Quiz 7 Flashcards
describe how acyl chlorides react with organocuprates to form ketones
(unicorn rxn #1)
- unlike grignard reagants, only one equivalent of organocuprate reacts with an acyl halide to produce a ketone as the product
describe the reaction of aldehydes and ketones with acetylide ions
a very weak acid reacts with a very strong base to produce a strong base and strong nucleophile
- the acetylide anions (like grignard reagants) are also organometallic reagants and thus react the same way with aldehydes, ketones, and carboxylic acid derivatives
name the unicorn reactions
unicorn rxn = very very specific to just these solvents
- acyl chlorides react with organocuprates to form ketones
- acyl chloride reduction as the aldehyde
what happens when an aldehyde reacts with an organometallic compound
gives a secondary alcohol with a triple bond still present
describe the reaction of aldehydes and ketones with a cyanide ion
- the overall reaction here is addition of HCN to the carbonyl to form a cyanohydrin.
- must be generated in solution using NaCN and HCl
describe the reaction of cyanohydrins in base, aqueous acid and heat, and with catalytic hydrogenation
Base:
- HO- and H2O (for ex.)
- reversible reaction
- the cyanohydrin reverts back ti the starting carbonyl compound since cyanide ion is also a good leaving group
Aqueous acid and heat:
- HCl and H2O –> AKA H3O+ and heat
- a cyanohydrin hydrolyzes to an alpha-hydroxycarboxylic acid.
Catalytic hydrogenation:
- H2 and Raney nickel
- catalytic hydrogenation reduces the cyanohydrin to a primary amine with an OH group on the neighboring carbon atom
describe the two reactions of aldehydes and ketones with a hydride ion
aldehydes:
- reacts with 1. NaBH4 and 2. H3O+ to produce a primary alcohol
- reduction of aldehydes with H- always gives primary alcohols
ketones:
- reacts with 1. NaBH4 and 2. H3O+ to produce a primary alcohol
- reduction of ketones with H- always gives secondary alcohols
describe reactions of acyl chlorides with hydride ions
- the acyl chloride is very reactive
- reacts with 1. 2NaBOH4 and 2. H3O+to produce a primary alcohol
- reduction of acyl chlorides with H- usually gives primary alcohols as the intermediate aldehyde also is reduced under the reaction conditions
describe how a weak and bulky hydride donor (DIBALH) stops the acyl chloride reduction as the aldehyde
(unicorn rxn #2)
- to stop the reduction of an acyl chloride at the aldehyde stage, you must use H- source that is a weaker nucleophile and sterically very bulky
- an acyl chloride reacts with 1. LiAl[OC(CH3)3]3H, -78 degrees celcius and 2. H2O
- LiAl[OC(CH3)3]3H contains 3 big groups so it can deliver one hydrogen. since only one H is attached, it can’t over reduce
- replacing three of the hydrogens of LiAlH4 with -OC(CH3)3 groups makes it a less reactive reducing agent than NaBH4
- to stop the reduction of an ester at the aldehyde stage, you must use an H- source that is a weaker nucleophile and sterically very bulky
describe reactions of esters with hydride ions
- an ester reacts with 1. LiAlH4 and 2. H3O+ –> esters are less reactive than aldehydes and ketones so it needs a strong solvent
- this reaction produces a primary alcohol
- like acyl chlorides, reduction of esters with H- usually gives primary alcohols as the intermediate aldehyde also is reduced under the reaction conditions
what does DIBALH do
- take an ester and stop it at the aldehyde, the big group helps it not over reduce
- only has one hydrogen separately attached
- the solvent used to stop the rxn
describe reactions of carboxylic acids with hydride ions (reduction)
like acyl chlorides and esters, reduction of carboxylic acids with H- gives primary alcohols as the intermediate aldehyde also is reduced under the rxn conditions
describe the different ways that reductions with H2 can happen
an organic compound is reduce when hydrogen is added to it
1. hydrogen can be added as a hydride ion and a proton (H- with two lone pairs)
2. hydrogen can be added as two hydrogen atoms (two H atoms with one electron each)
3. hydrogen can be added as two electrons and two protons (two electrons (dot-) and two H+’s that are all free floating separately)
what reactants are used for catalytic hydrogenation to reduce carbon-carbon double and triple bonds
solvents that can be used:
- Pd/C
- Raney nickel
- ramey nickel is a very reactive catalyst and is the preferred metal for reducing carbon-nitrogen triple bonds
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describe how aldehydes and ketones form imines with primary amines
an aldehyde or a ketone react with a primary amine and use a trace acid catalyst solvent to produce an imine and H2O
- reactions are reversible
- use Le Chatelier’s principle to remove H2O and drive rxn to the right
imines vs. amines
An amine is a functional group in which one or more hydrogen atoms are replaced by an alkyl or aryl group. An imine is a functional group containing a carbon-nitrogen double bond.
