Class 12 - Chemistry - Aldehydes Ketones Acids

Question 12.1.

What is meant by the following terms? Give an example of the reaction in each case.

  • Cyanohydrin
  • Acetal
  • Semicarbazone
  • Aldol
  • Hemiacetal
  • Oxime
  • Ketal
  • Imine
  • 2,4-DNP-derivative
  • Schiff’s base


  • Cyanohydrin

Cyanohydrins are organic compounds having the formula RR′C (OH) CN, where R and R′ can be alkyl or aryl groups.

Aldehydes and ketones react with hydrogen cyanide (HCN) in the presence of excess sodium cyanide (NaCN) as a catalyst to field cyanohydrin.

These reactions are known as cyanohydrin reactions.

RR’C=O + HCN --> RR’C (OH) CN   (In presence of NaCN)

Ketone                  Cyanohydrin

  • Acetal

Acetals are gem - dialkoxy alkanes within which 2 alkoxy teams groups attached to the terminal atom.

One bond is connected to associate degree alkyl whereas the opposite is connected to the hydrogen atom.

When aldehydes are treated with 2 equivalents of a monohydric alcohol within the presence of dry HCl gas,

hemiacetals are produced which are further reacted with one more molecule to alcohol to yield acetal as shown below:


  • Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction
  • between a ketone or aldehyde and semicarbazide.

Semicarbazones are useful for identification and characterization of aldehydes and ketones.


  • Aldol:

An aldol is a β-hydroxy organic compound. It's produced by the by the condensation reaction of

2 molecules of an equivalent or one molecule

every of 2 totally different aldehydes or ketones within the presence of a base.

The reaction is shown as below:


  • Hemiacetal

Hemiacetals are α−alkoxyalcohols.

Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas to from alkoxy alcohol, known as hemiacetal.

The following reaction shows the formation of hemiacetal:


  • Oxime:

Oximes are a class of organic compounds having the general formula RR′CNOH, where R is an organic side chain and R′ is

either hydrogen or an organic side chain.

If R′ is H, then it is known as aldoxime and if R′ is an organic side chain, it is known as ketoxime.


  • Ketals:

Ketals are gem−dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain.

The other two bonds of the carbon atom are connected to two alkyl groups.


General structure of Ketal

Ketones react with ethylene glycol in the presence of dry HCl gas to give a cyclic product known as ethylene glycol ketals.


  • Imine:

Imines are chemical compounds containing a carbon nitrogen double bond.


General structure of Imine

Imines are produced when aldehydes and ketones react with ammonia and its derivatives.


  • 2,4-DNP- Derivative

2, 4−dinitrophenylhydragones are 2, 4−DNP−derivatives, which are produced when aldehydes or ketones react

with 2, 4−dinitrophenylhydrazine in a weakly acidic medium.


To identify and characterize aldehydes and ketones, 2, 4−DNP derivatives are used.

  • Schiff’s base:

Schiff’s base (or azomethine) is a chemical compound containing a carbon-nitrogen double bond

with the nitrogen atom connected to an aryl or alkyl group-but not hydrogen.

They have the general formula R1R2C = NR3. Hence, it is an imine.

It is named after a scientist, Hugo Schiff.


Aldehydes and ketones on treatment with primary aliphatic or aromatic amines in the presence of trace of an acid yields a Schiff’s base.



Question 12.2.

Name the following compounds according to IUPAC system of nomenclature:

  • (CH3)3CCH2COOH
  • OHCC6H4CHO-p


  • CH3CH (CH3) CH2CH2CHO = 4-methylpentanal
  • CH3CH2COCH(C2H5)CH2CH2Cl =6-Chloro-4-ethylhexan-3-one
  • CH3CH=CHCHO = But-2-en-1-al
  • CH3COCH2COCH3 = Pentane-2,4-dione
  • CH3CH(CH3)CH2C(CH3)2COCH =3,3,5-Trimethylhexan-2-one
  • (CH3)3CCH2COOH = 3,3-Dimethylbutanoic acid
  • OHCC6H4CHO-p = Benzene-1,4-dicarbaldehyde


Question 12.3.

