 Now, next is sodium thiosulfate, Na2S2O3.5, the compound is not that important, okay. This reaction you must take care of, springs reaction, sodium sulphide with iodine and Na2SO3, sodium thiosulfate and Na8 forms, okay. The question is, they ask that what we use, like what compound we obtain in a springs reaction, so we use sodium thiosulfate we use to prepare for this, okay. And it is a salt of unstable acid, that is thiosulfuric acid S2S2O3, only one reaction, this is important, a spring reaction you must remember, use to prepare sodium thiosulfate, okay. Another method also we have, but not that important, like sulphur with NaOH forms this. Na2S sodium sulphide with sulphur forms Na2S5, pentahydrate. Heating effects, this reaction is important, like I have already told you, temperature dependent reaction in organic is very important. Pentahydrate form, when heated at 215 degrees Celsius, okay. Roughly temperature you must remember, 215 degrees Celsius forms thiosulfate and water dehydration takes place, water molecules goes out. Further you heated at 233 degrees Celsius, then form sodium sulphate and Na2S5, okay. Acidification, if you add acid to the solution, then sulphur dioxide gas evolves and sulphur precipitates. So what you have to memorize, sodium thiosulfate on acidification liberates SO2, sodium sulphur dioxide. Reduction with chlorine water forms sodium sulphate, it is not that important, okay. This complex color you must remember, like sodium thiosulfate with FECL3 forms a complex of iron of purple color and FECL2 of green color. Alkaline earth metal, it's group 2, 6 elements. Physical properties of these elements are electronic configuration we know, outermost configuration is NS2, atomic radius increases down the group. We all know, don't need to know to write on this. Density decreases down the group, because atomic radius increases, so density decreases, okay. Another thing is what density is larger than those of alkali metals. Left to right we are going, density is more than to that of alkali metals. This is due to a stronger metallic, okay. Melting and boiling point, the boiling point of these elements are low. Melting point is also low, but if you compare the boiling point and melting point from first group, that is alkali metals, it is more than that. The reason is what? They have two valence electron, which may participate in metallic bonding. And in alkali metal we have only one, hence group 2 elements are harder and have higher cohesive energy and have much higher melting and boiling point than alkali metals, okay. Melting point order you see, beryllium, maximum, and magnesium is the minimum, down the group if you go. So we have an irregular trend over here, must copy this down, melting point and boiling point order. Boiling point of stontium and beryllium is different here. Wherever you see irregular trend or some exceptions, you must remember that. Ironization energy if you see, it is obviously the ionization energy of alkaline earth metal is more than to that of alkali metal because of stable configuration, S2 configuration. But when you compare this with P block elements, then the alkaline earth metals because of their large size, because left to right if you go size decreases, large size have fairly low values of ionization energies as compared to the P block elements, okay. However, within the group, the ionization energy decreases as atomic number increases. Down the group you go, size increases, atomic number decreases. Ionization energy decreases, electropositive character increases, right? So down the group if you see, ionization energy decreases. Electropositive character increases because to lose electron will be more. Increases, electropositive character increases means metallic nature also increases because the energy to lose electron is more, also increases, okay. Oxidation number, it shows plus two oxidation state, we all know. And one more thing, it is less electropositive than alkali metals. Because of higher ionization energy. Okay, the oxidation state is plus two and M plus two ion possess a higher energy of hydration than M plus two ions are extremely hydrated, extensively hydrated and forms this complex, a hydrated ion, right? So first of all, what we require, we have to convert M into M plus two. So for this, we require ionization energy, IE one and IE two. And then if you dissolve this in water, then hydration energy evolves, okay. The total energy here is this plus this for this hydration process, okay. Reducing properties, as we go down the group, the metallic nature increases, tendency to lose electron is more, and hence the reducing property also increases. You see, down the group, reducing properties increases, okay. Flame coloration, this you remember, however it is not that important. Brick red, staunchium, crimson red, and barium, apple green. This also you see, beryllium and magnesium do not impart any color to flame. Because of the small size and high ionization energy. Chemical properties, alkaline earth metals are quite reactive due to their low ionization energy and high electropositive character. The reactivity of these elements increases with increase in atomic number, like ionization energy decreases. But it is less reactive than alkali metals, we know the reason, ionization energy. Hydrites of this, except beryllium, this is important here. You see, except beryllium, all alkaline earth metals forms hydrites of MS2 type. On heating directly with H2, M plus H2 gives MS2, where M is not equals to beryllium. Always, if you want to choose one different behavior of an element in a group, you should go with the first element because it shows abnormal properties. How do you prepare beryllium? Beryllium is prepared by action of LIALH4 on BECL2, direct hydration we cannot do. Direct, we cannot allow beryllium to react with H2. It won't form hydrides in that case. But if you want to prepare hydrides, we can take the chloride of beryllium with LIALH4, forms LICL, ALCL3, and BECL2. We know these hydrides, BECL2 and MGS2 are covalent, while other are hydrides. We know this fact, S-block hydrides are ionic in nature, except beryllium and magnesium, okay? The ionic hydrides of calcium, strontium, beryllium, liberate H2 at anode and metal at cathode. Hydrogen liberates at anode and metal at cathode. The stability of hydrides decreases from beryllium to beryllium. Because the size increases, bond length increases, stability decreases, okay? Hydrides having higher reactivity for water dissolves easily and produce hydrogen gas, okay? So, hydrides dissolves in water and evolves hydrogen gas. This is the reaction we have here. CAH2 plus H2O gives this, CAOH4 twice. What is this compound? You know the name of this compound? Yes. What is the name of this compound? Hydrolith, okay? Important in this name also, okay? Sometimes they ask this question that when hydrolith dissolve in water or cold water, which gas evolves? Hydrogen gas evolves, okay? Keep that in mind. Hydrolith is CAH2. All these alkaline metals reacts with water and evolves hydrogen gas. Reaction is given over here. Hydrogen gas evolves, H2 gas evolves. Magnesium decomposes hot water, but it also evolves H2. This also evolves H2. So, hydrogen liberates in all this. Beryllium does not react with water. Important this one is. Beryllium hydrides are, you know, strong. It's difficult to break the bond. Hence, it does not react with water. Order of reactivity. See, beryllium does not react, okay? So, beryllium we are not writing down here. To compare reactivity, molecule must react with another molecule. The reaction itself is not taking place. So, we are not considering beryllium here. It's not like we should write a beryllium in the here also. If you have this option and this option till here, then we'll go with this. Okay, don't get confused over here. Okay, that's the more precise one because beryllium does not react. Reaction with oxygen. All alkaline earth metals burns in oxygen to form oxides. Beryllium, magnesium and calcium form oxides, whereas beryllium and sodium forms peroxide. Okay, and that's why we also say tendency to form peroxide increases down the group. The hydrocytes of these elements can be formed either by resolving metal oxides in water or by reaction of elements with water like we have over here, this reaction. So, we can also use metal oxides or simply metal