydrogen straight away and oxygen steals a H from the base nearby.

Question 12

Halogenation reactions with enols

Answer 12

Form enol and then attack diatomic molecule (Br2)

Question 13

The Mannich Reaction

Answer 13

This reaction consists of extending the carbon chain and forming a (C-C).

Step 1) Forming Enol
Step 2) Forming Iminium Ion

Step 3) Mixing these these two together where alkene from enol attacks iminium alkene carbon.
Step 4) Deprotonate the carbonyl oxygen to then have a stable product.

Question 14

Formation of Enolates (Using LdeeA)

Answer 14

Use LDA, Li-N bond attacking acidic proton and C=O then attacking Li.

Conditions = -78 degrees celcius (dry ice).

Question 15

Formation of LDA

Answer 15

Li-R + H-NMe6 –> Li-NMe6 (using THF). (Dry ice conditions).

Question 16

Enolates with Alkyl Halides (making C-C)

Answer 16

See diagram on page 4 of master sheets. This reaction works with: esters, ketones and amides. NOT aldehydes.

Question 17

Enolates with Alkyl Halides (why not aldehydes?)

Answer 17

Because the enolate formed from that same aldehyde will rereact with the aldehyde due to aldehydes high electrophilicity to form an ALDOL instead of forming C-C bond by reacting with alkyl halide.

Question 18

Aldol Reaction (using enolate)

Answer 18

Enolate attacks ketone (O-Li electrons come down and then force alkene to go attack the ketone beta C), followed by acid workup.

Question 19

Aldol Reaction (using enols)

Answer 19

This reaction begins with protonating ketone and then reacting with its own enol version, see master sheet for clarity.

Question 20

Aldol Reaction (crossed aldols without sequential additon)

Answer 20

Use NaOH to deprotonate acidic alpha carbon H.

Then alkene end goes to other compound. Then adding water to give O- a hydrogen.

Question 21

The Diels-Alder Reaction (diene + dienophile)

Answer 21

This is the mechanism where it sort of goes in a circle, it is initiated by the alkene group on the compound with oxygens on them (hence called dienoPHILE)

This reaction forms a -cis isomer and -trans.

Question 22

Oxidative reactions of Alkenes (Dihydroxylation)

Answer 22

This mechanism uses osmium tetroxide to attach two oxygens SIMULTANEOUSLY. Initiation is from Os = O bond to electrophilic alkene.

Question 23

Oxidative reactions of Alkenes (Cleavage)

Answer 23

Com si com sa business. using osmium tetroxide and NaIO4 (with xs. H2O)

Question 24

Ozonolysis (Reaction Mech§anism)

Answer 24

Alkene pumped through ozone generator (O=O+-O-) and then this reacts to from ozonide to form com si com sa but com sa has extra oxygen attached therefore is an ion. PPh3 acts as nucleophile to take off O (remember P=O very stable fkn bond) and then com si com sa is formed :)

Question 25

Synthesis of tertiary alcohols

Answer 25

Using grignard addition on carboxylic acid

Question 26

Synthesis of ketones

Answer 26

Use a weinreb amide + grignard, note that O on N-OMe forms cylic intermediate with the MgBr.

int. is protonated then alcohol lonepair delocalises in to form C=O.

Question 27

Claisen Condensation (Forming B-Ketoesters)

Answer 27

Ester has acidic alpha proton attacked by strong base (NaOEt).. Another mole of ester present to then form other side (keto side of beta keto esters) and then base attacks again which takes middle proton to then induce a keto enol tautomer form to then make the ketone.

Question 28

Dieckmann Cyclisation (Making 5/6 membered rings)

Answer 28

Starts from the enol form and attacks carbonly electrophile by alkene in enol form and 1 OEt coming off.

Question 29

1, 3-dicarbonyl compounds (Alkylation reactions)

Answer 29

Using alkene (enol form) to attack an alkyl halide chain, halide comes off and the chain is attached forming C-C bond. This can be done for 1 or 2 groups. Reagents = NaOEt or NaH.

Question 30

1, 3-dicarbonyl compounds (Decarboxylation)

Answer 30

In this reaction we remove CO2. First turn ester into carboxylic acid via ester hydrolysis with acid and water. Then draw the molecule in a cyclic fashion where you can form the OH and kick off O=C=O.

Question 31

Alkenes with carbonyl groups gives them electrophilicity

Answer 31

Draw the resonance structures to end up on the positive carbon where alkene originally was. Electrophilic at beta carbon.

Question 32

Nucleophilic addition of a/B-unsaturated carbonyls

Answer 32

Alkylation reaction but with an alkene that has a carbonyl as we know they are electrophilic at the beta carbon.

Question 33

Acids from esters (Tricky steps)

Answer 33

Acidic conditions: The proton transfer to the OMe to get kicked out

Basic conditions: The attack on the carboxylic acid formed by the base formed when kicking off the ester (-OMe) so then you have to protonate as you get a carboxylate ion after -OMe deprotonates acid.

Question 34

Formation of Carboxylic Acid

Answer 34

Using CrO3 twice starting from a primary alcohol. There is proton transfer when adding water after C=O formed in 1st steps.

Question 35

Carbonylation of grignard reagents

Answer 35

Alkyl halide + Mg = grignard which then negative carbon attacks CO2 which kicks off MgBr+ then protonated with hcl to get your carboxylic acid.

Question 36

Forming acid chlorides

Answer 36

Using thionyl chloride and carboxylic acid. Forming SO2 gas in the process. S=O being the driving force of the mechanism. In the mechanism you for sort of a thiodiester.

Question 37

Forming esters from acid chlorides

Answer 37

Using pyridine and alcohol group. Pyridine comes in and kicks off Cl then alcohol comes in kicks off pyridinium ion, then compound deprotonated by pyridine.

Question 38

Reduction (Ester/Acid) LiAlH4

Answer 38

Keep attacking with hydrogens until each group comes off that is not R, then leaving O- and doing an acid work up.

Question 39

Reduction (Ester to Aldehyde) DIBAL-H

Answer 39

Lonepair on Oxygen carbonyl attacks the Aluminium to form positive oxygen and negatives aluminium. Hydrogen in DIBAL-H is attached to carbonyl alpha carbon. C=O broken, protonation, and then alcohol kicks off Ester bond OR and then deprotonate carbonyl.

Question 40

Reduction (Amide to Amine) LiAlH4/BH3

Answer 40

For LiAlH4, in intermediate negative oxygen forms interaction with AlH3 then kicked off by formation of N=C formation. Then another mole of LiAlH4 used to add hydrogen to C in C=N to then give you amide.

For BH3, Oxygen forms bond at attack of C=N in first step. H from OBH3- joins to alpha carbon and C=N goes back. This happens again and kicks off OBH2, this then causes BH3 to use another mole and attaches a H to C in C=N where spontaneously N also attaches to BH2, with water this can be turned into an amine.