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JEE MAIN & ADVANCE 11th PCM Physics -Circular Motion Demo Videos
Hello students welcome back to the chapter circular motion in the previous module we discussed about centripetal force which is necessary for a particle to move in a circular motion we need a uniform circular motion or non-uniform circular motion the value of the centripetal force is equal to mV square by r or M Omega square R we also discussed that the particle would be acted upon by a tangential force which is M alpha r non-uniform circular motion in this particular module will discuss about the examples of the same the statement of the first example says that the particle is projected to the speed u at an angle theta with the horizontal and we are required to find out the radius of curvature at the highest point of the trajectory for the project right now we know that in this particular case the particle is going in a projectile motion it is not a circular motion but every point in the path can be treated as a point of circle so we can find out the radius of curvature in this particular case as well so what we know is that the particle is initially thrown at an angle theta that will be horizontal with a speed u if we try to resolve the speed will get the value of horizontal speed to be ucos theta and the vertical speed here is usin theta the trajectory that the particle would follow would look something like this and at the highest point we know that the particle would have a horizontal speed the vertical speed vanishes and sends the horizontal speed is ucos theta it does not change if air resistance is neglected at the highest Point the particle would have a horizontal speed of ucos theta think that now we know that for particle moving in circle the velocity is always tangential so therefore at the highest point what we can say is the horizontal direction is a Tangential direction that is e theta a gap and the vertical direction is a radial direction that is er cap let’s say that the center of the circle here is C and the radius is R .we are required to find out this value of R now i will recall these steps that we have to follow to solve this kind of question is first we have to identify the forces and we do that by drawing the FBD in this particular case the particle is thrown as a projectile and therefore it is going under the influence of gravity that’s the only force that acts on the particle is mg and that is also what Vertically down that means it is towards the center and the net force which is required for a particle in circular motion towards the center is equal to mV square by r thus this mg is the only force and we can conclude that mg here is providing the necessary centripetal force so equating this we get mg is equal to mV square by r which on solving gives us the value of R to be equal to v square by G and we’re putting down the value of instantaneous speed at the highest point which is ucos theta so we get finally that the radius of curvature at the highest point as required in the question is u square cos square to theta divided by G this is the answer to this particular example let us move forward with the another example which says that there’s a bead which is placed the over a rod which is hinge at one end now this bead is at a distance of R from the hinge and the Rod is rotating with a constant angular acceleration of alpha we are also given the this coefficient of friction between the Rod and the bead and the values nil we have to find out the value of time and we angle after which the bead starts slipping and we have to neglect the gravity in this particular question let us try to visualize what is happening in this particular case the rod actually rotating horizontally in this fashion and we can conclude that The bead would be moving in a circle of radius R before it starts slipping so if we try to view this particular motion from the top end we’ll get the motion of the Rod looks like something like so we can conclude that the bead is moving in a circle of radius R and for a particle to move in a circle what is required is a radial force and this radial force should be equal to M Omega square R that means if it has a angular acceleration then we can conclude that the bead will also experienced and then tangential force which is equal to m into alpha are now what provides the bead this radial force there is no gravity and the normal reaction would provide the tangential force so the friction in this particular case provides the necessary centripetal force the radial force therefore we can write down and m omega square is equal to F which is equal to meu n even let us mark this as equation 1 also we discuss the tangential forces provided by the normal reaction and thus we can say that m alpha r is equal to n, this is a second equation now if we divided the equation one and two what we get is the value of meu the coefficient of friction is equal to omega square r / alpha r. now what we can conclude from here is that the value of omega is