Draw the structures of the following compounds.

  • 3-Methylbutanal
  • p-Nitropropiophenone
  • p-Methylbenzaldehyde
  • 4-Methylpent-3-en-2-one
  • 4-Chloropentan-2-one
  • 3-Bromo-4-phenylpentanoic acid
  • p,p’-Dihydroxybenzophenone
  • Hex-2-en-4-ynoic acid.


  • 3-MethylButanal


  • p-Nitropropiophenone


  • p-Methylbenzaldehyde


  • 4-Methylpent-3-en-2-one


  • 4-Chloropentan-2-one


  • 3-Bromo-4-phenylpentanoic acid

Class_12_Chemistry_Aldehydes_Ketones_&Carboxylic_Acids_3-Bromo-4-phenylpentanoic acid

  • p,p’-Dihydroxybenzophenone


  • Hex-2-en-4-ynoic acid

Class_12_Chemistry_Aldehydes_Ketones_&Carboxylic_Acids_Hex-2-en-4-ynoic acid


Question 12.4.

Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

  • CH3CO(CH2)4CH3
  • CH3(CH2)5CHO
  • Ph-CH=CH-CHO
  • Class_12_Chemistry_Aldehydes_Ketones_&Carboxylic_Acids_CHO
  • PhCOPh


(i) CH3CO (CH2)4CH3

IUPAC name: Heptan-2-one

Common name: Methyl n-propyl ketone


IUPAC name: 4-Bromo-2-methylhaxanal

Common name: (γ-Bromo-α-methyl-caproaldehyde)

(iii) CH3 (CH2)5CHO

IUPAC name: Heptanal

(iv) Ph-CH=CH-CHO

IUPAC name: 3-phenylprop-2-enal

Common name: β-Pheynolacrolein


  • Class_12_Chemistry_Aldehydes_Ketones_&Carboxylic_Acids_CHO

IUPAC name: Cyclopentanecarbaldehyde


IUPAC name: Diphenylmethanone

Common name: Benzophenone


Question 12.5.

Draw structures of the following derivatives.

(i) The 2, 4-dinitrophenylhydrazone of benzaldehyde

(ii) Cyclopropanone oxime

(iii) Acetaldehydedimethylacetal

(iv) The semicarbazone of cyclobutanone

(v) The ethylene ketal of hexan-3-one

(vi) The methyl hemiacetal of formaldehyde















Question 12.6.

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.

  • PhMgBr and then H3O+
  • Tollens’ reagent
  • Semicarbazide and weak acid
  • Excess ethanol and acid
  • Zinc amalgam and dilute hydrochloric acid













Question 12.7.

Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither?

Write the structures of the expected products of aldol condensation and Cannizzaro reaction.

(i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde

(iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone

(vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2, 2-Dimethylbutanal


Aldehydes and ketones having at least one α-hydrogen undergo aldol condensation.

The compounds (ii) 2−methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone, and

(vii) phenylacetaldehyde contain one or more α-hydrogen atoms. T

herefore, these undergo aldol condensation.

Aldehydes having no α-hydrogen atoms undergo Cannizzaro reactions.

The compounds (i) Methanal, (iii) Benzaldehyde, and

(ix) 2, 2-dimethylbutanal do not have any αhydrogen.

Therefore, these undergo Cannizzaro reactions.

Compound (iv) Benzophenone is a ketone having no α-hydrogen atom and compound

(viii) Butan-1-ol is an alcohol.

Hence, these compounds do not undergo either aldol condensation or Cannizzaro reactions.

Aldol condensation









Cannizzaro reaction








Question 12.8.

How will you convert ethanal into the following compounds?

(i) Butane-1, 3-diol (ii) But-2-enal (iii) But-2-enoic acid


(i) On treatment with dilute alkali, ethanal produces 3-hydroxybutanal gives butane- 1, 3-diol on reduction.


(ii) On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal.


(iii) When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid.



