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Solved Examples on Crabonyl Compounds

Question 1: Acid and acid derivatives although contain > C = O group, do not undergo the usual properties of carbonyl group explain.

Solution:  It is because of the possibility of resonance which compensates the electron deficiency of carbonyl carbon to some extent for example.

Question 2: The neutralization equivalent of an acid is 116. Fusion of sodium salt of the acid with sodalime gives a hydrocarbon of which butane monobromo substitution product is possible. What structural formula may be assigned to the acid.

Solution: The final product is an alkane derivative, this indicates the acid is mono carboxylic, RCOOH.

As neutralisation equivalent is 116, therefore, formula weight of alkyl group R is 116 – 45 (Formula weight of –COOH) = 71.

This corresponds to the pentyl group (C5H11). The formation of butane derivative suggests a branched alkyl radical containing four carbon atoms in the largest chain. Therefore, R-should be

and the acid would be

Question  3:  Compound A acidic in reaction, having the molecular formula C4H8O3, is oxidized with mild oxidizing agents to give B which is unstable, syrupy substance. It easily produces, C, C3H6O and carbon dioxide alone yield D, C4H6O2 having an acid neutralization equivalent of 104. What are A, B,C and D give their structural formulas too.

Solution: Compound A has molecular formula C4H8O3 as it is acidic in reaction, hence it should have –COOH group. The remaining group C3H7O contains oxygen. The product D, having formula C3H5COOH, an unsaturated acid is obtained by heating compound A. b-hydroxy substituted acids yield substituted acids on heating. Hence, A should be b-hydroxy substituted acid, i.e.,

B is keto substituted acid, hence, on further decomposition it will eliminate CO2 to give

Question 4: 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 

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

Question 5:  Compound A, C5H8O3, when heated with soda lime gives B which reacts with HCN to give C, C reacts with PCl5 to give D which reacts with KCN to form E. E, on alkaline hydrolysis gives a salt which is isolated and heated with soda lime to produce n-butane. A, on careful oxidation with K2Cr2O7 gives acetic acid and malonic acid. Give structural formulae of A to E.


On Oxidation only two –COOH groups can be introduced, i.e., one to each carbon undergoing C–C fission, but in the resulting products we have three –COOH groups. Hence, one –COOH group is already there in compound A, the remaining portion C4H7O–, resembles with Keto substituted alkyl group C3H4CO-

This indicates that the given organic compound A is Keto substituted acid. To assign position to Keto group in carbon chain, we know that keto acids on careful oxidation undergo C–C bond fission at a place where >C=O is situated, further Keto group is also converted into –COOH and remains with acid having small number of carbon atoms. From the above discussion it is clear that acetic acid is formed from.

CH3CO- arrangement C3H7CO – has 

structure and compound A is 

A, with soda-lime undergoes decarboxylation to give  CH3 – CO – CH2– CH3 (B)

B, being Ketone will give addition product with HCN to form

Question 6: An organic compound (A), C8H14O forms an oxime and gives positive haloform reaction. On ozonolysis it gives acetone and a compound (B), C5H8O2 (B) forms a dioxime and on subjecting to haloform reaction gives an acid (C), C4H6O4. On treatment with excess of ammonia and strong heating (C) gives a neutral compound (D), C4H5O2N (D) on distillation with Zn dust forms pyrrole. Suggest structures for (A), (B), (C) and (D). Give the IUPAC names of compounds (A) to (D).


A is (CH3)2 C = CH – CH2 – CH2 – COCH3

6 – methyl – 5 – hepten – 2 – one

B is CH3 – CO – CH2 – CH2 – CHO

4 – oxopentanal

C is HO2 CCH2 CH2 CO2 H

1, 4 – butandioic acid

 D is succinimide

Question 7:  

a)   Can the aromatic ring in benzoic acid stabilize the benzoate anion by  p-electron delocalization? Illustrate

b)   Discuss the electronic effect of the  p–NO2 group in p–NO2C6H4COO

Solution: a)   The only contributing structure I that delocalizes electron density from PhCOO has a positive charge on an O atom with only six electrons.

