Popular Chemistry Online: February 2023

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Monday, February 27, 2023

WORD PROBLEMS IN ORGANIC CHEMISTRY

1 Write down the functional isomers of carbonyl compound with molecular formula C3H6O, which isomer will react faster with HCN and why? Explain the mechanism of reaction also. Will the reaction lead to the completion with the conversion of whole reactant into products at reaction condition? If a strong acid is added to the mixture what will be the effect on concentration of the product and why?

 HINTS--- CH3CH2CHO AND CH3COCH3

2 An organic compound X having molecular formula C4H8O gives orange red ppt with 2,4- DNP reagent . It does not reduce Tollens Reagent but gives yellow ppt of iodoform on heating with NaOI. Compound X on reduction with LiAlH4 gives compound Y which undergoes dehydration reaction on heating with conc H2SO4 to form But-2-ene . Identify the compound X and Y and explain the reactions.

  HINT--- X = CH3CH2 CO CH3 y = CH3CH2CH(OH) CH3

3 An organic compound A (C3H4) on hydration in presence of H2SO4 / HgSO4 gives compound B (C3H6O) . Compound B gives white crystalline product D with NaHSO3 . It gives negative Tollens Test and positive iodoform test. On drastic oxidation B gives compound C (C2 H4O2 ) along with formic acid . Identify compound A, B, C , and explain all the reactions.

 HINT--- A-- CH3C=CH

4 An unknown aldehyde A on reacting with alkali gives ɮ--hydroxy aldehyde which loses water to form an unsaturated aldehyde 2- butenal. Another aldehyde B undergoes disproportionation reaction in the presence of conc alkali to form products C and D . C is an aryl alcohol with formula C7H8O. (a) Identify A and B (b) Write the reaction involved. (c) Name the product when B reacts with Zn-Hg and HCl

HINT---- A - CH3CHO B- C6H5CHO

5 An compound X (C2H4O) on oxidation gives Y ( C2H4 O 2) . X undergoes haloform reaction . On treatment with HCN , X forms a product Z which on hydrolysis gives 2-hydroxy propanoic acid. (a) Write down the structure of X and Y . (b) Name the product when X reacts with dil NaOH ( c) Write down the equation for the reaction involved.

HINT-- X- CH3CHO Y- CH3COOH

6 An alkene A molecular formula ( C5H10) on ozonolysis gives a mixture of two compounds B and C . Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling test but forms iodoform . Identify compound A, B, C giving suitable explanation and write the reaction of ozonolysis and iodoform formation either B or C .

  HINT-- CH3 CH = C(CH3) –CH3

7 An organic compound A has characteric odour. On treatment with NaOH it forms compound B and C . Compound B has molecular formula C7H8O which on oxidation gives back A . The compound C is an sodium salt of acid when C is treated with sodalime , it gives an aromatic compound D . Deduce the structure of A, B, C, D . Write the reaction involved.

 HINTS-- A - C6H5CHO

8 A ketone A which undergoes haloform reaction gives compound B on reduction. B on heating with conc H2SO4 gives compound C which forms monoozonide D . The compound D on hydrolysis in presence of Zn dust gives only CH3CHO. Write the structure and IUPAC name of A, B ,C . Write reactions involved.

  HINT-- CH¬3-CO-CH2-CH3

9 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b) .Compound (b) is reacted with HBr to give C which is isomer of (a) . When (a) is reacted with sodium metal it gives compound (d) , C8H18 WHICH IS DIFFERENT from the comound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equation for all reactions.

10 An aromatic compound A on treatment with aqueous ammonia and heating forms compound B which on heating with Br2 and KOH forms a compound C of molecular formula C6H7N. Write the structure and IUPAC names of the compounds.

11 An organic compound contains 69.77% acarbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens reagent but forms an addition compound with sodium hydrogen sulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

12 An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative , reduce Tollens Reagent and undergoes Cannizzaro reaction. On vigorous oxidation , it gives 1,2-benzenedicarboxylic acid . Identify the compound.

13 An organic compound A molecular formula C8H16O2 was hydrolysed with dilute sulphuric acid to give a carboxylic acid B and an alcohol C . On oxidation of C with chromic acid produced B . C on dehydration gives but-1-ene . Write equations for reactions involved.

