SET B | Chemistry Class 12 NEB Board | Full Paper Discussion 2081 | One Shot for A+
Firoj Malla
[{'start': '1.050', 'end': '5.720', 'text': '[Music]\nHello everyone and today we are going to discuss Set B of the last'}, {'start': '5.720', 'end': '11.160', 'text': "year that is 2081 under the NEB Board's A\nPast Year Choices."}, {'start': '11.160', 'end': '15.400', 'text': 'Before this,\nwe have already discussed Set A'}, {'start': '15.400', 'end': '20.760', 'text': 'and its MCQs. We have also discussed all\nfour sets at once.'}, {'start': '20.760', 'end': '26.000', 'text': 'That is why\nI will discuss only Group B and Group C'}, {'start': '26.000', 'end': '30.119', 'text': 'today. The MCQ has\nalready been done before us'}, {'start': '30.119', 'end': '35.760', 'text': "and in this Group B,\nwhat is given in our first number? Hoffman's"}, {'start': '35.760', 'end': '39.520', 'text': 'method has been given. This is\nbecause we have already discussed it before. So, I have not'}, {'start': '39.520', 'end': '43.520', 'text': 'discussed it here. Similarly, they asked about our name\nreaction. In our set, five'}, {'start': '43.520', 'end': '47.879', 'text': 'name reactions were also asked and here also\nour five name reactions were asked. In this,'}, {'start': '47.879', 'end': '51.360', 'text': 'only examples have to be given, like\ncarboxylation reaction,'}, {'start': '51.360', 'end': '56.760', 'text': 'carboxylation means that our carboxyl\ngroup should be D, like CO, group should be D.'}, {'start': '56.760', 'end': '60.719', 'text': 'Like we did\nin this example, we can also have a Coles reaction'}, {'start': '60.719', 'end': '65.560', 'text': 'of phenol or we have\ncarboxylated the Greek nitrogen,'}, {'start': '65.560', 'end': '71.320', 'text': 'we have R Mg X,\nwe have reacted with COt,'}, {'start': '71.320', 'end': '78.400', 'text': 'if it reacts with COt, it is plus or minus, the\nalkyl group goes with this carbon, this'}, {'start': '78.400', 'end': '84.840', 'text': 'Mg X goes with O and it forms\nRC, I put a double bond O,'}, {'start': '84.840', 'end': '89.400', 'text': 'what came with O, then Mg X came and when\nwe hydrolyze it,'}, {'start': '89.400', 'end': '94.720', 'text': 'EtO A+, what happens from here'}, {'start': '94.720', 'end': '99.720', 'text': 'in R C O, what comes like hydrolysis is the'}, {'start': '99.720', 'end': '105.000', 'text': 'H of water breaking into H and OH, the H comes and the\ncarboxylic acid is formed, plus Mg X'}, {'start': '105.000', 'end': '110.680', 'text': 'goes with MgOch X,\nthis is how our COt group is what'}, {'start': '110.680', 'end': '114.560', 'text': "we have, here the CO\ngroup is inserted, that's why"}, {'start': '114.560', 'end': '119.399', 'text': 'we can call it an example of\ncarbose ylation reaction, the example tells'}, {'start': '119.399', 'end': '125.399', 'text': 'us  Suppose we have\nCH3CO'}, {'start': '187.760', 'end': '192.599', 'text': 'Acidification was done, HCl was'}, {'start': '192.599', 'end': '197.519', 'text': 'added to it, what comes in place of N,\nthen OCH comes, the O of the phenol group comes'}, {'start': '197.519', 'end': '203.720', 'text': 'and says, here we have COOH and the other remaining\nwhich is ACL comes out as'}, {'start': '203.720', 'end': '209.439', 'text': 'N and CL remain, our A\nsalicylic acid is called'}, {'start': '209.439', 'end': '215.599', 'text': 'salicylic acid or whatever we say, then\northo hydroxy benzoic'}, {'start': '215.599', 'end': '222.080', 'text': 'acid is also called salicylic acid, the\nIPSC name of salicylic acid is the same, and in the same way, the other one has'}, {'start': '222.080', 'end': '227.080', 'text': 'given us a ozonolysis and\nozonolysis, I did Riemann time before,'}, {'start': '227.080', 'end': '233.079', 'text': 'why did Riemann time and this Coles reaction,\nthis Coles reaction is a little related,'}, {'start': '233.079', 'end': '237.400', 'text': 'like what we do in Riemann time, then we\ntake phenol, we take'}, {'start': '237.400', 'end': '244.159', 'text': 'phenol, we put phenol\nwith chloroform in it, then we walk with Kez, then'}, {'start': '244.159', 'end': '248.720', 'text': 'from here,\nours is now first like'}, {'start': '248.720', 'end': '256.160', 'text': 'this O our AX NOz is kept in it, it\nis O NA and here what it says like this'}, {'start': '256.160', 'end': '260.680', 'text': 'CO  It is called sodium\nsalicylaldehyde, it is'}, {'start': '260.680', 'end': '265.479', 'text': 'called sodium salicylaldehyde, the above is\nsalicylate, what happens in it, it is called'}, {'start': '265.479', 'end': '270.919', 'text': 'sodium salicylaldehyde,\nit is also called salicylaldehyde, after we acidify it,'}, {'start': '270.919', 'end': '278.240', 'text': 'it is called salicylaldehyde, here, here, CO comes, it is called salicyl aldehyde, what'}, {'start': '278.240', 'end': '285.039', 'text': 'happened in it, then salicylic acid is formed, CO,'}, {'start': '285.039', 'end': '289.919', 'text': 'here, CHO has come and that is why\nwe call it'}, {'start': '289.919', 'end': '294.280', 'text': 'carboxylation, just like we\ncalled it carbonylation group, it has'}, {'start': '294.280', 'end': '298.600', 'text': 'CO group, CO, this is\ncalled Riemann time reaction, that is why in Riemann time reaction'}, {'start': '298.600', 'end': '304.199', 'text': 'when phenol is treated with\nchloroform in presence of eq, then the'}, {'start': '304.199', 'end': '310.639', 'text': 'sodium salicylate is formed and on an\nacidification gives salicylaldehyde as a'}, {'start': '310.639', 'end': '314.960', 'text': 'major product, ortho hydroxy\nbenzaldehyde is formed  It is said that it also'}, {'start': '314.960', 'end': '319.800', 'text': 'forms a pair, but this is formed as a major product. Then\nwe have to write that in the'}, {'start': '319.800', 'end': '323.960', 'text': 'second we have\nozonolysis, what is ozonolysis?'}, {'start': '323.960', 'end': '328.319', 'text': 'When we say that ozone should be lysis, what is it? When we say that in alkyne, when we'}, {'start': '328.319', 'end': '333.800', 'text': 'do ozone, ozone breaks down like this and goes to this double'}, {'start': '333.800', 'end': '338.160', 'text': 'bonded carbon and it becomes d like this, the\naddition of ozone occurs. Thus, we'}, {'start': '338.160', 'end': '344.000', 'text': 'do three ozones like this, one is above and the other is in a\nseparate one like this. And this'}, {'start': '344.000', 'end': '349.080', 'text': 'carbon, this double bonded carbon, both of it has\nhydrogen with it, this hydrogen with it, this'}, {'start': '349.080', 'end': '354.840', 'text': 'also has both hydrogens with it, this is called\nozonoid. Now'}, {'start': '354.840', 'end': '360.199', 'text': 'what we do is decompose this ozonoid in the\npresence of Zn and EtO. From this,'}, {'start': '360.199', 'end': '364.759', 'text': 'one aldehyde is formed, the other one is formed from here. It is'}, {'start': '364.759', 'end': '376.280', 'text': 'said that it is formed from both.  Here we see that\nHCHO is formed like this. HCHO is formed'}, {'start': '376.280', 'end': '381.280', 'text': 'from here. From here, I have said that two HO are\nformed plus one oxygen. This is what is left.'}, {'start': '381.280', 'end': '386.360', 'text': 'This is what is combined with our water to\nform Et. And here, what is formed?'}, {'start': '386.360', 'end': '391.199', 'text': 'Methanol is\nformed. Two molecules of methanol are formed. This is what happened in'}, {'start': '391.199', 'end': '396.759', 'text': 'alkyne. What happened in\nalkyne is ethyne. What happened is'}, {'start': '396.759', 'end': '403.400', 'text': 'alkyne is treated with ozone. Then the ozonide is formed on\ndeition by treating with a'}, {'start': '403.400', 'end': '409.560', 'text': 'mixture of aldehyde and ketone. This is\ncalled an aldehyde reaction and it is called a talc test. The'}, {'start': '409.560', 'end': '414.280', 'text': 'talc test is like the\ntest that aldehydes give when we say that methanol is ethanol. For ethanol, we'}, {'start': '414.280', 'end': '419.479', 'text': 'can take an example. When'}, {'start': '419.479', 'end': '423.840', 'text': 'we write the talc reagent directly, we get'}, {'start': '423.840', 'end': '429.840', 'text': 'eth and diamine silver fus.