SOLVED PROBLEMS

 

Board Type Questions

Problem 1:  

Write chemical reactions to effect the following transformation.

(a) Butan-1-ol to butanoic acid

(b) Benzyl alcohol to phenylacetic acid

(c) Bromobenzene to benzoic acid

(d) p-methylacetophenone to benzene-1, 4-dicarboxylic acid

 

Solution:

 793_Chemical compounds.JPG

 

 

Problem 2:

Arrange the following compounds in increasing order of their boiling points:

acetic acid, methyl formate, acetamide, propan-1-ol

 

Solution:

(i) Out of all these compounds methyl formate does not undergo H-bonding therefore, its boiling point is the lowest.

 

(ii) Amongst the remaining three compounds intramolecular H-bonding is most extensive in acetamide, followed by acetic acid and least in propan-1-ol. Therefore, their boiling points decrease in the same order i.e. acetamide>acetic acid>propan-1-ol. Thus, the overall increasing order of their boiling points is methyl formate<propan-1-ol<acetic acid<acetamide.

Problem 3:

          2363_Chemical compound 60.JPG

 

 

Solution:

 

        2045_Chemical compound 61.JPG

 

 

Problem 4:

A dicarboxylic acid (A), C4H6O4, gave a compound (B), C6H10O4 upon treatment with excess of methanol and a trace of H2SO4. Subsequent treatment of (B) with LiAlH4 followed by usual work up gave C, 
C4H10­O2. Heating of A yielded D, C4H4O3. Assign structures to A, B, C and D.

 

Solution:

 

          1240_Chemical compound 62.JPG

 

 

 

IIT Level Questions

 

Problem 5:

Two organic compounds A and B have vapour densities 15 and 30 respectively. (A) reduces Fehling solution, but does not react with Na2CO3. (B) does not reduce Fehling solution but gives effervescences with Na2CO3. B when treated with a concentrated solution of base followed by prolonged heating gives a compound C(C3H6O). Identify A, B and C.

 

Solution:

 

2176_Chemical compound 63.JPG

Problem 6:

An organic acid A, C3H4O3 is catalytically reduced in presence of ammonia to give B, C3H7NO2. B reacts with acetyl chloride, hydrochloric acid and alcohols. It can also react with nitrous acid to give another compound C, C3H6O3, along with the evolution of nitrogen. What are A, B and C. Give reasons?

 

Solution:

Compound A is acid having one –COOH group only, the remaining part C2H3O can be

1861_Chemical compound 64.JPG

on catalytic reduction keto group is converted into secondary alcohol which with ammonia will give amino acid i.e.,

         O                                        OH                                  NH2
        ||              reduction              ||               HNH2             ||
CH3 – C – COOH ——————→ CH3 – CH – COOH ————→ CH3 – CH – COOH
                                                                    –H2O
 

with nitrous acid, B, react to give

         NH2                                       OH
          |                                          |
CH3 – CH – COOH + HNO2 ——→ CH3 – CH – COOH + N2↑ + H2O
         (B)                                       (C)


Problem 7:

Compound A (C6H12O2) on reduction yields 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 oxidised further to give F which was found to be a monobasic acid (mw = 60). Deduce the structures of A to E.

Solution:

2255_Chemical compound 65.JPG

From the molecular weight data F comes out to be CH3COOH. If we work back B is CH3CH2­OH and D is CH3CHO.

                OH                                      H2/Pt
CH3CHO ————→ CH3 – HC = CH – CHO —————→ CH3CH2CH2CH2OH
    (D)        Δ                    (E)                                       (C)

 

From here we can deduce that reduction of A with LiAlH4 gives two alcohols. So A must be an ester. Hence (A) comes out to be CH3CH2CH2COOC2H5.

 

Problem 8:

An organic compound (A) on treatment with acetic acid in the presence of sulfuric acid produces an ester (B). (A) on mild oxidation gives (C). (C) with concentrated KOH followed by acidification with dil. HCl regenerates (A) and produces (D). (D) with phosphorous pentachloride followed by reaction with ammonia gives (E). (E) on dehydration produces hydrocyanic acid. Identify compounds A, B, C, D and E.

