Hello, good day, welcome to the integration of the Organic Chemistry Apprentice 2.0, or Part Two, as you prefer to call it. The topics to be covered are alcohols, aldehydes, ketones, and carboxylic acids. With this, we will have covered all the topics included in the partial transfer exam, the main one being organic chemistry, as we mentioned before. The instruction tells us to name the following structures, so let's begin with structure number 1. Remember that these structures are in their linear form, similar to how they will appear on the exam. So, let's begin. In this case, we make our linear structure. Okay, we have our hydroxyl group there, a ring, an exile ring, and a proper ring. Okay, so we have our structure. Remember that the rule here is to look for the longest chain that includes the functional group. In this case, the longest chain that includes the functional group is this one we have right here. Also, remember that the rule tells us that the numbering starts from the end closest to the functional group. Here, the functional group is basically attached to carbon number 1, so it's carbon 1, 2, 4, 5, 6, 7, 8, and 9. 10, 11, 12, and 13 are very important; only carbons are counted, so please don't make the common mistake of counting oxygen. Oxygen is not counted, only the carbons, the little dots we have numbered. From there, we will identify the weapons by essences or branches. At carbon number 2, we begin with our first branch: 1, 2, 3, 4, 5, 6. For the modification of 6 carbons, carbon 3 and 4 are the same branch; it is a vent. Carbon 5 is also the same branch; treat it as below; it is a halogen, it is bromine. Carbon 7, the top part is its own base, the bottom part is a cycle. Carbon 9 is the same branch with two carbons, so it is a steel. And now we have the first letter, obviously it is the arsenic, so it will be bromine, the first branch to name: 55 of bromine, 55 and bromine. Okay, so we have already mentioned bromine. Let's see who comes next. Next would be the C for 21st century, so it would be 7 for 21st century. From there we would have what It's from till 9/10 diethyl 9 10 diethyl a b c d e f g h so starting the alphabet we see that the next one is 2 exile followed by seven and its profile and finally what comes to be the pair of neos penti the 34 money in till we have already named all the branches so now we indicate on which carbon up to the functional group hydroxyl is on carbon 1 it is 13 so it is tree decane remember in all the other names it will be left up to 3 dec but the hawks in alcohols the rule tells you not to men to the aldehyde so well with ending -gol now it is trick decane of canons 55 and bromo 7 cycles and the Nobel and steel 2007 made its own 34 and 91 tree decane we are going to compose the name are not appreciated correctly trini from here no well we continue now we go to our second structure we say an alcohol now we are going to make an aldehyde remember that the liquid is characterized Because of that functional group we have there at the ends, it's carbon, and in this case, carbon. Since it's the group of an aldehyde, it's called a methyl group. So, what will be the name of this figure? Remember, we have to look for the longest chain that has the functional group. After analyzing it a bit, it's evident that this chain would be the one shown here, where carbon 1 corresponds to the carbon of the functional group. The series of carbons is 1, 2, 4, 5, 6, 7, 8, 9, 10, 11, and 12, up to 12 carbons. Now let's identify the branches. The one from carbon 2 has one, two, three, so it's a proton. The one from carbon 3 has 12; it's a methyl group. The one from carbon 4 has 1, 2, 3, 4, 5; it's a pentyl group. The one from carbon 6 has two; it's a til group. The one from carbon 7 also has two carbons, so it's a reptile group. And so, let's name the structure. Remember, alphabetical order. So, the first letter It will obviously be useful 367 trigger til 367 tree destination then we have gotten rid of the three styles now it is between pen til and finally we are going to start with the letter p so we have to move on to the second letter and it would be against r obviously in band is that r so in alphabetical order now it is the turn of four fertile 2 own as it is of 12 remember important in aldehydes they have ending to the 12s hydrocarbon in this case a 2 of kano so it will be 2 of canal all of the final unlike alcohols here it is not indicated where the functional group is unless it has a double one as we saw in its time having only one it is understood that it is at the ends we will have it and in its time we talked about it if the carbonyl group is at the ends it is an aldehyde if the carbonyl group is on the intermediate carbons or the carbons in the middle it is a ketone as we are going to see next so let's go to it exercise number 3 well then we make our figure we have to be very careful Because with these lines, given that the presence of a double bond could be misinterpreted at some point, remember, a double bond will be indicated as it is there. We also have our basic structure. Now we are going to add substituents, and there it is. Okay, so we identify the main chain that encompasses all the constituents, sorry, that encompasses all the functional groups, which in this case is the carbon functional group and what is on the intermediate carbons. Therefore, it is a zone. Okay, the numbering runs from right to left because the carbon functional group is closest to the right end. So, yes, it is carbon 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18. Now, the name of the constituents: on carbon 5 there is a metal; on carbons 6 and 7 there is a alkyl group; on carbons 8 and 9 there is a cyclohexyl group; on carbon 10 there is a cyclohexyl group; on carbons 11 and 12 there is a cyclohexyl group; on carbon 12 there is a metal. 13 top part chlorine bottom part fluorine so now we have to name it in the first instance I would do the thermos with the checkmark that we already know that the rule tells us the prefix t is not counted and we start with the useful one so it would be 67 of chlorofluorocarbon 67 in the useful one from there we will go to the letter c we have a cyclohexyl cyclopentyl and chlorine at the time we already saw that in the it went before the l so the decision is between cyclohexyl and cyclopentyl and both start with the prefix cyclo so we go to the letter in this case that gives us the tiebreaker which would be the h of cyclohexyl against the p of cyclopentyl until h or before so the name is 89 chlorohexyl 10 if you ventilate it 13 chlorine a b c d e f in there is no style around here so e f fluorine 13 word I already mentioned the thermoses and the cyclohexyls and the cyclopentyls I am missing the neopentyls and the methyl phenyls which So it's 5 12, okay, then as I was saying, we've named the fluorine, now nobody, we're just between lying and lying, so it's 526 and inserted 11 12 dyn open till indicates with the carbons where there is carbon and the 215 is 18, so it's an opta from here of 12 15, opta of trucks, opta that refers to the 18 carbons of 212 carbons, and the ones that functioned were like ketones, okay, and now let's go for our last exercise number 4, in this case it's a carbolic acid, and so we have it there, this is our last structure, we enclose in blue the trunk main chain, the numbers will be given to me from the end closest to what is the functional group, carbolic, yes, it is very important, this functional group, carbolic, will go at the ends as well as happens with the aldehyde group, so it's carbon 1, it is counted from the carbon of the functional group 2 3 4 5 6 7 8 9 10 11 12 13 and then we name the carbons where there are enough 2 3 4 5 7 9 2 3 4 577 because there are 29 here as we say an error remember that between the number the number there are commas not hyphens so we check it again 2 3 4 5 779 we see how many there are they are 2 467 so the prefix is ​​captured I like me or in til we made a pretty classic mistake the name of carboxylic acids always starts with the word acid so its name is acid 2 john has been 2 3 4 7 7 9 this neo pentile we delete this from the top because it is obviously incorrect also til and as it is of 13 this train here not tribeca refers to 13 tribeca doctor and dean and with the ending today ccoo refers to carbolic acid so important what you have to remember is that carboxylic acids but you don't write the word acid and they go To have the ending today as if it had two functional groups, card, the carbon, chilo, then it would be 5. Well, this is our last exercise and with this we have finished the second integration of what we have learned, covering all the topics that will come in the next exam on Wednesday. I reiterate, as I said at the end of the previous video, if you have any questions, let me know through the subject's WhatsApp group. That's all from me. See you on Wednesday for your exam. Take care, goodbye.