\nHydroxide says from here, the'}, {'start': '429.840', 'end': '436.639', 'text': 'Ag+ of this complex is reduced, in\nthis way, the Ag+ is'}, {'start': '436.639', 'end': '441.400', 'text': 'reduced to metallic silver, which is positioned on\nthe wall of the test tube. What happened was that when'}, {'start': '441.400', 'end': '446.199', 'text': 'we warm it up, and\nthis is our aldehyde, what'}, {'start': '446.199', 'end': '450.720', 'text': 'happens is that the aldehyde is oxidized and comes in the form of a carboxylate ion.'}, {'start': '450.720', 'end': '455.280', 'text': 'What we do now is\nwrite A4, it is oxidized to carboxylic acid as an ammonium salt, and the'}, {'start': '455.280', 'end': '460.240', 'text': 'next thing is that our side product A3'}, {'start': '460.240', 'end': '465.960', 'text': 'comes out and this reaction is called a cold Talins\ntest or Talins reaction.'}, {'start': '465.960', 'end': '470.560', 'text': 'And now, next, our aldol condensation.\nAldol condensation is'}, {'start': '470.560', 'end': '475.199', 'text': 'our aldehyde and ketone\nhaving alpha hydrogen, so'}, {'start': '475.199', 'end': '482.319', 'text': 'I moved it down a little bit. In aldol\ncondensation, in a five, our'}, {'start': '482.319', 'end': '489.159', 'text': 'aldol condensation.\nAldol condensation reaction is an'}, {'start': '489.159', 'end': '493.960', 'text': 'aldehyde and a ketone. I brought an aldehyde.\nWhat should I do in the aldehyde? The'}, {'start': '493.960', 'end': '498.759', 'text': 'carbon after the carbonyl is\ncalled alpha carbon. If the alpha'}, {'start': '498.759', 'end': '502.680', 'text': 'carbon has hydrogen, then\nit is called alpha hydrogen. If the'}, {'start': '502.680', 'end': '507.240', 'text': 'aldehyde and ketone do not have alpha hydrogen, then\nwhat should we treat with'}, {'start': '507.240', 'end': '512.880', 'text': 'dilute NOH or dilute KOS?'}, {'start': '512.880', 'end': '517.519', 'text': 'Remember, here I have brought two molecules\nand from here an aldol is formed. When'}, {'start': '517.519', 'end': '521.560', 'text': 'making an aldol, how should I make it?\nWhich one has an alpha hydrogen? I'}, {'start': '521.560', 'end': '526.120', 'text': 'put these two, put H on it, and\nwhere should I attach this hydrogen? When making it, I take it and'}, {'start': '526.120', 'end': '530.839', 'text': 'attach it to O.\nWhat should I do? I'}, {'start': '530.839', 'end': '537.080', 'text': 'write from here CS is this Cich is this O\nin O. The hydrogen here came and'}, {'start': '537.080', 'end': '543.640', 'text': 'became Och and the rest is added from here.\nThis is what we call CIT and CO.'}, {'start': '543.640', 'end': '549.959', 'text': 'When counting from here,\nwhat we have in one to th four th is'}, {'start': '549.959', 'end': '555.480', 'text': 'hydroxy and then th is hydroxy and\nfour carbons buta and aldehyde is'}, {'start': '555.480', 'end': '560.399', 'text': 'butanal or this is\nalso called alpha this is what happened beta'}, {'start': '560.399', 'end': '566.320', 'text': 'like this carbonyl carbon which\ncarbon is alpha this beta this is called'}, {'start': '566.320', 'end': '572.760', 'text': 'betahydroxy bithydroxy what we call\naldehyde is formed or beta'}, {'start': '572.760', 'end': '578.760', 'text': 'hydroxy ketone is formed what\nwe call it aldol aldol'}, {'start': '578.760', 'end': '584.240', 'text': 'LD is aldehyde ol means this ol is\nalcohol ol this is called our'}, {'start': '584.240', 'end': '589.040', 'text': 'aldol and this reaction is called aldol\ncondensation reaction we have to'}, {'start': '589.040', 'end': '595.519', 'text': 'understand that next we have an\norganic compound a hang formula what is it'}, {'start': '595.519', 'end': '602.320', 'text': 'c4 a anith has given us this'}, {'start': '602.320', 'end': '608.680', 'text': 'compound called a on acidic hydrolysis\nit is hydrolyzed by adding EtOH'}, {'start': '608.680', 'end': '612.640', 'text': 'Give\ncarboxylic acid B from here'}, {'start': '612.640', 'end': '621.079', 'text': 'carboxylic acid is formed Run A is\nreduced This is what we reduced it to'}, {'start': '621.079', 'end': '627.079', 'text': 'LIALS F Lithium aluminum hydride\nWhat will be formed when we reduce it A'}, {'start': '627.079', 'end': '632.200', 'text': 'Alcohol is formed from here\nAlcohol is formed Re C becomes Re'}, {'start': '632.200', 'end': '638.839', 'text': 'And the alcohol C if\noxidized with K Mn A Ps gives back B It is'}, {'start': '638.839', 'end': '642.880', 'text': 'called again If we oxidize it again\nWhat should we do to'}, {'start': '642.880', 'end': '650.120', 'text': 'oxidize it We have to use K\nMno4 A+ What is formed Re Then the same B becomes'}, {'start': '650.120', 'end': '655.000', 'text': 'Re Draw the molecular structure of A B C\nWith the reaction involved'}, {'start': '655.000', 'end': '660.279', 'text': 'What should we understand When we say that this is\nour A A, it is like this A'}, {'start': '660.279', 'end': '664.560', 'text': 'What should we know is that\nthese three oxygens are our'}, {'start': '664.560', 'end': '668.880', 'text': 'one element compound That is A\nAnhydride That is why'}, {'start': '668.880', 'end': '672.000', 'text': 'we call Anhydride\nIf we do hydrolysis, carboxylic'}, {'start': '672.000', 'end': '676.440', 'text': 'acid comes out. If\nwe remove water from carboxylic acid, then'}, {'start': '676.440', 'end': '680.440', 'text': "anhydride is formed. That's why our\nanhydride is anhydride. What should happen if we"}, {'start': '680.440', 'end': '683.800', 'text': "hydrolyze to\nmake carboxylic acid? That's why what is"}, {'start': '683.800', 'end': '688.440', 'text': 'anhydride? When we make anhydride, we do\nlike this:'}, {'start': '688.440', 'end': '694.399', 'text': 'CH3O3. This is from acetic\nnitride, it has four carbons.'}, {'start': '694.399', 'end': '699.639', 'text': "That's why I divided two of them like this.\nWhat is its name? That's ethanoic"}, {'start': '699.639', 'end': '705.279', 'text': 'anhydride. This is an anhydride that comes from two carbons.\nWhat happens if we add ethanoic'}, {'start': '705.279', 'end': '709.040', 'text': 'anhydride? Here are two and there are two.'}, {'start': '709.040', 'end': '713.639', 'text': "That's why when we hydrolyze,\nwhat does our meaning of hydrolysis say? We will get"}, {'start': '713.639', 'end': '720.440', 'text': 'H. Now, taking this and adding O, what happens if we\nadd H? If we add CO, what happens if we'}, {'start': '720.440', 'end': '728.519', 'text': 'get H? The rest  What\nhappened to this is that we split it like this like we did'}, {'start': '728.519', 'end': '732.480', 'text': 'it in reverse, what did we do when we\ntook out H from here, oho from here, like when we made'}, {'start': '732.480', 'end': '735.760', 'text': 'anhydride, like when we hydrolyzed anhydride, what happens next is'}, {'start': '735.760', 'end': '740.760', 'text': 'carboxylic acid is formed and what is formed from both of these is\nethanoic acid, you'}, {'start': '740.760', 'end': '745.279', 'text': 'can check the formula, the formula\nalso has to be found and now we have to'}, {'start': '745.279', 'end': '750.760', 'text': 'reduce it, it is\nreduced by LELHO, what happens during reduction, what is'}, {'start': '750.760', 'end': '754.839', 'text': 'this ethanoic\nanhydride, what is ethanoic'}, {'start': '754.839', 'end': '759.560', 'text': 'acid, what is the anhydride that we made by removing water from ethanoic acid,\nthat is why the'}, {'start': '759.560', 'end': '765.079', 'text': 'reduction of ethanoic anhydride is\nwhat is the reduction of ethanoic acid, that is why the reduction of'}, {'start': '765.079', 'end': '768.720', 'text': 'carboxylic acid is what is formed when we reduce it, what is the\nprimary alcohol, what is the'}, {'start': '768.720', 'end': '772.760', 'text': 'primary alcohol formed, what is the\nalcohol of two carbons,'}, {'start': '772.760', 'end': '780.600', 'text': 'ethanol, S2, what does the other one say,\nch32  From this, what does our A say,'}, {'start': '780.600', 'end': '785.160', 'text': 'ethanol from here, ethanol from here, two ethanols are formed,\nnow two'}, {'start': '785.160', 'end': '789.839', 'text': 'ethanols are formed, then what do we do with ethanol now, is it\noxidized or put MnO2 and add H+, then what is formed'}, {'start': '789.839', 'end': '794.199', 'text': 'from here, then what is formed from ethanoic acid, then'}, {'start': '794.199', 'end': '799.079', 'text': 'what is obtained from these three things, then A means\nour ethanoic anhydride and our'}, {'start': '799.079', 'end': '804.680', 'text': 'B means ethanoic acid and C means\nwhat is our ethanol, then'}, {'start': '804.680', 'end': '809.839', 'text': 'we have to do this. Now, they\nask us what happens when B is heated'}, {'start': '809.839', 'end': '816.360', 'text': 'with PtO. They ask us\nwhat does it mean to put it with PtO? Then this is'}, {'start': '816.360', 'end': '821.040', 'text': 'dehydration. Then what does it mean to put it with PtO? Then'}, {'start': '821.040', 'end': '826.160', 'text': 'what do we do after putting it with PtO? If\nI remove H from here, then OH is'}, {'start': '826.160', 'end': '831.120', 'text': 'formed again. This is how it is\nrepeated from here to there and from there.'