Substitution Reactions and Their Products

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Treat chloroethane with alcoholic potassium cyanide, and you get one compound. Now use silver cyanide instead. Same cyanide group, same chloroethane, and a completely different compound! How can the same cyanide give two different products? Sounds strange, right? Today, one simple idea, a nucleophile pushing out the halogen, will explain it. These products come up often in JEE papers. So, let's begin.

0:32The idea: a nucleophile replaces the halogen

Look at the bond between carbon and a halogen. On screen, C is carbon, X is the halogen, and the dash is the bond. The halogen is more electronegative, so the bond is polar. Delta plus, the Greek letter delta with a plus sign, means a partial positive charge on carbon, and delta minus, a partial negative charge on the halogen. Okay? Now tell me something. This carbon is a little short of electrons. So who will attack it: a species rich in electrons, or one poor in electrons? Rich in electrons, yes, exactly! Such a species is called a nucleophile. We write it as N u minus, with the minus charge written high, or N u with two dots, which stand for a lone pair. Clear? Now the reaction itself. N u minus plus R, dash, X, where R is the alkyl group, and the arrow means gives. We get R, dash, N u, plus X minus. The nucleophile takes the halogen's place, and X minus leaves; we call it the leaving group. So, in one line, R X becomes R N u. With me?

1:41One rule, many products: put Nu in place of X

This one rule gives a whole table. Here is the trick: do not cram it, just put the nucleophile where the halogen was. First, aqueous potassium or sodium hydroxide; a q, with a dot, is short for aqueous. It gives O H minus, so R X becomes ROH, an alcohol. Simple, right? Next, N a O R prime, a sodium alkoxide. R prime, the R with a small dash written high, is a second alkyl group. The nucleophile is R prime O minus. Put it in place of X, and what do you get? An ether, ROR prime, exactly! The name in brackets marks it as the Williamson method of making ethers. Okay? Two more rows, same rule. Sodium iodide in acetone gives R I, an alkyl iodide. And R prime COOA g, a silver salt of a carboxylic acid, gives R prime COOR, an ester. Got it?

2:38More nucleophiles: nitrogen, sulphur, carbon, hydride

Now a nitrogen nucleophile, alcoholic ammonia, N H three; the small three written low counts the hydrogens. Nitrogen attacks with its lone pair, and we get RNH two, an amine. Okay? But wait, the nitrogen in RNH two still has a lone pair, right? So if more alkyl halide is around, what can this amine do? It attacks again, yes, exactly! The amine is itself a nucleophile. The small circle written high is a degree sign: two with it means a secondary amine, three with it a tertiary amine, and finally a quaternary salt. So ammonia can give a mixture. Remember this trap. Clear? With sulphur, N a S H, sodium hydrogensulphide, gives RSH, a thiol, and N a S R prime gives RSR prime, a thioether. Right? Now a carbon nucleophile. In R prime C triple bond C minus, the three lines are a triple bond, and the minus charge sits on the end carbon, with a sodium ion, N a plus, beside it. That carbon attacks, and we get R prime C triple bond C R, an alkyne. With me? And finally, L i A l H four, lithium aluminium hydride, gives the hydride ion, H minus, so the alkyl halide becomes R H, an alkane. Okay?

3:56Ambident nucleophiles: two ends to attack with

Now the answer to our puzzle. Some nucleophiles have two different donor atoms. We call them ambident. Cyanide can bond through carbon or nitrogen, and nitrite through oxygen or nitrogen. With me? Potassium cyanide, KCN, is ionic, so the cyanide ion, C N minus, is free. Which end will it attack with, carbon or nitrogen? It attacks through carbon, very good! Because the carbon to carbon bond formed is stronger than a carbon to nitrogen bond. So potassium cyanide gives RCN, a nitrile. Right? Now silver cyanide, A g C N, is mainly covalent: its carbon is already bonded to silver, shown as A g, dash, C, here. So which end is still free to attack? Only the nitrogen, exactly! So the bond forms through nitrogen, and we get RNC, an isocyanide. And there is our puzzle solved: same cyanide, but a different end attacks! Got it? Nitrite works the same way. Potassium nitrite bonds through oxygen and gives an alkyl nitrite, R, O, N, double bond, O; the two lines are a double bond. Silver nitrite bonds through nitrogen and gives a nitroalkane, RNO two. Notice the pattern: both silver salts bond through nitrogen. Clear?

5:19Example 1: chloroethane with alcoholic KCN and with AgCN

Time for Example one, our puzzle from the start. Write the product of chloroethane with, part a, alcoholic potassium cyanide, and part b, silver cyanide. Pause the video and try it first. Done? Okay, let's check. Chloroethane is C H three, C H two, dash, C l; the small numbers written low count the hydrogens. So it has two carbons, right? In part a, potassium cyanide is ionic, so the free cyanide ion attacks through carbon. The three lines between C and N are the triple bond of the nitrile. Okay? Now let us name it. Chloroethane had two carbons, so this is ethanenitrile, right? Oh no, careful, that is the trap! Count the carbons on screen: C H three, C H two, and the cyanide carbon. That makes three, so the name is propanenitrile. The chain has grown by one carbon. Clear? In part b, silver cyanide is mainly covalent, so the bond forms through nitrogen. We get C H three C H two, then N triple bond C, an isocyanide. The chain is still ethyl, so the name is ethyl isocyanide. Same chloroethane, same cyanide, two different products! With me?

6:37Example 2: bromoethane with AgCN, then LiAlH4 reduction

Now a harder one, Example two. Bromoethane with silver cyanide gives A, and then A with L i A l H four, lithium aluminium hydride, gives B, okay? The reagent of each step is written above its arrow. Identify A and B, and pause the video to try it first. Done? Okay, let's check. Bromoethane plus a cyanide, so A is propanenitrile, just like Example one, right? No, no, wait, look at the reagent again! It is silver cyanide, and silver cyanide bonds through nitrogen. So A is C H three C H two, dash, N triple bond C, ethyl isocyanide. Okay? In the second step, lithium aluminium hydride reduces an isocyanide to a secondary amine, with two carbon groups on the nitrogen. So where does the end carbon of N triple bond C go? It becomes a methyl group, C H three, on the nitrogen, yes! The slanted N in N methyl means the methyl sits on nitrogen. Clear? So B is C two H five, dash, N H, dash, C H three: N methylethanamine, a secondary amine. So always check the reagent first, potassium or silver, because it changes the whole answer. Got it?

7:57Quick revision: KCN, AgCN, KNO2 and AgNO2 products

So, quick revision. First, the nucleophile attacks the partially positive carbon, the halide ion leaves, and R X becomes R N u. Second, potassium cyanide gives a nitrile, through carbon, and silver cyanide an isocyanide, through nitrogen. Third, potassium nitrite gives an alkyl nitrite, through oxygen, and silver nitrite a nitroalkane, through nitrogen. Now go and try the practice problems on this page. All the best!