Chapter 12 Aldehydes, Ketones and Carboxylic Acids

The process of addition of water to ethyne in the presence of H₂SO₄ and HgSO₄ is known as hydration of alkynes.

When hydration of butyne takes place, the product is a ketone.

The OH group gets attached to the secondary carbocation (which is more stable due to the electron giving the property of alkyl group attached to it) after breaking of the triple bond.

The formation of an intermediate enol takes place. The deprotonation of oxygen takes place (i.e. hydrogen of oxygen is removed) to give a more stable final product which is ketone. Thus option (ii) is correct.

Thus, option (i) is wrong as primary carbocation is not stable, option (iii) is a carboxylic acid so it is also wrong and option (iv) is also wrong as the product is a carboxylic acid.

General reaction mechanism:

Aldehydes are more reactive than ketones due to the steric factor, the heavy alkyl groups on both the side of ketone hinder the approach of a nucleophile to the carbonyl carbon (carbon atom double-bonded to an oxygen atom).

Thus, option (ii) is wrong.

The polarity of the carbonyl group is reduced when it is attached to an aromatic ring due to its involvement in resonance. Thus their reactivity towards nucleophilic attack reduces.

Thus, option (iii) and (iv) are wrong.

Thus, option (i) is most reactive towards nucleophilic addition.

Phenols are more acidic than the alcohols due to to the more stability of phenoxide ion than alkoxide ion due to the resonance of electron pairs of oxygen with the aromatic ring.

Carboxylic acid is more acidic than phenol as the conjugate base of carboxylic acid, a carboxylate ion, is stabilized by two equivalent resonance structures in which the negative charge is at the more electronegative oxygen atom.

The conjugate base of phenol, a phenoxide ion, has non-equivalent resonance structures in which the negative charge is at the less electronegative carbon atom.

Thus, a carboxylate ion is more stable than a phenoxide ion.

When an electron-withdrawing group like chlorine is present on the carboxylic acid its acidity increase due to the fact that the O-H bond gets weaker due to the decrease in the negative charge of an oxygen atom.

Thus, the correct order of acidity in increasing order is:

Alcohols<phenols<acetic acid<choloroacetic acid.

The formation of the takes place when phenol and benzoyl chloride reacts in the presence of pyridine.

Phenol reacts with an aromatic acid chloride in the presence of the excess NaOH at room temperature to give ester. This reaction is called Schotten Baumann reaction. The presence of the base is to absorb acidic proton.

Phenoxide ion reacts rapidly with benzoyl chloride in comparison to phenol.

Fehling solution won’t react to both, acetone and benzaldehyde.

Fehling's solution is prepared by combining two separate solutions, known as Fehling's A and Fehling's B.

Fehling solution A is an aqueous solution of copper sulfate solution, which is deep blue.

Fehling's B is a colourless solution of aqueous of potassium sodium titrate (also known as Rochelle salt). The copper(II) complex in Fehling's solution is an oxidizing agent and the active reagent in the test.

Benzaldehyde does not contain alpha hydrogen so intermediate enolate formation does not take place. Thus, it does not react with Fehling's solution.

Ketone does not react with the Fehling's solution unless they are alpha-hydroxy ketones. Acetone is not alpha-hydroxy ketone so it will also not reduce the Fehling’s solution.

The Cannizzaro reaction is a redox reaction in which two molecules of an aldehyde are reacted to produce primary alcohol and a carboxylic acid under the presence of a hydroxide base. There should be no alpha hydrogen (hydrogen attached to alpha carbon) as there will be no enolization and aldol condensation.

Option (iv) will not give Cannizaro reaction as it contains alpha hydrogen.

Option (i), (ii) and (iii) will give Cannizzaro reaction as they do not have alpha hydrogen.

Step 1 Mechanism;

Step 2 Mechanism;

Step 3 Mechanism;

Benzaldehyde Cannizaro reaction

The above reaction is an example of a Cannizzaro reaction.