equal to under root mu into alpha so this is the value of Omega where the bead would actually start slipping why because after this Omega the friction would not be able to provide the necessary centripetal force the value of friction would be less than the required value so what we’ll do is now we have to find out the value of time and angle will use the fact that we can use the three equation of motion for the constant angular acceleration where the first equation is omega is equal to Omega naught plus alpha t which gives us the value of T to be equal to under root of meu by alpha now use the third equation we get omega square is equal to omega knot square plus 2 alpha theta which gives us the value of angle to be equal to under root of meu by 2 these two values are the final answer to this particular example i hope u understood the module will continue R discussion on circular motion thank u
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JEE MAIN & ADVANCE 11th PCM Physics-Real Fluids Demo Videos
Hello students we have discussed about the different phenomena viscosity we have talked about how exactly the liquids flow and so on right let’s try to understand about that other aspects of viscosity this in this particular topic we are going to discuss about poiseuilles equation of poiseuilles formula this is very, very important because this will tell us how exactly the fluid rate changes are how exactly the fluids move let’s try to understand about it so in this case what exactly the point this equation let’s try to understand about so this gives us this . this equation gives the pressure drop in the fluid flowing through a long cylindrical pipe let me explain how so just take it very long cylindrical pipe ok and if I just assume that both the ends have a uniform area of cross section there is some distance between the two ends there is going to be some pressure difference right because the fluid is moving from one into the other obviously that is something which is going to be different as a result the fluid is moving is the fluid would have been rest right so that tells us that is some pressure difference which makes the fluid to move how much is it how do we find it out what is the register’s with the flow that’s what we are going to understand in this particular topic so there are a few assumptions that we need to take care of here assumptions of the equation our first assumption flow is laminar that is it steady flow that is all the layers would have tangents being parallel each other or tangents are not at different angles then it’s a viscous force viscous fluid so obviously that is going to be viscous force that is going to be acting between the different layers then the liquid is incompressible that is you need to take into account that the density should not change from one point with other institution means constant from one into the other so it’s incompressible fine going further the flow is through a constant cross section they should not be it change in the cross section because if there is a change obviously the liquid is going to accelerate at some point it’s going to deselect some other right so the area of cross section has to be the same it is assumed that there is no acceleration of liquid taking place in the pipe it should not be accelerating there should be a constant velocity with which it should be moving so these are the assumptions that we are going to make you so let’s take it further from you so if we take the case of cylinder of ok the cylinder would have some length and there is a pressure difference between it so p1 is like that the pressure at one end and p2 is the pressure . in L be the length that liquid is entering all the fluid is entering from in and coming out from other r is the radius of the cylinder taking these things into account let’s go further and define everything so what is going to be the force the force due to the pressure is going to be defined as it’s going to be close to the pressure difference into the area, area of cross section that you have we are assuming that they remain same there’s also going to be a physical force because there are different layers of the liquid they are going to exert a force in the opposite direction so you have – ita 2piRL why 2pirl on it because it’s a cylinder so soon they will have a of cross section which is going to be given by a the circumference 2PI R into the length so 2pirL correct but we know that we assume that there is no acceleration right the net force has to be equal to zero so if I just add the pressure force and the viscous force we add the top so that gives the net force to be 0 so if you do that will get what – Delta P 2 pi r square is going to be equals to Ita2pir Dv divided by dr so this is what we have so from here dv by dr can be figured out and that comes out in the form of delta p divide 2ita nlr correct now we need to integrate this so using the empirical velocity gradient we can see that lets say from small are to capital R if you try to find out how exactly the velocity