JEE Advanced (Subjective Type Questions): Aldehydes, Ketones & Carboxylic Acids- 2 | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.16. An organic compound (A) on treatment with acetic acid in the presence of sulphuric acid produces an ester (B), (A) on mild oxidation gives (C), (C) with 50% potassium hydroxide followed by acidification with dilute hydrochloric acid generates (A) and (D), (D) with phosphorus pentachloride followed by reaction with ammonia gives (E), (E) on dehydration produces hydrocyanic acid. Identify the compounds A, B, C, D and E.              (1987 - 5 Marks)

Ans. 

Solution. TIPS/Formulae :

The above reactions lead to following conclusions.

(i) Reaction of A with CH₃COOH in presence of H₂SO₄ to form ester B indicates that A is an alcohol.
(ii) Reaction of C with 50% KOH followed by acidification to give alcohol A and the compound D seems to be the Cannizzaro reaction. Hence C must be an aldehyde and D must be an acid. The nature of C as aldehyde is again in consistent with the fact that it is obtained by the mild oxidation of A which has been established as an alcohol.
(iii) Structure of acid D is established by its given facts.

Formation of HCN by the dehydration of E establishes that E is HCONH₂ and hence D is HCOOH.

(iv) Thus the alcohol A produced along with HCOOH during Cannizzaro reaciton of C must be CH₃OH and hence C must be HCHO.

Thus the various compounds are as below :

Q.17. Complete the following reactions :

       (1988 - 1 Marks)

    (1988 - 1 Marks)
        (1988 - 1 Marks)

                 (1988 - 1 Marks)

Solution. 

(iv)  (i) fuse with NaOH 

Q.18. A hydrocarbon A (molecular formula C₅H₁₀) yields 2-methylbutane on catalytic hydrogenation. A adds HBr (in accordance with Markownikoff’s rule) to form a compound B which on reaction with silver hydroxide forms an alcohol C, C₅H₁₂O. Alcohol C on oxidation gives a ketone D. Deduce the structures of A, B, C and D and show the reactions involved.             (1988 - 5 Marks)

Ans. 

Solution. TIPS/Formulae :

For this type of problem, students are advised to summarise the whole problem in the form of reactions.

Let us draw some conclusions from the above set of reactions.

(i) The molecular formula C₅H₁₀ (CnH₂n) for A indicates that it is an alkene having one double bond.
(ii) Since the alcohol C on oxidation gives a ketone D, C must be a secondary alcohol and hence B must be a secondary bromide.
(iii) The structure of 2-methylbutane, the hydrogenated product of A, indicates that the secondary bromide must have following structure.

(iv) Thus the corresponding olefin A must have structure A which on Markownikoff addition of HBr gives the bromide B, the other possible alkene A' will not give B when HBr is addd on it according to Markownikoff rule.

Thus the reaction involved can be represented as below:

Q.19. A ketone ‘A’ wh ich undergoes haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives compound C, which forms monoozonide D, D on hydrolysis in presence of zinc dust gives only acetaldehyde. Identify A, B and C. Write down the reactions involved.  (1989 - 4 Marks)

Ans. 

Solution. The compound A, a ketone, undergoes haloform reaction.

Thus, it must contain CH₃CO group.

The compound C gives mono-ozonide D, which shows that the compound C contains a double bond. Since the hydrolysis of D gives only acetaldehyde, the compound C would be an alkene having four carbon atoms,

i.e. CH₃ – CH = CH – CH₃ (butene-2).

The compound B is obtained by the reduction of compound A (which contains CH₃CO group). Hence, the compound B would be an alcohol, which on heating with H₂SO₄ gives (C). Hence B and A would be

The reactions involved :

Q.20. The sodium salt of a carboxylic acid, A, was produced by passing a gas, B, into an aqueous solution of caustic alkali at an elevated temperature and pressure. A, on heating in presence of sodium hydroxide followed by treatment with sulphuric acid gave a dibasic acid, C. A sample of 0.4 g of acid C, on combustion gave 0.08 g of water and 0.39 g of carbon dioxide. The silver salt of the acid C weighing 1.0 g on ignition yielded 0.71 g of silver as residue. Identify A, B and C.  (1990 - 5 Marks)

Ans. TIPS/Formulae : 

The given set of reactions can be represented as below :

Calculation of molecular formula of C

By usual method, empirical formula of acid C = CHO₂

Mol. wt. of acid C = 45 × 2 = 90
∴ Mol. formula of C = C₂H₂O₄
Since it is dicarboxylic acid, it must have two –COOH groups.

