|Now let's look at the curly arrows for another NUCLEOPHILIC
SUBSTITUTION reaction, but this time a it's a little more involved.
Let's consider the SN2 reaction of isopropyl bromide and water.
First we need to draw in all the lone pairs. Once we have done that, we should work out which bonds have been made and broken.
Notice that these 2 curly arrows only show the formation of the C-O bond and the breaking of the C-Br bond and that the CHARGES BALANCE (2 neutral molecules give a negative and a positive). Since the neutral O in water gave away electrons to form the new C-O bond, the O becomes positive in the intermediate ion (an oxonium ion).
To complete the reaction we need to show the loss of a proton breaking the O- bond and the formation of the H-Br bond.
Do the charges balance ?
Now for the reverse reaction, i.e. the reaction of isopropanol with HBr to give isopropyl bromide.
This reaction must occur via the same pathway as we used above (the
Law of Microscopic Reversibility). However, we should be able to
deduce what happens without knowing that !
Do the charges balance ? Yes, we have 2 neutral molecules reacting
to give a +ve ion and a -ve ion.
Do the charges balance ? Yes, we have a -ve ion reacting with a +ve ion to give 2 neutral molecules.