varies between the different radii so we get into it this simple looking expression here correct this expression is related in terms of the pressure difference between the two ends the coefficient of viscosity ita the length of the tube l and the different radii right let’s go further we are now going to talk about in terms of the equation of continuity so using the equation of continuity let’s try to talk about the volume flux so volume flux this is going to be given by DV by DT that is the rate at which the volume is changing with time so that is going to be equal to the velocity into the area we are going to talk about the equation of continuity later but i’m just using that ok so what basically does that tell us we have the expression of velocity if you use that here we’ll get this particular expression in that just need to integrate it so if we integrate from the center to the radius R we’re going to get this particular expression of the flow rate so Q which is equal to the flow rate is going to be given by DV by DT which is dependent on the pressure difference the radius R coefficient of viscosity ita and the length of the tube right let’s try to understand and try to modify this expression to this particular equation is what is known as a poiseuilles of this equation it should be using now instead of delta p happy if he effectively write it down in terms of the pressure difference p 1 and p 2 so p1 and p2 for write it down so we can say that the expression Q can be given by p1 minus p2 and rest Pir4 if we take it to the denominator so we can write it in a very simple-looking form equal to p1 minus p2 upon Rp so what is Rp this term RP is known as the resistance that resistance to the flow of fluid that is what exactly stops the fluid from moving freely so this is what is known as that resistance to the flow of fluid we are going to do a numerical based on this so hopefully you have understood the concept of it thank you
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JEE MAIN & ADVANCE 11th PCM Maths -Sequence & Series-1 Demo Videos
Hello students welcome back so in this module will be discussing arithmetic geometric progression let us see what does arithmetic geometric progression means and shot it is called as AGP now students let us say i am given an AP and its terms are a1 a2 a3 till an And AGP term are b1 ,b2,b3 till bn then we can say the sequence which is obtained by multiplying the corresponding terms of an AP and AGP that is the fall term of AP into the first term of GP that is A1 b1second term of a AP second term of GP multiplied that is a2 B2 and then so on till AN BN would be called as arithmetic geometric Progression lets us from the AP is given as a ,a plus d a plus 2d and so on and GP is given that is B, BR& Br square till Br power of n minus 1 then multiplying the corresponding terms of these two sequence will give me a sequence which would look like this, this will become an AGP .students you can see over here and nTh term of AGP can be written as the multiplication of in nth term of AP and the end , term GP this one is the most important concept of here is to find the sum of N term in Agp let’s say I denote some of term in a AGP Sn then sn would be this some of these terms over here this students very important step of over here is I’m going to multiply complete series by the common ratio of GP as we used to do it in terms of in GP topic that is geometric progression this from multiplying complete series by R that is a common ratio so i have multiplied by R and shifted The terms like the this students no don’t worry i have written it in this fashion now i’m going to subtract these two series so when i’m going to subtract these two series this turns note over here the terms start cancelling out ABR cancels from the second and in the first expression similarly the second term third term and so on till interval Observe like this so what i’m going to subtract his students 1 – R into Sn would be written like this you can see over his students I obtained this expression and now i’m taking DB common from the second term and I obtained over you can see db into R plus R squared till R to the power n minus 1 and you can see this is AGP with first term R and common ratio also are so using the sum of N times in GP that is a into r to the power of n minus 1 upon r minus 1 we obtain this result and are now i’m dividing by 1 minus R to sum of n terms is open like this the students we don’t have to remember this formula we have to remember this method let us move further and find some of in finite terms in a AGP now students of common ratio better – 1to1 number of terms are in finite we have already found the sum of n terms in AGP and now it’s already given that R is between – 1to 1 number of times n is infinity therefore students R to the power n-1 when we were in understand this concept previously also that R to the power n -1towards zero and R to the power also towards zero using these two results in the above expression I’ll obtain our to the power n minus 