Nucleophilic attack on cabonyl group (SLOW)

Protonation of the anion that results (FAST) Leaving groups are anions that can be expelled from molecules taking their negative charge with them The first step is the most important - it forms a new C-C bond at the expense of a C=O bond If, however, the intermediate is unstable The polarisation gives the carbon atom and there are any good leaving groups When a nucleophile attacks some degree of positive charge, and this around it, they'll be expelled a carbonyl group, it forms a attracts the negative nucleophile tetrahedral intermediate The re-forming of a C=O double bond by far The attack involves the movement of electrons from the HOMO compensates the loss of the leaving group The attack at C is favoured of the nucleophile to the LUMO, which is the * antibonding This is a nucleophilic substitution mechanism orbital - the coefficient of which is greater on the C, meaning a better HOMO-LUMO interaction there Orbital explanations Mechanism The nucleophile approaches and the electron pair in its HOMO start to interact with the LUMO ( *) to form a new bond Mechanism Filling this antibonding orbital breaks the bond, the electrons of which end up on the electronegative oxygen Introduction The trigonal planar, sp2 carbon changes to a tetrahedral, sp3 carbon in the product

Nucleophile attacks at the Burgi-Dunitz angle , 107° from the C-O bond

The reaction goes Trigonal Planar Maximum overlap of the HOMO with * This is a compromise between --> Tetrahedral --> Trigonal planar Trajectory Minimum repulsion of the nucleophile HOMO ImPoRtAnT nOtE - this is NOT an Sn2 by the electron density in the carbonyl bond mechanism - the system prevents the correct Any group which hinders this trajectory will greatly reduce approach of the nucleophile. Therefore, all the the rate of addition - causing steric hindrance steps of this mechanism have to be drawn As the carbonyl carbon moves from sp2 to sp3, its bond angles get Radioisotope labelling studies have We know that the tetrahedral smaller - in other words, the substituents move closer together shown the "swapping" of the oxygens intermediate exists because... The least bulky the substituents, therefore, the easier this will Steric hindrance be, and the more favoured the reaction All have the formula RCOX This is why these reactions are usually faster Acid - X = OH for aldehydes than for ketones

Acid/Acyl chloride - X = Cl Small rings also favour these types of reactions Ester - X = OR Carboxylic acid Examples derivatives Bond angles change from 120 to 109 in the first step of the addition Amide - X = NH2 In a three-membered ring, bond angles are forced to Factors affecting Anhydride - X = RCOO Ring structures be 60, so the ring is very strained Nitrile - RC=N However, strain is involved when the carbon atoms are sp3 (ideal angle 109) than when they're sp2 (ideal angle 120) The nature of X is very important for substitution Similar results, to a lesser extend, apply to four and five ring structures A good leaving group must be a stable anion, and a convenient guide to the stability of anions is the pKa of their conjugate acids Electronegative groups make the carbon atom even more polarised, and makes the attack of the negative nucleophile even easier The lower the pKa of the conjugate acid, the more stable Electronegative groups the base and the better the leaving group The electronegative effects are greater than the steric hindrance ones pKa is a measure of stability only with Such effects can have a large influence on the equilibrium respect to the protonated form of the anion Here, we're looking for a measure of the The formaldehyde in water 13C NMR shows a peak at stability of the carbonyl compound plus However, this is only a rough guide 83ppm (C-O) rather than at 150-200 (C=O) leaving group with respect to the negatively Water has added to the carbonyl group to form charged tetrahedral intermediate Leaving groups a hydrate, or 1,1-diol R- 50 In the reaction, a proton has to be transferred from one NH2- 35 Addition of water to aldehydes and ketones oxygen atom to another, mediated by water molecules Learn these five values, of pKa RO- 16 of conjugate acid This is an equilibrium between starting materials and products RCO2- 5 Reaction of the hydrate in water with a trace of NaOH can produce the starting aldehyde (OH- acts as a base and deprotonates one of the hydrate OHs) Cl- -7 It is affected by the factors described above If the attacking neucleophile happens to be the best leaving group, then no reaction will occur H- is very small, and has a very high charge density

Again, pKa values are a good indication Addition The small 1s orbital is of an ideal size to interact with the H atom's contribution Good nucleophiles are poor leaving groups Nucleophilic attack by a simple hydride H- to the * orbital of an H-X bond A species that will readily form new bonds to hydrogen anion [for example, from NaH] does occur It only ever reacts as a base Much too small to interact easily with carbon's will probably also readily form bonds to carbons more diffuse 2p orbital contribution tot he Nucleophiles Therefore, the higher the pKa value of the LUMO ( *) of the C=O group conjugate acid, the the nucleophile The oribtal interactions that would be involved if it were to act NH3 9 as a nucleophile are very poor We add two values to the list above ROH -5 However, it would be rather useful to be able to add H- to a carbonyl group, because we would then have a reduction It depends on how good the We can class acids with the most reactive at the This is isoelectronic with methane, but contains a leaving group is top and the least at the bottom - transformations negative charge since boron has one less proton And on how good a nucleophile are always possible moving down the series However, the boron does have a lone pair, and it is WRONG the attacking thing is Carbonyl to draw an arrow showing the negative charge coming from the This is because the * orbital is raised boron - that would imply that the boron originally had 10 Chemistry electrons in its outer shell, impossible for a first row element Also because electron density donated to carbonyl carbon - less polarised Instead, all the electrons (including the one represented Delocalisation the acid by the negative charge) are in the C-H bonds (Ironically, this weakens the C=O bond) derivative, and makes it less Electrophiles (i.e.: It is then one of these bonds that interacts Amides show the greatest degree of reactive towards nucleophiles Substitution reactivity of the acid The most important is NaBH4, a water-soluble with the * MO - a much better match than H- delocalisation, esters less so, anhydrides less Examples derivative) Nucleophilic attack by "hydride" salt containing the tetrahedral BH4- ion so and acyl chlorides less so As a result, the boron becomes trivalent and neutral again, with an empty orbital This can clearly be seen in IR streching frequencies

