Name ______________________________________
1 Chlorination of alkanes gives mixtures of products where chlorine has replaced H at every
possible position. For example, chlorination of isobutane (CH3)3CH produces a mixture of tbutyl chloride and isobutyl chloride. (Although any one of the 10 Hs can be replaced by Cl, only
two products can be formed.) With this in mind, draw the monochlorination products of the
following alkanes, including stereochemistry. Be careful not to draw the same product twice!
(Experimentally, this reaction is difficult to control, and products with more than one Cl are
often produced. But for this problem we're only looking for the monochloro products.)
Cl2
h!
Cl2
h!
Cl2
h!
(think carefully about the
stereochemistry for this one)
2 Do you think this type of problem might show up on the final exam?
Lecture outline
Bromine atom is much more selective than chlorine atom in which H(s) it abstracts.
For example,
Cl2
h!
Br2
h!
+
Cl
+
Br
Cl
Br
1° vs 2° vs 3° Hs
for Br•
for Cl•
To simplify this, the relative reactivities mean that:
(1) Br• is very finicky and will abstract H to give the most stable radical
only;
(2) Cl• just grabs whatever random H it happens to run into, with (almost)
no regard for radical stability.
The key to understanding the difference in behavior of these two halogen atoms lies in the
energetics of the first chain propagation step.
For Br• + H–R, ΔH° ≈ +7 (3° H) to +13 (1° H) kcal/mol
For Cl• + H–R, ΔH° ≈ –8 (3° H) to –2 (1° H) kcal/mol
This, combined with the observation that exothermic free radical abstractions generally
have extremely low activation barriers, suggests the following rxn coordinate diagrams...
Chlorination — 1st prop step is exothermic
transition states resemble reactants
Cl•
H
Cl• + alkane
R
1° R• + HCl
3° R• + HCl
At the transition state, neither abstraction has gone very far, so the energy and structure of the
transition state still look like the reactants (Cl• + R–H) (Hammond postulate). The big
difference in energy of the final radicals, R•, doesn't matter much because there's not much
radical character on "R" at the transition state. Thus, the Eas are not very different, and the two
pathways go at similar rates.
Bromination — 1st prop step is endothermic
transition states resemble products
Br H
1° R• + HBr
•R
3° R• + HBr
Br• + alkane
At the transition state, both abstractions have gone almost all the way to completion, so the
energy and structure of the transition state look like the products (Br–H + •R) (Hammond
postulate). This time the big difference in energy of the final radicals matters a lot because the
radical character on "R" is almost fully developed at the transition state. Thus, the Eas are very
different, and the two pathways go at very different rates.
Free radical halogenation at allylic positions is especially facile —
Allylic and benzylic C–H bonds have relatively low BDEs
H
H
How do these values compare with a 1° C–H BDE of an alkane?
In practice, we can't brominate alkenes just by tossing in Br2 and turning on a light — the
Br2 would add to the double bond much faster than the desired radical rxn ever got going.
(Note that a light-initiated reaction won't go in the dark, but a thermal reaction — one that
doesn't require light — can go whether there's light or not. Stated another way, dark is the
absence of light, but light is not the absence of dark. There's still plenty of "dark" there, even if a
few photons go whizzing by once in a while. If there's enough thermal energy the "dark rxn" can
always go.)
We can get around this problem by using N-bromosuccinimide (NBS) instead of Br2.
The rxn works by the same free radical chain that we've already learned for alkane + Br2.
NBS reacts with the HBr formed in the rxn to generate Br2
O
only as fast as the rxn proceeds.
(The key is that the Br2 addition is more complicated than you
were led to believe — it's very dependent on the Br2
concentration; at very low concentration of bromine, the
reaction is extremely slow. In the NBS reaction the
concentration of Br2 stays extremely low, which slows the
competing addition to the π-bond, so the FR chain rxn wins.)
N Br
O
N-bromosuccinimide
(NBS)
An important practical difference between NBS brominations and simple brominations
with Br2 is that, like many free-radical rxns, the NBS rxn needs a separate initiator to get it
going. In the Br2 promination, Br2 serves as the initiator as well as the reactant. NBS can't
initiate the reaction because it doesn't absorb light and create radicals. Peroxides are good
initiators because the weak O–O bond breaks easily upon heating or when it absorbs UV light.
Azoisobutyronitrile (AIBN) is another popular initiator (see text section 21.4)
NBS
Br
h!
(or !)
and "RO–OR" (a peroxide)
or other f.r. initiator
Although it's unclear exactly how this gets started, the propagation sequence
has been established to be (complete the rxns)...
+
H–Br
+
Br•
Br2
NBS
+
+
"NHS" (i.e. just plain succinimide)
Br2
Now, predict the products of the rxn below. To do this, you'll need to draw the allylic radical
intermediate.
NBS
peroxide
h!
partial
mech
What products would be produced here?
NBS
peroxide
h!
(Here, 2 different radical intermediates are formed, and each leads to 2 bromo-alkene products,
so there are 4 products in all.)
In general,
— formation of delocalized radicals is much faster than formation of
localized radicals
— expect abstraction from all allylic positions (regardless of 1° vs 2°
vs 3° — Br• is not very picky when it comes to small variations
in the stability of delocalized radicals that are formed when it
abstracts H.
— expect addition of bromine to every C with some radical character
(i.e. every position with a δ•) — don't try to predict relative
amounts of the products from allylic radicals — there are too
many factors involved, and not enough info to figure it out.