The contribution from this extremely high energy resonance structure is nil.

b)   There is no direct resonance interaction between –COO and –NO2. The resonance interaction of the NO2 with the ring induces some positive charge on the ring C bearing the COO II, from which a strong electron with drawing inductive effect is generated. This effect is base stabilizing and thus acid strengthening.

Question 8:    An aromatic compound A gave a mixture of two isomeric compounds 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 oxidized with perbenzoic acid to G. Hydrolysis of G gave H and E. Anhydride of E and its sodium salt on condensation with PhCHO produced  cinnamic acid. H on reaction with pthalic anhydride in H­2SO4 gave phenolphthalein. Suggest structures for the compounds A to H

Solution:  The hydrolysis products (E) and (F) suggests the structure for (D). The phenyl group in (C) which is anti — OH migrates to give  N - phenyl substituted acetamide.

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

Question 9:  A compound X, C18H20O on ozonolysis gives C10H12O, A and C8H8O2, B.  A gives the iodoform reaction and with NH2OH gives the oxime C10H13ON, C. The oxime undergoes rearrangement to give an amide when treated with PCl5 in dry ether. The amide on hydrolysis produces CH3COOH and a compound C8H11N, D. Compound D with HNO2 at 0oC gives an aromatic alcohol C8H10O, oxidation of which gives phthalic acid. B on mild oxidation gives a carboxylic acid, C8H8O3 which is degraded by HI to CH3I and p-hydroxybenzoic acid. Identify X, A to D and give the reactions involved?

Solution:  The compound A must contain CH3CO— group as it shows positive iodoform test and forms oxime.The compound D which is obtained  after the  hydrolysis of amide must contain aliphatic —NH2 group as it gives alcohol on treatment with HNO2 at 0oC. Since the oxidation of alcohol obtained gives phthalic acid, the alcoholic substituent must be at ortho position to the another substituent. Guided by this and the given molecular formulae, we derive the structures as shown in the following.


Since the amide is obtained by the rearrangement of oxime, we may formulate, this reaction as

The intermediate compound C8H83 must be an ether. The compound B must be containing >C=O group as it is the ozonolysis product of X. Guided by these, we formulate the above reactions as

Question 10:    

a)   Convert 2-chlorobutanoic acid into 3-chlorobutanoic acid

b)   Prepare malonic acid from acetic acid

c)   Prepare 4-p tolylbutanoic acid from toulene and succinic anhydride. 

Question 11:    A ketone A which undergoes a haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives compound C, which forms mono-ozonide D. The compound D on hydrolysis in presence of zinc dust gives only acetaldehyde. Identify A, B and C. Write down the reactions involved

Solution:  We are given that 

The compound A gives a haloform reaction; it must contain CH3CO group. The compound C contains a double bond as it forms mono-ozonide D. Since, the compound D on hydrolysis gives only CH3CHO, the structure of C would be

Question 12:   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. The compound B on heating with Ca(OH)2 gives E (molecular formula C3H6O). E does not give Tollen’s test and does not reduce Fehling’s solution but forms a 2, 4-dinitrophenylhydrazone. Identify A, B, C, D and E.

Solution:   The given reactions are as follows.

The compound E must be ketonic compound as it does not give Tollens test and does not reduce Fehling’s solution but forms a 2, 4-dinitrophenyl-hydrazone. Therefore, its structure would be CH3COCH3 (acetone).

Since E is obtained by heating B with Ca(OH)2, the compound B must be CH3COOH (acetic acid).

Since B is obtained by oxidation of D with KMnO4, the compound D must be an alcohol with molecular formula CH3CH2OH (ethanol).

Since B and D are obtained by acid hydrolysis of C, the compound C must be an ester CH3COOC25 (ethyl acetate).

Since the compounds B (acetic acid) and C (ethyl acetate) are obtained by treating A with ethanol, the compound A must be an anhydride (CH3CO)­2O (acetic anhydride).

The given reactions are

Question 13:    Two moles of an ester A are condensed in the presence of sodium ethoxide to give a b-keto ester, B, and ethanol. On heating in an acidic solution, B gives ethanol and a b-keto acid, C. On decarboxylation C gives 3-pentanone. Identify A, B and C with proper reasoning.

Solution: The reaction of 2 mol  of an ester giving b-keto ester and alcohol in the presence of sodium ethoxide is known as Claisen condensation.

