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.

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

 JAWAHAR NAVODAYA VIDYALAYA RATIBAD BHOPAL

UNIT 12-ALDEHYDE KETONE AND CARBOXYLIC ACID

NOMENCLATURE

(1) Draw the structure of

i. 4-chloropentan-2-one.  (iii) 2-methylbutanal. (iv) 3-methylpentanal.

ii. p-Methylbenzaldehyde (v) 1-phenyl Propan-1-one

(2) Write the name of 

NAME REACTION

(1) Rosenmund reduction (02) Wolf kishner reduction (03) Etard reaction

(4) Aldol condensation (05) Cross aldol condensation (06) Ozonolysis

(7) Cannizzaro's reaction (08) Clemmensen reduction (09)   Decrboxylation

(10) Gatterman koach reaction (11) Kolbe reaction (12) Easterification

(13) Hell volhard zelinsky (HVZ)

CONVERSION

(1) Acetophenone to Benzoic acid (08) Benzoic acid to Benzaldehyde

(2) Benzoic acid from ethylbenzene (09) But-2-enal from ethanol

(3) Benzyl chloride to phenyl ethanoic acid (10) Butanoic acid from butanol

(4) Ethanoic acid to 2-Hydroxyethanoic acid (11) Propanone to propene

(5) Ethyne to Ethanal (12) Benzaldehyde from Phenol

(6) Benzaldehyde from toluene (13) Ethanol to acetone

(7) Benzene to acetophenone

REASONING

(1) Cl–CH2COOH is a stronger acid than CH3COOH. Why?

(2) Carboxylic acids do not give reactions of carbonyl group. Why?

(3) HCHO is more reactive than CH3–CHO towards addition of HCN. Why?

(4) pKa of NO2–CH2–COOH is lower than that of CH3–COOH. Why?

(5) The boiling points of aldehydes and ketones are lower than of the corresponding acids. Why?

(6) What is Tollens’ reagent? Write one usefulness of this reagent.

(7) Write the composition of Fehling solution?

(8) Aldehydes are more reactive than ketones towards nucleophilic reagents.

(9) Aldehydes and Ketones have lower boiling points than corresponding alcohols. Why?

DISTINGUISH

(1) Benzoic acid & ethyl benzoate (07) Benzaldehyde & acetophenone

(2) Benzoic acid & Phenol (08) Propanal & Propanone

(3) Butanal & Butan-2-one (09) Pentan-2-one & Pentan-3-one

(4) Ethanal and Propanal (10) Benzaldehyde & benzophenone

(5) Propanal and butanone (11) Benzaldehyde & benzoic acid

(6) Out of CH3CH2–CO–CH2–CH3 and CH3CH2–CH2–CO–CH3, which gives iodoform test?

COMPLETE FOLLOWING EQUATION

(1) Write the products formed when CH3CHO reacts with the following reagents

(i) HCN (ii) H2N-OH (iii) CH3CHO in the presence of dilute NaOH

(2) Write the structures of the products when Butan-2-ol reacts with the following:

(3) (i) CrO3 (ii) SOCl2

BRAIN STORMING QUESTION

1. An organic compound A, having the formula, C3H8O, on treatment with copper at 573 K, gives B. B does not reduce Fehling’s solution but gives a yellow precipitate of the compound C with I2/NaOH. Deduce the structure of A, B and C.

2. An organic compound (A) which has characteristic odour, on treatment with NaOH forms two compounds (B) and (C). Compound (B) has the molecular formula C7H8O which on oxidation with CrO3 gives back compound (A). Compound (C) is the sodium salt of the acid. Compound (C) when heated with soda lime yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D). Write chemical equations for all reactions taking place.

3. An organic compound with molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid. Identify the compound.

4. An organic compound (A) with molecular formula C8H8O forms an orange red precipitate with 2, 4-DNP reagent and gives yellow precipitate on heating with I2 and NaOH. It neither reduces Tollens’ reagent nor Fehling’s reagent, nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C7H6O2. Identify the compounds (A) and (B) and explain the reactions involved.

5. An organic compound with molecular formula C5H10O does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Identify the compound and write all chemical equations for the reactions.

6. An organic compound A has the molecular formula C8H16O2. It gets hydrolysed with dilute sulphuric acid and gives a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid also produced B. C on dehydration reaction gives but-1-ene. Write equations for the reactions involved.

7. An organic compound with molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-benzene-di-carboxylic acid. Identify the compound.

8. A ketone A(C4H8O), which undergoes a haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives a compound C which forms monozonide D. D on hydrolysis in presence of zinc dust gives only acetaldehyde E. Identify A, B, C, D and E. Write the reactions involved.

An organic compound A(C3H6O) is resistant to oxidation but forms compound B(C3H8O) on reduction. B reacts with HBr to form the compound C. C with Mg forms Grignard reagent D which reacts with A to form a product which on hydrolysis gives E. Identify A and E.