🔥 RENEET 2026 ALDEHYDE & KETONES | MOST EXPECTED QUESTIONS 💯 | Score 180+ in Chemistry 🚀

Added: 2026-05-29

[00:00:01]Hello, all my dear students. Welcome back again to HSC in world of chemistry.

[00:00:05]So, this is aldehyde and ketones.

[00:00:08]Newly framed questions, most expected for your re-need 2026. Let's try to solve it. Question is in our front.

[00:00:16]Try to get the answer. Definitely, they will help you in scoring good marks.

[00:00:21]Now, pause the video and check it out whether your answer is right or wrong.

[00:00:26]Baba HSC is going to solve it.

[00:00:28]Acetaldehyde dilute NaOH and heat. Aldol condensation. How will you get the product? Let me tell you those who don't know. So, write down CH3 CH=O like this and another molecule it's two alpha hydrogens like this.

[00:00:48]And now what you're supposed to do?

[00:00:51]Just simple thing.

[00:00:54]We have to eliminate out water. Once you eliminate out water your answer will be there in front of your eyes. Just check it out.

[00:01:05]Crotonaldehyde.

[00:01:07]Its name is crotonaldehyde.

[00:01:10]CH3CH=CH CH3CH=CH and CHO.

[00:01:19]Its name is crotonaldehyde. What kids?

[00:01:22]Croton aldehyde. Crotonaldehyde.

[00:01:29]Now, another part is NaBH4. So, NaBH4 will do the reduction.

[00:01:34]What I'm saying? Reduction. It will do reduction and your CHO will get converted into CH2OH. So, your C is CH3 double bond will not get affected, will remain same and CH2OH.

[00:01:56]You will get CH2OH. Is it clear? Now, thionyl chloride in place of OH, Cl will come. So, you will get CH3 CH double bond CH and CH2 Cl. Now, my dear student, this Cl will eliminate out. In place of Cl, hydrogen will come. This is given in class 11.

[00:02:20]So, your product will be but-2-ene. CH3 CH double bond CH and CH2H.

[00:02:31]That is CH2H is CH3. So, but-2-ene is your answer.

[00:02:37]Now, another question in front of your eyes, kids.

[00:02:41]Here is the question. Just see to it.

[00:02:43]And you have to reach up to H.

[00:02:47]Your target is to reach up to H.

[00:02:52]I hope that question is visible, very much clear to all of you. Now, let's it go.

[00:02:58]Baba HSP is going to solve this question. Just you see to it. What should be the answer? Now, you see CH3COCH3.

[00:03:07]Dilute barium hydroxide and heat. Aldol condensation. So, aldol condensation once again, what are you supposed to do?

[00:03:15]Take your acetone, CH3 CH3. Apply Baba HSP's short trick, another molecule of acetone. It's two alpha hydrogens will go like this. Mere pyare pyare bachcha log. Now, mare ko to pani nikalna. We have just to eliminate out water. Hey, the water has gone. And product B you must have obtained. So, product B is CH3 CH3 C double bond CHCO CH3. Aldol condensation product is always alpha beta unsaturated aldehyde or ketone. So, you see this is alpha and this is your beta alpha beta unsaturated ketone. Now, ozonolysis. What you're supposed to do? Break it from here.

[00:04:08]Ozonolysis will break the bond just like sandwich or burger. So, the bond will be broken from here. Now, one is acetone definitely left hand side one is acetone then and the right hand side is C. So, C is your C H double bond O and C O C H 3.

[00:04:29]This is your C and C is on Clemmensen's reduction. This is your Clemmensen's reduction.

[00:04:37]So, in place of carbonyl group you will get hydrocarbon like C H 3 C H 2 and C H 3. You got propane. So, then F R S R free radical substitution reaction. I hope it is very much clear.

[00:04:56]So, major product will be two chloro propane. Here you got two chloro propane like this, two chloro propane.

[00:05:04]Now, this one reaction with benzene will give you cumene, very well known product. Cumene, why are you so mean?

[00:05:14]This much mean cumene. Cumene oxygen cumene hydroperoxide cumene. You see this is benzylic hydrogen. This will eliminate out and you will get one unstable product C H 3 C H 3 O O H. Now, this one hydrolysis in acidic medium H3O+. How the bond will be broken up? Just see to it. Acetone will eliminate out as side product and worldwide production of phenol. So, this is the way by which people in the whole world they are producing phenol, right?

[00:05:58]Phenol. So, this is done and dusted. If you want to take a screenshot, you can have it for this question. I hope this is clear. Now, next question.

[00:06:09]Now, you see let me move aside away so that you can clearly see to it. I hope this is very much clear to all of you.

[00:06:17]Just see to it. You are supposed to reach up to J.