Let the given reactions may be depicted as shown in the following.

From these reactions, it is obvious that

 R’  \equiv–– CH2CH3

 R \equiv –– CH3

Hence, the compounds A, B and C are


Question 14:    An organic compound A(C6H10) on reduction first gives B(C6H12) and finally C(C6H14). Compound A on ozonolysis followed by hydrolysis gives two aldehydes D(C2H4O) and E(C2H2O2). Oxidation of B with acidified KMnO4 gives the acid F(C4H8O2). Determine the structures of the compounds A to F with reasoning.

Solution:  Since ozonolysis of A gives two aldehydes, the compound A contains the carbon-carbon double bond. In fact, the molecule of A contains two double bonds as it is successively ozonolyzed products will remain same as in the compound A. Hence, it may be concluded that the ozonolysis products include two molecules of D(CH3CHO) and one molecule of E(OHC –– CHO). From this, we derive the structure of A as shown in the following

Question 15:    An organic compound A (C6H12O) forms an oxime but does not reduce Tollen’s reagent. A on reduction with sodium-amalgam forms an alcohol B which on dehydration forms chiefly a single alkene C. The ozonolysis of C produces D and E. The compound D reduces Tollens reagent but does not answer iodoform test. What are the structures of the above compounds? Explain the reactions.

Solution:  The compound A must be a ketone as it forms oxime but does not reduce Tollen’s reagent.

The compound D must be an aldehyde. Its structure does not include the fragment CH3CO-– as it does not answer iodoform test.

The compound E must be a ketone containing CH3CO-– fragment.

 Let the compounds D and E be RCH2CHO and R’COR” respectively, where R, R¢ and R¢¢ are all alkyl groups. From these, we get

Question 16:    A compound A reduces Fehling’s solution and gives positive silver mirror test. On warming with dilute alkali, followed by dehydration it gives a product B which also responds to both the above tests. In addition, it decolourises the colour of bromine water. On treatment with hydrogen in the presence of nickel under pressure the compound B is converted to C which does not give any one of the above three test. Molar mass of the compound C is 74g mol–1. Deduce the structures of A and B giving the chemical equations involved.

Solution: The compound A must be an aldehyde as it reduces Fehling’s solution and gives positive silver mirror test. The compound A also contains a-hydrogen atom as it undergoes aldol condensation, the dehydration of which gives B. The latter contains unsaturation as bromine water is decolourised. Let the structure of A be RCH2CHO. The reactions involved are

Question 17:    When R glyceraldehyde, CH2OHCH(OH)CHO is treated with cyanide and the resulting product is hydrolysed, two monocarboxylic acids are formed. These acids are identical with the acids obtained by oxidation with Br2 - water of (-) threose and (-) erythrose. Assign single structure to (-) erythose and  to (-) threose.


Note: The erythro isomer is the one that is convertible (in principle at least) into a meso structure, whereas the threo isomer is convertible into a racemic modification. The names of these compounds are the basis for designations erythro and threo acid to specify certain configurations of compounds containing two chiral carbons.

Question 18:    How starting from (I and II), one can get (III)), 4, 4-dimethyl-2-cyclohexanone)

Solution:  The first reaction is the Micheal addition of the enolate anion to methyl vinyl ketone followed by intra molecular condensation.

Question 19:  Treatment of compound (A) C8H10O with chromic acid pyridine gives (B), C8H8O. Treatment of (B) with two equivalents of Br2 yields (C), C8H6OBr2, which on treatment with caustic soda followed by acidification gives a compound (D) C8H8O3. The latter liberates CO2 on treatment with NaHCO3 and is resolvable. Write structures for (A), (B), (C) and (D). Give mechanism of formation of (D) from (C).


Question 20:   An ester (A) is condensed in the presence of sodium methoxide to give a b-keto ester (B) and Methanol. On mild hydrolysis with cold conc. HCl, (B) gives methanol and 3-oxo-acid(C). (C) underwent readily decarboxylation to give cyclopentanone.

a)   Identify (A), (B) and (C)

b)   Name the reaction involved in conversion of (A) to (B)

c)   Give the mechanism of decarboxylation

b)  Diekmann condensation (intramolecular Claisen ester condensation)

c)  Mechanism of decarboxylation


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