ok so for this problem we’ve got a
substrate here which has a ketone functional group and a carboxylic acid
functional group and we’re gonna do two reactions with two different reducing
agents, we’ve got sodium borohydride and we’ve got lithium aluminum hydride so if
you don’t remember what these do go ahead and check out my lecture on oxidation
reduction in the organic chemistry playlist and we we these are going to
generate two different products based on reducing capacity of these two reducing
agents so go and see if you can draw the correct product for these two
transformations we’ve got our two different reducing agents, our sodium
borohydride and a lithium aluminum hydride, we know that those are
both sources of hydride, H- which is going to act as a nucleophile and
attack a carbonyl and reduce some carbonyl containing compound but the key
difference here is that sodium borohydride is a softer reducing agent, a
bit weaker whereas lithium aluminum hydride is a hard reducing agent, it’s a little
bit stronger, so there is a discrepancy in what kind of carbonyl
containing compounds they are able to reduce. as it happens sodium borohydride is
only able to reduce ketones and aldehydes, this is because the ketone
an aldehyde functional group, the partial positivity on this carbonyl carbon
differs from that of say a carboxylic acid because this carbon there, remember
that this adjacent oxygen atom is able to put electrons onto that carbon via
resonance and so that is the reason that this is not as partially positive and
therefore not as electrophilic as a ketone or aldehyde carbon so that means that
just the discrepancy in electrophilicity between this and this means that sodium
borohydride is able to reduce this ketone functional group but not the
carboxylic acid functional group so as a result on the product reducing with sodium borohydride we can
see that the ketone has been reduced to the secondary alcohol whereas the
carboxylic acid has been untouched it does not have the ability to reduce the
carboxylic acid so that won’t change however lithium aluminum hydride is able
to reduce ketones, aldehydes, carboxylic acids, esters, it is a stronger reducing
agent because it’s stronger it is able to reduce not just this but also
this and so as a result we can see in the product that the ketone has been
reduced to the secondary alcohol just as it has over here but in addition the
carboxylic acid has been reduced to the primary alcohol. so the key here is
noticing the difference in reducing strength of these two reducing agents and
using that information to draw the to correct products thanks for watching guys, subscribe to my
channel for more tutorials and as always feel free to email me