Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Glucose has a molar mass of one hundred and eighty. Nitric acid turns both ends of its chain into carboxylic acid groups, and each end picks up one oxygen atom. So the new acid should weigh one hundred and eighty, plus two times sixteen, which gives two hundred and twelve, right? But that is wrong! So where is the mistake? Hold that thought. This is Part A of the unit problem bank on biomolecules, at the JEE Main level. We will sharpen two tools, try three quick questions, and then solve two of its problems. Questions like these are typical of JEE Main, so let's begin.

0:44Tool 1: oxidising glucose

So, tool number one. On the right is the map of glucose reactions. In the figure, the box on its left side is D-glucose; the capital D in front names its configuration. Its formula has a C H two O H group at one end, then four CHOH units, shown by the bracket with a small four, and the aldehyde group, CHO, at the other end. Its molar mass, capital M, is one hundred and eighty. Okay? Now follow the arrow marked B r two water, that is bromine water, with a small lowered two. It is a mild oxidising agent, so it oxidises only the CHO group, into COOH, a carboxylic acid group; the arrow on screen means changes into. Right? The figure shows the product, gluconic acid. Now tell me, how much heavier is it than glucose? Yes, it is heavier by sixteen, the mass of one extra oxygen atom, and nothing else changes. So gluconic acid is one hundred and eighty plus sixteen, which gives one hundred and ninety six. Clear? Now the next arrow. The short form conc, with a dot after it, means concentrated, so this is concentrated nitric acid, HNO three. It oxidises the CHO group, and also the C H two O H group at the other end, the primary alcohol group. With me? So both ends become COOH groups, and we get saccharic acid, a dicarboxylic acid. Keep saccharic acid ready for Example one. Got it?

2:22Tool 2: enzymes and urease

Now, tool number two. Enzymes are biological catalysts. Look at the graph on the right of the figure: with the enzyme, the energy hill is much lower, so the activation energy is lower. Okay? And enzymes are highly specific, each one for one reaction and one substrate. In the figure on the left, the active site of the enzyme, E, fits only its own substrate, S, like a lock and its key. Right? In symbols, E plus S, with the two half arrows meaning a reversible step, gives E S, the enzyme substrate complex. Then the complex gives E plus P: the enzyme back again, plus the products. Clear? Now one enzyme to know well, urease. It breaks urea down into N H three, ammonia, and C O two, carbon dioxide. But notice, this line has no numbers in front, so remember it for Example two. With me?

3:16Three quick questions from Part A

Now, three quick questions from Part A, before the problems. Q five on screen means question five. Which reagent tells maltose from sucrose? The answer is Fehling's solution, or Tollens' reagent. Only maltose reduces it, because sucrose is a non-reducing sugar. Okay? Question nine: in what form does glycine exist at p H six, and what is its net charge? It exists as the zwitterion, a dipolar ion. On screen, N H three plus is at one end, C H two is in the middle, and COO minus is at the other end; the small plus and minus written high are charges, and the short dashes are bonds. One plus and one minus, so the net charge is zero. Right? And question fifteen: in RNA, uracil replaces which base of DNA, and which base does uracil pair with? Uracil replaces thymine, and uracil pairs with adenine. Got it?

4:20Example 1: formula and molar mass of saccharic acid

Now Example one, which is question one of the Part A set. Give the molecular formula and the molar mass of saccharic acid. Pause the video and try it first. Done? Okay, let's check. From tool one, saccharic acid has a COOH group at each end, and the same four CHOH units in the middle. Okay? Now count the atoms. Carbon first: two in the acid groups and four in the middle, so six. For hydrogen, each acid group has one and each CHOH unit has two, so two plus four times two, which is ten; the cross means times. For oxygen, each acid group has two and each unit has one, so four plus four, which is eight. Right? So the molecular formula is C six H ten O eight. Clear? Now the molar mass, and here is our puzzle from the start. Glucose is one hundred and eighty, and two extra oxygen atoms add thirty two, so two hundred and twelve, right? No, no, careful, that is the trap, so let us use the formula we just found. Six times twelve, plus ten times one, plus eight times sixteen. That is seventy two, plus ten, plus one hundred and twenty eight, which gives two hundred and ten grams per mole; g is grams, and mol to the power minus one means per mole. With me? So where did the two grams go? The CHO end only gains an oxygen, so plus sixteen. But the C H two O H end gains an oxygen and loses two hydrogen atoms, so only plus fourteen. One hundred and eighty, plus sixteen, plus fourteen, gives two hundred and ten, and that solves our puzzle. Got it?

6:15Example 2: mass of ammonia from 6 g of urea

Now Example two, which is question sixteen of the Part A set. Find the mass of ammonia formed when six grams of urea is completely hydrolysed by urease. Pause the video and try it first. Done? Okay, let's check. Urea is two N H two groups on a C O group; the small two after the bracket means two of them. So its molar mass is two times fourteen, plus four times one, plus twelve, plus sixteen, which gives sixty grams per mole. Okay? So the moles of urea are six over sixty, which gives zero point one mole; the slash means divided by, and m o l is short for mole. Got it? Now, from tool two, urea gives ammonia and carbon dioxide. So we get zero point one mole of ammonia, which weighs one point seven grams, right? That is exactly the trap, so balance the equation first. Urea has two nitrogen atoms, so with one water, H two O, it gives two N H three and one C O two. Check it: two nitrogens, six hydrogens and two oxygens on each side. With me? So zero point one mole of urea gives zero point two mole of ammonia. And the molar mass of ammonia is fourteen, plus three times one, which gives seventeen. Clear? So the mass of ammonia is zero point two times seventeen, which gives three point four grams. That is twice the trap answer, because each urea gives two ammonia molecules. Very good if you got it!

7:50Quick revision: glucose oxidation, saccharic acid, urease

So, quick revision. Bromine water oxidises only the CHO group, to gluconic acid, while concentrated nitric acid oxidises both ends, to saccharic acid. Saccharic acid is C six H ten O eight, with molar mass two hundred and ten, because the C H two O H end also loses two hydrogen atoms. And urea gives two ammonia molecules, so always balance the equation before you use the moles. Now try the other questions of Part A on this page as a test. All the best!