}, {'start': '831.120', 'end': '835.560', 'text': 'We got five\nmarks for this. Now, in the next'}, {'start': '835.560', 'end': '838.360', 'text': 'question, we are asked that\nformaldehyde is the first member of'}, {'start': '838.360', 'end': '843.079', 'text': 'aldehyde. It does not contain alkyl groups. Write\nthe molecular structure and'}, {'start': '843.079', 'end': '846.440', 'text': 'IPC name of formaldehyde. After\nsaying our first member of aldehyde, the'}, {'start': '846.440', 'end': '851.800', 'text': 'bond of aldehyde is CH.\nIf we add H to it, this is formaldehyde or'}, {'start': '851.800', 'end': '857.880', 'text': 'its IPSC name is one carbon.\nMethane is called methylanal. Thus, it becomes methylanal. Write'}, {'start': '857.880', 'end': '862.399', 'text': 'down the chemical reaction for the preparation of\nformaldehyde from methanol.'}, {'start': '862.399', 'end': '870.519', 'text': 'Methanol is called chooh ch3oh.\nWhat do we have to do to make ch3 a series of SO?'}, {'start': '870.519', 'end': '875.720', 'text': 'What do we have to do?\nThis is a primary alcohol. Methanol'}, {'start': '875.720', 'end': '881.639', 'text': 'is methanol. Methanol is a primary alcohol.\nWe have to change it into an aldehyde.'}, {'start': '881.639', 'end': '885.199', 'text': 'What do we have to do? The\nprimary alcohol is oxidized.  It is'}, {'start': '885.199', 'end': '888.320', 'text': 'formed as an aldehyde, but the aldehyde is then\nultimately oxidized to'}, {'start': '888.320', 'end': '892.440', 'text': 'form a carboxylic acid. For that,\nwe have to do controlled oxidation. In'}, {'start': '892.440', 'end': '897.160', 'text': "controlled oxidation, we either use\nCollins' reagent, which"}, {'start': '897.160', 'end': '902.600', 'text': 'we call a solution\nin the presence of'}, {'start': '902.600', 'end': '908.600', 'text': 'pyridine, or we\ncan use PCC, which is'}, {'start': '908.600', 'end': '913.560', 'text': 'pyridium chlorochromate. These\ntwo are controlled oxidation,'}, {'start': '913.560', 'end': '917.680', 'text': 'which just stops the primary alcohol from forming\naldehydes, or we have to understand'}, {'start': '917.680', 'end': '922.000', 'text': 'that they do controlled oxidation.\nThen in our'}, {'start': '922.000', 'end': '926.680', 'text': 'third question, we asked how is formalin prepared\nfrom formaldehyde? Formalin is what'}, {'start': '926.680', 'end': '933.839', 'text': 'we mean. Formalin is what is formaldehyde,\nwhat is formaldehyde? It is an'}, {'start': '933.839', 'end': '939.000', 'text': 'aqueous solution, a\nsolution made by adding water, which is  It'}, {'start': '939.000', 'end': '946.480', 'text': "says around 37%, but in some books it says 0%,\nthat's why we have made it 30% by mass."}, {'start': '946.480', 'end': '953.519', 'text': 'This is not by volume, but by mass. It\nmeans that we have dissolved formaldehyde in water and'}, {'start': '953.519', 'end': '959.279', 'text': 'made a solution. Where\nits percentage by mass is 37%,'}, {'start': '959.279', 'end': '964.920', 'text': 'it is called formalin. And the\nuses of formalin are also asked.'}, {'start': '964.920', 'end': '971.519', 'text': 'What is it used for? We have\npreservatives. What are preservatives for?'}, {'start': '971.519', 'end': '976.560', 'text': 'Laboratory equipment or'}, {'start': '976.560', 'end': '982.319', 'text': 'specimens are called specimens.\nAs they say, our biological things that are'}, {'start': '982.319', 'end': '986.560', 'text': 'used in the lab are not treated with ammonia.\nWe use formalin to prevent them from getting poisoned. And it happens that formaldehyde is'}, {'start': '986.560', 'end': '990.839', 'text': 'treated with ammonia.'}, {'start': '990.839', 'end': '996.880', 'text': 'We have read this in the chapter on aldehydes earlier.\nWe have called formaldehyde'}, {'start': '996.880', 'end': '1002.600', 'text': 'formaldehyde.  That is, A is and\nwhen we treat it with NH3,'}, {'start': '1002.600', 'end': '1006.600', 'text': 'our urotropin would be formed from here.\nHexamethylene'}, {'start': '1006.600', 'end': '1010.880', 'text': 'tetraamine is formed.\nWe put four on it. What else should we'}, {'start': '1010.880', 'end': '1016.079', 'text': 'put? Our six plus\nwhat do we call water? What do we call'}, {'start': '1020.480', 'end': '1026.039', 'text': 'hexamethylene and\ntetraamine? This is how we call it.'}, {'start': '1026.039', 'end': '1031.480', 'text': 'We call it urotropin. We\nalso ask about its uses. When we ask where it is used, we'}, {'start': '1040.640', 'end': '1046.199', 'text': 'say it is used as an antiseptic for urinary tract infections. We\nhave to write this. Sometimes this is also'}, {'start': '1046.199', 'end': '1050.640', 'text': 'asked more. Sometimes it is\nasked to make its structure. I have done this before'}, {'start': '1050.640', 'end': '1055.760', 'text': 'in the chapter on aldehydes. When we\ndiscussed the old is gold questions,'}, {'start': '1055.760', 'end': '1060.880', 'text': 'I have done this. And in our fit, it is called\nthe open hearth process for'}, {'start': '1060.880', 'end': '1064.799', 'text': 'manufacturing of steel. This is the\nsteel.  There are'}, {'start': '1064.799', 'end': '1069.600', 'text': 'two manufacturing processes which are our basic oxygen\nprocess and open earth process which is'}, {'start': '1069.600', 'end': '1073.960', 'text': 'very important from the exam point of view\nwhich I have discussed in the previous'}, {'start': '1073.960', 'end': '1077.880', 'text': 'chapter of iron and in the chapter of iron\nboth these processes are also known as'}, {'start': '1077.880', 'end': '1084.679', 'text': 'metal chloride'}, {'start': '1084.679', 'end': '1089.880', 'text': 'a and it is also called kk silver and\nname the metal m and write it c4 what is'}, {'start': '1089.880', 'end': '1094.320', 'text': 'metal m what is\nsilver we call it hg'}, {'start': '1094.320', 'end': '1101.159', 'text': 'mercury and its for is\nour cinnabar hg s and y is'}, {'start': '1101.159', 'end': '1106.480', 'text': 'the metal called quick silver like\nsilver silver like silver white and'}, {'start': '1106.480', 'end': '1111.080', 'text': 'quick means it is only a liquid\nmetal means it is liquid metal and its'}, {'start': '1111.080', 'end': '1115.360', 'text': 'movement is fast\nwe call it quick quick its'}, {'start': '1115.360', 'end': '1119.760', 'text': 'Because of the movement, it is white like silver, so'}, {'start': '1119.760', 'end': '1124.799', 'text': 'we have to write it as quick silver. Explain\nthe process of purification of metalium using its act'}, {'start': '1124.799', 'end': '1129.799', 'text': 'with its label diagram. The\nlast thing we did'}, {'start': '1129.799', 'end': '1135.080', 'text': 'during extraction was to purify it by adding it with ANO3 and\ndissolving it there.'}, {'start': '1135.080', 'end': '1138.960', 'text': 'We\nhad to write the reaction. I have also discussed'}, {'start': '1138.960', 'end': '1143.360', 'text': "this in Margari's one short video. You\ncan see it there. You"}, {'start': '1143.360', 'end': '1148.960', 'text': 'asked about\nelectrochemistry. Our A is z an plus'}, {'start': '1148.960', 'end': '1155.480', 'text': '2 Ag plus z an plus P 2 Ag is given like this.\nHere is the reaction and we have been'}, {'start': '1155.480', 'end': '1160.600', 'text': 'given the reduction potential of zinc and\nsilver. The above reaction is spontaneous.'}, {'start': '1160.600', 'end': '1166.320', 'text': 'First, what did we\ndo first? Zinc plus 2 Ag plus gives'}, {'start': '1166.320', 'end': '1172.120', 'text': 'z an plus plus plus 2 Ag solid.\nHere is the reaction.  Our zinc is'}, {'start': '1172.120', 'end': '1178.799', 'text': 'oxidized and Ag is reduced.\nWhere oxidation occurs, it is the anode'}, {'start': '1178.799', 'end': '1184.159', 'text': 'and where reduction occurs, it is the\ncathode. First, we'}, {'start': '1184.159', 'end': '1191.120', 'text': 'take out this E-not cell like this. E-not cathode minus\nE-not anode. We have'}, {'start': '1191.120', 'end': '1195.559', 'text': 'our silver at the cathode.