in water. Important property is BEH2 is amphoteric. BEH2 is amphoteric. Hydroxides of magnesium, calcium, strontium and beryllium are basic in nature and their strength from magnesium to beryllium increases, right? Means these hydrides, sorry, oxides are hydroxides are basic in nature and basicity increases down the group. Why it has more tendency as we go down the group, it has more tendency to release hydroxide ion because of larger size, okay? These hydroxides are less soluble in water as compared to the alkali metal hydroxides, okay? Not that important. Solubility of the hydroxides in water increases with increase in atomic number. BEH2, MGOH whole twice are almost insoluble. Why it is insoluble? So, lattice enthalpy is more than hydration enthalpy. Lattice enthalpy is more than, and why is that so? Why lattice enthalpy is more? It's covalent. It has, if you see the covalent, obviously it has more covalent character. Size is also small. So, because of that also it has high lattice energy. So, water molecule cannot penetrate into the lattice of this. Hence, it is insoluble, right? So, all these factors we have, covalent character, high lattice energy, right? Next, carbonates. Carbonates of alkali earth metals are insoluble in water. Important properties this one is. Carbonates are insoluble. This can be precipitated by addition of sodium or ammonium carbonate solution to the solution of salts of these metals. For example, we take calcium chloride and sodium carbonate. Form CSCO3 precipitate and NaCl. All carbonates decompose on heating to give carbon dioxide and metal oxides, MO and CO2. As the atomic number increases, how it should be Na2CO3? It's two is missing over here. Na2CO3, yes. Okay. As the atomic number increases, the stability of carbonates towards the heat increases. Okay. And why this happens, basically, as the atomic number increases. So, we have CaCO3, BeCO3, MgCO3, BacO3. Stability of these carbonates increases as we go down the group. Reason for this? So, it's that you have to look at polarizing power. Yes. Stability down the group increases. What is your high polarizing power? No, sir. Like there's a mean rate for polyatomic and monatomic. One of the famous polarizing power are the ones. Yeah, you can say like that also. But usually what happens, carbonates, sulphates, we consider these as larger anion. Okay. CO3, 2 minus, SO4, 2 minus. These are larger anions. So, we have the logic. You don't think about bond length. If you think about bond length and down the group bond length increases, then the stability should decrease, but that is not happening over here because bond length is not the major criteria. The criteria is the ions which stabilizes these anions. So, we have a concept here that we say that larger cations can stabilize larger anions. Okay. And carbonate sulphates are larger anions we have. Okay. So, if you go down the group, the size of cation increases and then the stability of carbonates also increases. Reason is larger cations can stabilize larger anions. Okay. Hence, you see that atomic number increases the stability of carbonates towards the heat also increases. Okay. Beryllium carbonate is unstable and can be kept only in atmosphere of CO2. Okay. It is also important. Beryllium carbonate can be kept only in the atmosphere of CO2. Okay. Sulfates, you see, forms by dissolving in acid, S2SO4, carbonates also if you dissolve, it forms sulphates. Property sulphates are what? Sulfates of alkaline earth metals are less soluble and the corresponding salts of alkali metals, right? Sulfates are lesser soluble than that of the alkali metals. The solubility decreases down the group. In water, the solubility decreases down the group because stability increases. On heating it, it involves sulphur trioxide. The thermal stability of sulphates increases on moving down the group. Thermal stability increases from top to bottom, right? Same logic I said just now that carbonates and sulphates are considered to be larger anion, which is stabilized by larger cation. Hence the thermal stability of sulphates increases on moving down the group from top to bottom. Lithopone is a mixture of BESO4 and ZNS. Many times they've asked this question. Is there any use for this Lithopone? Sorry? Doesn't Lithopone have any use? Yeah, we must have some use, but that use is not that important. Generally they ask, Lithopone is a mixture of what? Like BESO4 and ZNS. The composition you must remember. Okay, sir. Okay, done. Copy? Alta didn't get you. Yes, Alta, tell me. Okay, sulphates of alkaline earth metals are more stable than? Thermal stability, sir. Are more stable than the sulphates of group 1 metals? When you compare with the thermal stability, sir, I wanted to know. A thermal stability of sulphates of group 1 and group 2? Yes, sir, which is more greater? See, left to right if you go, the size decreases. Yes, sir. Correct, size decreases. Hence the stability, if you see, alkali metals will have slightly higher stability than alkaline metals. Okay, sir. Yeah, Lithopone, yeah, you can say it is also used as leather industry also we use it. Use are not that important here for Lithopone. Composition is more important, okay? Okay, nitrides you see, alkaline earth metal burns in nitrogen to form nitrides of N3 and 2 type, right? And these nitrides are ionic in nature. Direct combination. Nitrides when dissolved in water forms hydroxide, ammonia released in this. Carbides reaction you see, many a times you have seen this reaction, very important reaction. We get methane here, BE2C, CaC2 dissolved in a high, like water, it forms ethyne, acetylene, C2H2, okay? So BE2C, else methane on hydrolysis, whereas carbides of other metals is acetylene, okay? This line you must copy this up. Carbides for the question, both are example, J also, I've asked this. MC2 type carbides, group 2 elements forms, done? Okay, so carbides of group 2, group 2 metals are ionic carbides, ionic carbides, which may be in a methanoid on acetylide or allylide, like we can have anything, ions name it, it is methanide when one carbon atom, two carbon atom and allylide we have over here. So BE2C is methanide because it evolves methane, gives methane, calcium carbide, acetylide because it evolves acetylene, C2H2. Magnesium carbide, it calls allylide, reacts with water to gives allene, okay? CS3, C triple bonds in, methyl acetylene. This, you know, they've asked this one and this one also, I guess they have asked. Neat and JE both, okay? It dissociates like 2MG2 plus, plus C3, 4 minus, same thing. BEH2 is covalent, MGS2, it is partially covalent and partially ionic and the many hydrides are ionic in nature. BEH2 exist in, in polymeric form in which we have hydrogen abyssin. This bond you see, it is three-centered two electron bonded. Why it is three-centered two electron bond? Because BEH2, the bond is this and it has vacant orbital, barrier, right? 2P vacant orbital we have here. So what happens? This two electron, sigma electron, it donates into this vacant P orbital and this two sigma electron donates into this. It does not come out of the orbital. This here, we have a tri-junction, right? Orbital of this, orbital of this and orbital of this. These three will overlap and this two electron is now distributed among three atoms, two beryllium and one hydrogen atom. Two beryllium and one hydrogen atom. That's why we have three-centered, three nucleus, one, two and three, two beryllium nucleus and one hydrogen nucleus and among these three nucleus we have only two electron present. Hence the bond is called three-centered two electron bond. In vapor phase, BEH2 exists like this. Three-centered two electron bond and all the molecules are combined like this, right? So this is, we have the vapor phase molecules, two molecules we have here. If you have solid phase, then we have a polymeric structure. So keep that in mind. In vapor phase, beryllium hydrides exist as a dimer in which two molecules combines, but in solid state, it exists in polymeric form. That is this structure. All beryllium is sp3 hybridized here. Must remember this, very important. And here it is sp2 hybridized. So in vapor phase, hybridization is sp2. In solid phase it is sp3, halides. Two group metals when heated with halogens gives halides. Beryllium halides BEH2 and BECL2 are covalent due to the small size here, the covalent nature is. BEH2, BECL2 due to