Summary Chemical Reactivity - Nucleophilic Substitution and Elimination
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Organic Chemistry I
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The University Of Birmingham (UBir)
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Organic Chemistry
I got a 1st in my first year studying chemistry at the University of Birmingham using these revision notes that I have uploaded. They include detail on acidity, electron density and chemical reactivity, drawing reaction mechanisms, nucleophilic substitution reactions: the SN1 and SN2 reaction mecha...
Solution Manual For Organic Chemistry Second Edition Jonathan Clayden, Nick Greeves, and Stuart Warren
Mechanisms
Test Bank For Organic Chemistry 2nd Edition Klein 9780199270293 | All Chapters with Answers and Rationals
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Organic Chemistry I
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Voorbeeld van de inhoud
Chemical Reactivity - Nucleophilic Substitution and
Elimination
16 October 2017 18:01
PART 1: ACIDITY A Bronsted Acid (BA) is a proton donor. • H3O+ is the conjugate acid of water, water is the conjugate
• Water can solvate both cations and anions base of H3O+ The position of this dynamic
• Codeine is not soluble in water but contains basic nitrogen equilibrium depends on the
atom that can be protonated to give a more soluble salt relative stability of the species on
• Uncharged compounds can be separated by acid-base the left- and right-hand sides of the equilibrium i.e. the
extraction (such as in S1 practical) In dilute solutions of acids the concentration Bronsted acid HA and its conjugate base A-, and gives an
of water is effectively constant because it's in indication of how strong a proton donor the Bronsted acid is
Equilibrium constants vary over a wide range so pKa is used relative to water.
vast excess.
e.g. Ka of 10-40 gives a pKa of 40, When Ka is very large, equilibrium lies on the right hand side.
and a Ka of 109 gives a pKa of -9 • HA is a much stronger Bronsted acid than H2O and
dissociates readily. The acid is fully ionised in H2O.
• pKa tells us how acidic a given hydrogen atom in a • H2O is sufficiently strong a base to abstract a proton from
compound is Ka values are relative to the water molecule.
The solvent you record the pKa with is most of the molecules of HA that are in solution
• It is the pH at which a given solution contains exactly the When Ka is very small, the equilibrium lies on the left hand side.
same amount of protonated, acidic form as deprotonated, important as it'll change the pKa value since
everything is relative to the pKa of the solvent. • HA is a weak acid relative to H2O and does not readily
basic form (the acid is half dissociated) dissociate in H2O.
HCl(g) is not an acid but HCl(aq) is - the • H2O is functioning as a weak base, at equilibrium only a few
Strong acids difference is that isolated H+ is too unstable to protons are abstracted from a few HA molecules in solution.
• Strong acids essentially completely dissociate be encountered under normal conditions but in
(equilibrium lies completely on the right-hand side) water H of HCl is transferred to a water Factors the determine the position of equilibrium and hence
therefore high stability of the conjugate base molecule and not released as a free species. the pKa of these acids
• Stronger acid=easier to ionise as more readily donates
proton=stable conjugate base Acid pKa a. Electronegativity of the atom accommodating the charge in the conjugate base
• Stronger acid=easier to break HA bond • The more electronegative the element which is formally negatively charged in the
HCl -7 conjugate base, the better it will be at accommodating that negative charge (the charge
• The more negative the pKa value=the stronger the acid.
HBr -8 is more delocalised so the anion is more stable).
b. Charge density • Therefore expect HCl to be the strongest acid as chlorine is the most electronegative
HI -10
• There is increased stability when the charge can be element of the three. However looking at the pKa values HCl is actually the weakest acid.
distributed over a larger area or volume.
• The relative size of the anions is iodide>bromide>chloride c. Solvation effects
• Chloride ion will have the highest charge density • When talking about the Bronsted acidity, we are considering equilibria
• The larger iodide ion will have the same amount of charge in solution. Therefore should be comparing the stability of the solvated
distributed over a much larger volume resulting in a much (or hydrated if the solvent is water) acid and its conjugate base.
lower charge density. • Protic solvents such as water can stabilise an anion by hydrogen
• Therefore you would predict HI to be the strongest acid. bonding, electrostatic dipole interactions as well as through van der
Waals interactions.