Question 12.9.

Write structural formulas and names of four possible aldol condensation products from propanal and butanal.

In each case, indicate which aldehyde acts as nucleophile and which as electrophile.


  • Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.


  • Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.


  • Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.


  • Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.




Question 12.10.

An organic compound with the molecular formula C9H10O forms 2, 4-DNP derivative,

reduces Tollens’ reagent and undergoes Cannizzaro reaction.

On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.


It is given that the compound (with molecular formula C9H10O) forms 2, 4-DNP derivative and reduces Tollen's reagent.

Therefore, the given compound must be an aldehyde. Again, the compound undergoes Cannizzaro reaction and

on oxidation gives 1, 2-benzenedicarboxylic acid.

Therefore, the -CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted.

Hence, the compound is 2-ethylbenzaldehyde.


The given reactions can be explained by the following equations.



Question 12.11.

An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C).

Oxidation of (C) with chromic acid produced (B). (C) On dehydration gives but-1-ene.

Write equations for the reactions involved?


An organic compound A with molecular formula C8H16O2 gives a carboxylic acid (B) and an alcohol (C)

on hydrolysis with dilute sulphuric acid.

Thus, compound A must be an ester.

Further, alcohol C gives acid B on oxidation with chromic acid. Thus, B and C must contain equal number of carbon atoms.

Since compound A contains a total of 8 carbon atoms, each of B and C contain 4 carbon atoms.

Again, on dehydration, alcohol C gives but-1-ene. Therefore, C is of straight chain and hence, it is butan-1-ol.

On oxidation, Butan-1-ol gives butanoic acid. Hence, acid B is butanoic acid.


Hence, the ester with molecular formula C8H16O2 is butylbutanoate.

All the given reactions can be explained by the following equations.



Question 12.12.

Arrange the following compounds in increasing order of their property as indicated:

(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)

(ii) CH3CH2CH (Br) COOH, CH3CH (Br) CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)

(iii) Benzoic acid, 4-Nitrobenzoic acid, 3, 4-Dinitrobenzoic acid,

4-Methoxybenzoic acid (acid strength)


  • When HCN reacts with a compound, the attacking species is a nucleophile, CN.
  • Therefore, as the negative charge on the compound increases, its reactivity with HCN decreases.
  • In the given compounds, the +I effect increases as shown below.
  • It can be observed that steric hindrance also increases in the same.
  • Hence, the given compounds can be arranged according to their increasing reactivities toward HCN as:

Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

  • After losing a proton, carboxylic acids gain a negative charge as shown:

R-COOH --> R-COO- + H+

Now, any group that will help stabilise the negative charge will increase the stability of the carboxyl ion and as a result,

will increase the strength of the acid.

Thus, groups having +I effect will decrease the strength of the acids and groups

having −I effect will increase the strength of the acids.

In the given compounds, −CH3 group has +I effect and Br group has −I effect. Thus, acids containing Br are stronger.

Now, the +I effect of isopropyl group is more than that of n-propyl group. Hence,

(CH3)2CHCOOH is a weaker acid than CH3CH2CH2COOH.

Also, the −I effect grows weaker as distance increases.

Hence, CH3CH (Br) CH2COOH is a weaker acid than CH3CH2CH (Br) COOH.

Hence, the strengths of the given acids increase as:


  • As we have seen in the previous case, electron-donating groups decrease the strengths of acids,
  • while electron-withdrawing groups increase the strengths of acids.
  • As methoxy group is an electron-donating group, 4-methoxybenzoic acid is a weaker acid than benzoic acid.
  • Nitro group is an electron-withdrawing group and will increase the strengths of acids.
  • As 3, 4-dinitrobenzoic acid contains two nitro groups; it is a slightly stronger acid than 4-nitrobenzoic acid.
  • Hence, the strengths of the given acids increase as:

4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3, 4-Dinitrobenzoic acid


Question 12.13.

Give simple chemical tests to distinguish between the following pairs of compounds.

