for this problem I have drawn a few different carboxylic acids and we want
to be able to rank them in terms of increasing acidity. so recall that acid
strength is proportional to the stability of the conjugate base and so
go ahead and see if you can rank these properly
in order to tank these in order of increasing acidity we have to look at the conjugate base. we know that
the conjugate base of a carboxylic acid, the carboxylate anion over here, we're
going to have, this is the composite resonance structure so the two resonance
structures we could have an oxyanion on one of them and then by resonance on
the other one, so this is the composite resonance structure we know that we have
a formal negative charge and so that negative charge is stabilized by
resonance through pi bonds but it can also be stabilized through the sigma
bonds by induction and so induction is when we have some electronegative atom
remember that electronegative atoms are gonna pull electron density towards themselves
and so this excess electron density in the form of a negative charge can be
stabilized if there were some atom that through induction is able to take some
of that electron density and hang on to some of it so it's not the same as
resonance that's through pi bonds, induction is through sigma bonds but it is still a
stabilizing effect. if we understand that having electronegative atoms here is
going to stabilize the conjugate base then the question becomes which one
stabilizes it best. so first of all this one we know is going to be the least
acidic because there are no electronegative atoms on it anywhere so
there is no induction happening when this deprotonates but on the other ones
we have some chlorine atoms and we have some fluorine atoms so what we need to
understand is that the intensity of the
stabilizing effect via the induction increases in two ways
actually in three ways, it increases as the electronegative
element becomes more electronegative and more able to hold on to some of the
electron density, number two as that electronegative element gets closer to
the site of deprotonation, and number three if there are more more atoms of
them, so we have chlorine and we have fluorine, so putting this all
together over here we have a chlorine atom, this is less electronegative
atom than fluorine, so it is less able to provide induction and so this one is
going to be next because it is weaker and further away. now the next one if
we're comparing the rest of these three here we can see that we've got a
chlorine atom one away right there and then we've got a fluorine atom one away
and we know that fluorine is more electronegative than chlorine so it is
able to offer more induction and there is a greater stabilizing effect for the
conjugate base. so we're gonna call this one number three and then over here
we're comparing one fluorine and two fluorine atoms, obviously having two
fluorine atoms is going to generate even more induction and have an even greater
stabilizing effect so that means that this one is going to be next and then
this one is the most acidic. so once again we're seeing that these
electronegative elements are stabilizing the conjugate base by induction and then
the correct order is as follows
no induction, a chlorine atom that is two positions away from the carboxyl carbon
and then a chlorine atom that is one position away because it gets more
stabilizing as the chlorine atom gets closer. then if we switch this chlorine
atom to a fluorine atom there's going to be even more stabilization because the
fluorine is more electronegative and then lastly adding an additional fluorine
atom is going to make even more stabilization because now there are two
fluorines, so the correct order one two three four and five
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