Reactions of Carboxylic Acids

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Mix ethanoic acid with ethanol and a little concentrated sulphuric acid. You get an ester, and a molecule of water. Now, that water has one oxygen atom. Did it come from the acid, or from the alcohol? Chemists settled it by tagging the alcohol's oxygen as heavy oxygen, oxygen eighteen. We will crack it in a few minutes. Today, the reactions of carboxylic acids, which come up often in JEE, so let's begin.

0:35Breaking the O–H bond: salts

So, a carboxylic acid can break in two places. First, the O H bond breaks, as in any acid, giving salts. Active metals give hydrogen gas, H two, where the small two written low means two atoms. Alkalis give the salt. And carbonates and hydrogencarbonates give carbon dioxide, C O two. Okay?

0:57Esterification: the C–OH bond breaks

Now the second place, the bond between carbon and O H: esterification. RCOOH, the acid, plus R prime O H, the alcohol, where R prime, R with a small dash, is another group. The two half arrows mean equilibrium, giving the ester, RCOOR prime, and water. The conditions are concentrated sulphuric acid, H two S O four, or dry HCL, right? So how does it happen? The acid picks up a proton, and the alcohol attacks this protonated acid, replacing the O H group on the acyl carbon: a nucleophilic acyl substitution. Got it? So, back to our puzzle. The oxygen of the alcohol carries the oxygen eighteen tag. Does it end up in the ester, or in the water? Yes, exactly, it stays in the ester! The acyl oxygen bond of the acid breaks, so the acid loses its O H group, and the alcohol keeps its oxygen. The water gets the oxygen of the acid, and our puzzle is solved. Clear? Next, the speed. The greater than sign here means faster. Methanoic acid is the fastest, then ethanoic, then two methylpropanoic, and two two dimethylpropanoic acid is the slowest. With me? Why this order? Because bulky groups slow the reaction. So for alcohols too, primary is faster than secondary, and secondary faster than tertiary. Okay?

2:21Replacing the OH: acid chlorides, amides, anhydrides

The O H group can also be swapped for chlorine, to give an acid chloride. We can use PCL five, PCL three, or SOCL two, thionyl chloride. So which one would you pick, and why? Thionyl chloride is the one we prefer. Its other products, sulphur dioxide and HCL, are both gases, so they simply escape. Right? With ammonia, N H three, we first get the ammonium salt. Heat it strongly, it loses water, and we get the amide. Got it? And heating the acid with P two O five, phosphorus pentoxide, or with a mineral acid, gives the anhydride. Okay?

3:01The whole COOH group: reduction and loss of CO₂

Now the COOH group as a whole. Lithium aluminium hydride, L i A l H four, or diborane, reduces the acid. The arrow means gives, so RCOOH gives RCH two O H, the primary alcohol. Got it? But sodium borohydride, N a B H four, is also a reducing agent. So will it reduce the acid too? No, no, wait, this is a common trap. Sodium borohydride does not reduce the carboxyl group. And diborane does not easily reduce ester, nitro or halo groups. Clear? Next, decarboxylation, the loss of carbon dioxide. Heat the sodium salt with soda lime, N a O H plus C a O, sodium hydroxide with calcium oxide. RCOON a gives R H, with one carbon fewer. With me? Kolbe electrolysis is different. Electrolyse the aqueous salt, and at the anode two R groups join, giving R dash R, okay? And beta keto acids and malonic acids, beta being the Greek letter on screen, lose carbon dioxide on gentle heating, through a six membered cyclic transition state. Right?

4:06The hydrocarbon part, and two special cases

Now the hydrocarbon part. In the Hell-Volhard-Zelinsky reaction, chlorine or bromine, with a little red phosphorus, replaces an alpha hydrogen. Alpha is the Greek letter on screen, and the alpha carbon is the one next to the carboxyl group. Okay? Now, can two two dimethylpropanoic acid give this reaction? No, it cannot, because its alpha carbon carries three methyl groups and no hydrogen at all. The acid must have an alpha hydrogen. Right? In aromatic acids, the COOH group deactivates the ring and is meta directing. And no Friedel-Crafts reaction, because the ring is deactivated and the catalyst bonds to the carboxyl group. Clear? Heat the dicarboxylic acids, and oxalic and malonic acids lose carbon dioxide, while succinic and glutaric acids give cyclic anhydrides. Got it? Last, methanoic acid hides an aldehyde unit, H, C double bond O, inside it. So it reduces Tollens' reagent and acidified KMNO four, potassium permanganate. With me?

5:11Example 1: benzene from 14.4 g sodium benzoate and soda lime

Time for Example one. Find the mass of benzene obtained when fourteen point four grams of sodium benzoate is heated with excess soda lime, assuming complete reaction. Pause the video and try it first. Done? Okay, let's check. Sodium benzoate, C six H five COON a, has seven carbon atoms. So the product also has seven carbons, right? Oh no, careful, the carbon of the COON a group leaves. Here R is C six H five, so R H is C six H six, benzene, with six carbons. Over the arrow are the conditions, soda lime and heat, and one mole gives one mole. Okay? Now the molar mass, capital M, of sodium benzoate. Seven carbons give eighty four, five hydrogens give five, two oxygens give thirty two, and sodium gives twenty three. Total, one hundred and forty four grams per mole. Right? So fourteen point four over one hundred and forty four gives zero point one mole of salt, so zero point one mole of benzene. Benzene has a molar mass of seventy eight, and zero point one times seventy eight gives seven point eight grams. Easy, isn't it?

6:26Example 2: anode gas volume in Kolbe electrolysis at STP

Now Example two, a harder one. In the Kolbe electrolysis of sodium ethanoate, what total volume of gas at STP, standard temperature and pressure, is released at the anode when two faradays of charge are passed? Pause the video and try it first. Done? Okay, let's check. At the anode, two ethanoate ions, C H three COO minus, give C two H six, ethane, plus two C O two, plus two electrons. The small minus written up top is the negative charge, and e minus is the electron. Okay? Two faradays means two moles of electrons, so one mole of ethane. One mole of gas at STP takes twenty two point four litres. So the answer is twenty two point four litres, right? Oh no, careful, look at the equation again. The capital F here is the faraday. The same two moles of electrons also give two moles of carbon dioxide, and that gas comes off at the anode too. Clear? So the total gas is three moles. Three times twenty two point four gives sixty seven point two litres. And the hydrogen, H two, forms at the cathode, so it does not count here. Got it?

7:42Quick revision: esters, LiAlH4 vs NaBH4, Kolbe and HVZ

So, quick revision. When the O H bond breaks, we get salts; when the bond between carbon and O H breaks, we get esters, acid chlorides, amides and anhydrides. Lithium aluminium hydride reduces the carboxyl group, sodium borohydride does not; soda lime gives R H, and Kolbe electrolysis gives R dash R, okay? And the Hell-Volhard-Zelinsky reaction needs an alpha hydrogen, while the COOH group is meta directing. Now go and try the practice problems on this page. All the best!