A Cannizzaro reaction is a redox reaction in which two molecules of an aldehyde are reacted to produce primary alcohol and a carboxylic acid under the presence of a hydroxide base. There should be no alpha hydrogen (hydrogen attached to alpha carbon) as there will be no enolization and aldol condensation.

Benzaldehyde does not have an alpha hydrogen.

Thus, the product is a primary alcohol and potassium benzoate.

The process of addition of water to an ethyne in the presence of H₂SO₄ and HgSO₄ is known as hydration of alkynes.

When hydration of butyne takes place, the product is a ketone.

The OH group gets attached to the secondary carbocation (which is more stable due to the electron giving the property of alkyl group attached to it) after breaking of the triple bond.

The formation of an intermediate enol takes place. The deprotonation of oxygen takes place (i.e. hydrogen of oxygen is removed) to give a more stable final product which is ketone.

This process is called tautomerism .

Thus, the structure of A is pro-2-ene-1-ol and isomerization is tautomerism.

The reaction involves the :

Thus, A (propan-2-ol) and C (propan-1-ol) are positional isomers.

The given reaction takes place in the presence of the I₂ and NaOH. These two act as an oxidizing agent and convert ketones into a carboxylic acid (the number of carbon in parent chain reduces by one).

These reagents are most suitable reagents as they don’t affect the double C=C bond present in the molecules, whereas all the other reagent affects the double C=C bond.

Step 1 Mechanism;

Step 2 Mechanism;

Step 3 Mechanism;

This reaction is known as oxidation of methyl ketone by haloform reaction.

Primary alcohol like Butan-1-ol on reaction with a very strong oxidizing agent like KMnO₄ gives carboxylic acid. Thus, option (i) is wrong.

Secondary alcohol like Butan-2-ol gives ketone on oxidizing under the presence of KMnO₄.

Thus, only option (ii) will give butanone.

Clemmensen reaction is the conversion of the carbonyl group to CH₂. It is basically a reduction reaction.

The reaction takes place in the presence of the Zinc amalgam and HCl.

The carbonyl compounds which have alpha carbon undergoes the aldol condensation. Option (i) and (iii) have alpha hydrogen so they will undergo aldol condensation.

Option (ii) and (iv) will not undergo aldol condensation, as they do not have alpha hydrogen.

Thus, option (ii) and (iv) are the correct options.

Red color hydrogen is alpha hydrogen.

The hydrolysis of phenyl benzoate gives sodium phenoxide and sodium benzoate in the presence of strong base NaOH.

The intermediate product is phenol and sodium benzoate but due to the acidic character of phenol and presence of NaOH base, the reaction gives final product sodium benzoate and sodium phenoxide.

Clemmensen reaction is the conversion of the carbonyl group to CH₂. It is basically a reduction reaction.

The reaction takes place in the presence of the Zinc amalgam and HCl.

Thus, acetone gets converted into an alkane in Clemmensen reduction.

Thus, Cyclohexanone gets converted into cyclohexane and Benzophenone gets converted into diphenylmethane.

The other two options (i) and (iii) do not undergo Clemmensen reduction as they are not ketones.

Grignard’s reagent increases the number of carbons in the parent chain.

Increase in the number of carbon in parent chain on using grignard’s reagent.

The Cannizzaro reaction is a redox reaction in which two molecules of an aldehyde are reacted to produce primary alcohol and a carboxylic acid under the presence of a hydroxide base.

Thus, no increase in the number of carbon in the parent chain in a Cannizzaro reaction.

Aldol condensation increases the number of carbon atom in the parent chain.

HVZ reaction is a halogenation reaction it halogenates the carboxylic acid at alpha carbon.

Thus, Grignard reagent and aldol condensation increase the number of carbon atom in the parent chain.

The reaction (i) is also known as Friedel crafts acetylation. Electrophilic aromatic substitution takes place in this reaction.

The reaction (ii) also give benzophenone.

