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>> Solved Problems on Chemical Bonding Part I
Board Type Questions
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
Arrange the following compounds in increasing order of their boiling points:
acetic acid, methyl formate, acetamide, propan-1-ol
(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.
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.
IIT Level Questions
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.
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?
Compound A is acid having one –COOH group only, the remaining part C2H3O can be
on catalytic reduction keto group is converted into secondary alcohol which with ammonia will give amino acid i.e.,
CH3 – C – COOH ——————→ CH3 – CH – COOH ————→ CH3 – CH – COOH
with nitrous acid, B, react to give
CH3 – CH – COOH + HNO2 ——→ CH3 – CH – COOH + N2↑ + H2O
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.
From the molecular weight data F comes out to be CH3COOH. If we work back B is CH3CH2OH and D is CH3CHO.
CH3CHO ————→ CH3 – HC = CH – CHO —————→ CH3CH2CH2CH2OH
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.
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.
E ——————→ HCN So (E) must be H — C — NH2
H — C — OH ————→ H — C — Cl ————→ H — C — NH2
(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.
CH3OH CH3 — C — O — CH3 HCHO HCOOH HCONH2
(A) (B) (C) (D) (E)
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.
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.
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.
(A) and (B) gives tollen’s test. Hence must contain aldehyde group. But (B) shows phenolic test also. Hence (A) & (B) are
(C) shows Iodoform test show it must contain group. Therefore (C) is
(D) shows presence of carboxylic group. Hence (D) and (E) are
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.
Nature of reagents in the conversion of A to B indicates that the reaction must be Friedal – crafts reaction.
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.
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.
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.
But it must be ortho because on oxidation, it gives a highly volatile oil indicating that there is intramolecular hydrogen bonding.
% of bromine in E = 240×100/345 = 69.56%
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.
Anhydride of (E) (Perkin Reaction)
Formation of cinnamic acid proves that the anhydride is acetic anhydride. Hence E is acetic acid.