9 Organic Reagents in 9 Minutes / Quick Revision Series

[00:00:00]Hi students, in this video I'm going to teach you nine reagents in just 9 minutes. If you don't have a so much time and you want to learn more content in a small time period, then this video will be very much helpful for you. Let's say S4 C2 thionil chloride theiononyl chloride what is the trick is just you know see the trick tricks I'm going to teach you just use this and you know solve your question trick is remove O let's say remove O remove H and and keep CL that's it can you try your product remove O Keep CL remove O and keep CL in its place. Reagent number two when you see the phosphorus halids like PH3 PH5 in such cases remove O and keep halogen.

[00:01:03]Remove O. What do you need to do? Keep X in its place. Now can you try the product? Remove this O in its place.

[00:01:11]Keep Cl. Reagent number three. chlorine or a bromine in the presence of UV light or sunlight. In such case what you have to do bromine or a chlorine in the presence of UV or sunlight three types of reactions will happen. One is free radical substitution, one is ali substitution, one more is benzilic substitution. Let's say free radical substitution.

[00:01:42]Remove one hydrogen here. What you do? Remove hydrogen H and keep CL in this place. Or if you are using chlorine, you put CL. If you're using Br, you put Br. So I'll remove one hydrogen, I will add Br. And if you are doing alyic substitution like alaken if they give you alaken in exam if they give aline just remove hydrogen put a chlorine or a bromine. If they give alic position what do you mean by alic position double bonded carbon next carbon is alic this sp3 carbon is alic position. So in alic position, elic position, remove the hydrogen, remove hydrogen and keep cl.

[00:02:41]So in this place I am going to remove one hydrogen. I'm going to put CL.

[00:02:47]That's it. Instead of you know alaken if they give alleben.

[00:02:53]So what you have to do bingilic position you need to take and from there same remove hydrogen and keep chlorine. What do you mean by benzilic position?

[00:03:06]Benzene ring next connected carbon is benzylic carbon. There remove the hydrogen. This benzene ring next to carbon must be sp3. There remove one hydrogen. Put the halogen like chlorine or a bromine. That's it. Is it clear?

[00:03:25]Three reagents completed. Reag number four bromine over carbon tetra chloride.

[00:03:32]When you see this across double bond C double bond is there right to break the bond add Br on both the sides add Br in both the sets in anti-position anti-addition takes place anti-addition so here you see break the bond what you have to do students tell me one Br down side one Br upside Okay. Coming to the H. This will help you to cover in a short time like when you are walking or when you are you know you are in bus stop or when you are before sleeping or when you have very short time watch this video you will cover it. HBR or HI. Okay. When you see this it will follow Marikov's rule. Which rule? Tell me. Marikov's rule. What this rule will say double bonded carbon is there right? Take the double bonded carbon in that which carbon is having less hydrogens here one hydrogen here two hydrogens's this is having less hydrogen's right so break the bond you HBr how it will break H+ plus Br minus the negative part you add to the carbon having less number of hydrogens so negative part add to the double bonded carbon add to the double bonded carbon having less number of hydrogens less number of hydrogen. So double bonded carbon one two okay this is having less hydrogen's one portion so then you add that Br here and the hydrogen other side then CH2 one hydrogen you're adding it will become CH3 next even iodine also same place but HBr only HBr in the presence of peroxids it shows antimarony's rule what is the rule anti The maricovs rule antimarony's rule what you have to do HBR how it will break it will follow the free radical mechanism simply you can write the product reverse here Br minus is going to the carbon having less hydrogens here Br goes to the carbon having more number of hydrogens's see in this case 1 to let's say general numbering Which is having more hydrogen is this one you put Br there CH3 C H bond CH2 keep Br here and H here just reverse to that that's why it is antimaronic next eighth reagent sodium iodide over acetone it is fininkle stein reagent what is the reagent students tell mele stein reagent what this reagent will do is remove this remove the halogen like chlorine or bromine here and put iodine remove chlorine or bromine and keep iodine there. So what is our product?

[00:06:54]Can you tell me CH3 CH2 I and next one is either you can use AgF, Hg2, F2, CO F2 or SBF3 any of one reagent it is called SWAT reaction. What is the reaction? SWAT reaction. In this case, what is the trick? You know, remove chlorine or bromine and keep florine.

[00:07:21]What do you say? remove chlorine or bromine and keep florine its place. You know it it is very much helpful for you to revise in short time. Ninth one excess ammonia. When you see excess ammonia primary amines are the products remove the Brick is remove Br and keep NH3 keep NH2 NH2.

[00:07:48]So you will get CH3, CH2, NH2. That's it. Nine reagents you completed. What are those nine reagents? Students tell me S4 CL2, PX3 or PX5. Chlorine or bromine in presence of UV or sunlight.

[00:08:07]Bromine in carbon tetrachloride. HBr or HA maronic rule. HBr in peroxides antimaronic rule. Sodium iodide in acetone finger stain reaction. and AgF swarts reaction and excess ammonia gives primary amines. So in less than 9 minutes you learned nine different reagents. If you want me to make like these different reagents so that you can cover the syllabus in short time let me know in comment section and tell me how is this video. Thank you so much for watching this video. I'm your Kumali man chemistry mentor.

[00:08:45]Forgot to tell you try this question. This is from NCRT 10.5. Okay. Just try this and let me know your answer whether you are able to do it or not in the comment section.

[00:09:04]This is the question. Okay. O is there.

[00:09:07]SO2 is there. What is the product? You will keep remove O and keep Cl here.

[00:09:13]Same try mix questions and let me know how many you are able to do it.