Carbonyl compound is a planar molecule (all atoms lie in the same plane), hence two orientation regarding the attack of the nucleophile is possible. One will be front side attack and other will be rear side attack.

There is a large difference in the boiling points of butanal and butan-1-ol, i.e. aldehyde and alcohol of the same molecular mass. This is due to the presence of intermolecular hydrogen bonding in butan-1-ol, whereas there is no intermolecular hydrogen bonding in butanal. This bonding makes butan-1-ol slightly more stable at a higher temperature than butanal.

One can differentiate between pentan-2-one and pentan-3-one by iodoform test.

Pentan-2-one have a –CO-CH₃ group and therefore, forms a yellow precipitate of Iodoform. Pentan-2-one gives positive iodoform test, whereas, and pentan-3-one does not give positive iodoform test because they don’t have a –CO-CH₃ group.

IUPAC name: 3-Phenylprop-2-ene-1-al.

Here, the function group is aldehyde. The number of the longest carbon chain attached to the functional group is three. There is a double bond between the carbon atom at position 2 and position 3 with respect to the functional group and a phenyl group is attached to the carbon atom at 3^rd position. Thus, IUPAC name of the above compound is 3-Phenylprop-2-en-1-al.

(ii) IUPAC name: Cyclohexanecarbaldehyde

Here, the function group is aldehyde. The number of the carbon atom in the longest chain attached to the functional group is six and is present in a cyclic manner. Thus, IUPAC name of the above compound is Cyclohexanecarbaldehyde.

(iii) IUPAC name: 3-Oxopentan-1-al

Here, the functional groups are ketone and aldehyde. The number of carbon atoms in the longest chain is five. We know, the naming is done in an alphabetical order. Thus, IUPAC name of the above compound is 3-Oxopentan-1-al.

(iv) IUPAC name: But-2-enal

Here, the function group is aldehyde. The number of the longest carbon chain attached to the functional group is four. There is a double bond between the carbon atom at position 2 and position 3 with respect to the functional group. Thus, IUPAC name of the above compound is But-2-enal.

(i) 4-Nitropropiophenone

Here, Ketone group is the functional group in this compound and a nitro group is attached at a 4^th position with respect to the ketone group. Therefore, the structure of this compound is given below;

(ii) 2-Hydroxycyclopentanecarbaldehyde

Here, aldehyde and a ketone group are the functional group in this compound and the hydroxyl group is attached at the 2^nd position with respect to the aldehyde group. Therefore, the structure of this compound is given below;

(iii) Phenyl acetaldehyde

Here, aldehyde group is the functional group in this compound and acetaldehyde group is attached to a phenyl ring. Therefore, the structure of this compound is given below;

i) The number of carbon atoms in the longest chain is two and two aldehyde groups are present. Thus, IUPAC name of the above compound is Ethane-1,2-dial.

ii) Here, two aldehyde groups are attached to the benzene ring. Two –CHO groups are attached the carbon atom at position 1 and 4. Thus, IUPAC name of the above compound is Benzene-1, 4-dicarbaldehyde.

iii) Here, a bromine atom is attached at the 3^rd position with respect to the aldehyde group on the benzene ring. Thus, IUPAC name of the above compound is 3-Bromobenzaldehyde.

Benzaldehyde can be obtained from benzal chloride.

Toluene is first converted to benzal chloride by side-chain chlorination, in presence of Chlorine gas and light. Benzal chloride on hydrolysis at 373K gives benzaldehyde. The chemical reactions involved in the conversion of toluene to benzaldehyde are:

The electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl₃ is benzoylinium cation. The product formed in this reaction is benzophenone. This reaction is called Friedel Craft’s acylation reaction.

The chemical reaction involved when benzene reacts with benzoyl chloride in the presence of anhydrous AlCl₃:

The products formed on oxidation of 2, 5-dimethylhexan-3-one are the mixtures of ketone and carboxylic acids. Ketone is then further oxidized to carboxylic acids.

Overall the products formed in this reaction are 2-Methylpropanoic acid, 3-Methylbutanoic acid, ethanoic acid and methanoic acid.