[00:06:23]J J J J J because no. Ready? Let's it go and Baba HSP is going to solve it out and let's see what answer you are going to get. Now, see PCC. PCC is your MOA.

[00:06:37]MOA MOA MOA. MOA MOA means mild oxidizing agent. If you know what is MOA, mild oxidizing agent. So, this is your acetaldehyde CH3 CH double bond O. Acetaldehyde Grignard reagent. So, just this methyl add to this carbonyl carbon. Just add methyl to carbonyl carbon and oxygen to H. So, your B is CH3 CH CH3 OH. This is known as isopropyl alcohol.

[00:07:14]Concentrated H2SO4. Water will eliminate out OH from here and one H from this place. So, propene. C is your propene.

[00:07:24]Are you getting? I hope that everyone is able to understand. This is your propene.

[00:07:31]Propene ozonolysis will get formaldehyde and acetaldehyde.

[00:07:36]So, formaldehyde is already given. So, acetaldehyde CH3 CH double bond O. Now you must be getting dilute NaOH and heat. Aldol condensation reaction. So F is your crotonaldehyde.

[00:07:51]Earlier I have solved this question for crotonaldehyde. Now see to it. Now crotonaldehyde is alpha beta unsaturated aldehyde. This is your crotonaldehyde.

[00:08:02]Ozonolysis. Now see how to do ozonolysis. Break it from here. You will get glyoxal and acetaldehyde. So acetaldehyde also already given. So write down glyoxal. So G is your glyoxal CHO and CH double bond O. Reduction, you will get ethylene glycol CH2OH and CH2OH.

[00:08:25]Now PCC is going to go out. So once again you will get glyoxal.

[00:08:30]CH double reduction. You know very well.

[00:08:36]Well-known reaction, Wolff-Kishner reduction. So J is your ethane. What is J? J is your ethane. Ethane CH3 CH3 CH3 and CH3. Now let's see the whole reaction, the whole conversion. If you are really able to do such type of questions, then your revision is okay.

[00:09:02]And if you are not able to do such type of questions, then you have to work a lot. Work hard is required. Now see question. Very very simple question. If you see on your screen. Now you can go up to F. If you are able to reach up to F, it's okay. If you are not able to reach, more and more revision is required. So ethene plus HBr ethyl bromide I will get C2H5Br.

[00:09:28]Now alcoholic KOH once again ethene I will get simple simple very simple question. Now once again R2O2 is not going to affect because HBr and R2O2 is giving anti-Markovnikov's Kharasch effect and peroxide effect, but here only Markovnikov's rule. So, once again, you are going to ethyl chloride, C2H5Cl.

[00:09:47]Once again, alcoholic KOH, ethene, ethene. Again, I then I've given you such type of things so that you can be most aware of the such type of reactions. Ozonolysis is going to break from here two moles of formaldehyde.

[00:09:59]Now, remember that formaldehyde with ammonia excess, this reaction is missing in your NCERT, but given in Tamil Nadu board. So, urotropine, CH2 whole six, N4. How this reaction is going to form? That six moles of formaldehyde and four moles of ammonia is going to form urotropine. I have shown you one structure of urotropine somewhere. Now, if you're not able to get it, let me write down once again, NCH2 CH2 nitrogen, nitrogen, CH2. How many CH2 have consumed? I have consumed three CH2 and one nitrogen is still left. So, write down nitrogen in the middle, CH2N, CH2N, CH2N. This is urotropine, also known as hexamethylenetetramine. When you do the nitration of your urotropine, you will get RDX. RDX means explosive. On doing the nitration of this compound, you will get RDX, research and development explosive. Now, next question on your screen. Just see, it's very simple question.

[00:11:06]If you are not able to do this question, then remember that you have to do a strong hard revision, right? Now, see on your screen, you are supposed to find out F. Now, let's see F. The first reaction, if you remember in your brain, that is being Etard reaction, Etard reaction. What happens in Etard reaction?

[00:11:28]Etard reaction is going to give you benzaldehyde, benzaldehyde.

[00:11:33]So, A is your benzaldehyde. Now, another reaction is cross aldol condensation. If you don't know how to form the product of this cross aldol condensation, just see. Benzaldehyde is not having any alpha hydrogen, but acetaldehyde does have. So, just take it out and remove the water. We have to remove water, and your product will be C6H5CH=CH and CH=O.

[00:12:02]It's known as cinnamaldehyde.

[00:12:05]Cinna- mal- dehyde. Cinnamaldehyde. Now, my dear students, if you do the ozonolysis of cinnamaldehyde, it will break from here. Benzaldehyde and glyoxal. Glyoxal is already given, so C is your benzaldehyde.

[00:12:23]Benzaldehyde. Benzaldehyde, or we can say benzene carbaldehyde. NaBH4, so reduction reduction is benzyl alcohol, or you say benzyl alcohol. Benzyl alcohol. PCC is more of a mild oxidizing agent, so benzaldehyde.