\nWhat is the reduction potential of'}, {'start': '1195.559', 'end': '1202.760', 'text': 'silver? 0.80 minus what is the reduction\npotential of zinc? If we do 0.76, we get'}, {'start': '1202.760', 'end': '1211.960', 'text': '0.80 P, and if we do\n0.76 C, we get 0.56 F. If we do 0.66 T, we get'}, {'start': '1211.960', 'end': '1217.520', 'text': '1.56 F. If we do 0.66 T, we get 0.6'}, {'start': '1226.640', 'end': '1231.000', 'text': 'Anode and cathode are\nalways called what we do when we do cell notation. We'}, {'start': '1231.000', 'end': '1236.120', 'text': 'start from the anode. What\nhappened at the anode? Zinc is oxidized and'}, {'start': '1236.120', 'end': '1240.320', 'text': 'becomes Zn+. First we write the anode. Here\nwe also have to write the phase. Zinc has'}, {'start': '1240.320', 'end': '1247.280', 'text': 'gone from solid to Zn+ plus plus A. This is how it happens when we write a one molar. If we do it as a\nstandard, it happens'}, {'start': '1247.280', 'end': '1252.080', 'text': 'even if we do not write one molar.\nThis is our anode.'}, {'start': '1252.080', 'end': '1257.080', 'text': 'Oxidation occurs at the anode. Then our salt bridge\ncomes. After the salt bridge, we write the'}, {'start': '1257.080', 'end': '1262.520', 'text': 'cathode. What is our Ag+? What is\nour reduction to Ag? What is our reduction to Ag?'}, {'start': '1262.520', 'end': '1268.480', 'text': 'First we have to do. What did it\ngo from Ag+ to A? I did not'}, {'start': '1268.480', 'end': '1273.559', 'text': 'write one molar. Where did it go? It\nwent to Ag solid. This is'}, {'start': '1273.559', 'end': '1278.880', 'text': 'our cathode. This is how we\nhave to know how to write cell notation. Write down the'}, {'start': '1278.880', 'end': '1282.960', 'text': 'half cell reaction. Half cell reaction is called.\nThis is the anode half cell'}, {'start': '1282.960', 'end': '1288.039', 'text': 'reaction, Z and N went solid, Z and N plus plus A by'}, {'start': '1288.039', 'end': '1294.159', 'text': 'losing two electrons,\nwhat happened at the cathode, these two electrons are'}, {'start': '1294.159', 'end': '1301.000', 'text': 'two Ag plus gained and\nAg solid is formed, this is'}, {'start': '1301.000', 'end': '1306.440', 'text': 'what happened to us, half cell reaction happened,\ncalculate the EMF of cell,'}, {'start': '1306.440', 'end': '1311.960', 'text': 'I have calculated the EMF of cell here,\nwhat should I do with the cathode minus, anode has 1.56 fV, and'}, {'start': '1311.960', 'end': '1316.279', 'text': 'now we have'}, {'start': '1316.279', 'end': '1320.440', 'text': 'asked thermodynamics, define heat of\nformation, what is heat of formation, what is'}, {'start': '1320.440', 'end': '1324.880', 'text': 'the enthalpy change, what is the\nenthalpy'}, {'start': '1324.880', 'end': '1331.200', 'text': 'change, what is the enthalpy change or heat change, what is it that\nhappens when we write'}, {'start': '1331.200', 'end': '1336.640', 'text': '1 mole\nof substance is formed, what should we'}, {'start': '1336.640', 'end': '1343.080', 'text': 'say is formation, substance is formed\nfrom its form, what should we'}, {'start': '1343.080', 'end': '1347.720', 'text': 'say  It should be\nwritten as constituent elements, it should be'}, {'start': '1347.720', 'end': '1353.200', 'text': 'written as constituent elements,\nit is called our enthalpy of'}, {'start': '1353.200', 'end': '1359.520', 'text': 'formation or it is called heat of formation.\nNow in this, we have been given the enthalpy change of carbon plus the'}, {'start': '1359.520', 'end': '1365.880', 'text': 'heat'}, {'start': '1365.880', 'end': '1370.720', 'text': 'of formation of acetic acid, and from here, we have been told to extract the heat of formation of acetic acid. After\nthat, I put it in this reaction,'}, {'start': '1370.720', 'end': '1375.600', 'text': 'what is the enthalpy change of this\nproduct minus the reactant, the enthalpy of formation'}, {'start': '1375.600', 'end': '1383.320', 'text': 'of CO plus the\nenthalpy of formation of what should be done,'}, {'start': '1383.320', 'end': '1389.000', 'text': 'water should be done, EtO of this product\nminus the reactant, what do we have in the'}, {'start': '1389.000', 'end': '1396.080', 'text': 'enthalpy of formation of acetic acid\nCH3COH plus two times the enthalpy of'}, {'start': '1396.080', 'end': '1401.480', 'text': 'formation of oxygen, this is how we write\nit, and we have already been given its enthalpy'}, {'start': '1401.480', 'end': '1407.240', 'text': '- 21 is the enthalpy of formation of CO when\none mole of CO is formed.'}, {'start': '1407.240', 'end': '1413.919', 'text': 'How much enthalpy change has occurred -94 plus I think the\nreaction was balanced in this.'}, {'start': '1413.919', 'end': '1417.840', 'text': 'First we had to balance it.\nWe have two carbons here. Then'}, {'start': '1417.840', 'end': '1422.520', 'text': "we had to write two here. If we\nlook at hydrogen, how many hydrogens are there? It's 3 and"}, {'start': '1422.520', 'end': '1426.760', 'text': "one is four. I had to write two here. Is there\noxygen balance? Let's see if"}, {'start': '1426.760', 'end': '1432.679', 'text': "oxygen is balanced or not. If we look at oxygen, it's t fo 4 and\nt 6. Now, finally, it's balanced. Then"}, {'start': '1432.679', 'end': '1436.600', 'text': "we have to multiply it by t.\nThat's why the"}, {'start': '1436.600', 'end': '1441.400', 'text': 'equation is balanced or not.\nWe also have to look at 2 * ma-94 plus water.'}, {'start': '1441.400', 'end': '1446.640', 'text': 'How much is 2 int water? The\nenthalpy of formation is'}, {'start': '1446.640', 'end': '1453.360', 'text': 'ma-68. Minus the enthalpy of\nformation of ch3. This is ma-x. I have to'}, {'start': '1453.360', 'end': '1460.000', 'text': "figure it out. And what happens is that it's always\n- 2 * 0. It's in the elemental state"}, {'start': '1460.000', 'end': '1465.399', 'text': "or it's made of the same atoms. If it's a molecule, it's\nro. We have to figure it out from here."}, {'start': '1465.399', 'end': '1473.600', 'text': 'How much did Ma-208 come from here?\nMa-324 is one. I brought one here.'}, {'start': '1473.600', 'end': '1480.279', 'text': 'If\nMa-324 + 208,'}, {'start': '1480.279', 'end': '1486.279', 'text': 'Ma-16 is one. The above unit is given in\nkilocalories per mole.'}, {'start': '1486.279', 'end': '1490.880', 'text': 'We have to write it like this. Now, the\nnext question is the difference between'}, {'start': '1490.880', 'end': '1495.039', 'text': 'Brønsted Law and RNS concept. Acids are\ncalled'}, {'start': '1495.039', 'end': '1500.399', 'text': 'Brønsted acid and we are called Arsenic acid.\nWhat should be the Brønsted acid to be? Instead, we should say that it should be'}, {'start': '1500.399', 'end': '1505.760', 'text': 'a proton donor.\nWe call the proton donor a'}, {'start': '1505.760', 'end': '1512.520', 'text': 'Brønsted acid. What should be the anionic acid? It should give\nH plus ions. It should'}, {'start': '1512.520', 'end': '1518.159', 'text': 'not give protons. It should not dissolve in'}, {'start': '1518.880', 'end': '1525.159', 'text': 'water. What should be the water? We call it Arsenic\nacid. That is why now, for this, the'}, {'start': '1525.159', 'end': '1529.520', 'text': 'next difference is that the solvent'}, {'start': '1529.520', 'end': '1535.799', 'text': 'is not\nlimited'}, {'start': '1535.799', 'end': '1544.919', 'text': 'to water only. So, what should be the solvent in it?  It is\nnecessary to have water as'}, {'start': '1544.919', 'end': '1550.279', 'text': 'solvent, water is necessary, meaning that the\nconcept of acid is the same as the concept of base, if you put it in water, if you'}, {'start': '1550.279', 'end': '1555.279', 'text': 'give H plus, if you give acid, if you give O minus, then the\nconcept of base is there, then'}, {'start': '1555.279', 'end': '1559.679', 'text': 'whatever solvent is needed to be a bromine acid, it does not mean that it\ngives a proton, if it gives H plus,'}, {'start': '1559.679', 'end': '1563.640', 'text': 'we have to write it as bromine acid.