ALL ORGANIC TESTS in 10 min! 30 Marks Assured🔥NEET 2026

Added: 2026-05-09

[00:00:00]Today in this short video, we will be discussing all organic tests. So, without wasting time, let's begin. So, first organic test here is bromine water test or it is bromine test. Now, remember it's given by alkenes and alkynes. That means this is a test used for unsaturation. And reagents may be either what we call Br2 in water or it may be Br2 in CCl4. What will happen over here when alkene or alkyne is treated with bromine water or bromine in, I mean say CCl4, reddish brown of bromine is decolorized. So, with the help of this, you can distinguish whether the starting material is alkene or alkane. Similarly, alkyne and alkane. So, with the help of this, we can distinguish alkane from unsaturated hydrocarbons. After that, we will be having another test and that is known as Baeyer's test. Remember, this Baeyer's test is given, once again, by alkenes and alkynes. And the reagent used here is Baeyer's reagent. What is Baeyer's reagent? Baeyer's reagent is cold dilute alkaline KMnO4 solution.

[00:01:06]Here, what will happen? Alkene is converted to, I mean say, syn hydroxylation takes place. That means alkene is converted to 1,2-diol or what we call vicinal glycol. During this reaction, pink color of KMnO4 is changed to brown precipitate. So, this is used as a test for unsaturation under the name Baeyer's test. After that, we will be having another test that is known as Lucas test. With the help of this Lucas test, we can distinguish 1°, 2°, and 3° alcohols.

[00:01:35]So, with the help of which reagent?

[00:01:37]Reagent is known as Lucas reagent. And Lucas reagent means 1:1 mixture of anhydrous zinc chloride and HCl. What will happen over here? Primary alcohol on treatment with Lucas reagent does not react at room temperature. That means there will be no turbidity at all. But secondary alcohol can react with Lucas reagent to form alkyl chloride. That means turbidity appears but after 5 to 10 minutes. But tertiary alcohol immediately reacts because tertiary alcohol are more reactive alcohols. They immediately react with, I mean say, Lucas reagent and results in the formation of an corresponding alkyl halide, alkyl chloride, and hence turbidity appears very, very immediately. That means in seconds.

[00:02:18]After that, we have another test known as Victor Meyer test. This is also used to distinguish 1°, 2°, 3° alcohols. What are the reagents over here? Remember, if we have a, I mean say, doubt whether the test tube contains 1°, 2°, 3° alcohols, we add these reagents, iodine in presence of red phosphorus, then AgNO2, then HNO2, then NaOH. If a red color appears, that means at that time it's 1° alcohol. If blue color appears, at that time it's secondary alcohol. If it is colorless, then it's tertiary alcohol.

[00:02:49]Because of this color, we can also call this test as RBC test, red, blue, and colorless test. After that, we will be having another test known as ferric chloride test. This test is given by phenols. So, this test, with the help of this test, we can distinguish phenol and alcohols. The reagents used here are ferric chloride but anhydrous ferric chloride. What will happen? Phenol reacts with ferric chloride and results in the formation of ferric phenoxide.

[00:03:13]And the color of ferric phenoxide is violet color. So, with the help of this, we can distinguish phenol and alcohols.

[00:03:20]After that, we will be having another test known as azo dye test. This is given by either phenols or aromatic amines, for example, aniline. Now, reagents used here are benzene diazonium salt, for example, benzene chloride.

[00:03:32]What will happen? When benzene diazonium chloride is treated with phenol, reaction takes place in basic medium.

[00:03:37]That means in basic medium, coupling reaction takes place and this benzene diazonium group is, I mean say, coupled at para position of phenol and it results in the formation of 4-hydroxyazobenzene, known as an orange dye. But remember, aniline gives 4-aminoazobenzene but in acidic medium, it couples with this benzene diazonium ion at para position results in the formation of 4-aminobenzene, known as yellow dye. So, this can be simply used as a distinguishing test between phenol and rest of the compounds. Similarly, between aromatic amine, that means aniline, and rest of the compounds.

[00:04:10]After that, we will be having another test known as Liebermann's nitroso test.

[00:04:14]This test is given by phenol. So, with the help of this test, you can distinguish phenol and alcohol simply.

[00:04:20]Now, reagents used are NaNO2, H2SO4, or you can say HNO2 in excess of H2SO4.

[00:04:25]Now, what will happen over here? These are the reactions. Finally, this phenol results in the formation of indophenol but that upon, I mean say, basification gives sodium salt of indophenol and its color is blue. So, with the help of this, we can distinguish phenol from alcohols. After that, we will be having another test known as bromine water test. This can also be given by phenols and anilines. Phenols upon treatment with bromine water gives white precipitate of 2,4,6-tribromophenol but aniline gives white precipitate of 2,4,6-tribromoaniline.

[00:04:57]After that, we will be having another test known as Tollen's test. And this is given by all aldehydes or alpha-hydroxy ketones or what we call formic acid or N-phenylhydroxylamine hemiacetals or terminal alkynes. But remember, these five gives white, what we call, silver mirror in this test. But this terminal alkyne gives a white precipitate.