[00:12:41]Benzaldehyde. Now, benzaldehyde on reaction with hydrogen cyanide, you will get benzaldehyde cyanohydrin through um cyanide is nucleo- nucleophilic addition reaction, and then hydrolysis. So, the HCN will release cyanide. Cyanide will attack above the plane and below the plane. 50-50 Britannia 50-50 above and below the plane. That means you're going to get racemic mixture. So, finally hydrolysis, so you will get CH OH and COH.

[00:13:17]50% dextro and 50% levo. Britannia 50-50. It can be given as plus minus, or it can be given as dextro levo.

[00:13:29]So, that what is the name of this compound? It's from your biology, the father of uh what to say, father of in your bio what is there?

[00:13:39]A Pison sativum, like that. Father of genetics, Mendel, Mr. Mendel. Mandelic acid. Mandelic acid. So, this is your whole conversion for this question. I hope that everyone must have got it. Now, next question in front of your screen. Now, this question to definitely you have to do it, kids.

[00:14:04]Even if this question you are not able to do, I will say still time is left.

[00:14:08]Do proper preparation. Now, acetic acid if I take acetic acid, now you see B2H6 is very good reducing agent. It is doing very very fast reduction of RCOOH.

[00:14:19]LiAlH4 is taking too much time. It is given in an NCERT on the right hand page. Anyways, you will get ethyl alcohol, C2H5 OH. Now, PCC is more mild oxidizing agent once again. You will get acetaldehyde, CH3CHO.

[00:14:36]Now, NaOI, NaOI is sodium hypoiodite or iodate? Oh, sir, I don't know what is the name. Hypohalous halous salt.

[00:14:45]Halous will form halide. So, sodium iodite or you can say I2NaOH, iodoform test. Salt and water. So, C is your CHI3, yellow PPT, yellow papaya AG powder, silver reaction is giving acetylene.

[00:15:03]Acetylene with water, acetaldehyde, CH3CHO.

[00:15:09]If you don't remember the reaction, one side add oxygen and one side add two hydrogen. First, it will form enol then it will undergo tautomerization.

[00:15:18]Tautomerization to form acetaldehyde because enol is unstable and the keto or carbonyl group is highly stable. Now, sodium bisulfite, you will get white color PPT of acetaldehyde sodium bisulfite. White PPT white PPT is just a test also. So, you will get my dear students CH3 CH SO3 Na and OH. White PPT white PPT it is a distinguishing test. Now, another question on your screen. The question is very very awesome question, kids. See what the question is asking that how many compounds will show positive tolerance and failing test means those compounds we have to select which are showing both as positive the tolerance as well as failing. If I talk about the first benzaldehyde Oh, you want to do it? Just do it. Pause the video and do it. Try it out. Oh.

[00:16:21]A C D All the options you have ticked. Now, let me tell you. Benzaldehyde will give positive tolerance test but negative failing test. So, this is not answer.

[00:16:36]Formic acid both because it is only carboxylic acid which is having CHO group.

[00:16:42]And CHO group definitely will give you both positive tolerance failings. Now, this is hemiacetal and hemiacetal once go inside the water it will convert into acetaldehyde. So, it will give it will give white PPT. And now this is ketone but alpha hydroxy ketone. If I add basic medium it will show tautomerization. So, during Lobry de Bruyn van Ekenstein rearrangement due to that the keto group will change into aldehyde just like if I add fructose into the basic medium it will convert into glucose and mannose.

[00:17:19]So, this test is not actually given by ketone, it is actually given by aldehyde. So, yes, all alpha hydroxy ketones. So, this is alpha hydroxy ketone. All alpha hydroxy ketones of this world will show positive Tollens' and Fehling's test. Fructose, yes, my lord. Mannose, yes, my lord.

[00:17:35]Sucrose, no, no, no. Glucose, yes. So, how many 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 Got it?

[00:17:50]Now, the last question. Last, but not the least. Just see on your screen. If you are not able to do this question, you have to do thoroughly. You have to do very, very fantastic, awesome revision. Now, see how many compound many man no, man is written. How many compounds will show positive haloform test? Same, same, same, same.

[00:18:13]Uh ethyl alcohol will give propyl alcohol will give this will give this will give this will give this all will give. So, six. Yeah, first is the correct answer. Wow. So, my dear students, all the best. If you're new to this channel, subscribe it for 180 on 180. Bye-bye. Take care. This is your bye-bye to speak MSC topper MTech 2004.

[00:18:37]In between that GRF and gate qualified.

[00:18:39]I've been teaching for the last 27 years. Produced first rank, third rank, fifth rank, all top 10 rank along with huge mass selection. Bye-bye. Take care.