1 and r to the power of n from the last two terms cancels out gives me 0 and hence some of in finite terms you can see obtained as ab upon 1 minus r plus dbr upon 1 minus r square So students let us use these concepts in solving some question lets his students have to find the sum of these terms let’s check is this an AGP . look at the first term 1 into 2, 2nd 2*2 Square, 3 *2cube observe the terms of each and every term that is one two three and so on that is AP and the second terms that is 2,2 square, 2 cubed till to the power of hundred that is AGP there for students these terms are written in AGP and to find some of these term students we already understood we multiply complete series by the Common ratio of GP over here the common ratio of GPS students too some 1 to multiply the complete sequence by two so student multiply the complete sequence by 2 shift read the terms i’m going to subtract the two series students so i’m subtracting s minus 2’s becomes minus s and now the rest of the terms will be one into the first term as it is from second term onwards that’s where 2 Square, 2 cube, 2 power 4, two power of hundred minus the last term is written As it is now students you can see from first term to the hundredth term it’s AGP hence applying the sum of a hundred times in GP that is a into R to the power of n minus 1 upon r minus 1 that is two minus one minus the last term as it is that is hundred to two power of 101 its now simple calculations students doing the calculation i obtain this final result I hope his students you understood this concept thank you
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JEE MAIN & ADVANCE 11th PCM Chemistry -Redox Reaction Demo Videos
Hello students let us continue discussion on redox reaction. Now in a last module we will discussing about calculation of equivalent rates of oxidizing agent and reducing agent and we also defined 3 important term Z factor today we will continue to discuss some example on equivalent rate calculation and then we will see to particular special cases Fec2o4 and Cu2S. let us start some example Find equivalent weight of Kbro3and Br2 in the following reaction . now you can see in this reaction 10 electrons being exchange , Aur hume yaad hai Z factor kis tarah se calculate karna hai so let us begin with KBro3, so we can say that 2 moles of KBro3number of electrons exchange or a gained in this case equal to 10, so there for we can say that for 1 mol of KBro3 number of electrons gained so we can say that 1 mole of kbro3number of electrons gained will be equal to 10/2 that is five now Z factor KBro3 will become 5 and there for Equivalent weight Equal to molecular weight of kbro3 divided by 5 abhi me exact weight nahi nikal raha, me just symbolically represent kar raha hu and that is important for Br2 we see that number of electrons exchanged are 10 and since one mole of Br2 we need to see that nearly 10 electrons we have to use so Z factor of Br2 becomes 10 And equilent weight equals o molecular weight of bromine divided by 10. Let us see some example when Hno3 is oxidize into NH3the equilent weight of Hno3 will be how much? HNO3 changed In ammonia , sabse pehle hum inke oxidation no. likh lete hai Hno3 will be equal to 5 and you know the algebirc method to calculation of oxidation number , now in ammonia the oxidation number is -3 so we can see that we can calculate the change in oxidation number since there is 1 atom of nitrogen it will be equal to +5 –(-3) that is equal to 8, yaad rakhna change in oxidation is initial oxidation number minus final oxidation number so ye jo 8 hai so this is become the Z factor so the equivalent weight equal to molecular weight divided by 8, let us see the next example . The Equivalent weight of H2So4 in the following reaction is humme ek redox reaction given hai aur usme H2SO4 ka equivalent weight pucha jar aha hai now can I say the z factor for H2SO4 will be equal to 2 , it should be wrong because it is not acid base reaction . it should be analyze like a redox reaction , then we focus on loss and gained electrons , now what should be strategy , first we will find out the overall loss and gain of electrons in the redox reaction matlab jaise balanced karte hai uss process se loss and gained electrons nikalenge to do this we can focus on SO2 or na2Cr2O7, now we will use this information calculate Z factor for H2So4. Just remember there is 1 mole of H2SO4 involved in the reaction so let us focus on solution the oxidation no. of chromium K2cr2O7 is plus 6 and Cr2(So4)3 is plus 3 , since there are 3 atom of chromium we can see that number of electrons exchanged overall will be equal to six and there for we can say that for 1 mole of H2SO4 number of electrons exchanged will be also equal to six .Z factor for H2SO4 is 6 and equivalent weight will be equal to molecular weight divided by 6 so I remembered redox reaction me acid and base ka equivalent weight jo hota hai wo loss ya gained of electrons se nikalte hai . now this calculation could have be done if you just analyzed So2 and So4 