[00:05:17]Remember, ketones in general do not show this reaction. So, do not show this test. With the help of this, we can distinguish these compounds from rest of the compounds. And during this, remember, this aldehyde or terminal alkyne or these compounds on treatment with Tollen's reagent. Now, the question is what is Tollen's reagent? Tollen's reagent is ammoniacal silver nitrate gives either, what we call, silver mirror or white precipitate. But white precipitate is given by this and silver mirror is given by these all. After this, after this, we will be having another test that is Fehling's test and another test that is Benedict's test.

[00:05:49]Fehling's test and Benedict's test reactions are similar and Cu2O, that means red precipitate of Cu2O is, I mean say, obtained in this test.

[00:05:59]But this is given by either aliphatic aldehydes, alpha-hydroxy ketones, or formic acid. That means aromatic aldehydes do not show these tests. So, with the help of this test, we can distinguish aliphatic aldehyde from aromatic aldehyde or aliphatic aldehyde from ketones because ketones also do not show this reaction. Okay, reagent here in Fehling's solution is Fehling's solution A, which is aqueous copper sulfate, and Fehling's solution B, which is Rochelle salt, that means sodium potassium tartrate. And in, what we call, Benedict's test, it is aqueous copper sulfate and then sodium carbonate and sodium citrate. After that, we will be having another test that is iodoform test. Iodoform test is given by those aldehydes or ketones which contain C=OCH3 or it's it can also be shown by C=OCH2X, C=OCHX2. That means if alpha halo, I mean say, a carbonyl compound is there or it can be two halogens, three halogens at alpha position. Or it can also be shown by an alcohol in which OH is attached to a carbon. That means alpha carbon to which at least there must be one H and CH3. That means this group should be present. And remember, reagents generally used here are iodine in presence of basic medium or you can say sodium hypoiodide. So, what will happen during this reaction? Whether it is keto, what we call, aldehyde or ketone, whether it is alcohol, this CH3 group is changed to CHI3. That means a yellow precipitate of, I mean say, what we call, iodoform is obtained during this reaction. So, with the help of this reaction, we can distinguish these compounds from rest of the compounds.

[00:07:27]After this, we will be having another test. This is known as 2,4-DNP test. That means 2,4-dinitrophenyl, I mean say, test. So, this is given by all aldehydes, all ketones, and formic acid. That means reagent used over here is Brady's reagent. That means 2,4-dinitrophenylhydrazine.

[00:07:45]What will happen over here?

[00:07:46]2,4-dinitrophenylhydrazone will be prepared and its color will be either yellow, orange, or red. Remember, this reaction can also be used to distinguish these compounds from rest of the compounds. After that, we will be having Schiff's test. Schiff's test is given by aldehydes or formic acid. This test is not given by ketones. That means with the help of this test, you can distinguish aldehyde and ketone or you can distinguish aldehyde and formic acid from rest of the compounds. What will happen? First of all, we have Schiff's dye. Schiff's dye is basically pink in color. And Schiff's dye means aqueous solution of rosaniline hydrochloride. To this, we pass SO2 gas and it becomes Schiff's reagent. That means colorless.

[00:08:24]After when it becomes colorless, on treatment with aldehyde, pink color is restored. But this type of test is not given by ketones. That means ketones do not show, do not respond towards this test. After that, we will be having sodium bicarbonate test or sodium carbonate test. Let me tell you, this test is only given by those acids which are stronger acids than carbonic acid, except para-nitrophenol. As para-nitrophenol is lesser or weaker acid than carbonic acid, but still it can give this test. Now, here reagents are sodium bicarbonate in this one. And here reagents are sodium carbonate in this test. And during this test, what will happen? Brisk effervescence of carbon dioxide is obtained. I mean say, carbon dioxide is evolved with effervescence, brisk effervescence.

[00:09:07]After that, here another test is carbylamine test.

[00:09:11]It is given by primary amine. And the reagents are like, what we call, chloroform and what we call a base like KOH. And here, electrophile of this reaction is dichlorocarbene. And what will happen? This amine is converted to primary amine. Remember, primary amine, whether aliphatic or aromatic, it is converted to, uh, I mean say, isocyanide or carbylamine or isonitrile. This is having very bad smell. That's why with the help of this test, you can distinguish primary amine from rest of the amines. After that, we will be having another test that is Hofmann mustard oil test. This is shown by only primary aliphatic amine, not aromatic amines. And the reagent used are CS2 and HgCl2. With the help of this, you can form alkyl isothiocyanate having mustard oil smell. That's why this is known as Hoffman mustard oil test. And with the help of this reaction you can simply or with the help of this test you can simply distinguish aliphatic primary amines from rest of the amines or rest of the compounds. After that we will be having another test that is Hinsberg's test and it is used to distinguish 1°, 2°, and 3° amines. It is given by what we call benzene sulfonyl chloride, but nowadays we can use para what we call toluene benzene what we call para toluene sulfonyl chloride right in instead of benzene sulfonyl chloride. But what will happen over here? Primary amine reacts with this and results in the formation of N what we call solid, but that solid is soluble in what we call KOH. But if secondary amine is there, it results reacts with this results in the formation of solid, but that solid is insoluble in KOH and tertiary amine does not react with tertiary amine does not react with I mean say benzene sulfonyl chloride. So, this was all about these tests. Hope you like the test. Thank you and God bless you all.