 

Solution:

                                                       O
        –H2O                                        ||
E ——————→ HCN So (E) must be H — C — NH2

      

 

          O                                    O                                 O
        ||               PCl5               ||               NH3              ||
H — C — OH ————→ H — C — Cl ————→ H — C — NH2
      (D)                                                                       (E)

 

(A) reacting with acetic acid is giving an ester (B). So (A) must be an alcohol. (A) alcohol on mild oxidation gives (C). Hence (C) is an aldehyde. (C) with conc. KOH followed by acidification is yielding (A) alcohol and  H — C — OH (D). So this reaction is cannizzaro reaction. So A must be CH3OH.

                          O
                         ||
CH3OH       CH3 — C — O — CH3         HCHO        HCOOH       HCONH2
  (A)                   (B)                        (C)             (D)             (E)

Problem 9:

An aromatic compound A on treatment with CHCl3 and KOH gives B & C, both of which, in turn give the same compound D when distilled with Zn dust. Oxidation of D yields E of formula C7H6O2. The sodium salt of E on heating with soda lime gives F which can also be obtained by distilling A with Zn dust. Identify A, B, C, D, E and F.

Solution:

Molecular formula of (E) is C7H6O2 and reaction of its sodium salt with soda lime (decarboxylation) to form (F) indicates that (E) and (F) should be C6H5COOH and C6H6 respectively. Since (F) is also obtained from (A) by reaction with Zn dust, it indicates that (A) should be phenol. Nature of (A) as phenol is confirmed by the fact that it explains all the given reactions.

 

2155_Chemical compound 66.JPG

 

 Problem 10:

Five isomeric para – disubstituted aromatic compounds A to E with molecular formula C8H8O2 were given for identification. Based on the following observations, give structures of the compounds?

(i) Both A and B form a silver mirror with Tollen’s reagents; B gives a positive test with FeCl3 solution also.

(ii) C gives positive iodoform test.

(iii) D is readily extracted in NaHCO3 solution.

(iv) E on acid hydrolysis gives 1, 4 dihydroxy benzene.

 

Solution:

 

(A) and (B) gives tollen’s test. Hence must contain aldehyde group. But (B) shows phenolic test also. Hence (A) & (B) are

 

 

1690_Chemical compound 67.JPG

(C) shows Iodoform test show it must contain   group. Therefore (C) is

1810_Chemical compound 68.JPG

 

(D) shows presence of carboxylic group. Hence (D) and (E) are 

 

275_Chemical compound 69.JPG

 

 

Problem 11:

 An organic compound A, C8H4O3 in dry benzene in the presence of anhydrous AlCl3gives compound B. The compound B on treatment with PCl5, followed by reaction with H2/Pd (BaSO4) gives compound C, which on reaction with hydrazine gives cyclised compound D (C14H10N2). Identify A, B, C and D. 

 
Solution:

 Nature of reagents in the conversion of A to B indicates that the reaction must be Friedal – crafts reaction.

 

1767_Chemical compound 70.JPG


Problem 12:

An organic compound A, C6H10O on reaction with CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1- acetyl cyclopentene. The compound B on reaction with HBr gives compound D. Write the structures of A, B, C and D. 

Solution:

As B undergoes ozonolysis to form C, if must have a double bond. C in the presence of base gives a, b unsaturated ketone, so it is aldol condensation.

 

 

283_Chemical compound 71.JPG

 

 Problem 13 

 

A phenolic compound (A), C7H6O2 on mild oxidation gives a highly volatile oil (B). A forms (C) on reaction with dimethyl sulphate in alkali. Oxidation of (C) with hot KMnO4 gives (D), the silver salt of which reacts with bromine water followed by heating gives (E) containing about 72% bromine. Give structures A to E.

 

Sol.

 

765_Chemical compound 72.JPG

But it must be ortho because on oxidation, it gives a highly volatile oil indicating that there is intramolecular hydrogen bonding.

2359_Chemical compound 73.JPG

% of bromine in E = 240×100/345 = 69.56%

 

Problem 14
 

An aromatic compound (A) gave a mixture of two isomeric products B and C on reaction with NH2OH. C rearranged to D (C8H9NO) on heating with H2SO4. D on hydrolysis produced E and F. A was oxidised with per benzoic acid to G. Hydrolysis of G gave H and E. An anhydride of E and its sodium salt on condensation with PHCHO produced cinnamic acid. H on reaction with phthalic anhydride in H2SO4 gave phenolphthalein. Suggest structures for A to H.

 

Sol.

 

156_Chemical compound 74.JPG

Anhydride of (E)  (Perkin Reaction)

 

Formation of cinnamic acid proves that the anhydride is acetic anhydride. Hence E is acetic acid.

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