two negative, so lets us take a another example the equivalent of Hno3 molecualr weight is just 63 in the following reaction is so we have given the reaction between copper and Hno3 to give copper nitrate .aur hamare pass 4 option hai . we analyzed this reaction understand the Hno3 is here just not acting on a agent but it is also act oxidizing agent so Hno3 ke dual role hai, so it is act as a oxidizing agent and it is also provide acidic medium now find out the loss or gain of electrons to do this better copper because copper ka sirf oxidation ho raha hai Hno3 reduced bhi ho raha hai , Hno3 ke ander no ko jara dhayn se dekhenge so using this information we will be calculate the Z factor then we will gate the equivalent weight. Lets us see the solution so we can see that the oxidation number in Cu is zero and that in copper nitrate is +2. Since there is 3 copper involved we can see the change in oxidation number is -6. If we have to do NO remember we have only two NO corresponding to change in oxidation number . so we will find that is HNo3 +5 and that in NO is +2 . so the change in oxidation number is +6. To chahe me copper se karu , chahe Hno3 se karu dono sahi loss and gain in oxidation number is 6aur hona bhi chahiye so number of electrons overall changed is 6 so there for 8 moles of HNO3number of electrons is equal to six for 1 mole of Hno3it is 6/8 that is ¾ so equivalent weight will be equal to molecular weight divided by ¾ so we can say it will be 63*4/3 . now let us focus some special cases we focus on FeC2O4. FeC2O4 is a good reducing agent now fe^2 positive is oxidize to Fe^3 positive . ab iss reaction dekh sakte hai reactant side pe charges 2+, and product side pe Charges 3+ now to balanced the charge since atoms are already to balanced a charged we simply need to charged an electron to the product side now C2O4negative it is oxidize co2 and we can see that yah ape atoms pehel balanced karne padenge so just multiply the product side by 2 now to balanced the charged reactant side got two negative charged and product side is zero so to balanced the charge we simply add two electrons to the product side ab me dono reaction ke liye dekh raha hu overall fec2o4 pe dono component oxidize ho rahe hai so FeC2O4 overall losses the total of 3electrons and hence equivalent weight of FeC2O4 will be equal to molecular weight divided by 3 let us see another special case Cu2S. it is also good reducing agent now in this case Cu+ is oxidize to Cu^2+ again charge balanced karte hai reactant side pe charges 1 postive and product side pe charges 2+. So to balanced the charged we simply add an electron to the product side . now since there are two copper we can say overall 2 electrons lost for copper . now sulphide ions is oxidizes to SO2first we balanced the atoms we to do add 2 H2o on reactant side oxygen ko balanced karne ke liye and Add 4H+ to the product side yah ape assume kar liya medium in acidic in nature . so we add this and balanced the charge we see that reactant side has total charge 2 negative and product side 4+. To do this we will now add 6 electrons to product side and balances the charge so 2 electrons copper ke and six sulphide ions ke so CU2s losses 8 electrons and hence its equivalent weight Molecular weight divide by 8, Thank you
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JEE MAIN & ADVANCE 11th PCM Chemistry -Mole Concept Demo Videos
Hello students let us continue our discussion on mole and equivalent Concept now in this module first we will take an example on percentage purity and then we will discuss what is the percentage yield of a reaction let’s take the example first and impure six grams of NACL is dissolved in water and then treated with excess of silver nitrate the mass of AGCL precipitate is found to be 14 . 35 grams what is the percentage purity of NACL now Aise cases me sabse pehele reaction likh lete hai ho ky raha hai it is a double displacement reaction NACL plus silver nitrate gives yeild a AGCL plus NANO3 and now we will first develop the strategy here strategy ky honi chahiye hamari Jo bhi agcl precipitate hona hai to actual NACL se precipitate Hona hai so using the mass of the AGCL precipitate will first calculate the mass of NACL that is present and then using this information we find the percentage purity jitna actual hai therotically hume a given a sixth gram we can calculate so we write down the reaction and then we first do the shoso metric analysis one mole NACL it gives yeild one mole AGCL so using the molar mass data about atomic weights hume given hai silver ke liye hai 108, sodium 23, chlorine 35.5 so from here we can calculate the molar mass of any NACL 58.5 and that of AGCL ls 143.5 now from here converting the moles into mass we can write down 58.5 grams of NACL is producing 143.5 grams of AGCL up hum reverse ka answer analysis kr lete hai 143.5 grams of agcl is going to be produced by 58.5 grams of NACL so one gram of AGCL is going to be produced by 58.5 / 143.5 grams of NACL AB hume kitna mil raha hai 14.35 grams of AGCL it