OChem topic 4 - Acidity + Charge stabilisation

bronstead acid

a proton donor

what is Ka?

• an indication of how strong a proton donor a bronsted acid is relative to water

what determines the position of equilibrium

• the position of equilibrium depends on the relative stability of the species on the left-hand and right-hand sides of the equilibrium

• Ka gives an indication of how strong a proton donor the bronsted acid is relative to water

pKa values for strong and weak acids

• strong acid - large negative pKa

• weak acid - large positive pKa

the lower the pka the more stable the conjugate base and the stronger the acid

which equation describes the relationship between pKa and pH

Henderson-Hasselbalch equation

Henderson-Hasselbalch equation

pH = pKa + log ([A-]/[HA])

or

[A-]/[HA] = 10^[pH-pKa]

• used to estimate the extent of ionisation of an acid at a particular pH

examples + explanations of strong acids

• examples - sulfuric acid, nitric acid + hydrogen halide acids

• in water these are essentially completely dissociated

• eqm lies essentially completely on the right-hand side

• correlates well with the high stability of the conjugate bases of these acids

effect of solvation

• protic solvents such as water can stabilise an anion by H-bonding, electrostatic dipole interactions as well as through van der waals interactions

• solvation is exothermic therefore energetically favourable

• the more solvated an ion, the more stable it is - greater enthalpy of solvation

• effect of solvation effectively increases the ‘effective size’ of the anion which serves to reduce charge density

effect of electronegativity of atom accommodating the charge in the conjugate base

• the more electronegative the atom is that is formally negatively charged in the conjugate base the better it should be at accommodating that negative charge

effect of charge density

• any way by which charge can be distributed over a large area or volume tends to lead to increased stability

• eg. the chloride ion is much smaller than the iodide therefore iodide has the same charge distributed over a much larger volume resulting in a much lower charge density and increased stability therefore more acidic as it is better able to cope with losing a proton

how does a -I inductive effect stabilise negative charge?

• inductive effects operate predominately through sigma-bonding framework and are therefore always present in a molecule

• electron withdrawing group expert a -I indictive effect, the more electron withdrawing, the more they stabilise the charge

• delocalising the negative charge over the whole molecule stabilises it - distributing charge over a larger area reducing charge density

• if the conjugate base is especially stable then the acid is stronger and wants to give up a proton

examples of substituents with a strong -I inductive effect

• -halides

• -OH

• -OR

• -CF₃

• -NR₃⁺

• -C(O)R

• -NO₂

• -CN

• -sulfonly (-S(=O)₂R

examples of substituents with a weak -I inductive effect

• -NH2

• -NR2

• -NHC(O)R

• -SR

• -Phenyl

+I inductive effect

• electron donating groups - stabilise positive charge and destabilise negative charge

• more electron donating the stronger its +I effect

• need to look at the electronegativities

examples of substituents with a +I effect

• metals

• alkyl groups

distance dependence of inductive effects

• short range effects

• the more bonds through which they have to operate the weaker the stabilising/destabilising influence

• eg. the more remote a fluorine is from the negative charge the less stabilising it is

pka of ethanoic acid

4.76

π-resonance or mesomeric stabilisation

• sp3 orbitals ensures p orbitals on adjacent atoms can overlap with one another

• ability of aligned p orbitals to overlap provides a way of distributing charge over a greater area

• the delocalisation reduces charge density on the molecule creating added stabilisation

• negative charge often isn’t delocalised equally over all c atoms eg. in phenoxide the charge is mostly on o but some on 2, 4, 6 in 4 resonance forms

• generally more important that inductive when both are present

resonance forms - how to draw

• arrows always move from a source of electrons

• go from an are of high electron density/neg charge - push the neg charge around p orbitals from an atom to a bond and a bond to an atom

• only pi bonds or lone pairs are moved

• usually more resonance forms are generally indicative of greater stability we also need to consider the electronegativity of the atoms where the negative charge is located - better if neg charge is on an electroneg atom

• also need to pay attention to gaining or loss of aromatic character

resonance forms in benzene rings - with electroneg substituents

• usually get 4

• a second electroneg atom w a lone pair in the 2 or 4 position allows for a 5th resonance form

• p AOs are optimally aligned to donate lone pair into benzene ring through to the carbocationic centre

• lone pair on heteroatom allows 5th

• the lower the electronegativity of the atom the better it is at donating the lone pair - means that the 5th resonance form has greater weighting

• effectiveness of heteroatoms to donate electron density also depends on the effectiveness of orbital overlap - similar size and energy is better

examples of substituents which stabilise negative charge by π-resonance ie. -M effect

• nitro -NO2

• sulfonic acid -SO3H

• carbonyl substituents

• Nitriles

examples of substituents which stabilise positive charge by π-resonance ie. +M effect

• -NH2

• -NR2

• -NHC(O)R

• -OH

• -OR

• -SR

• less effectively: -F (too electroneg), -Cl, -Br, -I (poor orbital overlap)

examples of substituents which stabilise both positive + negative charge by π-resonance

• unsaturated carbon substituents

hybridisation effects

• the more s character the more stable the conjugate base, the lower the pka the stronger the acid

• The more s-character an orbital has, the closer the electrons are held to the nucleus, making the conjugate base more stable. - electrons experience a greater stabilising effect from the positively charged nucleus

• results in an sp hybridised c being more electronegative than an sp2