The Unimolecular Mechanism: S_N1

Transcript

0:00Introduction

Hello everyone! Here is a strange result from the lab. You react a halide with hydroxide ions, then double the concentration of hydroxide, the attacking reagent. Twice the attackers, so twice the speed, right? But for some halides, the rate does not change at all. So how is that possible? Today it is the S N one mechanism, which comes up often in JEE questions. Let's find out.

0:27SN1: two steps, and one is slow

So, first the name. In S N one, the S stands for substitution, the small N written low for nucleophilic, and the one for unimolecular. Three words, three letters. With me? Step one is slow. The carbon halogen bond breaks by itself: R X, the alkyl halide, with the dash standing for the carbon halogen bond, gives R plus and X minus; the arrow means gives. R plus, with the plus written high, is a carbocation, a positively charged carbon. In the figure, R one, R two and R three are any three groups, and B r minus leaves in the slow step. Okay? Step two is fast. The nucleophile, N u minus, attacks the carbocation. The two red arrows in the figure show it coming in. Right? Now let me ask you this. We have one slow step and one fast step. Which one decides how fast the whole reaction goes? Yes, exactly, the slow step sets the pace, like the slowest counter in a queue. Only one species, R X, takes part in this slow step, and that is why it is called unimolecular. So rate equals k times the concentration of R X; the square brackets mean concentration, and k is the rate constant. The rate law is first order. Clear? And look at what is missing: the nucleophile. The rate is independent of the nucleophile. That is the answer to our opening puzzle, and we will check it properly in Example one. Got it?

1:58Which halides react fastest?

The slow step makes a carbocation, so the more stable the carbocation, the faster the reaction. Reactivity follows carbocation stability. So tell me: tertiary, secondary, primary or methyl halide, which reacts fastest by S N one? Yes, the tertiary halide, absolutely! The small circle written high is read as degree, so three degree means tertiary, two degree secondary and one degree primary, and C H three is the methyl halide. The greater than signs show the order of reactivity. Okay? In allylic halides, the halogen is on a carbon next to a carbon carbon double bond; in benzylic halides, it is on a carbon next to a benzene ring. Both are very reactive. With me? Now here is a trap. Allyl chloride is a primary halide, and primary halides are slow by S N one. So allyl chloride must be slow too, right? No, wait, allyl chloride is hydrolysed much more readily than one chloropropane, another primary halide. Look at the allyl cation. The equals sign is the double bond, and the double headed arrow means resonance: the double bond shifts, and the plus charge moves to the other end. Clear? So the positive charge is spread over two carbons, and the cation is stabilised by resonance. The benzylic cation spreads its charge into the ring in the same way.

3:21The carbocation can rearrange

Got it? Now, one more thing about this carbocation. Before the nucleophile attacks, R plus may rearrange. Right? In a hydride shift, a hydrogen on the next carbon moves over with its pair of electrons; in an alkyl shift, an alkyl group such as methyl moves the same way. The result is a more stable carbocation, for example secondary to tertiary, as in Example two. Okay?

3:45Stereochemistry: racemisation

Look at the middle of the figure. The carbocation is planar, flat like a table top, and its carbon is s p two hybridised, with the small two written high. In the figure, the solid wedge is a bond pointing towards you, and the dashed line is a bond pointing away. With me? So let me ask you. The carbocation is flat. From which side can the nucleophile attack it? From either face, exactly, as the two red arrows show. Attack from the face the halide left keeps the arrangement, which is retention. Attack from the opposite face flips it, which is inversion. Both products are drawn on the right. Clear? So an optically active halide gives both, in nearly equal amounts, and the product is largely racemised: most of its optical activity is lost. S N two, remember, gives inversion, so do not mix them up, okay?

4:37Conditions that favour SN1

Now, the conditions. S N one is favoured by polar protic solvents, such as water, alcohols and ethanoic acid. They solvate both ions, the carbocation and the halide ion, so ionisation becomes easier. Right? S N one is also favoured by weak nucleophiles. Even a weak one is enough, because the nucleophile acts only in the fast step. And the silver ion, A g plus, helps by pulling off the halide, so the carbocation forms more easily. Got it? Now a quick question. Chlorobenzene is an aryl halide. Can it react by S N one? No, it cannot, and not by S N two either! Vinylic halides, with the halogen on a double bonded carbon, and aryl halides, with it on a benzene ring, react by neither mechanism. With me? This is because their cations are very unstable, so S N one is blocked. And the pi cloud, pi being the Greek letter on screen, keeps the nucleophile away from the back of the carbon. So S N two is blocked too. Clear?

5:39Example 1: double OH- and the SN1 rate stays the same

Now, let's finish our puzzle. Write the rate law for the hydrolysis of two bromo two methylpropane, and say how the rate changes when the concentration of hydroxide ion is doubled. Pause the video and try it first. Done? Okay, let's check. C H three in brackets with a small three outside means three methyl groups, all on the carbon that carries the bromine. That carbon is joined to three carbons, so this is a tertiary bromide, and it reacts by S N one. Okay? So the rate law is: rate equals k times the concentration of the bromide. Now we double the concentration of the hydroxide ion, O H minus. Does the rate double? No, it does not change at all! Hydroxide is not in the rate law; it acts only in the fast step. Putting hydroxide into the rate law, as you would for S N two, is the trap. And that solves our puzzle. Clear?

6:39Example 2: hydride shift gives 2-methylbutan-2-ol

Now a harder one, closer to JEE level. Two bromo three methylbutane is warmed with water. Name the major alcohol formed. Here is a hint: watch what the carbocation does. Pause the video and try it first. Done? Okay, let's check. Bromine is on carbon two of a four carbon chain, and C H three in brackets with a small two means carbon three carries two methyl groups, one of which is carbon four itself. In water the bromide ion leaves, and the plus charge sits on carbon two: a secondary carbocation. Okay? So water attacks carbon two, and the answer is three methylbutan two ol, right? No, wait, look at carbon three first! Its hydrogen moves over to carbon two with its pair of electrons, which is a hydride shift. Now the plus charge is on carbon three, joined to three carbons: a more stable, tertiary carbocation. With me? Water attacks this tertiary carbon, giving the alcohol on screen, with O H in brackets. The longest chain has four carbons, and both the O H and a methyl group are on carbon two. So the major alcohol is two methylbutan two ol. Three methylbutan two ol was the trap, because the carbocation moved first. Got it?

7:59Quick revision: rate law, reactivity order, racemisation

So, quick revision. S N one is slow ionisation to the carbocation, then fast attack, with rate equal to k times the concentration of R X only. The order is tertiary, secondary, primary, methyl, and allylic and benzylic halides are very reactive. Watch for rearrangement, and optically active halides give largely racemised products. Now try the practice problems on this page. All the best!