Going back, compound C must be produced from sodium oxalate which in turn is produced from sodium formate.
Hence A is formic acid and B is CO₂. Thus the complete series of reactions can be written as below.

Q.21. Compound A (C₆H₁₂O₂) on reduction with LiAlH4 yielded two compounds B and C. The compound B on oxidation gave D, which on treatment with aqueous alkali and subsequent heating furnished E. The latter on catalytic hydrogenation gave C. The compound D was oxidized further to give F which was found to be a monobasic acid (molecular weight = 60.0). Deduce the structures of A, B, C, D and E.              (1990 - 4 Marks)

Ans.  C₃H₇COOC₂H₅, C₂H₅OH, C₄H₉OH, CH₃CHO

Solution. TIPS/FORMULAE :

Molecular weight of the monobasic acid (RCOOH) indicates that the R– should be CH₃– i.e., acid F should be acetic acid (CH₃COOH, mol. wt. 15+45). Thus compound D must be acetaldehyde CH₃CHO, and compound B which on oxidation gives CH₃CHO must be ethanol, CH₃CH₂OH.
Acetaldehyde (D) on treating with aqueous alkali will undergo aldol condensation.

Nature of A. Thus it is evident that reduction of A with LiAlH4 gives two alcohols; B (ethanol) and C (butanol). Hence A must be an ester i.e., ethyl butanoate (CH₃CH₂CH₂COOC₂H₅).

Q.22. An organic compound containing C, H and O exists in two isomeric forms A and B. An amount of 0.108 g of one of the isomers gives on combustion 0.308 g of CO₂ and 0.072 g of H₂O. A is insoluble in NaOH and NaHCO₃ while B is soluble in NaOH. A reacts with conc. HI to give compounds C and D. C can be separated from D by ethanolic AgNO₃ solution and D is soluble in NaOH. B reacts readily with bromine water to give compound E of molecular formula, C₇H₅OBr₃.
 Identify, A, B, C, D and E with justification and give their structures.             (1991 - 6 Marks)

Ans. 

Solution. Empirical formula of A and B.

∴ Empirical formula of A and B = C7H₈O
Nature of (A) : Since A is insoluble in NaOH and NaHCO₃, it can’t have –OH and –COOH groups. Further the reaction of A with conc. HI to give compounds C and D separable by means of ammonical AgNO₃ and solubility of D in NaOH indicates that C and D are alkyl halide and phenol respectively. Thus A is an ether i.e. it is C₆H₅.O.CH₃ which explains all the given reactions.

Nature of (B) : Solubility of B (C₇H₈O) in NaOH indicates that it is a phenol which is further confirmed by its reaction with bromine water to give compound E of molecular formula, C₇H₅OBr₃. Further bromination of B to give tribromo product indicates that it is m-cresol.

Q.23. (i)     

Identify C, D and E.                                                                       (1991 - 2 Marks)

Identify F, G and H.                                         (1991 - 2 Marks)

Solution.

Q.24. Compound ‘X’, containing chlorine on treatment with strong ammonia gives a solid ‘Y’ which is free from chlorine. ‘Y’ analysed as C = 49.31%, H = 9.59% and N = 19.18% and reacts with Br₂ and caustic soda to give a basic compound ‘Z’. ‘Z’ reacts with HNO₂ to give ethanol. Suggest structures for ‘X’, ‘Y’ and ‘Z’.      (1992 - 1 Mark)

Ans. 

Solution. For empirical formula of (Y)

∴ Empirical formula of (Y) is C₃H₇NO.

(Y) reacts with Br₂ and NaOH to give (Z) and (Z) reacts with HNO₂ to give ethanol and thus (Y) seems to have —CONH₂ group.

Y is formed from (X) having Cl on treatment with NH₃ and so (X) is CH₃CH₂COCl i.e. propanoyl chloride.