\nSimilarly, in our question,'}, {'start': '1563.640', 'end': '1570.640', 'text': 'the given coupling in regular\nsequence, use of a first reagent,'}, {'start': '1570.640', 'end': '1577.360', 'text': 'so that it does not take it from another, in which\nwe are given what is given here, here, ether is given, here, another ether is given,'}, {'start': '1577.360', 'end': '1582.640', 'text': 'two\nethers, one methoxy, this methoxy ethane,'}, {'start': '1582.640', 'end': '1588.080', 'text': 'here, ethoxy ethane is\ngiven, another Cth C2 I is given, another is'}, {'start': '1588.080', 'end': '1592.520', 'text': 'our ethanol is given, another is given,\nnow how do we make it from this,'}, {'start': '1592.520', 'end': '1597.080', 'text': 'hey, just what do we do,\nfirst of all, we go from amines,'}, {'start': '1597.080', 'end': '1603.399', 'text': 'CH3 CH2 A2 is this  1N degree amine is in this,\nwhat will happen if we put Ano in it?'}, {'start': '1603.399', 'end': '1612.480', 'text': 'C2 OoH means our ethyne amine has\nbecome ethane amine has become ethanol from this.'}, {'start': '1612.480', 'end': '1617.720', 'text': 'Now what do I do with ethanol? I\ndehydrate it by putting it with H2SO4'}, {'start': '1617.720', 'end': '1624.760', 'text': 'but when the temperature is 140,\nwhat happens from here? CH3 or I wrote S2 A5 and said'}, {'start': '1624.760', 'end': '1630.960', 'text': 'Sot A5O C2 H5 I say ethoxy ethane has become.\nWhat happened to this?'}, {'start': '1630.960', 'end': '1636.120', 'text': 'Ethoxy ethane has become. It\ndehydrates from here. From two molecules of off'}, {'start': '1636.120', 'end': '1642.320', 'text': 'ethanol, ethoxyethane has become. Now this is ours.\nThis is also gone. Now ethoxy ethane has also'}, {'start': '1642.320', 'end': '1646.399', 'text': 'gone. Now I say what do I put in ethoxy ethane? I\nput H here and'}, {'start': '1646.399', 'end': '1653.000', 'text': 'had to put H and heat it. I could not cool it. What does it\nsay about this?'}, {'start': '1653.000', 'end': '1658.240', 'text': 'With this, I take this access.\nWhat happens if I take this access? Both C2 A5'}, {'start': '1658.240', 'end': '1664.000', 'text': 'I become. In this, one Sot A5O and\none  C2A5I is formed and that C25OH'}, {'start': '1664.000', 'end': '1668.799', 'text': 'reacts with Access HI to form this,\nthen C2HFI is formed. Here we have C2A5I. Then we have iodoethane. Now we have to make it from'}, {'start': '1668.799', 'end': '1674.919', 'text': 'this iodoethane. Now we have'}, {'start': '1674.919', 'end': '1679.919', 'text': 'methoxyethane. Then we have to make\nit into an ether. To'}, {'start': '1679.919', 'end': '1683.880', 'text': 'make ether,\nwhat do I take? Etherification. We have to'}, {'start': '1683.880', 'end': '1690.080', 'text': 'give it to Williamson etherification. In which I\ntake CH3O which has one carbon, so'}, {'start': '1690.080', 'end': '1698.640', 'text': 'N and I combine to form NA and form CH\nand CTH. What is formed is our'}, {'start': '1698.640', 'end': '1703.919', 'text': 'methoxyethane. This is how we were able to do\nthis conversion. Here'}, {'start': '1703.919', 'end': '1707.480', 'text': 'we have to write these reagents.\nHere we have to write Access.'}, {'start': '1707.480', 'end': '1712.760', 'text': 'Similarly, in our text, there is a\nlong question like this, where'}, {'start': '1712.760', 'end': '1718.440', 'text': 'Titson is a common volumetric method,\nwhere it is called Distance between Titan and'}, {'start': '1718.440', 'end': '1723.720', 'text': 'Titan. Now this is  Titan is\nour tight'}, {'start': '1723.720', 'end': '1731.039', 'text': 'trend and tightend. This tight trend is\ndifferent from D and T. That subsense whose'}, {'start': '1731.039', 'end': '1736.159', 'text': 'concentration is what we have,\nwhen we say a non, what is non, what is'}, {'start': '1736.159', 'end': '1741.960', 'text': "called Titan, which means what\nwe don't know, means what we don't know."}, {'start': '1741.960', 'end': '1746.200', 'text': "We remember from D that we don't know,\nfrom T that we remember that we know, when we"}, {'start': '1746.200', 'end': '1750.799', 'text': "remember that we know, we don't know. The solution whose"}, {'start': '1750.799', 'end': '1755.919', 'text': "concentration is non, what we\ncall Titan, what we call Don't is not known,"}, {'start': '1755.919', 'end': '1760.640', 'text': 'what we call Titrant,\nand similarly, there are other differences. When we'}, {'start': '1760.640', 'end': '1765.440', 'text': 'say Titan, usually\nwhat we mean is this primary standard, what'}, {'start': '1765.440', 'end': '1770.120', 'text': 'we mean is this\nsecondary standard, usually what we'}, {'start': '1770.120', 'end': '1773.960', 'text': 'mean is sometimes we can use the secondary\nstandard as a Titan.'}, {'start': '1773.960', 'end': '1777.960', 'text': 'That is also something to keep in mind.'}, {'start': '1778.240', 'end': '1782.919', 'text': 'Similarly, the major requirement\nfor a substance to be a primary'}, {'start': '1782.919', 'end': '1787.720', 'text': 'standard is that it should be\npure, dry, readily available,'}, {'start': '1787.720', 'end': '1794.840', 'text': 'non-toxic,'}, {'start': '1794.840', 'end': '1801.080', 'text': 'non-\nhygroscopic, non-'}, {'start': '1801.080', 'end': '1806.919', 'text': 'deliquescent, non-absorbing, non-water-forming, and'}, {'start': '1806.919', 'end': '1811.159', 'text': 'most'}, {'start': '1811.159', 'end': '1817.720', 'text': 'importantly, it should\nhave a high molecular weight and an effective weight.'}, {'start': '1817.720', 'end': '1824.360', 'text': 'Finally, it should not have\nvariable'}, {'start': '1824.360', 'end': '1833.519', 'text': 'compositions, both in the solid state and in the\nsolution state.'}, {'start': '1833.519', 'end': '1837.159', 'text': 'We consider such substances as primary standards'}, {'start': '1837.159', 'end': '1842.600', 'text': 'and the other is that\nwe have to calculate the normality equation.  We'}, {'start': '1842.600', 'end': '1847.159', 'text': 'need to know how to derive. Similarly, now\nour numerical question is to calculate the'}, {'start': '1847.159', 'end': '1853.000', 'text': 'molality of a solution obtained by\ndissolving 2.5 grams of salt in'}, {'start': '1853.000', 'end': '1858.559', 'text': '1 liter of solution. Given that the concentration of the\nsolution is 0.85 5 grams per cc,'}, {'start': '1858.559', 'end': '1862.559', 'text': 'and for this, first we need to\nknow the formula for what molality is.'}, {'start': '1862.559', 'end': '1867.399', 'text': 'Molality is the number of moles\ndivided by the'}, {'start': '1867.399', 'end': '1874.000', 'text': 'number of moles of solute. First, we need to\nknow the formula to tell what is needed.'}, {'start': '1874.000', 'end': '1877.840', 'text': 'What is the weight of\nsolvent'}, {'start': '1877.840', 'end': '1883.559', 'text': 'in kg? That is why we\nneed the weight of solvent and'}, {'start': '1883.559', 'end': '1887.159', 'text': 'we need the number of moles of solute.\nThat is why first,'}, {'start': '1887.159', 'end': '1891.320', 'text': 'what do we do? For the weight of solvent,\nwe have given the weight of solute. For the weight of salt, we have'}, {'start': '1891.320', 'end': '1896.519', 'text': 'given the weight of our solute.\nFor that, we have given the weight of'}, {'start': '1896.519', 'end': '1900.679', 'text': 'salt.  We\ncalculate the weight of the solution. First, the'}, {'start': '1900.679', 'end': '1906.720', 'text': 'volume of the solution is given to us. We are given the volume of solution. One liter\nis 100 ml.'}, {'start': '1906.720', 'end': '1912.120', 'text': 'Since it is given in CGS units, I\ntook the density in ml and did the same. The'}, {'start': '1912.120', 'end': '1919.399', 'text': 'density of the solution is given to us. The\ndensity of the solution is given to us. The density of the solution is given to us. It is given to us as 0.85 grams per'}, {'start': '1919.399', 'end': '1925.159', 'text': 'cc. It is given in grams per cc. So\nwhat do I calculate from here? I'}, {'start': '1925.159', 'end': '1931.960', 'text': 'calculate the mass of the solution. If the mass is equal to 0.85 grams, then\nwhat is the mass? Density into volume. Density'}, {'start': '1931.960', 'end': '1937.080', 'text': 'into volume. Density is 0.85 grams per volume. We have\n100. Then we say how much we got.'}, {'start': '1937.080', 'end': '1942.399', 'text': 'Then we say how much its weight is 850 grams. Now, how do we calculate the weight of solvent? We'}, {'start': '1942.399', 'end': '1951.880', 'text': 'subtract the weight of the solution from the weight of the solution. We subtract the weight of the solute. The'}, {'start': '1951.880', 'end': '1957.279', 'text': 'weight of the solution is 850 grams. And the weight of the\nsolute is given to us.  