The chemical reactions involved are:

The decreasing order of their acidic strength:

FCH₂COOH > ClCH₂COOH > C₆H₅CH₂COOH > CH₃COOH > CH₃CH₂OH.

EXPLAINATION:

We know carboxylic acids are more acidic than alcohols of comparable molecular masses. So, CH₃CH₂OH is least acidic among the given compounds.

Electron withdrawing groups increase the acidity of carboxylic acids by stabilizing the conjugate base through resonance effects, whereas, electron-donating groups decrease the acidity of carboxylic acids by destabilizing the conjugate base through resonance.

Thus, FCH₂COOH and ClCH₂COOH are highly acidic due to the –I effect of halogen. FCH₂COOH is most acidic among the given compounds due to the higher electronegativity of Fluorine atom.

C₆H₅CH₂COOH is more acidic than CH₃COOH due to the resonance in C₆H₅CH₂COOH.

Thus, the decreasing order of their acidic strength is:

FCH₂COOH > ClCH₂COOH > C₆H₅CH₂COOH > CH₃COOH > CH₃CH₂OH.

When propanal react with 2-methylpropanal in the presence of NaOH, the mixture of aldehydes are formed.

Both the reactants have an alpha-hydrogen and hence, can undergo cross aldol reaction in the presence of NaOH.

The products formed in this reaction are

This is called cross aldol reaction.

Compound ‘A’ belongs to the carboxylic acid.

Compound ‘B’ belongs to alcohol.

Compound ‘C’ belongs to an ester group.

EXPLAINATION:

When compound A is heated with compound B in the presence of H₂SO₄, it produces fruity smell (compound C). Thus, compound ‘C’ belongs to an ester group as they are known for their fruity smell.

Compound ‘B’ is oxidized to compound ‘A’ by alkaline KMnO₄. Thus, compound ‘A’ belongs to the carboxylic acid as alkaline KMnO₄ is a strong oxidizing agent.

Compound ‘A’ on reduction with lithium aluminium hydride gets converted back to compound ‘B’. Thus, compound ‘B’ belongs to an alcohol as lithium aluminium hydride is a strong reducing agent.

The decreasing order of their acidic strength:

NO₂CH₂COOH > FCH₂COOH > C₆H₅COOH.

EXPLAINATION:

Electron withdrawing groups like -NO₂, increases the acidity of carboxylic acids by stabilizing the conjugate base through resonance effects, whereas, electron-donating groups decrease the acidity of carboxylic acids by destabilizing the conjugate base through resonance.

We know that the negative inductive effect of nitro group is greater than that of the fluorine atom.

Thus, NO₂CH₂COOH is most acidic among the given three compounds.

Hence, the decreasing order in terms of their acidic strength is:

NO₂CH₂COOH > FCH₂COOH > C₆H₅COOH.

In both alkenes and carbonyl compounds, carbon atom is attached to an oxygen atom by a double bond. We know, the electronegativity of oxygen is greater than that of carbon atom. Therefore, oxygen atom in carbonyl compound pull more shared pair of electron towards itself and so, carbon acquires partial positive charge and oxygen acquires partial negative charge in carbonyl compounds. So, carbon in carbonyl atom is attacked by a nucleophile.

Hence, alkenes show electrophilic addition reactions whereas carbonyl compounds show nucleophilic addition reactions.

We know, oxygen atom in carbonyl compound pull more shared pair of electron towards itself and so, carbon acquires partial positive charge and oxygen acquires partial negative charge in carbonyl compounds. So, the carbon in carbonyl atom is attacked by a nucleophile.

Though carboxylic acids contain carbonyl group they do not show the nucleophilic addition reaction like aldehydes or ketones and this is due to the resonance. The partial positive charge on the carbonyl atom gets reduced because of the resonance in carboxylic acids.

