Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Butane has ten hydrogens: six on its two end carbons, and only four on its two middle carbons. Bromine replaces just one of them, in a free radical reaction. Six against four, so most of the product should have bromine on an end carbon, right? But in the flask, almost all of it has bromine on a middle carbon! How can that be? We will settle it in the last problem today. This is the JEE Main part of our hydrocarbons problem bank: three quick problems first, and then two full ones. So, let's begin.

0:39Quick problem 1: air is not pure oxygen

Quick problem number one. What volume of air, twenty percent oxygen by volume, is needed to burn one litre of propane completely? The sign after twenty means percent, and the capital L stands for litre. Okay? First, the equation. Propane is C three H eight, where the small numbers written low count the atoms. It burns in O two, oxygen, to give C O two and H two O, and the arrow means gives. The big numbers in front count the molecules: three C O two and four H two O, so we need five O two on the left. With me? At the same temperature and pressure, gas volumes follow the mole ratio, so one litre of propane needs five litres of oxygen. So the answer is five litres, right? No, no, careful, the question asks for air, not oxygen! Only twenty percent of the air is oxygen, so we take five, times one hundred over twenty. That gives twenty five litres of air.

1:42Quick problem 2: work back from ozonolysis

Clear? Quick problem number two. An alkene gives cyclohexanone and methanal, HCHO, on ozonolysis, and we have to name it. Remember, in ozonolysis, ozone adds to the double bond to form an ozonide, and splitting it with zinc and water cuts the molecule there, giving aldehydes and ketones: each carbon of the double bond becomes a carbonyl carbon. Okay? So, how do we go backwards, from these two products to the alkene? Yes, exactly, we simply reverse it! Remove the two oxygen atoms, and join the two carbonyl carbons by a double bond. With me? On screen, the two short lines are a double bond. Cyclohexanone is a ring carbon double bonded to oxygen, plus methanal, O double bond C H two. Take out both oxygens and join the carbons, and the arrow leads to the alkene: the ring carbon, double bonded to a C H two group outside the ring. Got it? So the alkene is methylidene cyclohexane, also called methylene cyclohexane, the name in brackets. Not cyclohexene, careful, because a cyclic alkene would give just one compound with two carbonyl groups. Okay?

2:54Quick problem 3: which step comes first?

Quick problem number three. We want one chloro three nitro benzene from benzene, and the arrow shows we go from benzene to this product. Positions one and three on the ring are meta to each other, with one carbon in between. So we need one chlorination and one nitration. Clear? Now, which step comes first? Chloro is written first in the name, so we chlorinate first, right? No, no, wait, let us check that first! Chlorine on the ring, the dash meaning it is joined to the ring, is ortho, para-directing, though deactivating. So the nitro group would go ortho or para to the chlorine, and that is the wrong product. Okay? Now the nitro group, N O two. It is meta-directing, and deactivating too, because the nitrogen joined to the ring carries a multiple bond to the more electronegative oxygen. With me? So the answer is nitration first, then chlorination, and the nitro group sends the chlorine to the meta position. Remember the rule: in a synthesis, put in first the group whose directing effect you need. Got it?

3:57Example 1: propane mass from sodium butanoate and soda lime

Alright, example one, an easy one. What mass of propane is obtained by heating eleven grams of sodium butanoate with soda lime? Small g is grams, and capital M, one hundred and ten, is the molar mass of the salt. Pause the video and try it first. Done? Okay, let's check. N a O H with C a O is called soda lime, and heating the sodium salt of an acid with it is decarboxylation. RCOON a, the sodium salt, with N a O H, gives R dash H, the alkane, plus N a two C O three, sodium carbonate. Here R stands for the alkyl group, and C a O and heat are written over the arrow. Okay? Now, sodium butanoate has four carbons. So how many carbons does our alkane have? It has three carbons, very good! The alkane has one carbon less than the acid, because the carbon of the COO group ends up in sodium carbonate. So C three H seven COON a gives C three H eight, propane, and one mole gives one mole. Right? The moles of salt are eleven over one hundred and ten, which is zero point one mole. Clear? Propane, C three H eight, has a molar mass of forty four grams per mole. So its mass is zero point one times forty four, which gives four point four grams. Count the carbons first, and the rest is just moles. With me?

5:22Example 2: percent 2-bromobutane from bromination of butane

Now example two, the harder one, and it is our opening puzzle. Find the percentage of two bromo butane in the monobromination of butane. The small circle written high is read as degree, so one degree is primary and two degree is secondary. Their reactivities per hydrogen are one to eighty two, and the colons mean a ratio. Pause the video and try it first. Done? Okay, let's check. Butane is C H three, C H two, C H two, C H three, and the dashes are single bonds. The two end methyl groups give six primary hydrogens, and the two C H two groups give four secondary hydrogens. Okay? Two bromo butane comes from the four secondary hydrogens. Four out of ten, so forty percent, right? Oh no, careful, that counts the hydrogens but forgets their reactivity! The amount of a product is proportional to the number of equivalent hydrogens, times the reactivity per hydrogen, and the curly sign means proportional to. Clear? For primary, six hydrogens times one gives six. For secondary, four hydrogens times eighty two gives three hundred and twenty eight. With me? The total is six plus three hundred and twenty eight, which is three hundred and thirty four. So two bromo butane is three hundred and twenty eight over three hundred and thirty four, and then times one hundred, which gives about ninety eight point two percent; the wavy equals sign means approximately. Only four hydrogens, yet almost all the product, because bromine is less reactive and so far more selective, and that solves our opening puzzle!

7:02Quick revision: air, ozonolysis, ring order, soda lime, Br2

So, quick revision. When a hydrocarbon burns, balance the equation first, then turn oxygen into air, twenty percent oxygen here. For ozonolysis backwards, remove the oxygen atoms and join the carbons by a double bond. On benzene, put in first the group whose directing effect you need. And soda lime gives one carbon less, while in halogenation, the amount goes as equivalent hydrogens times reactivity per hydrogen. Now try the other problems on this page yourself, before reading their solutions. Keep going, and all the best!