Overall the most important factor in determining relative acidity of HCl, HBr and HI
• Solvation is an exothermic process (as you are making bonds - electrostatic or H bonds). seems to be the size of the halide conjugate base as this best reduces charge density.
The more solvated an ion, the more stable it is (greater enthalpy of solvation). Solvation
increases the effective size of the anion as it is surrounded by H2O molecules, which in The strength of acids or bases we can use in any solvent is limited by acidity/basicity
turn reduces charge density and therefore increases stability. of the solvent itself.
• Enthalpy hydration of Cl-, Br- and I- is not significantly different. These large anions are • If trying to deprotonate a compound with pKa 30, cannot do with water as the
not particularly good hydrogen bond acceptors, so only van der Waals and relatively strongest base you can use is -OH. Since pKa 30 is a stronger base,
weak electrostatic interactions are involved in hydrating these ions. deprotonates water instead of compound.
Case study - why hydrogen fluoride is such a weak Bronsted acid (pKa=3): • Acids stronger than H3O+ cannot exist in water, just protonate water
• Fluorine is the most electronegative element in the Periodic Table, therefore on these completely to make H3O+
grounds it should be best able to accommodate negative charge, and would predict it to be Inductive effects
acidic.
• The fluoride anion is much smaller than other halides and has a high charge density,
therefore might explain a bit why HF is not very acidic
• However the high charge density also means it forms relatively strong hydrogen bonds with
water molecules, reflected in the high enthalpy of hydration for the fluorine anion (-560)
compared to the enthalpy of hydration for the chloride anion (-364). para-toluenesulfonic acid Methanesulfonic acid Trifluoromethanesulfonic
• Bond dissociation enthalpy of H-F bond is high (+562, compared to +431 for HCl and +299 acid
for HI). So whilst fluoride anion has a high enthalpy of hydration, this favourable enthalpic Trifluoromethanesulfonic acid is much more acidic
term is compensated by the large amount of energy required to break the HF bond. than the other acids. The three fluoro substituents in
• There is high polarisation (large difference in electronegativity) in the HF bond, therefore the triflate anion exert a -I inductive effect.
form strong hydrogen bonds with water molecules. The dissociation of HX requires • The -I inductive effect can be represented by
desolvation, so this will be energetically more costly for HF than for other hydrogen halides. drawing an arrow on the bonds, with the arrow
Therefore HF less acidic than other hydrogen halides as less enthalpically favourable to pointing in the direction of electron flow i.e.
ionise HF than other hydrogen halides. towards the electronegative element.
• When looking at the calculation △G = △H - T△S, in the case of HF, T△S is strongly negative, • Each fluoro substituent withdraws electron density from the sulfonate
as there is strong electrostatic attraction between H3O+ and highly charged F- anions, and group
also as there is an increase in order associated with the strong degree of hydration of these • Negative charge distributed over greater area
ions in solution. Resulting value of △G is small, helping to explain why HF is a weak acid. • Increased stability of the triflate conjugate base compared with the other
two strong acid conjugate bases.
• In general electron-withdrawing substituents exert a -I inductive effect, and are therefore Inductive effects operate predominantly through the σ-bonding framework and
able to stabilise negative charge (and destabilise positive charge) are therefore always present in a molecule.
• The more electron-withdrawing the substituent, the stronger the -I effect. Therefore a
fluoro substituent will exert a stronger -I effect than a chloro substituent etc. Common substituents that exert -I inductive effects:
• Electropositive elements and electron-donating substituents exert a +I inductive effect. Relatively strong -I effect:
These substituents therefore stabilise positive charge (cations) and destabilise negative
charge (anions).
• The more electron-donating the substituent, the stronger will be its associated +I
inductive effect.
• The +I inductive effect can be represented by drawing an arrow on the bonds, with the Weaker -I effect:
arrow pointing in the direction of electron flow, which is away from the electropositive
element.
Common substituents that • Inductive effects operate over relative short ranges, thus the (S and N are less electronegative than O, F, Cl)
exert +I inductive effects: more bonds through which the effect has to operate, the weaker Case study - acidity of water:
• Metals will be a substituent's ability to influence the stability of charge. • The pKa of water is approximately 15.7, the pKa of ammonia is
• Alkyl groups (R)
Organic Chemistry I Page 1
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