Compound A = CH₃-MgBr

Compound B = CH₃-COOH

Compound C =

EXPLAINATION:

CH₃-Br reacts with the Grignard reagent and form CH₃-MgBr.

CH₃-MgBr then reacts with CO₂ and water to form the corresponding carboxylic acid, CH₃-COOH.

CH₃-COOH is then heated in presence of CH₂-OH/H⁺ to form.

Carboxylic acids are more acidic than alcohols or phenols, although all of them have a hydrogen atom attached to an oxygen atom (—O—H) because the conjugate base of carboxylic acids or the carboxylate ion is stabilized by resonance. Due to the resonance in carboxylic acids, the negative charge is at the more electronegative atom (oxygen atom), whereas, in alcohols or phenols, the negative charge is on less electronegative atom. Thus, carboxylic acids can release proton easier than alcohols or phenols.

Structure of compounds A, B and C in the above reaction sequence are:

Acetone reacts with Grignard reagent followed by hydrolysis to form 2-Methylpropane-2-ol.

2-Methylpropane-2-ol then reacts with sodium metal in the presence of ether to form sodium tert-butoxide.

Sodium tert-butoxide is then allowed to react with CH₃-Br to produce 2-Methyl-2-methoxypropane. This is called Williamson’s synthesis of ether.

Ethylbenzene is generally prepared by acetylation of benzene followed by reduction and not by direct alkylation. The reason behind this is the formation of polysubstitution products.

To avoid the formation of polysubstituted products Friedel-craft’s alkylation reaction is not used for the preparation of ethylbenzene but one can use Friedel-craft’s acylation reaction.

Yes, one can consider Gatterman-Koch reaction similar to Friedel Craft’s acylation reaction.

In Friedel Craft’s acylation reaction, an aryl group or benzene is treated with acid chloride in presence of anhydrous AlCl₃ and corresponding aldehyde or ketone is formed. In Gatterman-Koch reaction, benzene is treated with CO and HCl in presence of AlCl₃ and CuCl to yield benzaldehyde.

If the acid chloride in Friedel Craft’s acylation reaction is H-CO-Cl, then the product will be benzaldehyde.

(i) Cinnamaldehyde - (d) 3-Phenylprop-2-enal

(ii) Acetophenone - (e) 1-Phenylethanone

(iii) Valeraldehyde - (a) Pentanal

(iv) Acrolein - (b) Prop-2-enal

(v) Mesityl oxide - (c) 4-Methylpent-3-en-2-one

EXPLAINATION:

i) The structure of Cinnamaldehyde is:

Here, a phenyl group is attached to a carbon atom at 3^rd position with respect to the functional group. A double bond is present between the C₂ and C₃ atoms. Thus, the IUPAC name of Cinnamaldehyde is 3-Phenylprop-2-enal.

ii) The structure of Acetophenone is:

Here, the functional group present in the above structure is ketone. This ketone group is attached to one methyl group and one phenyl ring.

Thus, the IUPAC name of Acetophenone is 1-Phenylethanone.

iii) The structure of Valeraldehyde is:

CH₃CH₂CH₂CH₂CHO

The functional group in this compound is aldehyde. The number of carbon atoms in the longest chain is five. Thus, the IUPAC name of Valeraldehyde is Pentanal.

iv) The structure of Acrolein is:

CH₂=CH-CHO

The functional group in this compound is aldehyde. The number of carbon atoms in the longest chain is three with one double bond between C₂ and C₃ atoms. Thus, the IUPAC name of Acrolein is Prop-2-enal.

v) The structure of Mesityl oxide is:

(CH₃)₂C=CHCOCH₃

The functional group in this compound is ketone. The number of carbon atoms in the longest chain is five with one double bond between C₃ and C₄ atoms. Thus, the IUPAC name of Mesityl oxide is 4-Methylpent-3-en-2-one.