(i) Propanal and Propanone (ii) Acetophenone and Benzophenone

(iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate

(v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone

(vii) Ethanal and Propanal


(i) Propanal and propanone can be distinguished by the following tests.

(a) Tollen’s test

Propanal is an aldehyde. Thus, it reduces Tollen’s reagent.

But, propanone being a ketone does not reduce Tollen’s reagent.


(b) Fehling’s test

Aldehydes respond to Fehling’s test, but ketones do not.

Propanal being an aldehyde reduces Fehling’s solution to a red-brown precipitate of Cu2O,

but propanone being a ketone does not.

CH3CH2CHO + 2Cu+ + 5OH- --> CH3CH2COO-  + Cu2O       + 3H2O

Propanal                              Propanoate ion  Red-Brown ppt.

(c) Iodoform test:

Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom respond to iodoform test.

They are oxidized by sodium hypoiodite (NaOI) to give iodoform.

Propanone being a methyl ketone responds to this test, but propanal does not.

CH3COCH3 + 3NaOI --> CH3COONa     + CHI3      + 2NaOH

Propanone                 Sodium Acetate   Iodoform

                                                         (Yellow ppt)

(ii) Acetophenone and Benzophenone can be distinguished using the iodoform test.

Iodoform test:

Methyl ketones are oxidized by sodium hypoiodite to give yellow ppt. of iodoform.

Acetophenone being a methyl ketone responds to this test, but benzophenone does not.

C6H5COCH3 +           3NaOI                 --> C6H5COONa     + CHI3      + 2NaOH

Acetophenone     Sodiumhypoiodite     Sodium Acetate    Iodoform

                                                                                (Yellow ppt)

C6H5COCH3 + NaOI --> No yellow ppt. of CHI3


(iii) Phenol and benzoic acid can be distinguished by ferric chloride test.

Ferric chloride test:

Phenol reacts with neutral FeCl3 to form an iron-phenol complex giving violet colouration.

6C6H5OH + FeCl3 --> [Fe (OC6H5)6]3- + 3H+ + 3Cl-

Phenol                       Iron-phenol complex

                                (Violet colour)

But benzoic acid reacts with neutral FeCl3 to give a buff coloured ppt. of ferric benzoate.

6C6H5OH       + FeCl3 --> (C6H5COO) 3Fe + 3HCl

Benzoic Acid                       Ferric benzoate

                                          (Buff’s coloured ppt)

(iv) Benzoic acid and Ethyl benzoate can be distinguished by sodium bicarbonate test.

Sodium bicarbonate test:

Acids react with NaHCO3 to produce brisk effervescence due to the evolution of CO2 gas.

Benzoic acid being an acid responds to this test, but ethylbenzoate does not.

C6H5COOH + NaHCO3 --> C6H5COONa + CO2 (gas) + H2O

Benzoic Acid                 Sodium benzoate (Buff coloured ppt)

C6H5COOC2H5 +NaHCO3 --> No effervescence due to evolution of CO2 gas

(v) Pentan-2-one and pentan-3-one can be distinguished by iodoform test.

Iodoform test:

Pentan-2-one is a methyl ketone. Thus, it responds to this test.

But pentan-3-one not being a methyl ketone does not respond to this test.


(vi) Benzaldehyde and acteophenone can be distinguished by the following tests.

(a) Tollen’s Test

Aldehydes respond to Tollen’s test.

Benzaldehyde being an aldehyde reduces Tollen’s reagent

to give a red-brown precipitate of Cu2O,

but acetophenone being a ketone does not.

 C6H5CHO +      2[Ag (NH3)2] + + 3OH- --> C6H5COO-   +   2Ag +    2H2O + 4NH3

Benzaldehyde   Tollen’s reagent          Benzoate ion      Silver Mirror


(b) Iodoform test

Acetophenone being a methyl ketone undergoes oxidation by sodium hypoiodite (NaOI) to give a yellow ppt. of iodoform.

But benzaldehyde does not respond to this test.