so that is going to produce the other that is going to be produced by 58.5 by 143.5 *14 .35 grams of NACL it it comes out as five point eight grams of NACL ye tha actual amount theoretically kitna given hai given six grams so percent a purity is simply what 5.85/6 *100 that is equal to 97.5 now let us discuss a very important percentage yield what is percentage yeild sometimes the reaction may not get completed according to the initial amount taken , matlab aapne job hi amount liya use aap jo sochte hai banega wo banta hi nahi What are the possible reason is the first possible reason establishment of equilibrium now this chapter you will read later in the portion of chemistry but let me tell you jitna reactant aap lete ho woe k time ke baad step stabilized ho jata hai wo aage badh jata hai And that is called as equilibrium and therefore a poora reaction the corresponding ke product expect krte hai wo aata hi nahi hai now for the second reason some of the reactants are not able to react to the formation of a protective layer suppose karo ek solid uske upar humne kuch liquid dala that reacts with the surface and then cover set up not that we meaning so it cannot react and therefore we will not get the actual amount that we are thinking now the third possible case would be when we take gaseous reactants they may simply escape out of the reaction chamber and will not give you the products and the fourth possible reason can be other reactions happening in parallel for the same reactants Jaise abhi tak aapne padha n2 and h2 can give you ammonia NH3 but they can also combine to give you N2h 4 which is called as hydrazine so you may not get ammonia in total that you’re thinking now in such cases the efficiency of the process is measured using percentage yield what is percentage yeild it is the product mass that you actually get divided by the product mass that you theoretically predict times hundred now let us take an example for this A on controlled oxidation gives x according to the reaction hume so-so metric given it they’re saying some of the reactant a oxidizes and only seventy percent of maximum yield is obtained what mass excess is produced if 200 grams of A is taken now what should be a strategy sub se pehle normal calculate krte hai Hum maan ke chalet hai hundred percent yeild there and let us find out how much mass A that we can produce and then using the mass of X that we have calculated we will calculate the actual mass by using the percentage yeild seventy percent so let us see how we do it right down the So-so metric of the reaction to 2a plus 9b gives yeild 2 x + 4 y plus 5 z it but interesting parts hamare liye ky hai 2A and 2X two mole so we write down 2a is going to produce two mole X so one mole A is going to produce one mole X and the first convert to moles into mass so 128 gram of a will give you 128 grams of X based on the molecular mass data that was given to us so from here we can say 128 gram of a produces 140 grams of x and therefore one gram of a is going to produce 148 x 128 grams of X hundred-person yield Mante hue weigh 200 grams of is going to produce 148 x 128 times 200 grams of X now since percentage yield is equal to product mass actually / product mass Theoretical times hundred this equation re arrange kr lete hai kyu ki mass actual aata hai a so we can write it down as mass actually equal to percentage yield by hundred times mass actual now putting in the value yeild the key value Hai seventy percent so 70 by hundred times this much mass and that is going to give us 161 .875 grams so first we calculate the actual amount or rather the theoretical amount and then we calculate the actual amount in that yeild’ thank you
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JEE MAIN & ADVANCE 11th PCM Chemistry -General Organic Chemistry Demo Videos
Welcome back students so since we have learned quite a lot about the phenomenon of resonance now this will be the last module we will learn the application of resonance and proceed to the next electronic effect that is the introduction of hyper Conjugation alright so when you talk about the application of a resonance is basically the phenomena which helps us to understand the stability in the compound so here we are going to learn the acid strength of carboxylic acid as we know the fact any components acid if its conjugate base is stable so let’s look at the structure of carboxylic acid and if it donates a proton it forms a conjugate base which is called as a carboxylate ion as we have seen early well as now we’re having 2 structure one is a carboxylic acid and others carboxylate ion we will try and draw the resulting structure of both of this because both of them are having a loan ordered by condition present then which is one of the condition which makes the compound eligible to show phenomenal resonance so look at the carboxylic acid structure the one visit to the left if we should delocalization over here we’ll get a resonating structure which can be simply drawn like this so since the lone pair from the