(i) Phthalic acid - (b) Benzene-1,2-dicarboxylic acid

(ii) Oxalic acid - (e) Ethane-1,2-dioic acid

(iii) Succinic acid - (d) Butane-1,4-dioic acid

(iv) Adipic acid - (a) Hexane-1,6-dioic acid

(v) Glutaric acid - (c) Pentane-1,5-dioic acid

EXPLAINATION:

(i) The structure of Phthalic acid is:

The functional group in this compound is carboxylic acid. Two carboxylic groups are present on the adjacent carbon atom on the benzene ring. Thus, the IUPAC name of Phthalic acid is Benzene-1,2-dicarboxylic acid.

(ii) The structure of Oxalic acid is:

HOOC-COOH

The functional group in this compound is carboxylic acid. Two carboxylic groups are present on the adjacent carbon atoms. Thus, the IUPAC name of Oxalic acid is Ethane-1,2-dioic acid.

(iii) The structure of Succinic acid is:

HOOC -(CH₂)₂-COOH

The functional group in this compound is carboxylic acid. Two carboxylic groups are present on the first and the last carbon atom. Thus, the IUPAC name of Succinic acid is Butane-1,4-dioic acid.

(iv) The structure of Adipic acid is:

HOOC -(CH₂)₄-COOH

The functional group in this compound is carboxylic acid. The number of carbon atoms in the longest chain is four. Thus, the IUPAC name of Adipic acid is Hexane-1,6-dioic acid.

(v) The structure of Glutaric acid is:

HOOC -(CH₂)₃-COOH

The functional groups in this compound are carboxylic acids. The number of carbon atoms in the longest chain is five. Thus, the IUPAC name of Glutaric acid is Pentane-1,5-dioic acid.

(i) Benzophenone → Diphenylmethane - (c) Zn(Hg)/Conc. HCl

(ii) Benzaldehyde → 1-Phenylethanol - (d) CH₃MgBr

(iii) Cyclohexanone → Cyclohexanol - (a) LiAlH₄

(iv) Phenyl benzoate → Benzaldehyde - (b) DIBAL—H

EXPLAINATION:

(i) Benzophenone is converted to Diphenylmethane by using Zn(Hg)/Conc. HCl. This is called Clemmensen reaction.

(ii) Benzaldehyde is converted to 1-Phenylethanol by using CH₃MgBr. This conversion basically uses grignard reagent followed by hydrolysis.

Benzaldehyde + CH₃MgBr + H₃O⁺

1-Phenylethanol

(iii) Cyclohexanone can be converted to cyclohexanol by a reducing agent like LiAlH₄.

LiAlH₄ is a strong reducing agent.

(iv) Phenyl benzoate can be converted to Benzaldehyde by reducing agent DIBAL—H.

(i) → (e) Rosenmund’s reduction

(ii) → (d) Cannizaro’s reaction

(iii) → (a) Friedel Crafts acylation

(iv) → (b) HVZ reaction

(v) → (f) Stephen’s reaction

(vi) → (c) Aldol condensation

EXPLAINATION:

(i) The given reaction is known as Rosenmund’s reduction.

Acid chloride reacts with Pd-C/ BaSO₄ to form the corresponding aldehyde.

(ii) The given reaction is known as Cannizaro’s reaction.

Benzaldehyde undergoes self-oxidation and reduction reaction when heated with concentrated NaOH to form benzyl alcohol and sodium benzoate.

(iii) The given reaction is known as Friedel Crafts acylation reaction.

When benzene is treated with CH₃-CO-Cl in presence of anhydrous AlCl₃, acetophenone is formed.

(iv) The given reaction is known as the HVZ reaction.

Carboxylic acids having an α-hydrogen when treated with bromine in the presence of small amount of red phosphorus gives bromocarboxylic acids.

(v) The given reaction is known as Stephen’s reaction.

Here, nitriles are reduced to imine and then imine is hydrolyzed to the corresponding aldehyde.

(vi) The given reaction is known as Aldol condensation. Here, ethanol having one α-hydrogen atom undergoes a reaction in the presence of dilute NaOH to form 3-Hydroxybutanal and is further converted to but-2-enal.

