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A good leaving group is a weak base, the conjugate base of a strong acid: iodide leaves best of the halides and fluoride worst, hydroxide hardly at all, and an alcohol substitutes only once its –OH is protonated or converted.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will rank leaving groups, choose the better of two from their conjugate acids, judge the forms of an –OH group, measure the pKa gap between leaving groups, and explain why alcohols need acid.
Unit 4 ranked acids by pKa and their conjugate bases by stability. A leaving group is a conjugate base in the making, so the same table ranks leaving groups.
A leaving group departs with the electrons of its bond to carbon. A tosylate, –OTs, is an alcohol converted into a sulfonate ester, an excellent leaving group. Protonation adds H+ to an atom.
When the leaving group departs it takes the bond's electron pair and a negative charge (or leaves neutral). It departs readily if it is comfortable holding that pair — if it is a weak base, the conjugate base of a strong acid.
| Leaving group | Conjugate acid | pKa | Verdict |
|---|---|---|---|
| I− | HI | −10 | excellent |
| Br− | HBr | −9 | very good |
| Cl− | HCl | −7 | good |
| H2O | H3O+ | −2 | good |
| F− | HF | 3 | poor |
| HO− | H2O | 16 | very poor |
| NH2− | NH3 | 38 | never leaves |
So alkyl iodides and bromides substitute readily, alkyl fluorides barely, and alcohols and amines not at all as they stand.
Another way: steps
To judge a leaving group:
Alcohols are common and useful, but hydroxide is a strong base and will not leave. Two fixes turn it into a weak base. Protonate it: in strong acid the –OH becomes –OH2+, which leaves as neutral water, so butan-1-ol reacts with hydrobromic acid but not with sodium bromide. Convert it: turning the alcohol into a tosylate makes a leaving group better even than chloride, without needing strong acid. Either way the chemistry is the same idea: make the departing group a weak base.
A leaving group must be a strong base. It must be a weak one.
Fluoride is the best halide leaving group because it is most electronegative. It is the poorest: HF is the weakest acid.
An –OH leaves as hydroxide in acid. In acid it leaves as water.
Any group attached to carbon can leave. Only weak bases leave well.
HCl has pKa −7; HF has pKa 3.
Compare conjugate acids.
HCl is ten billion times stronger, so chloride is the far weaker base.
Ten units, ten powers of ten.
So chloride leaves far more readily, and alkyl chlorides substitute where alkyl fluorides do not.
Weak base, good leaver.
Conjugate acids and pKa values?
Methanol about 16; HBr about −9.
Which is the better leaving group?
Bromide, the far weaker base.
Rank the halide leaving groups from best to worst: bromide, chloride, fluoride, iodide.
Number the steps in order (write the number in the box):
Which is the better leaving group: fluoride, F− (conjugate acid HF, pKa about $3$) or bromide, Br− (conjugate acid HBr, pKa about $-9$)?
An alcohol's –OH must leave in a substitution. Match each form of that oxygen group to how well it leaves.
| very poor | good | excellent | |
|---|---|---|---|
| –OH, leaving as hydroxide | |||
| –OH2+, leaving as water | |||
| –OTs, leaving as a sulfonate ion |
By how many pKa units is the conjugate acid of bromide, Br− (HBr, about $-9$) stronger than that of the amide ion, NH2− (NH3, about $38$)?
Answer:
Butan-1-ol does not react with sodium bromide, but it does react with concentrated hydrobromic acid to give 1-bromobutane. Why does the acid make the difference?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Say whether each is a good or a poor leaving group.
| leaving group | |
|---|---|
| bromide, Br− | |
| hydroxide, HO− | |
| water, from –OH2+ | |
| the amide ion, NH2− |
You can say why a leaving group leaves. Tell someone why butan-1-ol needs hydrobromic acid rather than sodium bromide. Next: choosing among substitution and elimination for a given substrate and reagent.
8. Your turn: methoxide against bromide, step 3