IIT-JEE Questions: Polyhalogen Compounds

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. An old IITJEE paper gives just three clues about a compound: twenty four point two four percent carbon, four point zero four percent hydrogen, and one gram of it gives two point nine zero grams of silver chloride. From only these numbers, can you draw two complete structures? Sounds like detective work, right? Today we solve two old IITJEE questions, from nineteen ninety two and nineteen eighty nine, on dihalides, compounds with two halogen atoms. Both test ideas worth knowing well. So, let's begin.

0:39Idea 1: geminal or vicinal?

So, idea number one. On screen, capital X stands for any halogen atom, such as chlorine or bromine. When both halogens sit on the same carbon, the dihalide is called geminal, like twins. For example, C H three, dash, C H Cl two; the dash is a single bond, and the small numbers are written low. Both chlorines are on carbon one, so this is one, one dichloroethane. Okay? Now look at this one: Cl C H two, dash, C H two Cl. Same formula, C two H four Cl two, but now each carbon carries one chlorine, on neighbouring carbons. So tell me, is this one geminal too? No, it is not, very good! Halogens on adjacent carbons make a vicinal dihalide, from vicinity, the neighbourhood. So this is one, two dichloroethane, an isomer of the first one: same formula, different structure. It comes back in Example two. Right?

1:43Idea 2: boiling with aqueous alkali

Now idea number two: boil a dihalide with an aqueous alkali, such as aqueous KOH, potassium hydroxide. This is hydrolysis, and each halogen atom is replaced by an O H group. Clear? For the vicinal one, each O H lands on its own carbon. The arrow means gives, so we get HOCH two, dash, C H two O H, which is a diol, a dihydroxy compound, with two O H groups on two different carbons. With me? Now the geminal one. Here both O H groups land on the same carbon. So the product is a diol again, and we stop there, right? No, no, careful, that is the trap! This geminal diol, with two O H groups on one carbon, loses a molecule of water at once. So that diol is never the final product. Got it? And what is left depends on where that carbon sits. At the end of the chain, written dash CHX two, we get an aldehyde. In the middle of the chain, the angle sign before C X two shows two bonds to two other carbons, and we get a ketone.

2:55Idea 3: a formula from analysis data

Okay? Now idea number three, for analysis data. Chlorine is often given as a mass of silver chloride. So how do we get the mass of chlorine from it? We use Carius' method, from earlier in the course. One mole of silver chloride, A g Cl, contains one mole of chlorine atoms. Its molar mass is one hundred and forty three point five, of which thirty five point five is chlorine. So the mass of chlorine is thirty five point five over one hundred and forty three point five, times the mass of silver chloride. The long line is a fraction bar, and the cross means times. Okay? Next, take each percentage, with the little percent sign, and divide it by the atomic mass of that element. Turn the results into the simplest whole number ratio, and that gives the empirical formula. Right? And the molecular formula is n times the empirical formula, where n is a whole number. Usually the molar mass gives n, but in Example two, watch where n comes from. Clear?

3:58Example 1: CH3CH2CHCl2 with aqueous alkali gives propanal

Time for Example one, from the IITJEE paper of nineteen ninety two. What is formed when C H three, dash, C H two, dash, C H Cl two is boiled with aqueous alkali? Pause the video and try it first. Done? Okay, let's check. Where are the two chlorines? Both on the end carbon, carbon one. So this is one, one dichloropropane, a terminal geminal dihalide. Okay? Now boil it with aqueous alkali. Each chlorine is replaced by an O H group, both on that same carbon. On screen, O H in brackets with a small two written low means two O H groups. So we get a geminal diol. Clear? So the answer is this diol, with two O H groups on one carbon, right? Oh no, careful, remember idea two! The geminal diol loses water at once: plus H two O, the water molecule on screen. What is left is C H three, C H two, CHO, okay? So the answer is propanal, an aldehyde, exactly as we expect from a terminal carbon. Writing the diol here is the classic mistake, so don't do that. With me?

5:19Example 2: Carius data to C2H4Cl2, then isomers Y and Z

Now the harder one, Example two, from the nineteen eighty nine paper: our puzzle from the start. Compound X has twenty four point two four percent carbon and four point zero four percent hydrogen, and one point zero gram of it gives two point nine zero grams of silver chloride. Careful, here capital X is the compound, not a halogen. X has two isomeric structures, Y and Z, okay? With aqueous KOH, Y gives a dihydroxy compound, and Z gives ethanal. Find the molecular formula of X, and the structures Y and Z, alright? Pause the video and try it first. Done? Okay, let's check. Chlorine comes from the silver chloride, by idea three. Thirty five point five over one hundred and forty three point five, times two point nine zero, gives zero point seven one seven grams; the small g means grams. That is in one gram of X, so chlorine is seventy one point seven percent. Okay? Now divide each percentage by its atomic mass. The colons on screen mean a ratio. Twenty four point two four over twelve gives two point zero two. Four point zero four over one gives four point zero four. And seventy one point seven over thirty five point five gives two point zero two. With me? Divide all three by two point zero two, and we get one is to two is to one. So the empirical formula is C H two Cl. Clear? So the molecular formula of X is simply C H two Cl, and we are done, right? No, no, wait, count the bonds on that carbon: two hydrogens and one chlorine, which makes only three. Carbon needs four, so n equals one fails, and the formula must be at least doubled to satisfy valency. Doubling gives C two H four Cl two, with two carbons, just what we need, because Z gives ethanal, a two carbon compound. So X is C two H four Cl two. Got it? Now, C two H four Cl two. We met its two isomers in idea one. Which one gives a dihydroxy compound with aqueous KOH? The vicinal one, yes, exactly! Each O H lands on its own carbon, giving a diol. So Y is Cl C H two, dash, C H two Cl, which is one, two dichloroethane. Right? And Z, which gives ethanal, must be the geminal one: C H three, dash, C H Cl two, which is one, one dichloroethane. Its geminal diol loses water to give the aldehyde, ethanal. So, from three numbers we got two complete structures, and the detective work is done. Clear?

8:21Quick revision: diol, aldehyde or ketone, and formula steps

So, quick revision. Boil a vicinal dihalide with aqueous alkali, and you get a diol. For a geminal dihalide, the diol loses water: a terminal one gives an aldehyde, and a non-terminal one gives a ketone. And for the formula: chlorine by Carius' method, then the mole ratio, then fix n by valency or by molar mass. Now go and try the practice problems in the earlier lessons of this chapter. All the best!