Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle from Set three. In cis one chloro three methylcyclopentane, the chlorine and the methyl group sit on the same face of the ring. Hydroxide ion replaces the chlorine, and the methyl group is not touched at all. So the product must still be cis, right? Most students say yes without thinking. Hold that thought for now. Today we warm up with quick problems from Set three, then solve two in full, including this one. These are JEE Main pattern problems, so let's begin.

0:36Quick fire: name the product

So, first a quick fire round. Toluene, which is methylbenzene, is heated with NBS, that is N bromosuccinimide. Where does the bromine go: on the ring, or on the side chain? On the side chain, yes, exactly! NBS brominates the benzylic carbon, the carbon joined to the ring, giving benzyl bromide, C six H five C H two B r; the small numbers written low count the atoms. The plus sign means reacts with, and the arrow means gives. Okay? Next, a solution of benzenediazonium chloride is warmed with K I, potassium iodide. What is the organic product, and which gas comes off? Iodobenzene and nitrogen gas, that is correct! The N two plus group is replaced by iodine, and no copper is needed. The plus and minus written high are charges, and the small upward arrow means nitrogen escapes as a gas. Right? Now look at the third problem. Iodomethane, C H three I, is heated with sodium phenoxide, C six H five O minus, N a plus. The phenoxide ion pushes out iodide, giving the ether methoxybenzene, C six H five OCH three, which is anisole. Clear?

1:55Bonds and carbocations

Now look at the bonds of the methyl halides. The carbon fluorine bond, C F, has the highest bond enthalpy, four hundred and fifty two kilojoules per mole, and the carbon iodine bond, C I, the lowest, two hundred and thirty four kilojoules per mole. The short dash is the bond, and k J mol with a minus one written high means kilojoules per mole. Okay? So let me ask you this. Which methyl halide reacts fastest with nucleophiles? Fluoromethane, with the strongest bond, right? No, no, careful, it is the other way round! In a substitution, the carbon halogen bond has to break, and the weakest bond breaks most easily. So iodomethane, C H three I, is the most reactive. Got it? Now three chlorides are hydrolysed in aqueous acetone, with one, two and three phenyl groups. Which one reacts the fastest? The one with three phenyl groups, exactly! So the order is triphenylmethyl chloride, then diphenylmethyl chloride, then benzyl chloride. C six H five is the phenyl group, the small number after the bracket counts them, and the greater than sign means faster than. Right? And the reason is the carbocation, the positive carbon left behind when the chloride ion leaves. Each phenyl group adds to the stability of the carbocation, so more phenyl groups mean faster hydrolysis.

3:20Haloarenes: nitro groups help

Clear? Now to the haloarenes: chlorobenzene resists nucleophiles. To make it react with sodium hydroxide, you need six hundred and twenty three kelvin and three hundred atmospheres; capital K is kelvin, and ATM means atmospheres. The figure on the right shows how a nitro group changes this. Okay? Now look at the figure on the right. Hydroxide ion, O H minus, adds to the carbon carrying the chlorine in four nitrochlorobenzene, in the slow step, and the ring takes a negative charge, the minus sign. The double headed arrow shows resonance: that charge moves onto an oxygen of the para nitro group, N O two. Then chloride ion leaves fast, giving four nitrophenol. So a nitro group at ortho or para takes up the charge; at meta it cannot. Right? Now here is the problem from Set three. One chloro two four dinitrobenzene is warmed with sodium methoxide, N a OCH three, in methanol. The numbers in the name give the positions on the ring. What is the product? The product is two four dinitroanisole, that is correct! The methoxide ion replaces the chlorine. With one nitro group ortho and one para, gentle warming is enough. Got it?

4:38Example 1: mass of 1-bromopropane from 0.1 mol PBr3

Now, Example one, a medium problem from Set three. Find the mass of one bromopropane made from zero point one mole of P B r three, phosphorus tribromide, and excess propan one ol. Bromine is eighty, its atomic mass, and m o l means mole. Pause the video and try it first. Done? Okay, let's check. Zero point one mole of P B r three, so zero point one mole of bromopropane, right? Oh no, careful, look at the equation first! Three ROH plus P B r three gives three R B r, plus H three P O three. In this equation the alcohol is written ROH, and R stands for the propyl group. The big three in front means three molecules. So one P B r three converts three alcohol molecules. Okay? So zero point one mole of P B r three gives three times as much: zero point three mole of one bromopropane. Right? Now the molar mass, capital M, of one bromopropane, C three H seven B r. Three carbons give three times twelve, seven hydrogens give seven times one, and bromine gives eighty. Thirty six plus seven plus eighty gives one hundred and twenty three grams per mole. Clear? So the mass is zero point three times one hundred and twenty three, which gives thirty six point nine grams; the cross means times, and g means grams. Write the balanced equation first, and this trap disappears. With me?

6:14Example 2: SN2 turns cis chloro compound into trans alcohol

Now, Example two, the hard one, and it is our puzzle from the start. cis one chloro three methylcyclopentane reacts with hydroxide ion by the S N two mechanism, written as a capital S with a small N low, then two. Name the product, with its stereochemistry. The figure on the right will help us in a moment. Pause the video and try it first. Done? Okay, let's check. The figure shows S N two on bromoethane. Hydroxide ion, H O minus, comes in along the orange arrow, opposite to the bromine; the solid wedge points towards you, and a dashed bond points away. In the square brackets with the double dagger sign is the transition state: the dashed lines to H O and to B r are bonds being made and broken, and delta minus, the Greek letter delta with a minus sign, is a small negative charge. On the right, bromide ion, B r minus, has left, and the other three bonds have turned inside out. That is inversion. Okay? Now back to our ring compound. In S N two, the hydroxide ion, O H minus, attacks the carbon from the back. Right? So the O H group enters on the face opposite to the chlorine, C l, on the other side of the ring. Clear? In the starting compound, the chlorine and the methyl group, C H three, are on the same face of the ring; that is what cis means. The methyl group is not touched. So the product is cis as well, just like we guessed at the start, right? No, no, wait, that is the trap! The O H came in opposite to the chlorine, and the chlorine was on the methyl group's face. So now the O H and the methyl group are on opposite faces. Got it? And opposite faces means trans. The product is trans three methylcyclopentanol; the carbon with the O H is carbon one, so the methyl group is at three. Puzzle solved: nothing changed sides on its own; the new group simply came in from the back. Clear?

8:16Quick revision: PBr3 ratio, SN2 inversion, C-X bonds

So, quick revision. Balance first: one P B r three converts three alcohol molecules. S N two inverts the carbon, and here cis became trans. The weakest C X bond, with X the halogen, reacts fastest, and more phenyl groups mean faster hydrolysis. And for nucleophiles to replace the halogen, haloarenes need a nitro group at ortho or para, or harsh conditions. Now try the rest of Set three on this page. You handled some real traps today, well done. All the best!