C6H5COCH3 +           3NaOI                  --> C6H5COONa     +        CHI3      + 2NaOH

Acetophenone     Sodiumhypoiodite     Sodium Benzoate           Iodoform

                                                                                          (Yellow ppt)


(vii) Ethanal and propanal can be distinguished by iodoform test.

Iodoform test

Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom responds to the iodoform test.

Ethanal having one methyl group linked to the carbonyl carbon atom responds to this test.

But propanal does not have a methyl group linked to the carbonyl carbon atom and thus, it does not respond to this state.

CH3CHO + 3NaOI --> HCOONa + CHI3 + 2NaOH

Ethanal                   Sodium       Iodoform

                              Methanoate (yellow ppt)



Question 12.14.

How will you prepare the following compounds from benzene?

You may use any inorganic reagent and any organic reagent having not more than one carbon atom

  • Methyl benzoate
  • m-Nitro benzoic acid
  • p-Nitrobenzoic acid
  • Phenylacetic acid
  • P-Nitrobenzaldehyde














Question 12.15.

How will you bring about the following conversions in not more than two steps?

(i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde

(iii) Ethanol to 3-Hydroxybutanal (iv) Benzene to m-Nitroacetophenone

(v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol

(vii) Benzaldehyde to 3-Phenylpropan-1-ol

(viii) Benazaldehyde to α-Hydroxyphenylacetic acid

(ix) Benzoic acid to m- Nitrobenzyl alcohol?


  • When propanone reacts with NaBH4 it will form propan-2-ol.This alcohol is dehydrated to form propene.




  • When ethanol is treated with Cu at 573 k, it will oxidize to ethanal.
  • When ethanal is treated with Dilute NaOH it will form 3-Hydroxy butanal





  • When benzaldehyde is oxidized with dichromate and treated with calcium carbonate it forms calcium salt.
  • And after dry distillation it gets converted into benzophenone.


(vi)    When bromobenzene reacts with Mg in dry ether it forms Grignard reagent, further treating with ethanal

          in acidic condition it forms 1-phenyl ethanol.












Question 12.16.

Describe the following:

(i) Acetylation (ii) Cannizzaro reaction

(iii) Cross aldol condensation (iv) Decarboxylation?


  • Acetylation

The introduction of an acetyl functional group into an organic compound is known as acetylation.

It is usually carried out in the presence of a base such as pyridine, dirnethylaniline, etc.

This process involves the substitution of an acetyl group for an active hydrogen atom.

Acetyl chloride and acetic anhydride are commonly used as acetylating agents.

For example, acetylation of ethanol produces ethyl acetate.

CH3CH2OH + CH3COCl -->      CH3COOC2H5 + HCl (in presence of Pyridine)

Ethanol        Acetyl Chloride   Ethylacetate

(ii) Cannizzaro reaction:

The self-oxidation-reduction (disproportionation) reaction of aldehydes having no αhydrogens on treatment

with concentrated alkalis is known as the Cannizzaro reaction.

In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.

For example, when ethanol is treated with concentrated potassium hydroxide, ethanol and potassium ethanoate are produced.


(iii) Cross-aldol condensation:

When aldol condensation is carried out between two different aldehydes,

or two different ketones, or an aldehyde and a ketone, then the reaction is called a cross-aldol condensation.

If both the reactants contain α-hydrogens, four compounds are obtained as products.

For example, ethanal and propanal react to give four products.


(iv) Decarboxylation:

Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolyzed.

This electrolytic process is known as Kolbe’s electrolysis.



Question 12.17.

Complete each synthesis by giving missing starting material, reagent or products.



  • When 1- phenyl ethane is oxidized with a strong oxidizing agent like KMnO4, it forms a benzoic acid ion.


  • When phthalic acid is treated with SOCl2 it chlorinates both carboxyl group to form phthaloyl chloride.


  • When benzaldehyde is treated with semicarbazide to form benzaldehyde semicarbazone.


  • When benzene is mixed with benzoyl chloride in presence of Anhyd.AlCl3 to give benzophenone.


