AIEEE and Earlier JEE Main Questions: Elimination and Metals

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take one alkyl halide, two chloro two methylpentane, and treat it with one reagent, sodium methoxide in methanol. How many products do you expect? Just one, right? Hmm, maybe not, so hold that thought, and we will come back to it at the end. Today we solve older A I triple E and JEE Main questions on elimination and on reactions with metals. Questions like these come up often, so let's begin.

0:33Elimination: losing H and X to make an alkene

So, first, elimination. In the lower picture, X is the halogen, on the carbon marked alpha, a Greek letter. The next carbon is marked beta, and a hydrogen on it is a beta hydrogen. With alcoholic KOH and heat, the halogen and a beta hydrogen are lost together. B minus, the base, takes the hydrogen, and the arrow means gives: B H, the alkene with its carbon carbon double bond, and X minus. The small angle marks are just other bonds. With me? Now the top of the picture: two bromobutane, where the small numbers written low count the hydrogens. Its bromine carbon has beta hydrogens on both sides, so the double bond can form in two positions. The alkene but two ene, with the double bond in the middle, is eighty percent, mostly the trans form, and but one ene only twenty percent. This is Saytzeff's rule: the more substituted alkene, with more alkyl groups on its double bond carbons, is major. In fact, A I triple E two thousand and five asked exactly this. Okay? And how does it happen? With a strong base, by E two, in one step: the base pulls off the beta hydrogen while the halide leaves, as the curved orange arrows show. And remember one condition: the H and the X must be on opposite sides, which we call anti. One of the hard questions on this page depends on that anti rule. Got it?

2:01Substitution or elimination?

Now, here is the big question in every halide problem. The reagent that attacks carbon as a nucleophile can also pull off a hydrogen as a base. So substitution and elimination compete with each other. Okay? So tell me this. Treat one alkyl halide with aqueous KOH in one flask, and with alcoholic KOH in another flask. Which one gives the alkene? Alcoholic KOH, yes, exactly! In water you get substitution, an alcohol. In alcohol you get elimination, an alkene, because the base there is the alkoxide ion, a stronger base. Clear? Now the tertiary halides. The small circle written high is read as degree, so three degree means tertiary. With a strong base, such as an alkoxide, R O minus, a tertiary halide gives E two. Right? But there is one more case. When a tertiary halide sits in water or an alcohol as the solvent, you get both S N one and E one products, with the N written low: substitution and elimination side by side. Remember this line, because our puzzle from the start depends on it. With me?

3:12Halides with metals

Now the second half: halides with metals. An alkyl or aryl halide, R X, with magnesium in dry ether, written over the arrow, gives R Mg X, the Grignard reagent. The short dashes are bonds, okay? Look at the carbon to magnesium bond. The small delta minus on carbon and the delta plus on magnesium are partial charges. So that carbon behaves as a carbanion, a carbon with a negative charge, and a strong base. Right? So now think. This strong base meets water, or an alcohol. What will it do? It grabs the acidic hydrogen, exactly! R Mg X plus H A gives R H, plus magnesium A X, with the A in brackets. Here H A stands for any compound with an acidic hydrogen, and the Grignard reagent is destroyed. Okay? So here is a topper's habit. In any Grignard question, first scan for an acidic hydrogen: water, alcohols, amines, carboxylic acids, terminal alkynes. That is also why the ether on the first line must be dry. Got it? One more metal reagent: R two, copper, lithium, a lithium dialkylcuprate. With R prime X, where the small mark called prime means a second alkyl group, it gives R R prime, an alkane, plus R copper and L i X, the lithium halide. This is the Corey House synthesis, and A I triple E two thousand and five asked just this.

4:43Example 1: PhMgBr with methanol gives benzene, not toluene

Clear? Let us try Example one, from A I triple E two thousand and six. What does phenylmagnesium bromide give with methanol? Pause the video and try it first. Done? Okay, let's check. Phenylmagnesium bromide is C six H five, M g B r, a Grignard reagent, and C six H five is the phenyl group. Methanol, C H three O H, has a hydrogen on oxygen, and that is an acidic hydrogen. Right? Now, suppose a student writes this: the phenyl group attacks the methyl carbon of methanol, and gives toluene. Is that right? No, no, careful, that is the trap! Methanol has an acidic hydrogen, and the Grignard reagent simply takes it. C six H five M g B r, plus C H three O H, gives C six H six, plus M g, OCH three in brackets, B r. Okay? So the products are benzene, C six H six, and M g OCH three B r. It is the same pattern we just saw, with phenyl as the R group. Clear?

5:57Example 2: 2-chloro-2-methylpentane with NaOCH3, all 3 form

Now Example two, from JEE Main two thousand and sixteen, and it is our puzzle from the start. Two chloro two methylpentane reacts with sodium methoxide in methanol. Which of these are formed: two methoxy two methylpentane, two methylpent one ene, or two methylpent two ene? Pause the video and try it first. Done? Okay, let's check. C H three in brackets with a small two outside means two methyl groups, on the carbon that carries the chlorine, C l in brackets. That carbon is joined to three carbons, so this is a tertiary halide. Right? Now, methoxide is a strong base, and the halide is tertiary. We saw that this gives E two, so the answer is only the two alkenes, right? No, not so fast, there is a catch here. Yes, E two gives the alkenes, and they are the main products. But the question asks which are formed, not which is major. The solvent is methanol, and a tertiary halide in methanol also gives substitution, as we saw before. Substitution puts a methoxy group, OCH three, in place of the chlorine: two methoxy two methylpentane. So that one forms too. Okay? Now elimination, and it can take a hydrogen from either side. Take a hydrogen from one of the methyl groups, and the double bond goes to the end: C H two, double bond, C, as on the screen. That is two methylpent one ene. With me? Or take it from C dash three, which means carbon number three, the C H two of the chain. Then the double bond sits inside the chain: two methylpent two ene. It is the more substituted alkene, so by Saytzeff's rule it is the major alkene. Got it? So the answer is all three! And the trap here is picking only the Saytzeff alkene, so do not forget the minor alkene or the substitution product. Puzzle solved: one halide, one reagent, three products.

7:59Quick revision: Saytzeff, tertiary halides, RMgX, R2CuLi

Clear? So, quick revision. With alcoholic KOH, the halide loses a beta hydrogen and X, and the Saytzeff alkene is major. Tertiary halides in an alcohol give both substitution and elimination. And R Mg X with an acidic hydrogen gives R H, while R two copper lithium with an alkyl halide gives an alkane. Now go and solve the other past questions on this page yourself. You can do this. All the best!