phagansal, hence have comparatively low melting point and boiling point. The chlorides and fluorides of other metals of the group are ionic solids. BECL2 in the solid state is polymeric containing chains. Okay, BECL2 in solid state is polymeric. So we have similar kind of structure like we seen in BEH2. In the vapor phase, at higher temperature, BECL2 exist as the monomer, which is BECL2 itself. This also you copy. Solubility of halides in water decreases down the group from calcium to beryllium. BEH2 vapor phase, see. Vapor phase we have dimer, right? If you look at the configuration of this, one is to two is to two p is vacant, vacant, right? That's what two p orbitals will have on both sides. Subcell we can say two p and in two p, which orbital we have that we cannot say. Here you see. Oh sir, I meant two orbitals of p. Yes, you see hybridization is sp2, right? So one s and two p orbital overlaps. Sir, vapor phase. What? Your voice is breaking. Sorry, vapor phase. Oh, this is the vapor phase. See, I'll tell you it. Beryllium has four electrons. So one is to two s2 and it combines with two hydrogen. BEH2 you consider now, right? Combines with two hydrogen. So this one electron will be here and one will be in two p1 like this in excited state. And the other two are p orbitals are vacant. Correct. So when this two orbitals forms sp because here we have dimer. So this two electrons, it donates into the p orbital. But in the bonding, what happens? We have one s, one p, which has one, one electron and one p also, which has supposed to two px. It is two p y I'm taking, which has no electron vacant p orbital. It takes part in hybridization and forms sp2 hybridized orbital. Okay, so I thought bottom one is vapor phase. I meant to ask for the bottom structure. Oh, this one you're talking about? Yes, sir. That's right. This is solid state. That's right. In this also it is same. You see, we have four bonds now. So one s and three p will take part in hybridization. So one h and one h, one c. Suppose this is h here, I'll take the middle one. This is the actual molecule you understood, this one. And we have two sp3 hybridized orbital here. So this electron donates into this and this electron from here, it donates into this. Two times the same thing is happening. Okay, sir, I understood. Correct, yeah. One second, Pradhup, Karbait, I'll just go back. This one, okay. So these are the questions I have taken from Bansal classes module, okay. Only few questions, but important one, okay? So I did not get the unsolved one, otherwise I would have taken that. But yes, we can go, like we can do the practice for this kind of questions. You see, NAOH gives disproportionation, reaction with what? Reaction is given with sulfur it reacts, you see. It forms sodium thioculfate, sulfide and H2O, right? We have seen this reaction just now. And this further reacts with sulfur, converts into Na2S5, okay? With excess pentaculfide forms, right? Pentaculfide forms. So disproportionation reaction with sulfur it gives, okay? CO2 gas along with solid Y is obtained when sodium salt X is heated. X is again obtained when CO2 gas is passed into Y, X and Y are. Okay, so we know what all things can eliminate CO2 gas. We can have carbonate or bicarbonate, two things we can think of, right? CO2 gas along with solid Y is obtained when sodium salt is heated, right? So we have sodium salt, suppose sodium bicarbonate we have options are given. Sodium bicarbonate when it is heated forms carbonate H2O and CO2 goes out, right? And we also know when this two combines with carbonates, it forms metal hydrogen carbonate, that is NAHCO3 or any metal hydrogen carbonate, okay? So that's what the question is. Answer for this question would be NAHCO3 and NA2CO3, okay? This is also factual, you see number 14, which of the following is used as barium meal for getting the X-ray spectrum of the human digestive system, okay? BSO4 is both insoluble in water and opaque to X-rays and hence is used to get X-rays spectrum of the digestive system, okay? Barium sulfate is correct over here. Potassium is kept in, kept in kerosene. Why it is in kerosene reason? Because it is highly reactive, okay? Alkali metals are highly reactive metals. They react with alcohols, water, ammonia, like the reaction given. Hence we kept this in kerosene, okay? To reduce the reactivity of potassium. Yeah, that's what we can say. That is also we can say. Highly reactive so it can reacts at normal condition also, normal temperature also, okay? Oh, it's the same thing I have taken. Anyways, this one we'll see. Which liberates ammonia when treated with, okay? CaCn2, mg3n2 and Li3n. All nitrides reacts with H2 to liberate ammonia, okay? So nitrides on reaction with water gives ammonia. This also is a nitride reaction with water gives ammonia. Calcium cyanamide on hydrolysis gives Ns3 with water, produce carbonate and Ns3. Hence the answer would be all here. The cation which gives yellow precipitate with potassium chromate, yellow precipitate with potassium chromate, okay? This is also factual. Barium gives a yellow precipitate of barium chromate. This definition, this color of this, you must remember, BACRO4. It is yellow in color, okay? Okay. So K2CRO4 plus BA2 plus gives barium chromate and 2K plus yellow precipitate of it. Na plus is larger than mg2 plus ion. S minus 2 ion is larger than Cl minus ion, which is the following least soluble in water. Sodium chloride, sodium sulfide, magnesium chloride and magnesium sulfide. Reason is you see the magnesium sulfide has higher lattice energy. We have seen this, higher lattice energy lower the solubility. Out of the four combination possible, the lattice energy of MGS bivalent cation ionic solid is higher than those of Na2S, MgCl2 and NaCl, right? So the one which has the higher lattice energy, it is difficult for water molecule to penetrate into its lattice and hence it is least soluble, right? So answer for this one is magnesium sulfide. MGS, MgCl2, NaCl and Na2S, these are the four compounds, just a second. And write down these four possibilities of the compounds like magnesium sulfide, magnesium chloride and the data is given Na plus ion is larger than mg2 plus, S2 minus is larger than this, okay? So based on the size which is given already, we can compare which one has the, no, like least lattice energy or maximum lattice energy and accordingly you can decide, okay? Okay, so these are the few questions. I think one of the slide I have missed a few questions. Okay, fine, just let it be. I'll share some PDF on this, okay? Some PDF we have, I have for this thing, four, five doc file I have, okay? I'll share that after the class, you can just go through it. Ansaki is also given into that and solution also if it requires, okay? So we can finish that on your own product blog, okay? So we'll share with you, okay? Next we are going to start biomolecules, okay? All these chapters you see one questions, like mostly you will get from this chapter also, biomolecules and polymers. So all these chapters are, we cannot say it is not important, you never know what questions they're gonna get, you're gonna get here. So like whatever theory is there, that you should know, you never know like which portions they'll ask, okay? So questions definitely you saw, you'll get the idea of which one is important topic or not, but theory you must go through for the entire chapter, okay? Like you see, carbohydrates, it is nothing but biomolecules, okay? So as the name suggests, you see it is the hydrates of carbon. We know it initially the definition was given like this, it is the hydrates of carbon and the general formula is C-X-H2-O-Y, which was later on proved to be wrong because there are many compounds which follows this particular form, but are not biomolecules. Like for example, any example of this could you give me? Lactic acid, what is the formula for lactic acid? Lactic acid, the formula is C-H3-C-H-C-O-O-H, and here we have OH, carol carbon, right, optically active. Lactic acid, this also we can represent in this form that is C-3-H2-O-3, same form. It is a carbohydrates, sorry, okay, these are not, lactic acid is not a carbohydrates, but the formula representation is same. Similarly, formaldehyde, H-C-H-O, obviously it is not a carbohydrate, but the formula representation is same, we can represent it this way, for which this definition was wrong later on. And