How much is given? 2.5 grams is'}, {'start': '1957.279', 'end': '1963.039', 'text': 'given. 47.5 grams is given. Then whose weight is given?\nWe need the weight of the solvent in'}, {'start': '1963.039', 'end': '1967.080', 'text': 'kg. So I divided it and it'}, {'start': '1967.080', 'end': '1972.639', 'text': 'came to 0.8475 kg. We got this. Now\nwhat we need to get is the'}, {'start': '1972.639', 'end': '1977.519', 'text': 'number of moles of a solute. What is the\nnumber of moles? The number of moles is given by the'}, {'start': '1977.519', 'end': '1983.240', 'text': 'molecular mass. The weight is given to us as 2.5.\nAnd how much is the molecular mass? The'}, {'start': '1983.240', 'end': '1989.240', 'text': 'molar mass is given to us here as 138. When we divide it, we'}, {'start': '1989.240', 'end': '1995.639', 'text': 'get 0.181 1812 moles. Now how do we directly\nget the molality? The number of'}, {'start': '1995.639', 'end': '2002.200', 'text': 'moles of solute is\n0.1812 divided by the kg of solvent is'}, {'start': '2002.200', 'end': '2005.840', 'text': '0.8475. When we say how much is ours, it'}, {'start': '2005.840', 'end': '2013.159', 'text': 'is 0.21. 0.21\n1 37 per mole per kg.'}, {'start': '2013.159', 'end': '2017.200', 'text': 'We have to do this and\nthen in our next question, it is'}, {'start': '2017.200', 'end': '2023.039', 'text': 'given as Draw the Energy Profile.  Diagram\nfor catalyzed and uncatalyzed chemical'}, {'start': '2023.039', 'end': '2029.200', 'text': 'reactions. For this,\nwhat happens when we show an energy profile diagram?'}, {'start': '2029.399', 'end': '2033.519', 'text': 'Here on the x-axis is'}, {'start': '2033.519', 'end': '2039.360', 'text': 'our progress of the reaction,\nwhich is also called time. On the other side is'}, {'start': '2039.360', 'end': '2045.480', 'text': 'our time. On the y-axis is\nour energy. This is'}, {'start': '2045.480', 'end': '2052.079', 'text': 'our graph of energy versus time. An\nenergy profile diagram is where'}, {'start': '2052.079', 'end': '2056.240', 'text': 'we have one of our\nreactants. Its reactant has this much energy.'}, {'start': '2056.240', 'end': '2061.639', 'text': 'Now, in\norder for the reactant to change into a product, it has to cross one of our'}, {'start': '2061.639', 'end': '2066.280', 'text': 'energy barriers.\nHere is our product. And the'}, {'start': '2066.280', 'end': '2072.639', 'text': 'energy from here to here is what it\nneeds. That energy comes from our'}, {'start': '2072.639', 'end': '2077.440', 'text': 'kinetic energy, which is our collision.\nAfter the collision, the kinetic energy that comes out of'}, {'start': '2077.440', 'end': '2082.480', 'text': 'it\novercomes this energy barrier. And this'}, {'start': '2082.480', 'end': '2088.359', 'text': 'energy we call activation energy.\nAnd if we say that'}, {'start': '2088.359', 'end': '2093.639', 'text': 'it is uncatalyzed, it is'}, {'start': '2093.639', 'end': '2098.800', 'text': 'For an uncatalyzed reaction, this energy profile\ndiagram is done, but in this case,'}, {'start': '2098.800', 'end': '2102.000', 'text': 'what happens if we use a catalyst? It is faster to\nget the product from the reactant.'}, {'start': '2102.000', 'end': '2108.359', 'text': 'How is it faster?\nIt gives an alternative path like this.'}, {'start': '2108.359', 'end': '2113.800', 'text': 'And here, what happens is the\nactivation energy, I'}, {'start': '2113.800', 'end': '2119.440', 'text': 'called it E1, this is the\nactivation energy of the catalyzed reaction. This is'}, {'start': '2119.440', 'end': '2124.040', 'text': 'called uncatalyzed. In this, there is a small energy barrier\nand the reactant is easily crossed and the'}, {'start': '2124.040', 'end': '2127.880', 'text': 'product is changed. In\nthis way, what does our catalyst do? It makes the'}, {'start': '2127.880', 'end': '2132.320', 'text': 'rate of reaction\nfaster by providing an'}, {'start': '2132.320', 'end': '2137.040', 'text': 'alternative pathway that has lower\nactivation. We need to understand this and in'}, {'start': '2137.040', 'end': '2142.079', 'text': 'its double eye, we are asked how does a\nbundle of sticks burn faster than a big'}, {'start': '2142.079', 'end': '2147.079', 'text': 'single log. For this,\nwe have to say the rate of  The reaction'}, {'start': '2147.079', 'end': '2149.839', 'text': 'depends on what we call the surface\narea. We have to talk about it like'}, {'start': '2149.839', 'end': '2156.480', 'text': 'a big single lock, one like this, one like this,'}, {'start': '2156.480', 'end': '2160.960', 'text': 'one like this, one like this, one like this, and one like this,\nor its single pieces are'}, {'start': '2160.960', 'end': '2165.920', 'text': 'made like this, a bundle. Which one\nburns faster?'}, {'start': '2165.920', 'end': '2169.760', 'text': 'Why does it burn? When we burn, the reaction happens'}, {'start': '2169.760', 'end': '2174.079', 'text': 'only on the surface. The reaction area is there.\nIf it is only this surface, the'}, {'start': '2174.079', 'end': '2179.280', 'text': 'molecules inside it\ncannot react or burn, but the'}, {'start': '2179.280', 'end': '2183.440', 'text': 'individual molecules here\ncan react like this. Why does it increase the'}, {'start': '2183.440', 'end': '2187.079', 'text': 'surface area of \u200b\u200bthe reacting particles?'}, {'start': '2187.079', 'end': '2192.240', 'text': 'For that reason, we understand that a bundle of\nsticks burns faster than a big single lock.'}, {'start': '2192.240', 'end': '2196.560', 'text': 'For this, we need to calculate the rate of\nreaction.  We need to understand the proportional to surface'}, {'start': '2196.560', 'end': '2200.880', 'text': 'area. Then we\nare asked to define the order of reaction.'}, {'start': '2200.880', 'end': '2206.200', 'text': 'What is the order of reaction? If we say that\nany rate we have is where the'}, {'start': '2206.200', 'end': '2214.400', 'text': 'power of a is equal to 1 and the power of b is equal to y, then\nwhat do we call the'}, {'start': '2214.400', 'end': '2219.119', 'text': 'order of reaction? That is why the order\nof reaction is the sum'}, {'start': '2219.119', 'end': '2225.720', 'text': 'of the powers of the powers of the concentration\nterm in the rate law equation. We'}, {'start': '2225.720', 'end': '2230.640', 'text': 'have to write the order of reaction as the sum of the powers of the concentration term in the rate law equation. And how is it determined? What is the\norder?'}, {'start': '2230.640', 'end': '2236.359', 'text': 'What is our experimentally determined? What is the experimentally'}, {'start': '2236.359', 'end': '2242.200', 'text': 'determined? What is the determination? And how is it'}, {'start': '2246.000', 'end': '2252.119', 'text': 'determined? By'}, {'start': '2252.119', 'end': '2256.319', 'text': 'changing the concentration of one of the reactants\nand keeping the concentration of another. And what happens in the reaction?  The'}, {'start': '2256.319', 'end': '2261.000', 'text': 'rate is how many times does it\nchange, we look at it,'}, {'start': '2261.000', 'end': '2265.440', 'text': 'what do we do, we find out the\norder of reaction of any reaction,'}, {'start': '2265.440', 'end': '2270.400', 'text': 'and the next one asks us numerically,\nnow the order of reaction is the concentration'}, {'start': '2270.400', 'end': '2275.920', 'text': 'of reactant is reduced by 5%,\nwe assume that the'}, {'start': '2275.920', 'end': '2282.960', 'text': 'initial concentration is 10%, now\nhow much is left, say a-x,'}, {'start': '2282.960', 'end': '2289.319', 'text': 'it says that it is 25% left, in how much\ntime, it is one hour, now we'}, {'start': '2289.319', 'end': '2293.800', 'text': 'call the calculator half-life period,\nfor the half-life, what do we'}, {'start': '2293.800', 'end': '2299.359', 'text': 'do first, we calculate the value of the rate constant, what is\n2.303 33'}, {'start': '2299.359', 'end': '2306.280', 'text': 'by log a ba a ba-x, where\nour a 2.303'}, {'start': '2306.280', 'end': '2313.760', 'text': '33 time is how much is our, one hour is\nlog a ba a ba-x a is our'}, {'start': '2313.760', 'end': '2322.880', 'text': '1st% a ba-x is our 25 is our\n25  So I will solve this 1.38 38 sec is what is the'}, {'start': '2322.880', 'end': '2327.599', 'text': 'value of what and\nfor the first order what is our per'}, {'start': '2327.599', 'end': '2333.000', 'text': 'hour but the unit is time is a\nrate constant