University of Warwick
Department of Chemistry
Advice for successful arrow pushing (suggestions from MW).
1) Don’t push arrows from protons – they have no electrons.
CH3
O
Avoid:
H
CH3
Instead push arrow to proton:
O
H
H
wrong - no electrons on a proton.
H
correct
2) Substitution at unhindered saturated (sp3) C atoms is via SN2 mechanism (and inversion
when there is a chiral centre):
OMe
d- OMe
OMe
H
Et
Me
via transition state:
H
Et
H
Me
Et
Me
Br
d- Br
Br
Note that the reaction proceeds via a transition state, as illustrated.
3) Substitution at tertiary sp3 atom is via SN1 mechanism and a cation intermediate.
MeOH
H
OMe
OMe
Me
Me
Ph
-H
Me
Me
Et
Ph
Me
-Br
Br
Ph
Me
Me
Me
cation intermediate
(i.e. it is not a transition state).
4) Substitution at a sp2 C atom is via a tetrahedral intermediate:
MeO
Ph
Cl
OMe
Cl
Ph
Ph
-Cl
O
O
OMe
O
Tetrahedral
intermediate
5) The two steps above can be condensed into one:
1
MeO
MeO
Ph
OK
Cl
Ph
OMe
Ph
Cl
OR
-Cl
Ph
OMe
-Cl
O
O
OK
O
O
Me O
Ph
But you MUST show the
arrows on the C=O bond, i.e. this is INCORRECT:
Cl
WRONG
-Cl
Ph
OMe
O
O
6) Bromine addition goes via the bromonium ion:
Br
Br
Br
Br
Br
Br
(although not illustrated, note that the bromonium ion opening proceeds via an SN2
mechanism, and the bromide atoms in the product end up trans to each other).
7) The addition of ‘HX’ goes via a cation (the most stable one if there is a choice):
Br
Br
H
Br
H
H
Note – there is no cyclic intermediate in this case because the proton is too small to bridge the
bond (in contrast the Br+ is big enough).
8) Elimination – the E2 reaction is a concerted reaction:
Br
Base: alkoxide, LDA etc. (do you know what these are?).
H
:base
9) Protons at the a-position to a carbonyl group are acidic because an enolate is formed, with
the charge on O, not C. This is how it is formed:
You may have seen an enolate drawn like this.
But remember that is doesn't really have a charge
on the C atom:
O
O
H
O
:base
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10) The aldol reaction is a remarkably important and valuable synthetic reaction. Learn the
mechanism:
First deprotonate:
O
O
+
H-base
:base
H
(could be from the
protonated base,
or another source)
Then react with an aldehyde or ketone and finish by protonating:
O
O
O
H
OH
O
O
R
R
R
In some cases an elimination can take place:
O
O
OH
H
R
R
:base
11) Grignard Reagents have partial ionic and partial covalent character. Draw one form or the other in
mechanisms:
MeMgBr
=

Me

MgBr
or
O
Me
MgBr
O
but AVOID:
Me
MgBr
So if you draw the ionic version, the arrow comes from the bond. If you draw the ionic one then the arrow comes
from the negative group.
12) Lithium aluminium hydride is a more powerful reducing agent than sodium borohydride. LiAlH4 will reduce
virtually all C=O containing substrate, whereas NaBH4 will generally reduce only aldehydes and ketones
(although there are some exceptions). LiAlH4 will instantly ignite into flames if you splash water on it whereas
NaBH4 can be used as a reagent in a mixture of methanol and water!
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