Reactions and Mechanisms in Organic Chemistry Pt. 2
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what are the ways to reduce other carbonyl derivatives (not aldehydes/ketones)
• Reduction of acid chlorides/anhydrides to alcohols using sodium borohydride (NaBH4) • Reduction of esters to alcohols using lithium aluminium hydride (LiAlH4) • Reduction of amides to amines using lithium aluminium hydride • Reduction of Carboxylic acids using Borane
How can we determine which species will/won’t react with sodium borohydride
• using the reactivity series of carbonyls Acyl Chlorides > acid anhydrides > aldehydes/Ketones > esters > amides Aldehydes/Ketones are the lowest reactivity species that can be reduced by sodium borohydride anything with a higher reactivity will be reduced anything with a lower reactivity will not be
what is the mechanism for anhydrides/ acyl chlorides with sodium borohydride to form alcohols (same mechanism for both)
• the borohydride ion attacks the carbonyl carbon as before • But this time the oxygen attacks back into the molecule and the chlorine or alcohol like ion (for anhydrides) • this reforms a Ketone • this ketone is attacked again by a borohydride ion and is then protonated as in previous mechanisms to form an alcohol
what do we use to reduce esters to alcohols and why
• lithium aluminium hydride (LiAlH4) • this is because esters are LESS reactive than ketones/aldehydes so a stronger reducing agent is needed
what is the mechanism for the reduction of esters to alcohols
• very similar to previous mechanisms for reductions that used NaBH4 but just with LiAlH4 • AlH4(-) ion attacks carbonyl • O(-) attacks back into the structure and the OR group is lost as a leaving group • this forms an aldehyde • this aldehyde is then attacked again by AlH4(-) and the mechanism continues as before to form an alcohol
Why, in the reduction of esters to alcohols, does the mechanism not stop at aldehydes
two reasons: - LiAlH4 can be the source of more than one hydride (as in BH4(-) in previous booklet) • the aldehyde is more reactive than the starting material so it reacts in preference to the ester with LiAlH4
Why, in the reduction of esters to alcohols, is the reducing agent not added at the same time as a protic solvent (to protonate the resultant oxyanion in the final step)
• LiAlH4 will react with a protic solvent • so the protonation is done during a work-up, after the reaction itself is complete
Give a brief overview (no mechanism required) of how carboxylic acids can be reduced to alcohols
• they can be reduced using Borane (BH3) • 3 eq’s are required in the first step • it forms a triacylborate intermediate which reacts further with a BH3 to form an alcohol
what is used to reduce amides to amines and what is the mechanism
• AlH4(-) attacks the carbon of the carbonyl and the C=O bond breaks by heterolytic fission • the O(-) attacks the spare p-orbital on the resultant AlH3 molecule (AlH3 acts as a Lewis acid) • the lone pair on the nitrogen attacks back into the structure and now OAlH3 is removed as it’s an ok leaving group • this forms an Iminium ion containing a C=N(+)R1R2 section • the AlH4(-) ion attacks the carbon of the C=N bond on the iminium ion forming an amine, this is analogous to the reduction of a ketone/aldehyde
Summarise which reducing agents can be used to reduce which species
(OOO) = (NaBH4, LiBH4, LiAlH4) Acid Chlorides (YYY), Acid Anhydrides (YYY), Ketones/Aldehydes (YYY), Esters (Slow YY), Amides (NNY) Carboxylic Acids (NN Slow so use BH3)
as a general rule, which reducing agent should be used
• the mildest agent which will still reduce the substance
What are the main two reactions that are possible when an organometallic reagent (Grignard or Organolithium) react with carbonyls
• Ketones can form from a single addition of the nucleophile • Alcohols can form from a double addition of the nucleophile
Which species can be reacted in the single addition of an organometallic reagent to carbonyls to form ketones, give the mechanism
• Acid Halides and Acid Anhydrides react with 1eq of RMgBr or RLi to form ketones • the organometallic attacks the carbon of the carbonyl, the C=O bond breaks by heterolytic fission • the lone pair/ -ve charge on the oxygen attacks back into the structure and now the chlorine or OCOR (anhydride thing) is removed as chlorine/OCOR are good leaving groups
Which species can be reacted in the double addition of an organometallic reagent to carbonyls to form alcohols, give the mechanism