later on, we have given a new definition that these are the polyhydroxy, aldehyde or ketone, which can be broken down into smaller polyhydroxy, aldehyde or ketone through hydrolysis, okay. So carbohydrates, these are the classification, but the definition of carbohydrates is what? These are the polyhydroxy, aldehyde or ketone, which can also be broken down into another, like a smaller polyhydroxy, aldehyde or ketone through hydrolysis, okay. Like we have glucose, fructose, other examples we have. So later on, these carbohydrates also classified into different categories, okay. We have monosecrites, we have oligosaccharides, we have polysaccharides. This term based on the number of monoseccharides we get on hydrolysis, right. Like monoseccharides are those carbohydrates, which cannot be broken down into small carbohydrates by hydrolysis, no hydrolysis possible for this. Right. General formula is this, oligosaccharides are what? These includes two to 10 monoseccharides linked together. Means when you hydrolyze this, you'll get two to 10 monoseccharides in this, after hydrolysis. Polysaccharides in this on hydrolysis will get more than 10 units of monoseccharides, greater than 10. So based on the number of monoseccharides we are getting after hydrolysis, we have done this classification. Mono, oligo and polysaccharides. Okay, monoseccharides, the examples are what? Okay, one more thing I'll tell you here. In oligosaccharides, what happens? Just a second. In oligosaccharides, we have two to 10, right. So if this oligosaccharides gives you two monoseccharides, if it gives two monoseccharides, then it is called as disaccharides, okay. It's called as disaccharides. And then if it is three trisaccharides, four heteroseccharides, et cetera, okay. Example of disaccharides are, example, the only important thing I've written on here, we have sucrose, we have maltose, and we have lactose. All these, one of the component is glucose. Whether it is sucrose, maltose or lactose, one of the component is glucose in this. But we have two monoseccharides possible. One is glucose, other one with the name you can understand, like if you write sucrose, one is glucose, other one is fructose, sucrose and fructose. You can understand it looks similar name. Maltose, one is glucose, other one is glucose itself here. Galactose, one is glucose, other one is lactose, other one is galactose, similar name you see. In short, I've written it, right. You should know this, that what is the component present in disaccharides? Try and tetra are not important. We're not going to do this. Second thing, like we should know like how these two molecules are combined, which carbon atoms are there in the, which are joined between the two, like glucose, what carbon atom, fructose, what carbon atom have been joined, okay. That we'll discuss later, but this is what that you need to know, the combination of these disaccharides, okay. Monosaccharides you see, it is polyhydroxy aldehyde or ketone, like I said. This one is he, polyhydroxy means more number of hydroxy group present, but the primary functional group is hydride. That's why it is polyhydroxy aldehyde. Primary functional group is ketone and hence it is polyhydroxy ketone, right. So if you have this N value, what is the possible value of N here, could you tell me? Number of carbon atom here in monosaccharides, the number of carbon atom, yes, in monosaccharides the number of carbon atom could be minimum three to maximum nine, right. Minimum three to maximum nine, total carbon atom. So N value could be, N value could be one to seven. So if N value is one here, this is the compound we get, three carbon atom. So it is aldotryos, because we have three carbon atom, try. Why aldo? Because aldehyde is the primary functional group. If you have two carbon atom, then it is total four here, N value is two total four here, aldotetrose, okay. Like this we do, aldopentose, aldohexose, right. Aldohexose is nothing but glucose. This name also you should know, it is not written here. Aldohexose, when N is equals to four, this is nothing but glucose, okay. Six carbon atom totals. Similarly, when you have ketone as the primary functional group, then ketotryos, ketotetrose, ketopentose like that, okay. So if N is equals to one here, it is ketotryos, N is equals to zero, it is no. N is equals to zero. N is equals to zero, it is ketotryos. N value could be here from zero to six. For it is zero, ketotryos, it is one, ketotetrose, it is two, ketopentose, okay. It is three, N value three, ketohexose. And ketohexose is nothing but fructose. I've done all these things in the class, but still if you forget, these things are important. Ketohexose is nothing but fructose. So fructose, the primary functional group is ketone and glucose, the primary functional group is aldehyde, okay. They have asked this name also, like what is glucose, aldehyde. So this also they have asked once, okay. These two name you must remember, okay. This is the cyclic structure, Havarth structure also we call it, right. Glucose is also known as dextrose. Okay, before going into this, glucose is also known as dextrose, because in nature, it exists as the optically active dextrorotatory form, dextrorotatory isomer. Hence it is, hence it is also called dextrose, okay. Found in sweet fruits, whatever sweet fruits you see, now all those glucose are present, okay. Now the thing is, why this is alpha, what is this alpha and beta we have, okay. Alpha and beta, it is defined with the reference of a compound called glyceridehyde, okay. So we took an example of glyceridehyde and its formula is this. We have OH on the right, H on the left, CH2 OH on the bottom and C double bond OH on the top. This compound is glyceridehyde. Glyceridehyde is the smallest unit of carbohydrate, okay. Smallest carbohydrate we have, okay. So based on the position of this OH on this carbon, the carol carbon, the position of OH, we say it is D or L, correct. D or L. So when OH is on the right side of this carbon atom, then it is D-glyceridehyde when it is on the right. Similarly, if you draw the another one in which OH is on the left and H is on the right, obviously it is isomer, CH2 OH. This we call it as L-glyceridehyde, glyceridehyde. D and L we define like this with respect to glyceridehyde, okay, OH on the right side of the carol carbon, it is D-glyceridehyde, okay. Now, if you talk about glucose here, then we have so many carol carbons, one, two, three, four carol carbon we have, okay. If you look at the structure here, glucose, this one is a structure, one, two, three, four, four carol carbon we have. So which carbon we should consider, right? So we consider the carbon atom, which is at the maximum distance from the primary functional group. Primary functional group is aldehyde, it is not alcohol, but it is aldehyde primary functional group. So carol carbon, which is at the maximum distance from the primary functional group, which is this carbon. So on this carbon, if OH is on the right, it is D. When OH is on the left, it is L glucose, like this with D and L we compare, okay. Now, when you talk about Havarth structure, so what happens in this, you see, the cyclic structure, this lone pair, it attacks onto this carbonyl carbon and shift this electron pair to this, okay. So this carbon atom, you see, it is sp2 hybridized here. Initially, it was sp2 hybridized, okay. This attack is possible from any side, from the top or from the bottom, both ways possible. If it is on the top, which means if one side attack will lead to OH on the right side and another one will lead to OH on the left side. So if it is OH on the right side, OH is present on the right side, then it is called as alpha D glucose. It is already D glucose because OH is here. Why it is alpha? Because this OH is on the right side, correct. So alpha when OH on the right, beta when OH on the left, we have both possibilities possible, right. Important thing here is right down in solid state, beta form is more stable. Solid state, this is more stable, beta form, okay. These two are, if you consider this one and two, these two are di-stereomers of each other. Alpha and beta form are di-stereomers of each other, correct. There's a term here called anomeric carbon. Anomeric carbon is the carbon atom which configures, okay, anomeric carbon is the carbon which has both OH and OR group attached to it. Like you see