unit and calculate the'}, {'start': '2333.000', 'end': '2339.920', 'text': 'half is called period what is half what is\n0.693 ba for the second order what is our per hour what is'}, {'start': '2339.920', 'end': '2348.200', 'text': '0.693 c93 ba what is 1.386 ba\nwhile doing this 0'}, {'start': '2348.200', 'end': '2353.560', 'text': 'pat f na na na na\nna hour what is our per'}, {'start': '2353.560', 'end': '2358.280', 'text': 'hour what is our per hour what is its half life what is it\nwe have to show and then'}, {'start': '2358.280', 'end': '2364.000', 'text': 'our how much concentration of return is\nconverted into product in a t hour is'}, {'start': '2364.000', 'end': '2372.520', 'text': 'given a time time its formula is\n2.303 33/ what is log a ba a ma- x'}, {'start': '2372.520', 'end': '2379.520', 'text': 'time gives us to hours 2.303\n33 and what is our value of what is it so we have'}, {'start': '2379.520', 'end': '2387.839', 'text': 'come to 1.386 and log a ba a ma- x'}, {'start': '2387.839', 'end': '2396.359', 'text': 'From here we have log a ba a ma- x\nis 2 * 1.38'}, {'start': '2396.359', 'end': '2404.680', 'text': 'divide ba 2.303\n33 this divide is 1 pa 2036'}, {'start': '2404.680', 'end': '2410.480', 'text': 'and from here what I do now is I take\na ba a ma- x e this log over there is not log'}, {'start': '2410.480', 'end': '2420.160', 'text': 'shift log throw\n1.36 by doing this how much does it come to 5.98 na et'}, {'start': '2420.160', 'end': '2425.800', 'text': 'we are asked what percentage\nreactant consumed is converted in'}, {'start': '2425.800', 'end': '2432.599', 'text': 'that now what we do now is\nI take 1n by this a'}, {'start': '2432.599', 'end': '2445.440', 'text': 'I bring down a ba a ma- x ba a e 5.98 it comes to\n1 ba1 ma- x ba a e 5.98'}, {'start': '2445.440', 'end': '2454.839', 'text': '8 what I do is I do 1\nba1.98 e 1 ma ams x ba a and put x ba a'}, {'start': '2454.839', 'end': '2465.119', 'text': 'here I do 1n minus a 1 ba1.98 by\ndoing this how much does it come to'}, {'start': '2465.119', 'end': '2468.119', 'text': '0.9374.93'}, {'start': '2468.400', 'end': '2475.079', 'text': '3374 = x/ a  Here x is how much\nreactant was consumed or'}, {'start': '2475.079', 'end': '2480.839', 'text': "changed in the product. A is the total. That's why\nwe wrote it in percentage."}, {'start': '2480.839', 'end': '2486.040', 'text': 'If we did 0.9374 E100, then how much will we get?\n93.74%J4'}, {'start': '2486.040', 'end': '2490.800', 'text': 'Percentage. What happens is that\nthe product is converted in how many hours?'}, {'start': '2490.800', 'end': '2495.599', 'text': 'We have to understand that in the\nsame way, in our T1, we asked N'}, {'start': '2495.599', 'end': '2502.319', 'text': 'alcohol with molecular formula C3Ht\norder, vector Mertus T give blood red'}, {'start': '2502.319', 'end': '2508.400', 'text': 'collation at the of the reaction after adding\nwhat solution is called. What is our blood red'}, {'start': '2508.400', 'end': '2513.160', 'text': 'coloration? When we\nsaid that it is a primary alcohol, then I'}, {'start': '2513.160', 'end': '2522.359', 'text': 'had to make a primary alcohol with three carbons. I had to make a primary alcohol with three carbons.\n3CH2 CHtooh. It is propane one ol'}, {'start': '2522.440', 'end': '2528.319', 'text': 'and its structural formula is IPSC.\nThis is what we did. Write down the complete Victor'}, {'start': '2528.319', 'end': '2533.160', 'text': 'Mayer test. In the Victor Mayer test,\nwhat do we do first? Add P4 plus eight and get an'}, {'start': '2533.160', 'end': '2543.200', 'text': 'iodo alkane.  We make a C2 SO2 solution.\nIn this, we add Ano and'}, {'start': '2543.200', 'end': '2549.760', 'text': 'make CH3 TH CH2\nCH2 AOT. To make a nitro alkane,'}, {'start': '2549.760', 'end': '2556.280', 'text': 'we add AOT. When we'}, {'start': '2556.280', 'end': '2560.119', 'text': 'add Ano, what we get is nitro\nacid. We learned how to make nitro acid.'}, {'start': '2560.119', 'end': '2567.480', 'text': 'We made a double bond of HO\nand from here, we get ATO. From here, we get'}, {'start': '2567.480', 'end': '2575.440', 'text': 'ATO and we get AOT on C2 C. And what we get is a'}, {'start': '2575.440', 'end': '2582.640', 'text': 'double bond of NH.\nThis is what we call nitric'}, {'start': '2582.640', 'end': '2588.520', 'text': 'acid. And when we\nfinally add an alkaline solution like a KOS'}, {'start': '2588.520', 'end': '2592.960', 'text': 'solution, what we\nget is a'}, {'start': '2592.960', 'end': '2599.760', 'text': 'blood red color. This is our\npictoremio test. Similarly,'}, {'start': '2599.760', 'end': '2607.040', 'text': 'starting from CTH CTH MgB and you can\nsay that now we'}, {'start': '2607.040', 'end': '2612.359', 'text': 'have CTH CTH MgB and to\nmake primary alcohol.  Whenever we'}, {'start': '2612.359', 'end': '2617.040', 'text': 'put what we have to put H CHO from here\nour'}, {'start': '2617.040', 'end': '2623.400', 'text': 'this is plus in it and it has a negative charge the\nplus one is with oxygen the negative'}, {'start': '2623.400', 'end': '2630.160', 'text': 'alkyl part goes with carbon and the\nseat this seat is connected to who this C with'}, {'start': '2630.160', 'end': '2636.400', 'text': 'C has two H and the other axis is\noxygen what comes in the oxygen what comes Mg B R comes like this'}, {'start': '2636.400', 'end': '2641.079', 'text': 'adduct now\nwe hydrolyze it'}, {'start': '2641.079', 'end': '2649.720', 'text': 'if we hydrolyze it then seat C2 with STO H comes\nMg B R goes Mg Och and makes'}, {'start': '2649.720', 'end': '2656.359', 'text': 'Mg Och B R so we make Mg Och B R so we'}, {'start': '2656.359', 'end': '2662.440', 'text': 'have to show that t happens no p is\nheated with copper at t0 degrees centigrade so'}, {'start': '2662.440', 'end': '2669.960', 'text': 'we heat CH3 CH2 CH2\nCH2 with copper at 300 then'}, {'start': '2669.960', 'end': '2675.160', 'text': 'dehydrogenation takes place from here H from here a\nhydrogen molecule is released from here a hydrogen molecule is released'}, {'start': '2675.160', 'end': '2680.079', 'text': 'one hydrogen molecule is\nreleased and becomes seat C2 and what we have'}, {'start': '2680.079', 'end': '2685.040', 'text': 'left here is two  After the hydrogen is gone,\nonly H is left and the oxygen is gone. In this way,'}, {'start': '2685.040', 'end': '2690.800', 'text': 'we make an aldehyde. What is the meaning of this? We have to understand that\nour aldehyde is formed by dehydrogenation. We'}, {'start': '2695.920', 'end': '2701.400', 'text': 'have to write that our propane is formed. In the next last one, we are asked how do you convert P into propanone. In order to make'}, {'start': '2706.200', 'end': '2710.720', 'text': 'propanone, what should we\ndo first? We have to make PSL or PSL'}, {'start': '2710.720', 'end': '2715.359', 'text': 'fi to make\nchloro propane.'}, {'start': '2715.359', 'end': '2721.720', 'text': 'And what should we do with this? We\nput alcoholic KOH'}, {'start': '2721.720', 'end': '2727.839', 'text': 'and make\nCH3C double bond CH. Dehydro'}, {'start': '2727.839', 'end': '2732.079', 'text': 'halogenation occurs and it becomes propene. In\npropene, what should we put in it?'}, {'start': '2732.079', 'end': '2740.160', 'text': "Then HCl is added. It follows Markopani's\nreaction and with CTH CH comes CL and with C2 comes CTH."}, {'start': '2740.160', 'end': '2746.559', 'text': "Thus, from Markopani's reaction, two\nchloropene is formed. What should we do with this? E2KOH  When we add"}, {'start': '2746.559', 'end': '2751.920', 'text': 'KOH or KOH,\nour'}, {'start': '2751.920', 'end': '2758.559', 'text': 'CO2 is\nconverted into propane.'}, {'start': '2758.559', 'end': '2764.800', 'text': 'If it is a secondary alcohol, now we simply\noxidize it by adding KNO3 + H2O.'}, {'start': '2764.800', 'end': '2768.920', 'text': 'What will be formed when the secondary alcohol is oxidized is ketone.'}, {'start': '2768.920', 'end': '2774.760', 'text': 'In this way, our propane is formed from what, but we have to'}, {'start': '2774.760', 'end': '2778.920', 'text': 'show that it is from propane alone.\nSimilarly, we are also asking about this in other'}, {'start': '2778.920', 'end': '2782.680', 'text': 'questions. Write down the\nchemical equation for the preparation of aniline'}, {'start': '2782.680', 'end': '2786.839', 'text': 'from nitrobenzene. It is said that\nwe have to make nitrobenzene into aniline.'