If the attached group is electronegative, it It turns out that this can be done using compounds increases the +ve charge on the carbonyl There is, however, also an that contain nucleophilic hydrogen atoms carbon andmakes it a better electrophile inductive effect

Thus, the attached group has two effects - inductive and conjugative The oxyanion produced is them likely to stabilise the electron A base is used to remove the deficient BH3 by donating a pair of electrons into the empty p orbital proton from the alcohol as it Usually, a water/ethanol attacks the carbonyl group mixture is used as the solvent Acid anhydrides and acid This is very weak - it can even be used in water chlorides react with alcohols to make esters Much stronger H- donors, such as lithium Pyridine is often used aluminimum hydride can be used Reacts violently with water, produced flamable H2

Aldehydes Ketones (slower) If this is done under basic Sodium borohydride reacts with Good selectivity if the molecule conditions, it is irreversible This reaction is reversible, the opposite NOT with less reactive carbonyl compounds such as ester or amides contains several groups This is because the final being called ester hydrolysis deprotonation step is irreversible Li-C or Mg-C bonds are highly polarised towards carbon The lone pair on the carboxyl group can be protonated They are therefore very powerful nucleophiles, and attack the by strong acids (see before) protonated carbonyl carbonyl group to give alcohols, forming a new C-C bond groups are then extremely powerful neucleophiles Addition of organometallic reagents Carboxylic acids react with It is important that water not be present in the first step, as it It also makes the leaving group much better alcohols to make esters, destorys the organometallic compound Proton protonates carbonyl under acid catalysis the second step, in which water is added, in known as the work-up Alcohol attacks Hemiacetals are the result of the addition of Interconversions between Proton transfer from alcohol to solvent The mechanism is as follows alcohols to carbonyl compounds acid derivatives Proton transferred to the leaving OH Hemiacetal formation This really only ever occurs if the product is H2O leaves cyclic, which gives it extra stability

The position of the equilbirum is determined The first step is the protonation of the carbonyl oxygen's lone pair by which reagents are in excess The positive charge makes the carbon much more electrophilic Acid This is very similar to ester formation and so the reaction is faster A second molecule of amine has to The proton is regenerated Catalysis act as the base, to remove the extra The first step is deprotonation of the alcohol by the base hydrogen on the amine Base (for reactions It results in a Amine+ Cl- ionic by-product involving alcohols) This makes the reaction faster by making the alcohol more nucleophilic Amines react with A way round this is the Again, the hydroxide is regenerated at the end acid chlorides to Reactions Schotten-Baumann method produce amides The NaOH and organic layers are NaOH can't be used as the base kept separate using a heavy solvent because it also attacks the acyl in which the organic stuff can chloride to give carboxylic acid Formed between two sp2 hybrids on carbon and oxygen dissolve, but the NaOH can't The two other sp2 hybrids on carbon form two bonds to the substituents The HCl released by the reaction system The two other sp2 hybrids on the oxygen contain lone pairs can diffuse into the NaOH layer, and be neutralised there The sp2 hybridisation means that the carbonyl group has to be planar, with bond angles nearly 120° All that happens is that the acid is deprotonated The bond results from the overlaping of the resulting two p orbitals on C and O Once that happens, substitution is Carboxylic acids do not prevented because (almost) nothing can The bond is skewed towards oxygen, because undergo substitution reactions attack the (stable) carboxylate anion oxygen is more electronegative than carbon under basic conditions This means that if we want to make esters from Notes Similarly, the * antibonding orbital is skwed alcohols and carboxylic acids, we use acidic towards the carbon, with a larger coefficient there conditions, because alcohols by themselves just Bonding in aren't reactive enough, and we can't use a base carbonyl system compounds The mild reducing agent NaBH4 will not reduce esters First by substitution of -OR for a hydrogen to make an aldehyde The stronger LiAlH4 reduce esters all Reduction of esters Then by the reduction of the aldehyde the way to the corresponding alcohol all the way to an alcohol by addition

Esters react twice with organometallics The bond is rather shorter than a typical C-O First time in a substitution to make an aldehyde or ketone single bond, and over twice as strong Esters and Consequences Second time to add to that one and make an alcohol organometallics The polarised C=O gives the carbon Note that the initial product is more electrophilic than the atom some degree of positive charge starting material, so even with one equivalent of organometallic, the final product will be formed

(C) Daniel Guetta, 2006