this carbon has one OH and one OR group attached to it. This is the anomeric carbon. This is the anomeric carbon. This carbon, anomeric carbon, this carbon, anomeric carbon. So you see on this carbon atom, OH is on the right and on this carbon atom, OH is on the left. So obviously the configuration of these two carbons are different. So anomers are what? Anomers are the carbohydrates which differs only in the configuration at anomeric carbon. Like these two carbon you see, in these two carbon, the configuration differs only at anomeric carbon. All other carbon atoms the configuration is same. Hence these two carbon atoms are what? These two carbon or molecules are anomers to each other, okay? So understood anomeric carbon is a carbon atom which contains both OH and OR group. This ether linkage should be there and hydroxy linkage should be there. Configuration differs at anomeric carbon are called anomers, okay? So this you see here, it is a cyclic structure. If you observe this, if I number this one, one, two, three, four, five, six. This is a cyclic structure. And this is what we're talking about here. Each carbohydrate molecules contains both a carbonyl group and a hydroxy group. If these two functional groups are properly positioned in the same molecule, then there can an intermolecular rearrangement forming a cyclic hemiacetyl structure. Why it is called as hemiacetyl structure? Because it has both OH and OR group present over here. If both OR group is there, then what is the structure called? If both group are OR, what is the structure called? Acetyl structure, correct? When we have OH and OR present the same carbon, it is hemiacetyl structure. If both are OR, then it is acetyl structure, okay? There are a few things that you must understand. Havarth projection, like I said, it is the same thing. We have six-member ring carbon atom, which we have to represent. I'll show you this here. It is not, this is a Havarth projection. This was a cyclic structure of this molecule. It is ribose formula, you should know. It has one carbon less, five carbon atom present, and both all the OH on the right side. One second, I forgot to tell you one thing. This kind of question they have asked in need here. If they ask you the, how do you represent glucose here? Like in this one, how do you represent glucose? So primary functional group is on the top, this one. So we'll number the carbon atom as one, it is two, it is three, it is four, five, and six, okay? So we can write the configuration of all the carol carbon. Like the second carbon, we have OH on the right side, so it is 2D. Third OH on the left, 3L. Four OH on the right, 4D, and five OH on the right, 5D. This represents glucose. So if sometimes they give this kind of four options, okay? Just you need to see OH on the right or left. Accordingly, we can say D or L configuration here. Okay, now Havarth projection you see, we have five carbon atom, okay? This OH will, this lone pair on this oxygen attacks onto this carbon atom, this goes up, and this H will get attached over here. Right, so we'll get this kind of structure. Okay, this we call it as D ribose, Puranose five-membered ring. Okay, so the five-membered ring. So in Havarth projection, what we do, there's one more term. See, if you talk about glucose, on the second, OH on the right, OH on the left, OH on the left, OH on the right. This is not glucose, right? In glucose, this OH is on the right side, we have. Okay, this is another molecule. We can draw the cyclic structure this way and then this, okay? The Havarth structure of glucose is based on a molecule called pyrene, okay? It's based on a molecule called pyrene. Pyrene, the formula is this and Havarth structure is based on this only of glucose. This is pyrene, okay? And the structure of glucose, Havarth structure, if you see, you'll get this. If it is alpha, then OH will be on the top, H will be here. It is a first carbon atom, second, third, fourth, fifth, and sixth. OH on the top, it is, it means it is on the right, it is the alpha form of it. Then we'll write down OH on the bottom, then OH on the top, alternate, then OH on the bottom, and then the last one we have CH2, OH on the top, right? OH on the top, yes, OH on the top, it is the beta form, yeah, that's right, beta form it is, okay? So when you have this, two things here you have to keep in mind that when this oxygen, this position is the first carbon, one, two, three, four, five, sixth carbon atom is not the part of the ring. That is one thing. Second thing, on the first carbon, if you change the configuration, means when you have OH on the top, then it is the beta form we have, beta form. Similarly, if you write this OH bottom and H on the top, then it is alpha form, right? Like I said, in solid state, beta form is more stable, because when you place OH over here on the bottom and H here on the top, then we have more repulsion in these two hydroxy group, right? And because of this repulsion, that structure is unstable, right? Hence, the stable structure is this one when both OH on the first and second carbon are opposite to each other, correct? This repulsion, when we have OH here, this repulsion we call it as diaxial lip. I'll draw that structure also, otherwise you will get confused in this only. So another one, the alpha form would be this. We have OH on the bottom, H on the top, then OH on the bottom, H on the top, OH on the top, H on the bottom, and the bottom, H on the top, and then we have H over here, H over here, and CH2, OH on the top, okay? You see here, OH on the first carbon atom, it is down. So it is alpha form, okay, alpha form. In alpha form, you see, we have these two OH, obviously we have more repulsion than this OH and this OH over here. This repulsion, we call it as diaxial repulsion, the technical term we have here, diaxial repulsion. And because of this only, the alpha form is slightly lesser stable than the beta form, okay? Since this structure is given on the reference of Pyrene, hence the name of these two structure is glucopyrinose. So when they have given this structure, then you should consider, when they have given this structure, then you should consider the Havod structure of glucose, not the simple one, okay? Mutarotation, important on this also, they have asked question, when an open chain monocyclic site recyclizes to a furanose or a pyranose form, a new carol center at the carbon, at the carbonyl carbon is created, the two diastereomers produced are called anomers, we have discussed it. So what happens in this? This one, furanose we haven't done, I'll discuss this, just give me one minute, after this we'll discuss. So it is a spontaneous change in that specific rotation, okay? So we have alpha D glucose and beta D glucose. So what happens if you have alpha D glucose, alpha D GL right down here, if you dissolve this in water, okay? So it converts into D glucose simply, open chain. This was the cyclic chain because we have alpha over here, D glucose, and when this again try to form the bond, the cyclic structure, then this could be beta as well as alpha also. Both ways possible. If it attacks from the top, when OH on the right, it is alpha, when OH on the left, it is beta, right? So after D glucose, we can get beta D also over here. So basically in solution, we have both form exist, alpha and beta both form exist. If you look at the specific rotation of this, it is somewhere around 111 degree. And for this one, it is around 19, 19.5 or 19 degree, you can say, it is specific rotation. So this two exist in equilibrium and the equilibrium value of this specific rotation is 15 to 0.5 degree, right? So what is beta rotation? It is the spontaneous change in the specific rotation before the equilibrium established, spontaneous change in the specific rotation. Alpha is converting into beta, beta is converting into alpha. So this is spontaneous change, continuous change that we have. This phenomenon is beta rotation, right? So what is written here, you see? If either alpha or beta form of glucose is dissolved in water, the optical rotation changes eventually a value of this is obtained. Thus alpha form changes to beta form or vice versa and eventually a solution containing 36% of alpha form and 64% of beta form is obtained. This phenomenon is known as muta rotation. So all these spontaneous changes happening before the equilibrium established is called