Q.25. An organic compound ‘A’ on treatment with ethyl alcohol gives a carboxylic acid ‘B’ and compound ‘C’. Hydrolysis of ‘C’ under acidic conditions gives ‘B’ and ‘D’. Oxidation of ‘D’ with KMnO₄ also gives ‘B’. ‘B’ on heating with Ca(OH)₂ gives ‘E’ (molecular formula, C₃H₆O). ‘E’ does not give Tollent’s test and does not reduce Fehling’s solution but forms a 2,4-dinitrophenylhydrazone. Identify ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’.An organic compound ‘A’ on treatment with ethyl alcohol gives a carboxylic acid ‘B’ and compound ‘C’. Hydrolysis of ‘C’ under acidic conditions gives ‘B’ and ‘D’. Oxidation of ‘D’ with KMnO₄ also gives ‘B’. ‘B’ on heating with Ca(OH)₂ gives ‘E’ (molecular formula, C₃H₆O). ‘E’ does not give Tollent’s test and does not reduce Fehling’s solution but forms a 2,4-dinitrophenylhydrazone. Identify ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’.                            (1992 - 3 Marks)

Ans. 

Solution. TIPS/Formulae :

(i) Since E (C₃H₆O) forms a 2, 4-dinitrophenylhydrazone but does not reduce Tollen’s reagent and Fehling solution, it must be a ketone, CH₃.CO.CH₃.
(ii) The compound E (established as ketone) is obtained by heating compound B with Ca(OH)₂, B must be CH₃COOH.
(iii) Compound B is obtained by the oxidation of D, the latter must be ethyl alcohol, C₂H₅OH and hence C must be ethyl acetate, CH₃COOC₂H₅.
(iv) Since compound A when treated with ethyl alcohol gives acetic acid (B) and ethyl acetate (C), it must be acetic anhydride.

Q.26. Arrange the following in increasing order of expected enol content      (1992 - 1 Mark)

CH₃COCH₂CHO₃ CH₃COCH₃, CH₃CHO, CH₃COCH₂COCH₃

Ans.  CH₃CHO < CH₃COCH₃ < CH₃COCH₂CHO < CH₃COCH₂COCH₃

Solution. CH₃CHO < CH₃COCH₃ < CH₃COCH₂COCH₃ < CH₃COCH₂CHO

Q.27. In the following reactions identify the compounds A, B, C and D.                (1994 - 1 × 4 = 4 Marks)

Ans. 

Solution. 

Q.28. When gas A is passed through dry KOH at low temperature, a deep red coloured compound B and a gas C are obtained. The gas A, on reaction with but-2-ene, followed by treatment with Zn/H₂O yields acetaldehyde. Identify A, B and C.              (1994 - 3 Marks)

Ans. 

Solution. TIPS/Formulae :

The reaction of gas (A) with but-2-ene followed by treatment with Zn/H₂O gives CH₃CHO. This shows that the gas (A) is ozone (O₃).

Reaction of ozone with KOH.

Q.29. An organic compound A, C₈H₆, on treatment with dilute sulphuric acid containing mercuric sulphate gives a compound B, which can also be obtained from a reaction of benzene with an acid chloride in the presence of anhydrous aluminium chloride. The compound B, when treated with iodine in aqueous KOH, yields C and a yellow compound D. Identify A, B, C and D with justification. Show how B is formed from A.                       (1994 - 3 Marks)

Ans. 

Solution. TIPS/Formulae :

(i) Formation of (B) from benzene and acid chloride in presence of anhydrous AlCl₃ (Friedel-Craft reaction) indicates that it is a ketone, C₆H₅COR.

(ii) Further th e keton e (B) reacts with alkaline iodine forming yellow compound (D) (haloform reaction). This indicates that one of the alkyl groups in ketone B is – CH3. Hence it should be C₆H₅.CO.CH,.

(iii) Since ketone (B) is also formed from the hydrocarbon C₈H₆ (A) by reaction with dil. H₂SO₄ and HgSO,, the hydrocabon (A) must have an acetylenic hydrogen atom, i.e. ≡ C – H grouping. Hence (A) must be C₆H₅C ≡ CH.

Thus compounds (A) to (D) are

Q.30. Which of the following carboxylic acids under goes decarboxylation easily? Explain briefly.               (1995 - 2 Marks)

(i) C₆H₅–CO–CH₂–COOH
 (ii) C₆H₅–CO–COOH

Solution. (i) β-Keto acids are unstable and undergo decarboxylation most readily.