  • When 4-oxocyclohexanecarbeldehyde is treated with Tollen’s reagent it gets oxidized to carboxylate anion.
  • As it is aldehyde it reduces Tollens reagent.


  • When 2-formyl benzoic acid is treated with NaCN it produces 2-[1-hydroxycyanomethyl] benzoic acid.


  • When benzaldehyde and propanal mixed equally in presence of Dilute NaOH it forms 2-methyl-3-phenyl-prop-2-enal.


  • When Ethyl 3-oxobutanate is treated with sodium borohydride it converts oxo to hydroxyl


  • When cyclohexanol is oxidized with CrO3 it forms cyclohexanone.


  • When methylene cyclohexane undergoes hydroboration oxidation reaction it will form alcohol.
  • Further treating with oxidizing agent PCC it converts into aldehyde.


  • When cyclohexylidenecyclohexane undergo ozonolysis it will form cyclohexanone.



Question 12.18.

Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6- trimethylcyclohexanone does not.

(ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst,

the water or the ester should be removed as soon as it is formed.


  • Cyclohexanone forms cyanohydrin in good yield.


In this case, the nucleophile CN can easily attack without any steric hindrance.

However, in the case of 2, 2, 6 trimethylcyclohexanone, methyl groups at α-positions offer steric hindrances and as a result,

CN cannot attack effectively.For this reason, it does not form a cyanohydrin.


(High steric hindrance)

  • Semicarbazide undergoes resonance involving only one of the two −NH2 groups, which is attached directly to the carbonyl-carbon atom.


Therefore, the electron density on −NH2 group involved in the resonance also decreases. As a result, it cannot act as a nucleophile.

Since the other −NH2 group is not involved in resonance; it can act as a nucleophile and

can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.

  • Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.


RCOOH     +        R’OH <-->RCOOR’ + H2O (In presence of H+)

Carboxylic acid    Alcohol      Ester      Water

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible.

Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed.


Question 12.19.

An organic compound contains 69.77% carbon, 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 hydrogensulphite and give positive iodoform test.

On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.


% of carbon = 69.77 %

% of hydrogen = 11.63 %

% of oxygen = {100 - (69.77 + 11.63)} %

= 18.6 %

Thus, the ratio of the number of carbon, hydrogen, and oxygen atoms in the organic compound can be given as:

C: H: O = (69.77)/ (12): (11.63)/ (1) :( 18.6)/ (16)

= (5.81) :( 11.63) :( 1.16)

= 5:10:1

Therefore, the empirical formula of the compound is C5H10O.

Now, the empirical formula mass of the compound can be given as:

= (5 × 12 + 10 ×1 + 1 × 16)

= 86

Molecular mass of the compound = 86

Therefore, the molecular formula of the compound is given by C5H10O.

Since the given compound does not reduce Tollen's reagent, it is not an aldehyde.

Again, the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test.

Since the compound is not an aldehyde,

it must be a methyl ketone.

The given compound also gives a mixture of ethanoic acid and propanoic acid.


Hence, the given compound is Pentan-2-one.

The given reactions can be explained by the following equations:


Question 12.20.

Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol.



Resonance structures of phenoxide ion are:


It can be observed from the resonance structures of phenoxide ion that in II, III and IV, less electronegative carbon atoms carry a negative charge.

Therefore, these three structures contribute negligibly towards the resonance stability of the phenoxide ion.

Hence, these structures can be eliminated. Only structures I and V carry a negative charge on the more electronegative oxygen atom.

Resonance structures of carboxylate ion are:


In the case of carboxylate ion, resonating structures I′ and II′ contain a charge carried by a more electronegative oxygen atom.

Further, in resonating structures I′ and II′, the negative charge is delocalized over two oxygen atoms.

But in resonating structures I and V of the phenoxide ion, the negative charge is localized on the same oxygen atom.

Therefore, the resonating structures of carboxylate ion contribute more towards its stability than those of phenoxide ion.

As a result, carboxylate ion is more resonance-stabilized than phenoxide ion. Hence, carboxylic acid is a stronger acid than phenol.

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