oxygen atom as delocalized it is experiencing loss of a lone pair so we carry a positive charge and the double bonded oxygen gaining one additional electron pair so caring and negative charge which means now this newly opting resonating structure carries a positive and the negative charge which means this structure is the charge separated Structure now try to work on the phenomenal for resonance in this structure right here which is nothing but the carboxilate ion so we again have a loan alternate by condition present over here so if we sure do localization will get one more Resorting structure which is equivalent of this carboxylic ion and you can see that there’s only negative charge which is getting dispersed delocalized so i will say that in this structure 1and 2 which are the equal and structure you will find that this equal and structures are charged the localized structure and whenever you get a chance temperature structure it is relatively a less stable structure on the other hand charge delocalized which means if only one charge is getting dispersed the locals or several items then that delocalize a structure is always a most stable structure than the other and now you think about why are we calling this application we see that any compound acid if its conjugate base is stable so in case of carboxylic acid we saw that the conjugate base was a carboxylate ion which is giving us charge delocalized structure which is more stable more stable the conjugate bases more stable is the acid from which their conjugate bases obtained so likewise student and if you try and work on this sell phonic acid or if you try and work on the phenol you will find that the charge delocalized structures are going to be more stable due to phenomenal for resonance so phenomenal resonance can simply justify the acid strength of the Compounds well having known this now we are going to learn or see a simple sequence of acid strength but the most important one so remember if you talk about the several acid the sell phonic acid it is the strongest among all because is reasoning structures are more stable and after so sulphuric acid in the next acid is their logs leak acid followed by we have a carboxylic acid and then we have phenol by after phenol all the structure of the compound which comes most acid is methanol check out here we have a methanol all then relatively weaker acid is a water and then we again have any other alcohol student please look at this methanol is the only exceptional alcohol which is more acid than water and rest alcohols are weaker acid than water and after this we have alkynes then we have ammonia and then we get alkenes followed by alkenes so this is a simple order of acid strength of several compound and you got to remember this all right students now having said that we are going to learn application of resonance and this is what was the application of a resonance now we’ll see introduction to the phenomena of a hyper conjugation and there are several more news coming up right there will justify each one of this name but for the timing let’s look at this reaction so we have tertiary butyl halide highlight when this tertiary butyl and is brought in presence of a polar solvent like water again we learn what it means a polar then the six bond here to literally breaks and which means the bond pair between can have been completely taken by X so the carbon has lost born pair and therefore produces a positively charged species like this positively charged species were positive charge is carried bicarbonate Called as carbo cation now let’s take that carbon cation over here and remove all this methyl group will get the structure like this so this is a general representation of carbo cation and let’s take a replicate that so we are taken 4 such general representation of a carbo cation a time and now instruction number one right there we are fulfilling all the valances let’s see by methyl groups so now that carbo cationic which is carrying a positive charge is attached to three more carbon so the nature of the carbon carrying positive charge is Tertiary we call it a Tertiary carbo cation same way if we take 2 methyl group to such carbon you will call it as the secondary carbo cation since it is attached 2 the carbon when so if you move further if I’m fulfilling to valances with edges and one with methyl the nature of carbonic cation is primary and if I fulfill all the valances with edges I will call it as the methyl carbo cation since it looks like methyl so the nature of carbon carrying positive charge decides the nature of carbon cation and it is said that among all this the order of stability is found to be Tertiary the most then secondary then primary and methyl is stable carbo cation and why is the order so can be proven by the phenomenon of hyper conjugation so in the very next model will try and prove this order and with respect to this we will learn the phenomenal hyper conjugation but we’ll stop here in this module thank you
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