describe the bonding in an imine
- the pi bond is formed by side-to-side overlap of a p-orbital of carbon with a p-orbital of nitrogen
- the pi bond is perpendicular to the orbitals
describe the formation of imine derivatives
in all instances, imine formation replaces C=O with C=NR
describe how tetrahedral compounds in imine formation are unstable intermediates
- imine formation is reversible because there are two protonated intermediates in the mechanism that can eliminate a group
- imine formation can be pushed to completion by removing water as it’s formed (use of Le Chatelier’s principle)
describe how the pH must be controlled for imine formation
- pKa of ammonium ion is 6 and the maximum reaction with acetone is at pH 4.5
- this difference suggests that for imine formation, the pH should be 1.5 units less than the pKa of the amine nucleophile
- most amines have a pKa of 10.5, so that means the pH of the solution should by 9.0
- there must be enough acid to protonate the oxygen atom of the tetrahedral intermediate so that water can be expelled
- if too much acid is present, however, the amine will be protonated and will not be a nucleophile
describe imine hydrolysis
- acid-catalyzed hydrolysis (H3O+) converts the imine back ti the carbonyl compound and the amine
- the amine co-product is protonated in the acidic solution (not a trace amount), so it is no longer nucleophilic and thus cannot react with the carbonyl compound
- more importantly, the hydrolysis reaction is performed with a huge excess of water, so Le Chatelier’s principle favors the carbonyl compound
describe the reaction of aldehydes and ketones with secondary amines
- a ring with a double bond to an O and a secondary amine react with the trace acid catalyst solvent to produce an enamine
- use le chatelier’s principle to remove H2O to drive rxn to right
what is an enamine
ring with the double bond in the ring attached to the carbon that is attached to N, the N is bonded to R groups
enamine vs imine
imine consists of a C=N double bond while enamine consists of a C-N single bond.
describe enamine hydrolysis
- acid-catalyzed hydrolysis (H3O+) converts the enamine back to the carbonyl compound and the amine
- the amine co-product is protonated in the acidic solution (not a trace amount), so it is no longer nucleophilic and thus cannot react with the carbonyl compound
- more importantly, the hydrolysis reaction is performed with a huge excess of water, so Le Chatelier’s Principle favors the carbonyl compound
describe another way to convert a carbonyl group to a methylene group
(reducing O double bond to just CH-)
- use a Wolff-Kishner reaction
- reagants: NH2NH2 KOH and heat
describe the mechanism for the wolff-kishner reduction
- the ketone forms a hydrazone for the imine formation
- hydroxide removes a proton from the weakly acidic NH2 group
- reprotonation occurs at carbon
- another deprotonation, loss of N2, and reprotonation forms the product
describe how water reacts with aldehydes and ketones
- forms a hydrate
- most hydrates are not stable and usually revert back to the carbonyl form
- the acid protonates the carbonyl oxygen
- the nucleophiles adds to the carbonyl carbon
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