muta rotation, okay? So like I said, gluco-paranos is based on pyrene, the molecule that we take to represent the structure of gluco-paranos. Similarly, we also have defined the Havarth structure of fructose. And Havarth structure of fructose is defined by a molecule called furan, okay? So if you look at this molecule, furan, and hence we call it as fructo-furanose. Furan, the molecule is this, this one, okay? Based on this, the structure of Havarth structure of fructose has been given and hence the name of the structure is fructo-furanose. Okay, so beta rotation is done. You will have, you will get some questions also and it's numerical question, we'll see that. We'll see that after this discussion. Okay, what is reducing and non-reducing sugar? Yes? Attached by both the anomeric carbon when it's non-reducing because there's no group which can lead you. They attach at both anomeric carbon. Yes. Two monomers, like we can say two monosecrites. They combines to form a disaccharides, correct? So we can have any carbon atom attached of the two monosecrites. So reducing sugar is the one in which, first of all, definition of reducing sugar is which can reduce tolerance and tolerance reagent and failing solution, okay? Non-reducing sugar, right? Non-reducing sugar is the one which cannot reduce these reagents, tolerance reagent and failing solution, right? All those molecules or carbohydrates in which there are free aldehyde or ketone group present, free aldehyde or ketone group present are called reducing sugars, okay? Basically in which anomeric carbons are not involved, are called reducing sugars. It's reducing sugars are easily oxidized to give carboxylic acid, okay? They reduce tolerance reagent. Here it is written you see. They reduce tolerance reagent. They reduce, you know, failing solution, right? Benedict reagent or Benedict solution to red precipitate of cuprous oxide. So all those carbohydrates which can reduce these reagents are called reducing reagents, reducing sugars. All aldoses are reducing sugars, but some ketoses are reducing sugars as well. For example, fructose reduces tolerance reagent. Even it contains no aldehyde group. We know the reason the automerism and all the mechanism we have, but fructose has have the functional group that is ketone. Still it can reduce tolerance reagent. We as you know, aldehyde can reduce tolerance reagent. Okay, but fructose can do because of that equilibrium we'll discuss that later. This occurs because fructose is readily isomerized to an aldose in basic solution. This is the example given over here. It is fructose double bond O, right? Isomers of this, you'll get this tautomerase. And then again, this converts into this molecule. C-H-O-H is a double bond comes over here and H will go on to this carbon atom. Okay, you see here, this group you see. It is C double bond C H-O-H, the top two. This is O-H, let it be as it is and other things we have here. What happens? This pi will shift here, tautomerism and this H will come over here in basic media. So you get C-H-O-H here. Okay, C-H-O-H here on the top we have C-H-O. You see this keto converts into aldehyde finally. And when the aldehyde forms, it can reduce the tolerance reagent solution, correct? That's why we usually say all ketone cannot reduce tolerance reagent except fructose, okay? Because of this tautomerism. Okay, this is the reaction we have. So reducing sugar, you have to keep in mind. If the molecule has free aldehyde or ketone group, then it is reducing sugars. Important one, all monosecrites you can say. Monosecrites in which we have aldehyde as a major functional group. In ketone, we have only one, we can think of fructose here. All aldoses are reducing sugars. Many a times they've asked this question, which one is reducing or not reducing sugar? Now these are the few reaction-based thing we have, like what happens in what reaction that you must understand here. Okay, so sugar is completely degraded with H-I-O-4. You have to keep this in mind that when we use H-I-O-4, all the C-H-O molecule converts into H-C-O-H, C-S-T-O, C-S-2-O-H converts into aldehyde H-C-H-O. If you have carbonyl group, converts into CO2, C-double bond law, right? Based on these, we have this reaction given, okay? C-S-2-O-H, you see, it converts into C-H-O. This is C-H-O, this is C-H-O. So at C-H-O, we got C-H-O-H we have, C-H-O-H converts into C-O-O-H. We have CO group converts into CO2, okay? With H-I-O-4, strong oxidizing agent, phenyl hydrazine. One minute, can you go back? Yeah, one second. Hadn't we earlier seen that H-I-O-4 was only for consecutive diodes? Yes, sir, thank you, sir. So what happened, Luchey? Sir, earlier, like in early age, ketones or alkyl receptor, we saw H-I-O-4 was only for consecutive diodes or triodes. Over here, we can do with adiados, sir. Yes, it's possible, no? It's consecutive diol only. We have like this now, O-H, like O-H, O-H. But it's a strong reducing agent, oxidizing agent. So it can convert C-H-O also, aldehyde also, it's possible. Okay, okay. It's fine, we are happening over here. But with aldehyde also, it's possible. It converts first, like aldehyde, it will oxidize into acid. Alcohol, obviously we know, it converts into acid. So it converts into C-H-O-H also. Like this is for fructose volathing because the second carbon, you'll get C-H-O-H. So whether you have aldehyde or alcohol, this aldehyde will convert into, this is just for you to memorize mechanism because we are not doing. So if you have C-H-2-O-H present, then you'll get an aldehyde oxidation. If you have C-H-O acid, C-H-O-H acid, if you have keto, then carbon dioxide you'll get. Okay, sir. Sir, in the haward structure, it doesn't open that oxygen bond, sir. That will happen in solution. Yes, sir. But if they give the haward structure reaction, then only three of the things will react, sir. Three of the things, as in which one? So only the second, third, and fourth carbon will react. The other two will still stay. Yeah, you are talking about with H-I-O-H, no? Yes, sir, yes, sir. Yeah, yeah, right, right. Okay, sir. Okay, so in this one, you see phenylhydrazine, on this, they have asked question in J also, like in previous year, phenylhydrazine reaction. I have done this reaction in the class also, if you just go back and check your notes, I've done this reaction with mechanism I have done. Okay, so in phenylhydrazine, what happens? What is the name of that reaction we call? Something formation, if you remember. What is that? Remember that osazone formation? Yes, it is osazone formation. So important reaction it is, osazone formation. Sugars form characteristics, osazones with phenylhydrazine. Okay, so when you have phenylhydrazine, you have to just memorize this that what happens here, we'll get this C double bond N-N-H-P-H, C double bond N-N-H-P-H, and this lower part will be as it is. Okay, here they have written this, O-H, this side, this is for what? This is for ketose. So this is for, just a second, we have this, there's some other compound they have taken, not a problem. So what we need to do, when you take a big glucose here, in the first two carbon atom, we have discussed the mechanism, tautomerism takes place, and finally with glucose, you will get with phenylhydrazine, the name of the compound is glucosazone. In glucosazone, what happens? First two carbon atom will have double bond N-N-H-P-H, H2O will go out. Along with H2O, what forms? Along with H2O? Along with H2O, we get these three bonds, just let it be. Along with H2O, we'll get primary, we'll get aniline, phenyl with NH2. This is one compound we get. We also get ammonia, we also get H2O in this reaction. Okay. If you look at the structure with glucose and fructose, since the reaction takes place at first two carbon atom, and if you see the difference in the structure of glucose and fructose, it is only at the first and second carbon atom. This part from the third onwards, it is exactly same in glucose and fructose. Since the reaction takes place here is same, so the formula of glucosazone and fructosazone is exactly similar. There's no change in that, correct? If you take fructose, we'll say fructosazone. If you take glucose, we'll take glucosazone, okay? This is the important thing. We have both have the same structure, because the last lower part is exactly same. Okay, oxidation reaction you see will take aldose. This one we have, it's