Acid Derivatives

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Ethanoyl chloride fumes in moist air, and reacts violently with water to give ethanoic acid. But ethanamide, from the same family, needs hot acid or hot alkali before it hydrolyses. Same C double bond O group, so why is one so wild and the other so calm? Hmm, hold that thought. One mechanism and one order explain it all, and this order comes up often in JEE papers. So, let's begin.

0:33Nucleophilic acyl substitution

Look at the top row of the figure. On the left is an acyl compound: R, then C double bond O, the two lines being the double bond, and a group, L, that can leave. The nucleophile, N u with a minus sign for its negative charge, attacks along the red arrow and adds to the carbon. The double bond opens, and oxygen takes the charge: this is the tetrahedral intermediate. With me? Now, the C double bond O can form again only if some group leaves the carbon. Which group goes? The leaving group goes, yes, exactly! The C double bond O re-forms, and L leaves as L minus. So the nucleophile takes the place of L, okay? Addition, then elimination: this is nucleophilic acyl substitution. Clear? But aldehydes and ketones stop at addition, and here is why. They would have to throw out H minus or R minus, and neither of these is a leaving group. Got it?

1:29The reactivity order and its two reasons

Now the row of boxes; the pointed sign means greater than. The acid chloride, RCOC l, reacts fastest with nucleophiles. Then the anhydride, where the brackets and the small two mean two RCO groups on one oxygen. Then the ester, RCOOR prime, the little mark being a prime. Then the amide, RCONH two. Last, in red, the carboxylate ion. Okay? Why this order? The first reason is the leaving group. Which one leaves more easily: chloride, C l minus, or N H two minus? It is chloride, exactly right! The weaker the base L minus, the better it leaves, and chloride is a weak base. Right? The red arrow under the boxes gives the basicity: chloride, less than carboxylate, RCOO minus, less than R O minus, less than N H two minus; the open sign means less than. So the leaving group gets worse as we go right, in step with the reactivity. With me? The second reason: the lone pair on L is delocalised into the C double bond O, so the carbonyl carbon becomes less electrophilic. This is greatest in amides. And our puzzle is solved! Ethanoyl chloride has the best leaving group, and ethanamide the worst, with the most delocalisation. Clear? And here is one useful result, from the bottom of the figure: a derivative is easily converted into any derivative to its right, but not directly into one to its left. Got it?

2:57Acid chlorides and anhydrides

Alright, now the reactions of acid chlorides. The plus sign adds a reagent, and the arrow means gives. Water, H two O, gives the acid. An alcohol, R prime O H, or a phenol gives an ester, with pyridine or N a O H, sodium hydroxide, in the brackets. Okay? Now use the mechanism. Ammonia, N H three, attacks, and chloride leaves. What is the product? It gives an amide, very good! N H two takes the place of chlorine, and amines do the same. And a carboxylate salt gives an anhydride. All these lie to the right in our order, right? Three more are useful for conversions. Hydrogen with palladium on barium sulphate, P d, dash, B a S O four, gives the aldehyde. Dialkylcadmium, R two C d, gives a ketone. And an arene with aluminium chloride, A l C l three, gives an aryl ketone. With me? Anhydrides react in the same ways, less vigorously, and the second product is one mole of the acid. Clear?

4:01Esters and amides

Now esters. Hydrolysis with an acid is reversible; the double arrow, a half arrow each way, means the reaction goes both ways. Okay? So with an alkali, it is reversible too, right? No, no, careful, that is the trap! In alkali, the acid is removed at once as its salt, the carboxylate ion, which the alcohol does not attack. So alkaline hydrolysis, saponification, is irreversible, and a simple ester, RCOOR prime, with one ester group of an alcohol, takes one mole of sodium hydroxide per mole. Keep this for Example two, okay? Now amides. Phosphorus pentoxide, P two O five, turns an amide into a nitrile, and lithium aluminium hydride, L i A l H four, gives an amine with the same number of carbons. With me? But bromine, B r two, with sodium hydroxide, the slash meaning together, turns the amide RCONH two into the primary amine RNH two, with one carbon fewer: the Hoffmann bromamide reaction. So count the carbons carefully. Got it?

4:59Example 1: rank four propanoic acid derivatives towards NH3

Time for Example one. Arrange in order of decreasing reactivity towards a nucleophile such as ammonia: propanamide, methyl propanoate, propanoyl chloride and propanoic anhydride. Here is a hint: look at the leaving groups. Pause the video and try it first. Done? Okay, let's check. Propanoyl chloride loses chloride, the weakest base, so it comes first. Propanamide would lose N H two minus, the strongest base, so it comes last. With me? Now the middle two. Both of them lose an oxygen anion, so the anhydride and the ester are equally reactive, right? No, no, wait, they are not equal at all! The anhydride loses a carboxylate ion, RCOO minus, while the ester loses methoxide, C H three O minus, which is a stronger base and so a poorer leaving group. So the anhydride is second. Okay? So the answer is propanoyl chloride, greater than propanoic anhydride, greater than methyl propanoate, greater than propanamide, exactly as in our figure. Clear?

6:01Example 2: ester formula from a saponification numerical

Now a harder one, Example two. One point four eight grams of an ester of a saturated monocarboxylic acid and a saturated monohydric alcohol needs twenty millilitres of one molar sodium hydroxide for complete saponification; capital M means molar. Give the molecular formula and the possible esters. Pause the video and try it first. Done? Okay, let's check. Twenty millilitres is zero point zero two litres, and one molar is one mole per litre, mol L to the power minus one. So the sodium hydroxide is zero point zero two times one, which is zero point zero two moles. With me? Now tell me, how many moles of ester did this saponify? Also zero point zero two moles, yes, exactly! This ester has one ester group, so saponification takes one mole of sodium hydroxide per mole of it. Right? So the molar mass, the italic M here, is one point four eight over zero point zero two, which gives seventy four grams per mole. Okay? Such an ester has the formula C n H two n O two, where n counts the carbons. Its mass is fourteen times n, plus thirty two. Set this equal to seventy four, so fourteen times n is forty two, and n is three. The formula is C three H six O two. Clear? Be careful here: the question asks for the possible esters. How many esters have the formula C three H six O two? There are two of them, very good! Methyl ethanoate, C H three COOCH three, and ethyl methanoate, HCOOC two H five. Both are possible answers. Got it?

7:47Quick revision: acyl substitution, order, saponification

So, quick revision. Acid derivatives react by nucleophilic acyl substitution: add, then expel the leaving group. The order is acid chloride, anhydride, ester, amide, and a derivative converts easily only into those to its right. And saponification is irreversible, with one mole of sodium hydroxide per mole of a simple ester, RCOOR prime. Now go and try the practice problems on this page. All the best!