 So, now that you are familiar with the reducing agents, lithium aluminium hydride and sodium borohydride that are used to synthesize alcohol from carbonyl compounds, let us look at a few reactions. Okay, in the first reaction that we have here, we will try to learn or understand the mechanism of this reduction. So here we need to figure out the product that is obtained in the given reaction. We are reacting an ester, more specifically a cyclic ester with lithium aluminium hydride. And we know that lithium aluminium hydride is a source of the hydride ion, H minus ion which helps us bring about the reduction reaction. So the first step obviously involves the addition of H minus to the carbonyl group. Now here the electron pair in the ALH bond attacks the C double bondo group because the CO bond is polar as carbon has a partial positive charge and oxygen has a partial negative charge and hence the hydride ion attacks the electrophilic carbon atom. And this gives us the following intermediate. Now in the next step elimination takes place, that is the C double bondo is restored and this CO bond gets broken. And as a result we have O minus for me here and here we get an aldehyde group. And aldehyde is much more reactive as compared to an ester. Now this is because if you look at the structures, an aldehyde group has an alkyl group attached to one end and a hydrogen atom on the other end. Whereas an ester has an alkyl group attached to one end and an alkoxy group attached to the other end. And if it is an acid you have OH group here. Now if you compare the structures you can see that the carbonyl carbon here is much more electron deficient as compared to the carbon in this case. Here the lone pair of electrons on the oxygen atom of the alkoxy group or in the case of acid the OH group will delocalize with the pi electrons of the C double bondo group. Now because of this continuous delocalization the carbon here is less electron deficient as compared to the carbon here. Now here you have no such electron donating group right? Here you have a weak electron donation via inductive effect by the alkyl group. But here the electron donation is much more stronger due to the resonance or delocalization. Now greater the electron deficient carbon or more electrophilic the carbon center is more ideal it is for a nucleophilic attack. And it is not just for the hydride ion attack but for any nucleophilic attack. More electrophilic a center is more tempting muridbi for a nucleophile to attack. Now the reason I'm telling this entire story is to substantiate why aldehydes and ketones are much more reactive than esters and acids. And if lithium aluminium hydride could reduce an ester it can easily reduce an aldehyde group as well correct? Exactly. So this is why another hydride attack takes place where the electron pair in this ALH bond attacks the C double bondo group of the carbonyl carbon and causes the delocalization of the pi electrons. Now once the first step or the reduction is completely done then we add water or mild acid to obtain the final alcohol. Here the O minus ions abstract proton from water molecules and give us the final product. Now the mechanism is quite similar even in the case of sodium borohydride reduction where the electron pair from HB bond or hydrogen boron bond attacks the C double bondo group and gives us the primary alcohol in the case of aldehyde and the secondary alcohol in the case of ketones. Let's look at the second question it says how can the following alcohols be synthesized by reducing carbonyl compounds? Okay so what carbonyl compounds should we reduce using lithium aluminium hydride or sodium borohydride to obtain these alcohols right? So let's draw the structure of these alcohols first. First one is one pentanol so the structure would be yeah this is one pentanol. Now as you can see this is a primary alcohol. Now we can get a primary alcohol by reducing an acid, an ester and even an aldehyde group. We can obtain the primary alcohol from an acid and an ester via lithium aluminium hydride and from an aldehyde by using sodium borohydride. So what acid should be the starting reactant here? So to obtain one pentanol our starting acid should be pentanoic acid or an ester where you have OH replaced by an OR group alright? Now pentanoic acid on reduction with lithium aluminium hydride would give us pentanol but not sodium borohydride. We can also pin the same primary alcohol by reducing an aldehyde like pentanol by reacting it with either lithium aluminium hydride or sodium borohydride that is either of the two reducing agents can be used to reduce this aldehyde group to the primary alcohol group. So let's look at the second one here we have 2 methyl 3 hexanol okay so let's draw the structure so this is 2 methyl 3 hexanol so OH group is at the C3 position now this as we know is a secondary alcohol now the only way we can get the secondary alcohol is if we had a keto group at this position at C3 so our starting reagent should be this ketone that is 2 methyl 3 hexanol and this on reduction with either lithium aluminium hydride or sodium borohydride would give us this secondary alcohol alright so let's look at one more question before wrapping up this video now here we need to identify the final products that are formed when this compound is subjected to reduction so now you know how lithium aluminium hydride and sodium borohydride differ in their selectivity correct so take a moment for yourselves and answer this question now sodium borohydride reacts selectively with aldehydes and ketones but not with less reactive ester group and it reduces aldehydes to primary alcohols and ketones to secondary alcohols so the product here would look like let's see so this is what we get with sodium borohydride reduction let's look at the product obtained in the second case now lithium aluminium hydride being highly reactive will reduce all of these groups aldehydes to primary alcohols ketones to secondary alcohols and again esters to primary alcohols