a general thing they have written. So what happens? C-H-O-H will be as it is with bromine water, but C-H-O will get oxidized into C-O-H. Okay, bromine water is what? It is a mild oxidizing agent for what? Right, bromine water is a mild oxidizing agent, not very strong, right? So this will oxidize aldehyde into acid. HNO3 is a strong oxidizing agent, and it will oxidize both this alcohol here in the bottom and C-H-O into acid. Name of the compound is aldatic acid here. This is aldonic acid, okay? Have done all these reactions in the class as well, so you can go through it. Okay, so next is in this one. I have to discuss disaccharide. This is important, and on this, you get many questions in the exam, okay? So what is important here? Disaccharides, like we know, disaccharides are made up of the combination of two monosaccharides, so which all carbon atoms are involved in this combination? Like you see, if you talk about glycosidic acetyl bond we have here, so OH and OH, both combines, H2O molecules eliminates, and both monosaccharides are combined with a oxygen bond here, ethyl bond here, we can say, okay? So in this one you see, we have this molecule given, right? If you talk about, because this is some other example we have, the important one is this, I'll tell you. When you talk about sucrose, I won't write down the entire structure. We know sucrose is made up of glucose, and fructose, so glucose and fructose, we also have this kind of bond present involved in this bonding that is important here to memorize, okay? So in glucose and fructose, for glucose we have C1 carbon atom involves, C1, and from fructose we have C2 carbon atom involves. So what we say, sucrose is formed by the condensation of alpha D, glucopyrinose, glucopyrinose, and fructose we say, beta D, fructofurinose. So you can draw the haworth structure of these two, H2O you have to eliminate out, and you will get the combined structure similar to this kind of, that structure is called sucrose, right? So which carbon atom is involved here, C1 of glucose and C2 of this fructose, right? So here what happens you see, this bond, C1 and C4 we have, the bond is called glycosidic bond, can be either alpha or beta, the end monosecarates on the right is known as glycone, which is the common dissect rights are as follows, cello bios it is not important, okay? In which we have one for glycosidic linkage, but here in sucrose which is important here, we have glycosidic linkage present between C1 and C2 carbon atom, and hence we say fructose forms, sorry sucrose forms by the glycosidic linkage between C1 and C2 carbon atom of glucose and fructose, okay? And this is structured by the combination of this, the structure that we get, we call it as acetyl structure, acetyl structure. So glycosidic linkage present between C1 and C2 carbon atom, this is important and you must memorize this, okay? Next is maltose, in maltose we have glucoparanose we take and glucoparanose, two different structures we have, but the bond that combines here it is C1 of glucoparanose, C1 of glucoparanose and C4 of the other glucosin, because both are glucose only we know maltose are made up of two glucose unit G plus G. So first C1 and second C4, C1 and C4 we have glycosidic linkage present here. So H2O eliminates this kind of bond is present. Now this one is reducing sugar or not, could you tell me? This one is reducing sugar or not? It is the reducing sugar, why? It is the reducing sugar because in which one of the carbon atom which contains OH and OR group is not involved in the bonding. See, this one is OH and OR group it is involved in bonding, but for this molecule, this carbon atom which contains OH and OR group is not involved in bonding. And when this carbon atom is free, it is a reducing sugar. If you look at the example of the previous one, which I have taken sucrose, in sucrose what happens? Both C1 and C2 carbons are involved, right? So both molecule of glucose and fructose, anomeric carbon is involved in bonding, in sucrose. If you draw the structure, you will see anomeric carbon is involved in bonding and hence sucrose is a non-reducing sugar. But this anomeric carbon is free here. Anyone we want, anomeric carbon free. So it is a reducing sugar. This value is not required, just let it be. You just need to know what is the combination of this two units, which carbon atom is involved, C1, C4. We have glycosidic linkage between the two. And it is a reducing sugar because the anomeric carbon is free. And when it is reducing, it shows meter rotation. This property you should know. Just a second, I'm coming. Oh, I wasn't mean. OK. So we have C4 of glucopyrinose and C1 of galactopyrinose. We take beta form of it, glucose here, and beta form of galactose here, both beta form we have. In the previous one, we take alpha form of glucose. Maltose, both will have the alpha form. It is the beta form, C1 and C4 carbon. C1 of galactose and C4 of glucose is involved. Now, again, you see for glucose, C4 carbon is involved. So obviously C1 carbon is left here. It is not involved in the reaction. Hence, it is again a reducing sugar. It is a reducing sugar. And when it is reducing sugar, it shows meter rotation. So all the three different saccharides, you should know where we have the glycosidic linkage present, which carbon atom involved, and whether it is reducing or not. OK. A few more things we have left into this, and then we'll go to the numericals. OK. So, because we have discussed already, glucose and fructose, C1, C2 carbon involves, and it is a non-reducing sugar. OK. We have discussed this already. Next, we have polysaccharides. What is polysaccharides? It is a polymer of monosecarides. It is the polymer of monosecarides. OK. Example, we have starch, cellulose. OK. So what happens in this? First of all, you see these are non-reducing sugars. Key point. Do not exhibit muta rotation. Cellulose and starch are the two common polysaccharides we have. Cellulose formed from D-glucose via 1,4-beta glycosidic linkage. OK. Cellulose is formed from D-glucose via 1,4-glycosidic linkage. So first and fourth carbon are involved in glycosidic linkage. It is non-reducing all this property, remember. OK. Beta form. Keep that in mind. Beta glycosidic linkage. Beta glucose we are taking here. Starch. The starch we have, 1,4-alpha glycosidic linkage. OK. Present in wheat, rice, potatoes, et cetera. OK. So starch has two components. We'll discuss that a bit later. Two components it has. That is amylose and amylopectin. OK. So sugar is separated in a cold water soluble fraction called amylopectin and a cold water insoluble fraction called amylose. So what is the property here you have to memorize? Amylose is water insoluble, this one. Amylose is water insoluble. Fraction you don't like this. Again, ignore that. But water insoluble, amylopectin is water soluble. OK. These two things you must remember. OK. Amylose is a straight chain. Amylose is a linear polymer. Right. Straight chain. Alpha D glucose. I'll write down a few properties here. What? Amylopectin it says insoluble. NCRT. Because could you check once quickly? Sir, I checked, sir. It says amylose is soluble and amylopectin is insoluble. Amylopectin. OK. We'll follow NCRT. We'll follow NCRT. Just maybe some, you know, correction will be here. Error will be here in this media. Possible. OK. So we'll follow NCRT. Amylopectin, guys, just to correct it here. It is water insoluble. Water insoluble. Do you have the note that I have done in the class? Check that also once. Yes, sir, given insoluble. OK. Then maybe some correction over here. And it is soluble. OK. So keep that in mind. Important property. This one is. OK. So in amylose, what happens? The bonding we have between C1 and C4 carbon atom. C1 and C4 glycosidic linkage we have present. And it is alpha D glucose we are taking for this. Because we have N number of monomer. All are alpha D glucose. Right. So the repeating unit also sometimes they ask, what is the repeating unit here? Repeating unit here, if this oxygen you are taking in, then this will be out of the repeating unit. This is a repeating unit. You see, this is getting repeated. So sometimes they also ask you, what is the repeating unit we have? Important. OK. Amylose is this. Amylopectin is highly branched polymer. This one is linear. One after the other, you see it is arranged. Amylopectin is highly branched