}, {'start': '2786.839', 'end': '2791.240', 'text': 'What do we do in this? We'}, {'start': '2791.240', 'end': '2796.079', 'text': 'reduce nitrobenzene by adding what to\nreduce it.'}, {'start': '2796.079', 'end': '2800.440', 'text': 'What is formed when we add SNAC and reduce it in acidic medium? In'}, {'start': '2800.440', 'end': '2807.880', 'text': 'this way, Et+ water is formed. Thus, aniline is formed.\nSimilarly, from benzamide, it is said that it has to be made from amide.'}, {'start': '2811.720', 'end': '2817.920', 'text': 'C2 is what we just do, we\nput a bit of KOH in it,'}, {'start': '2817.920', 'end': '2825.520', 'text': 'this CO group that does the Hoffmann-Bohm reaction is added and it becomes\nour Et. From here,'}, {'start': '2825.520', 'end': '2832.839', 'text': 'what we have is another KBr plus KetCo C3 plus\nwater is formed as a side product'}, {'start': '2832.839', 'end': '2839.000', 'text': 'and Y is aniline is more\nbasic than ammonia. We have already talked about this before,'}, {'start': '2839.000', 'end': '2845.280', 'text': 'which is to\ncompare our ammonia and our NH.'}, {'start': '2845.280', 'end': '2850.920', 'text': 'NH is this.\nBoth of them have a lone pair, but where is this lone pair'}, {'start': '2850.920', 'end': '2854.960', 'text': 'involved? It is involved in the\nresonating structure of the benzene ring.'}, {'start': '2854.960', 'end': '2858.960', 'text': 'For that reason,\nthis lone pair of electrons is less'}, {'start': '2858.960', 'end': '2865.640', 'text': 'available for donation. For that reason,\naniline is less basic than'}, {'start': '2865.640', 'end': '2870.200', 'text': 'ammonia, but what happens to ammonia? The\nlone pair on nitrogen is'}, {'start': '2870.200', 'end': '2876.000', 'text': 'easier.  It is available for protonation\nor for donation to others,'}, {'start': '2876.000', 'end': '2881.119', 'text': 'that is why we have to understand that our ammonia\nis more basic than aniline. It happens that'}, {'start': '2881.119', 'end': '2886.079', 'text': 'aniline\nis treated with chloroform and alcohol.'}, {'start': '2886.079', 'end': '2891.800', 'text': 'This means that our aniline is a\nprimary amine. When we treat primary amine with'}, {'start': '2891.800', 'end': '2897.960', 'text': 'chloroform in the presence of\nalcohol,'}, {'start': '2897.960', 'end': '2904.720', 'text': 'our phenyl isocyanate comes from here. The\noffensive smell of isocyanide is formed.'}, {'start': '2904.720', 'end': '2910.359', 'text': 'Plus KCl plus our KCl is formed. This is how\nwater is formed. We'}, {'start': '2910.359', 'end': '2915.359', 'text': 'have to write that we have to make a sequence\nwhere A is HCOAT and'}, {'start': '2915.359', 'end': '2922.440', 'text': 'AC is kept in ice-cold conditions. When AC is kept in ice-cold conditions,\nACCA5N is formed. So what'}, {'start': '2922.440', 'end': '2927.040', 'text': 'is our A? What is\nneeded to become a body?'}, {'start': '2927.040', 'end': '2935.319', 'text': 'What is aniline? In aniline, we have\nANO or AOT ACL 0 T5, which means if we'}, {'start': '2935.319', 'end': '2938.640', 'text': 'put it in isothermal conditions, what will be formed? Body C will be formed.'}, {'start': '2938.640', 'end': '2945.440', 'text': 'N CL N double bond N SAL. If we say A,\nwhat should have happened to us? Aniline should have been formed.'}, {'start': '2945.440', 'end': '2952.520', 'text': 'This means that Body C has become and when we\nput X in Body C, then C C A5'}, {'start': '2952.520', 'end': '2957.440', 'text': 'CL is formed. When we say that chlorobenzene has been formed, there\nare two ways to say chlorobenzene.'}, {'start': '2957.440', 'end': '2961.680', 'text': 'One is the Senme reaction. The other is the Gutterman\nreaction. In the Senme reaction, C2Cl is a'}, {'start': '2961.680', 'end': '2968.799', 'text': 'concussion cell or C2 in the presence of a concussion cell. This is how'}, {'start': '2968.799', 'end': '2973.559', 'text': 'chlorobenzene was formed. Chlorobenzene was formed.\nThis chlorobenzene is called'}, {'start': '2973.559', 'end': '2978.119', 'text': 'CLTH CO plus\nConcussion. With chloral, what is'}, {'start': '2978.119', 'end': '2983.480', 'text': 'formed by chlorobenzene is DT.\nWe learned to make DDT, DDT is now'}, {'start': '2983.480', 'end': '2989.520', 'text': 'B. Now I\nwill make DDT.  For which I have a'}, {'start': '2989.520', 'end': '2995.480', 'text': 'Clth CO plus what we do two\nI'}, {'start': '2995.480', 'end': '3003.400', 'text': 'write like this chlorobenzene is two\nchlorobenzene a here is H here is another I did'}, {'start': '3003.400', 'end': '3009.079', 'text': 'chlorobenzene\nhere is H and seal is from here like this'}, {'start': '3009.079', 'end': '3016.520', 'text': 'H TO like this H TO comes out in the\npresence of what we put is the compound A4'}, {'start': '3016.520', 'end': '3024.359', 'text': 'and it becomes DT like this C salt CH CH\nNow with two phenyl groups like this we'}, {'start': '3024.359', 'end': '3031.720', 'text': 'make a bond here CL and another phenyl group\nbelow CL this is called'}, {'start': '3031.720', 'end': '3038.880', 'text': 'dichloro diphenyl trichloro ethane\nwhich we have to write as DT is the'}, {'start': '3038.880', 'end': '3043.839', 'text': 'major use of B is DD where is\nused'}, {'start': '3043.839', 'end': '3047.359', 'text': 'DT should be used as'}, {'start': '3048.520', 'end': '3054.000', 'text': 'ins A similarly\nin our 22 asked given example of each of the'}, {'start': '3054.000', 'end': '3059.799', 'text': 'following is said a homo polymer is like a homopolymer\nmade of the same monomer unit like'}, {'start': '3059.799', 'end': '3065.720', 'text': 'ours can be called a polyethylene  Ethylene is\nmade up of ethylene, polyethylene is number'}, {'start': '3065.720', 'end': '3073.240', 'text': 'two, what is a polymer? Nylon is made up of\ntwo different monomer units. It is made up of two different monomer units. If we'}, {'start': '3073.240', 'end': '3079.079', 'text': 'say antibiotics, we\ncan say penicillin.'}, {'start': '3079.079', 'end': '3083.640', 'text': 'Penicillin is an example of\nantibiotics. Now, in antipyretics,'}, {'start': '3083.640', 'end': '3087.559', 'text': 'what can we take? We can take paracetamol.'}, {'start': '3087.559', 'end': '3094.319', 'text': 'This is what we have mentioned. Mention the rattles for\nmaking paper. Now, if we'}, {'start': '3094.319', 'end': '3099.319', 'text': 'look at the materials of paper, we have\ncalled paper and materials.'}, {'start': '3099.319', 'end': '3104.839', 'text': 'Now, for paper and materials,\nwhat do we need? First, we can have wood. Similarly, we can have'}, {'start': '3110.960', 'end': '3117.240', 'text': 'our grasses.\nSimilarly, we can have our starches. Similarly, we can have'}, {'start': '3117.240', 'end': '3124.559', 'text': 'cotton lints. Similarly, we can have recycled\npaper. Similarly, we need'}, {'start': '3124.559', 'end': '3130.400', 'text': 'other chemicals.\nWe'}, {'start': '3130.400', 'end': '3134.280', 'text': 'can also take them as materials.\nSimilarly, we can'}, {'start': '3134.280', 'end': '3139.359', 'text': 'call it paper.  Paper is what\nwe call Nepali paper or'}, {'start': '3139.359', 'end': '3143.920', 'text': 'Nepali paper, Lokta paper is the paper that is extracted from the\ntrees called Lokta'}, {'start': '3143.920', 'end': '3149.359', 'text': 'and which we\ntraditionally use.'}, {'start': '3149.359', 'end': '3154.480', 'text': 'Another distinction is between\nnatural radioactivity and artificial'}, {'start': '3154.480', 'end': '3158.720', 'text': 'radioactivity. Where\nour natural radioactivity is what'}, {'start': '3158.720', 'end': '3163.160', 'text': 'happens naturally, what happens'}, {'start': '3182.920', 'end': '3187.960', 'text': 'naturally, what happens spontaneously'}, {'start': '3187.960', 'end': '3193.400', 'text': 'If we could write a difference like this, what can we do about natural radio?'}, {'start': '3193.400', 'end': '3198.960', 'text': 'It cannot be controlled. If it cannot be'}, {'start': '3198.960', 'end': '3203.040', 'text': 'controlled, then we can control it. Why is it that we have'}, {'start': '3203.040', 'end': '3207.200', 'text': 'induced this disintegration?\nThat is why'}, {'start': '3207.200', 'end': '3214.040', 'text': 'we had to write that it can be controlled and if it cannot be controlled,'}, {'start': '3214.319', 'end': '3219.599', 'text': 'then we\ndiscussed the questions of set B. Okay, thank you'}, {'start': '3219.599', 'end': '3223.119', 'text': 'for watching this video.'}]