• esters react with 2 eq’s of RMgBr or RLi followed by aqueous acid to form alcohols • the organometallic reagent attacks the carbon of the carbonyl and the C=O breaks by heterolytic fission • the lone pair/-ve charge on the oxygen attacks back into the molecule and the OR group of the ester is removed as it’s an ok leaving group • this forms a Ketone • this ketone is then further attacked by the organometallic reagent and an oxyanion forms as expected • this oxyanion is protonated by aqueous acid (HCl) which is only added AFTER the first part • this forms an alcohol
Why can we not make alcohols from acid anhydrides/acyl chlorides and organometallics but we can make ketones Why can we not make ketones from esters and organometallics but we can make alcohols
For Acid Anhydrides/Acyl Chlorides: - the starting material is more reactive than the ketone - this means that the starting material will react quicker with the organometallic than the ketone so if 1eq of organometallic is added then the major product is the ketone For Esters: - the starting material is less reactive that the ketone - so the ketone (formed after the first part of the reaction) will react quicker with the organometallics than the starting material and hence the reaction is forced through to completion - if 1 eq is added then the alcohol is the major product and about 50% of the starting material remains
what occurs in the reaction of organometallic reagents with carbon dioxide
step 1) add RMgBr or RLi to CO2, nucleophilic attack step 2) add HCl, protonation carboxylic acid formed 1) C-Mg sigma protons attack delta+ve carbon on CO2, one of the CO2 bonds breaks by heterolytic fission, this forms RCOOMgBr 2) the lone pair on the single bonded oxygen attacks a proton on the H3O+ ion (formed from the acid) this forms a carboxylic acid
Summarise what each carbonyl type (and carbon dioxide) forms on reaction with organometallic reagents then protonation
Acyl Chloride and ester: ketone (with 1eq) alcohol (with 2eq) Ketone: Alcohol Ester: Alcohol (with 2eq) Carboxylic acid: Carboxylate anion metal salts Carbon dioxide: Carboxylic acid
What is a generalised reaction of water with a carbonyl and what do the specifics depend on
• oxygen on water attacks carbon on C=O bond • O(-) formed attacks back into molecule • proton from water group is removed, X group is removed to form carboxylic acid • the exact order in which this happens depends on the carbonyl
give the reaction conditions for each of the carbonyls (4) with water, what does this agree with
• Acyl Chloride: Fast at 20 degrees • Acid Anhydride: Slow at 20 degrees • Ester: only on heating with an acid or base catalyst • Amide: only on prolonged heating with an acid or base catalyst • this confirms the three factors of strength of nucleophile, reactivity of carbonyl and leaving group ability for nucleophilic substitutions
explain how acid-mediated hydrolysis of esters and amides speeds up the reaction
• it protonates the carbonyl oxygen making it more susceptible to attack from a nucleophile • it can protonate the leaving group which lowers the pKa and makes it a better leaving group
give the mechanism for the acid-mediated hydrolysis of an ester (and effectively an amide too)
1) the carbonyl oxygen is protonated 2) the H2O attacks the carbonyl carbon and the C=O pi bond breaks by heterolytic fission 3) +- H(+) to form an O(Me)(H)+ group and an OH group 4) let one of the lone pairs on one of the oxygens on one of the OH groups attack back into the molecule and the MeOH group is lost as it’s a good leaving group 5) Let an H2O deprotonate the product to form a carboxylic acid this also reforms the catalyst so it is truly catalytic in acid
in what way is the acid-mediated hydrolysis of an amide different to the acid-mediated hydrolysis of an ester
• in the ester case the acid catalyst is regenerated so it is truly catalytic in acid • in the amide case the amine (which was the leaving group) is protonated under meaning that one equivalent of acid is used up in the reaction so it is not catalytic in acid (but still acid-mediated)
explain how base-mediated hydrolysis of esters and amides speeds up the reaction
• by creating a more reactive negatively charged nucleophile • by deprotonating the carboxylic acid product ‘pulling the equilibrium over irreversibly to the hydrolysis products
give the mechanism for the base-mediated hydrolysis of an amide (and effectively an ester too)
• an OH(-) nucleophile is formed from the reaction of water with a base • this attacks the carbonyl carbon forming a tetrahedral product with an OH, O(-), OR/NH2 and R • this step is reversible with the majority returning to the starting product as OH(-) is a better leaving group than NH2(-) • in the small number of cases where the NH2 group is removed, the resulting carboxylic acid is immediately deprotonated in an irreversible reaction to form a carboxylate anion • this can then be protonated in a work up to form the carboxylic acid