14 An organic compound A with molecular formula C8H8O forms an orange red precipitate with 2,4- DNP reagent and gives yellow ppt on heating with iodine in the presence of sodium hydroxide. It neither reduce Tollens or Fehling reagent, nor does it decolourise bromine water or Bayers reagent. On drastic oxidation with chromic acid , it gives a carboxylic acid B having molecular formula C7H6O2. Identify the comound A and B and explain the reaction involved.


Wednesday, February 22, 2023

Important Distinguish between organic pairs

Class XII Chemistry Distinguish between organic pairs

Lucas test

Used to distinguish primary, secondary and tertiary alcohols.

Lucas reagent – equimolar mixture of conc. HCl and anhydrous ZnCl2

Alcohol is treated with Lucas reagent, turbidity due to the formation of insoluble alkyl chloride is observed.

If the turbidity appears immediately, the alcohol is tertiary.

If the turbidity appears in about five minutes, the alcohol is secondary.

A primary alcohol does not react with Lucas reagent at room temperature and hence no turbidity is formed.

Iodoform test

Iodoform test is a test for the CH3-CO group and is characteristic for alcohols and methyl ketones

Reagents- compounds containing CH3-CO or CH3-CH(OH) group, sodium hydroxide and iodine 

Product formed- iodoform,(CHI3) which is a pale yellow in colour


Tollens’ test

Given by aldehydes and formic acid.

Tollen’s reagent- ammoniacal silver nitrate solution 

The aldehydes are oxidised to corresponding carboxylate anion.The reaction occurs in alkaline medium.

A bright silver mirror is produced due to the formation of silver metal.

Also known as Silver mirror test.

Fehling’s test

Fehling reagent comprises of two solutions,

Fehling solution A and Fehling solution B

Fehling solution A is aqueous copper sulphate 

Fehling solution B is alkaline sodium potassium tartarate (Rochelle salt).

On heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained

this test is not given by benzaldehyde

Azo dye test

It involves the reaction of any aromatic primary amine with nitrous acid followed by treatment with an alkaline solution of 2-naphthol, when a brilliant yellow , orange or red coloured dye is obtained.

Used to distinguish between aliphatic primary amine and aromatic primary amine.

Hinsberg’s test

Used for distinguishing primary, secondary and tertiary amines.

Hinsberg’s reagent- benzenesulphonyl chloride (C6H5SO2Cl)

In this test amine is shaken with Hinsberg’s reagent in presence of excess of aqueous KOH solution.

A primary amine gives a clear solution which on acidification gives an insoluble N-alkylbenzenesulphonamide.10

A secondary amine gives an insoluble N,N-dialkylbenzenesulphonamide which remains unaffected on addition of acid.

A tertiary amine does not react at all. Therefore, it remains insoluble in the alkaline solution but dissolves on acidification to give a clear solution.

Carbylamine test-

only Primary aliphatic and aromatic amines are treated with chloroform and KOH . they produce isocynide (RNC) compound having foul smells.

Sodium bicarbonate test-

Carboxylic acids react with NaHCO3 with Brisk effeverscene and librates CO2 gas passed lime water becomes milky due to formation of calcium carbonates.futher more passed through it. its milkiness disappears due to formation of calcium bicarbonate 

while phenol being acidic it does not give this test because phenol is weak acid than carboxylic acids(carboxylic acids having most stable resonating structures)

Ferric chloride tests FeCl3.- phenol reacts with FeCl3 ,forms purple/violet coloured complex 

The given organic samples are teated with ferric chlorides If the sample turns to red, green, purple, or blue colouration then it indicates the presence of phenols and its derivatives.

Ferric chloride will not react with any aliphatic alcohol



Differentiate 

1. Phenol and benzoic acid

2. Butan-1-ol and 2-methylpropan-2ol

3. CH3CH2NH2 and(CH3)2NH

4. Ethyl amine and diethyl amine

5. Propan-1-ol and propan-2-ol

6. Methyl ethanoate and ethyl ethanoate

7. Aniline and N-ethyl aniline

8.  N methyl propane-2 amine and N -ethyl-N methyl ethanmine

9. Aniline and benzylamine

10. Ethylamine and Aniline



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CARBON AND ITS COMPOUNDS-Functional Group

1. Name the functional groups present in the following compounds. a)     CH 3  – CO –CH 2 –CH 2  –CH 2 –CH 3 b)    CH...