Aldehydes and Ketones Flashcards
C=O with neutral nucleophile?
Produces an intermediate containing both a negative and a positive charge empale of a zwitterion, transfer of a proton from Nu to O in two subsequent steps is facilitated by the nucleophile to form a product containing an OH group, all steps reversible, acid catalyst often added to increase rate of addition of neutral nucleophile to C=O bond, net result is addition of NuH across C=O bond
C=O with negatively charged nucleophile?
produces an intermediate alkoxide ion, addition of H- or R- is irreversible but additional of CN- is reversible since it is a better leaving group than the other and the C=O bond can be reformed from the intermediate alkoxide ion, in all cases alkoxide ion is protonated in second step
Nucleophilic addition reactions of aldehydes and ketones?
Aldehydes react faster than ketones explained by steric and electronic facts in ketones two alkyl or aryl groups hinder approach of nucleophile and rescued the size of partial positive charge on the carbon atom in the C=O bond exerting +I effect making atom less electrophilic
Reaction with hydrides?
Aldehydes and ketones react with hydride ions to form alcohol and the C=O bond is reduced
Using complex metal hydrides?
Eg NaBH4 or LiAlH4 followed by treatment with acid in a separate stage. Reduction of aldehydes from primary alcohol and the reduction of kenos produces secondary alcohols
LiAlH4 vs NaBH4?
LiAlH4 much stronger reducing agents than NaBH4and reacts more rapidly with aldehydes and ketones, Al-H bonds are more reactive and weaker than the B-H bonds in NaBH4 because aluminium is more electropositive than boron, electropositive aluminium ion in [AlH4]- rapidly donates a hydride ion to the C=O bond and so loses a negative charge
End of the reaction?
Aqueous acid is added to protonate the intermediate alkoxide ion, the reaction picture is cooled to 0 degrees and the aqueous acid added drowse,
Na(CN)BH3?
Another useful hydride less reactive than NaBH4, electron withdrawing CN group stabilises the negative charge on the bon atom and reduces the rate of transfer of a hydride ion, mild reducing agent with molecules containing both an aldehyde and ketone group it can selectively recuse the more reactive aldehyde group
Cannizzaro reaction?
hydride ion generated from aldehyde by reaction with hydride ions. Aldehyde must not contains any alpha hydrogen atoms because hydroxide ion mist reacts with the C=O bond of the aldehyde in the first step. Example of a disproportionation reaction
Reaction with cyanide to form cyanohydrins?
formation of the cyanohydrin is reversible and the position of the equilibrium depend on size of R groups attached to C=O bond, ketones with large R groups equilibrium favours ketones rather than the cyanohydrin in the ketone the two R groups are separated by a bond angle of 120 degrees but this angle is reduced to 109 in the cyanohydrin and the R groups pushed closer together, if R groups close enough the resulting steric strain causes the equilibrium to favour the ketone, for aldehydes equilibrium favours the cyanohydrin, steric strain formed from an aldehyde is less than that formed from a ketone because a very small hydrogen atom replaces one of the R substituents
Reactions with organometallic to form alcohols?
Metals more electropositive than carbon so covalent C-metal bond is polarised with a partial negative charge on the carbon atom, hence why it can reactive with nucleophile or base, Grignard reagents
Wittig?
Aldehyde and ketones react with phosphonium ylides to form alkenes in the wittig reaction
Mechanism of reaction with metal hydrides?
Involves breaking a B-H or Al-H in the [BH4]- or [AlH4]- ion to donate a hydride ion to the carbon atom of the C=O bond to form a tetrahedral alkoxide ion, protonation of the alkoxide ion forms an alcohol
Mechanism of reaction with organometallic to form alcohols?
Organometallic reagents react with aldehydes and ketones to form alcohols, the organic groups acts as a nucleophile and track the carbon atom of the C=O bond to form an alkoxide ion, the alkoxide ion then reacts with H+ in a separate second stage
Preparation of phosphonium ylides?
In first step reaction of halogenolkane with a trialkyl or triply phosphine typically triphenylphosphine which is a strong nucleophile produces a phosphonium salt in an SN2 reaction, in the phosphonium salt the positively charged phosphorous atom is the electron withdrawing and this makes the hydrogen atom on the adjacent carbon relatively acidic, treatment with a strong base removes H+ to form the phosphonium ylide