Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a tertiary alkyl halide, where the carbon holding the halogen is joined to three other carbons. In one kind of substitution, it is the fastest of the alkyl halides. In the other kind, it is the slowest, and practically does not react. So the same compound is both the fastest and the slowest? That is strange, right? Hold that thought, because Example two answers it. This is Part C of the unit problem bank: short concept checks, where every answer needs a reason. Reasoning questions like these come up often in JEE, so let's begin.

0:41Check 1: why Finkelstein needs acetone

Let us begin with check number one, the Finkelstein reaction. On screen, R X is an alkyl halide, the short dash is the carbon halogen bond, and X here stands for chlorine or bromine. It reacts with sodium iodide, N a I, and the arrow means gives. So we get the alkyl iodide, R I, and N a X, the sodium salt. Okay? Now, this reaction is run in dry acetone, not in water. Let me ask you this: why does the solvent matter so much here? Look at the salts. Sodium iodide dissolves in dry acetone, but sodium chloride and sodium bromide do not. So once they form, they cannot stay in solution. With me? So they precipitate, coming out as a solid. And taking a product out of the mixture pulls the equilibrium forward, towards R I, the alkyl iodide. That is what drives the reaction. Clear? And in water, all the salts stay dissolved, so nothing leaves the mixture, and there is no such pull. So it is all about which salts dissolve. Got it?

1:50Check 2: why methanol slows SN2

Now check number two, about the solvent. On screen, S N two is written with a small N set low, and N u minus is the nucleophile, with its minus sign written high. In the figure, H O minus, the hydroxide ion, attacks bromoethane. The orange arrow shows it coming at the carbon from the back, opposite the bromine. The middle structure, in square brackets with a double dagger, is the transition state, with half made dashed bonds; delta minus, the Greek letter delta, is a partial negative charge. On the right, the product forms, and the bromide ion, B r minus, leaves. With me? Now run this reaction in methanol, and it slows down. So tell me, what is methanol doing to the negative ion? Yes, it holds on to it! Methanol is a protic solvent: its O H hydrogen can form hydrogen bonds. So methanol molecules surround the anion and hold it by hydrogen bonds. Okay? Before the anion can attack the carbon, this shell of solvent molecules must be removed. That is the extra work, so the attack is slower. Right? That is why polar aprotic solvents, such as acetone, DMSO and DMF, favour S N two. They do not hydrogen bond to the anion, so it stays free to attack. And notice, acetone again, the same solvent as in check one.

3:15Check 3: vinyl chloride versus allyl chloride

Clear? Now check number three, and it hides a trap. Vinyl chloride is very unreactive towards nucleophiles, but allyl chloride is very reactive. Both have a double bond, so you might expect them to behave alike. Then why are they so different? Look at where the chlorine sits. Vinyl chloride is C H two, double bond, C H, C l. The equals sign is the double bond, and the small two written low counts hydrogens. So chlorine is on a carbon of the double bond itself, an s p two carbon, with the small two written high. Okay? And a lone pair of chlorine is in resonance with the double bond; the double headed arrow means resonance. In the second structure, the double bond has moved between carbon and chlorine, so chlorine carries a plus and the end carbon a minus, both written high. Right? So the carbon chlorine bond, shown with a dash, gets some double bond character, and becomes stronger. It is very hard to break, and that is why vinyl chloride is so unreactive. With me? Now allyl chloride: C H two, double bond, C H, C H two, C l. Here chlorine is on the C H two carbon next to the double bond, an s p three carbon, with the small three written high. Clear? And here is the key. When this chlorine is replaced, the double bond next door stabilises the cation or the transition state that is formed. Same double bond, but where the chlorine sits makes all the difference. Got it?

4:46Example 1: which atom of the C-X bond a nucleophile attacks

Now the two examples. Example one is an easy one. Which atom of the C X bond does a nucleophile attack, and why? Here the dash is the bond, and X stands for any halogen. Pause the video and try it first. Done? Okay, let's check. The halogen is more electronegative than carbon. So it draws the bonding pair of electrons towards itself. Okay? Now, a nucleophile is electron rich, and it looks for a positive centre. So which end of this bond becomes partly positive? Yes, it is the carbon, exactly! The bond is polar. On screen, delta plus, the Greek letter delta with a small plus written high, marks a partial positive charge on carbon, and delta minus on the halogen marks a partial negative charge. Right? So the nucleophile attacks the carbon, because it is the partly positive end of the bond. Got it?

5:45Example 2: tertiary halide fastest by SN1, slowest by SN2

And now Example two, a step harder, and it is our opening puzzle. Here is the question: a tertiary halide is the most reactive in S N one reactions, but the least reactive in S N two reactions. So why is that? Here is a hint: ask what each mechanism needs. Pause the video and try it first. Done? Okay, let's check. Start with S N one. Its slow step is the ionisation of the halide, giving the carbocation, R plus, with the plus written high. Okay? Now, this slow step is easier when the carbocation is more stable. So tell me: tertiary, secondary, primary or methyl, which carbocation is the most stable? Tertiary, yes, exactly! The small circle written high is read as degree, so three degree means tertiary, two degree secondary, and one degree primary. The greater than signs give the order, with methyl last, and reactivity follows this cation stability. So by S N one, the tertiary halide is the fastest. Right? Now let us look at S N two. Here the nucleophile must reach the carbon from the back, and in the transition state, five groups crowd round that carbon. With me? But a tertiary carbon carries three alkyl groups. What do they do to a nucleophile coming in from the back? They block it, almost completely! So the order flips: C H three X, the methyl halide, comes first, then primary, secondary, and tertiary last. Oh no, careful here: many students write the same order for both mechanisms. Don't do that! Clear? So here is the full answer. The tertiary halide gives the most stable carbocation, which is what S N one needs, but its three alkyl groups block back side attack, which S N two needs. Puzzle solved: one compound, two mechanisms, two different needs. Got it?

7:47Quick revision: the five haloalkane reasons in one place

So, quick revision. In the Finkelstein reaction, sodium chloride or bromide precipitates in dry acetone, and that drives it. Protic solvents hold the anion, while polar aprotic ones favour S N two. Vinyl chloride is held by resonance, while in allyl chloride the double bond helps. Nucleophiles attack the partly positive carbon of the carbon halogen bond. And tertiary halides are the fastest by S N one, but the slowest by S N two. Now go and try the practice problems on this page. All the best!