polymer. It is composed of highly branched complex structure composed of alpha D glucose again, in which C1 and C4 carbons are involved in glycosidic linkage. Insoluble in water. We have already done. OK. And number of units here is around 300 to 600. You're not important. Just to ignore this. OK. This is, you see, because we say it is a highly branched. So if you talk about the linear one, then C1 C4 carbon is involved in this, like we have in amylose. In branching what happens? C6 and C4 carbon is involved. Right. So right now in branching, C6 and C4 carbon atoms of various units are involved. Branching C4 and C6 carbon of various units are involved. Cross-linking will be there between C4 and C6 carbon. OK. So we don't have much to understand here. These are the informations of these, you know, polysaccharides that we should understand or we should memorize. OK. Now we see this question. Try this one by one. This chapter is done. Acetylation of aldohexose with excess of this. I've done this reaction in the class. Done. All. OK. So the first one, 34th, we are doing acetylation of aldohexose. Acetylation gives you what aldohexose is. Glucose, which is C-H-O-H-O-H on the left. On the right. On the right and C-H-2-O-H. H here, H here, H and H here. OK. So the reaction we have aldohexose with AC2O. So reaction is this. We have C-H-3 C double bond O. O-C double bond O-C-H-3. Right. So all these OH converts into O-A-C and C-H-3-C-O-H forms. So the product would be C-H-O. And here we have C-H-O-A-C-4 and C-H-2-O-A-C. All these H will take this form C-H-3-C-O-H and C-H-3-C-O-H will get attached to it. So this is the product we get. What is the option given here? O-A-C we are getting. Penta acetate of aldehyde. Are we getting penta? Yes. We are getting penta acetate 4 and plus 1-5. So we are getting penta acetate of aldehyde. Answer will be option D, 34th one. Furanose and parenose are respectively contained. Furanose and parenose. Five-membered and six-membered ring. Okay. Six-membered and five. Five and seven. Seven and six and seven. Option A is correct in 35th one. The specific rotation of alpha and beta anomers of glucose is this. If the specific rotation of the equilibrium mixture is this, then the mole fraction of alpha and beta anomers. How do we do it? If one is X, other one would be 1-X. X into, how did you do? X into 1-1-2 plus 1-X into 1-9 equals to this. Have you done this? Yes. What is the answer you are getting? C. C. That's right. Okay. Mole fraction is given. The specific rotation of alpha and beta anomers is given. Is this equilibrium also mole fraction of alpha and beta. So one you assume is X. So other one would be 1-X mole fraction. So X into its specific rotation, 1-1-12 plus 1-X into its specific rotation, 19 equals to the equilibrium one. Answer is option C in this one. Similarly, it's 37 also or what? It's exactly the same, no? 37. What is the answer? B you are getting? Yes. If you want me to do just let me know. Okay. Both questions are exactly same. Okay. And these kinds of questions nowadays they're asking in need also. Okay. Treatment of glucitol with pariotic acid. Just now we did glucitol with pariotic acid. What happens in this? CH2OH, CHOH and CH2OH. So all these bond you continue to break this bond, you need to break this bond and it forms HCHO with this, HCHO with this and for CHO OH convert into COOH. Two HCHO formaldehyde and four HCHO OH option B is correct. Treatment of glucuric acid with HIO4. Glucuric acid gives you what? COOH. We won't have any effect. One second. We have CHOH, CHOH and CHOH. Okay. So we have four CHOH we have. So the last two will convert into CHO and the between two will convert into COOH. See in this one, I'll write down this molecule as this, this one, 39th one. The molecule is HOOC, CHOH, CHOH. And we have COOH. This bond you need to break. So from this, this CHOH converts into CHO. This gives you COOH here too. And this gives you again CHO. So one product would be COOH, CHO, two of this molecule and two of COOH we get. So we'll get two of these two options C we are getting here. Last one, treatment of glucuric acid consumes. Have done this reaction. What is the answer? 40th one. Glucuric acid with obviously HIO4 reaction we have. So one, two and three molecules we have. Option three is correct. See in one question only, they can frame that question in different way. Okay. If you have done this, you can do this also. That's the same thing. Okay. Next question you see. Take care of alpha and beta as well. So when we have both alpha and beta, which one do we put? No. We have done this. No. Like you should know this. What is the unit we have? Like what is the monomer we are using? Monosaccharides we are using. Yes. Like I went like if one monomer is alpha, other is beta. Which one do we put? So we cannot change it because it is already mentioned. No. Like if I tell you, if you think of sucrose, right? So gross. What is the monomer we have in sucrose? Alpha beta. Alpha glucose beta fructose. That you cannot change. Alpha glucose and beta fructose. Like you see 44th one. We have alpha glucose and HIO. You are saying only one thing is given here. So we have one and four. One and four. One and two. We have no in sucrose. We have one and two because it's not reducing sugar. So one and two glycosidic linkage. What is your doubt? 44th is simply simple. It's the option. Most of the other options. So what is glycosidic linkage? So what? So like in the glycosidic linkage, whether it is alpha or beta. No, glycosidic linkage is not alpha or beta. It is the mono saccharides that we are taking alpha form of that or beta form of that. Yes. Yes. Alpha beta. Alpha beta is because of the mono saccharides that we are taking. Like for example, you see, if you talk about sucrose, right? In sucrose, we'll take alpha form of glucopyrinose and beta form of fructofyrinose. Right. So that alpha and beta is associated with the mono saccharides that we are taking. Glycosidic linkage is just a bond between a carbon atom, you know, which contains hydroxy group means two molecule we have. That is what the glycosidic linkage is. So here you see the first question that you have maltose. Maltose is both glucose. What form alpha beta? Maltose is alpha, right? Both alpha we are taking. So 43rd one, the answer will be option A. If suppose you must be like confusing, like if alpha and beta both feel like in case of sucrose, one is alpha and one is beta. It's given, right? So they haven't mentioned, right? You see here in sucrose, we have one, two glycosidic linkage, right? So one, two glycosidic linkage we have here. Beta it is given. And it is just one, two glycosidic linkage we have here. Right? It's not like we have only beta in monomer. So we won't go with this. Here we have what maltose we have only alpha in the monosaccharides. So we can go with option A. It is alpha glycosidic. Sir. Sir, I think on the slide it was written beta. I'm not sure if I took it down wrong. Beta for which one? Sir, for maltose, I think it was written one, four beta glycoside on the slide. Okay, just a second. Maltose you are saying. I'll take one second. Should not be beta in this one? Yes, sir. At the top it's written beta. It might be misprinted. One, four beta glycoside. Beta should be alpha, I guess. I'll cross check this. Just give me some time. I'll cross check this in the break. Okay. Maltose, I'll check that. Okay, so next is this we have done. So first one, I'll cross check this whether it is alpha or beta. I guess it should be alpha. Okay, but we'll cross check this, but in A and B we'll cross check. Okay. So 45th one, invert sugar is sucrose. Sucrose we know it is a non-reducing sugar. That's how we discuss this. Next one you see we have, we'll have to finish this one. Before the break. Yeah. Simple one. Tell me. Maltose is formed by the union of two forms of glucose. Option C. Sucrose we have one glucose, one fructose. And galactose we have one molecule of glucose and one molecule of lactose. Option B. Okay. Simple one. Fine. So we are done with this biomolecules. Next we have to start after the break. I'll check that question in the break. And we'll start a polymers after the break. Okay. And probably polymers also we can finish and then we will move into alcohol chapter. Okay. We'll do that. Fine. So we'll take a break now. We'll resume at 625. 625 will resume. Take a break.