According to Molecular orbital theory (MOT), the carbon in the carbonyl groups (C=O) is actually sp² hybridized (consisting one s and two p orbitals) and therefore it contributes 3 sp² orbitals for ∂- bonds (sigma i.e.single bonds) and the 4^th valence electron (as carbon has a valency of 4) remains in its (C) one unhybridized partially filled p-orbital.

• This unhybridized p orbital forms a π bond (pi i.e. double bond) by collateral or sidewise overlapping with the partially free p-orbital of the oxygen atom.

• Thus the carbonyl atom of formaldehyde and the other 3 atoms attached to it ( 2 H atoms and one O atoms through sigma bonding) lie in the same plane while the π electron cloud (pi bonds are not actual bonds, they are more like electronic attractions) lies above and below of this planar structure.

•

• In addition the lone pairs left on the oxygen atom also lie in the same plane results in a planar structure for formaldehyde (HCHO) with bond angles of almost 120◦(angle for triagonal planar structures).

• The reason does not provide any explanation to the assertion statement given; as the two different reaction leading to different kinds of products and reasons one is about oxidation of aldehydes (-CHO) and other is about reductions of carboxylic acids (-COOH).

• Compounds containing —CHO group do easily oxidised to corresponding carboxylic acids, but the reason behind it is that aldehydes get easily oxidised even with mild oxidising agents as the H atom(which is more acidic) on carboxyl carbon(-CHO) changes easily to hydroxyl group (–OH) without any cleavage of any other bond.

• Whereas, the reason statement Carboxylic acids can be reduced to alcohols by treatment with LiAlH₄ has a different reason behind it.

• The carbon next to any functional group (here carbonyl group) is called the α-carbon and the H group attached to it is called the α-hydrogen atom. The reason is a wrong statement because the α-hydrogens in carbonyl compounds are generally acidic in nature and in chemical reactions there will be an easy loss of these in forms of H⁺ ions.

• The main reason behind this is electron withdrawing (attracting) effect of the carbonyl group which is powerful enough to remove α-hydrogens by any base (as H is acidic).

• Another fact also contributes to this acidity which is the resonance stabilisation of the anion (i.e. the conjugate base of the acid) formed (called enolate ion) after the removal of the α-hydrogens which is stated in the reason, which would have been explained the assertion if it was correct.

•

enolate ion

• Cannizaro is actually a type of disproportionation reaction (a reaction involving both oxidation and reduction). And in this case of aldehydes which do not have any α-H atoms and therefore undergoes self oxidation in the presence of concentrated alkali solutions (NaOH or KOH) to form acid and reduction to form alcohols.

• Aromatic aldehydes also do not have α-H so they also undergo this reaction.

• But aromatic aldehydes cannot be almost reactive as the aliphatic aldehydes like formaldehyde so the reason statement is wrong because the aromatic aldehydes are usually very stabilised by the more resonance stabilization of the aromatic ring attached with it which making its carbonyl carbon less acidic or electrophillic unlike formaldehyde where there is no such ring present so the carbonyl carbons will be more acidic or electrophillic(electron loving nature) .

• Reaction with Tollen’s reagent (which is actually the ammoniacal silver nitrate solution) is actually is an oxidation reaction.

• But only the aldehydes respond to Tollen’s reagent which oxidises aldehydes to carboxylic acids and simultaneously aldehydes reduce the Ag⁺ ions of this reagent to metallic silver and a brightly appeared silver mirror is formed on the sides of the test tube(silver mirror test )

• Ketones do not give reactions with Tollen’s reagent, therefore the assertion statement is wrong because they are not oxidised by weak oxidising agents as they are less acidic than aldehydes (absence of any α-H atoms hence inability to reduce silver ions).

• And the reason statement cannot explain the reactions of tollen’s reagent because in spite of having carbonyl group, aldehydes and ketones reacts to it differently.

The compounds A, B , C are

,

2 Methyl but-2-ene, Acetaldehyde and acetone respectively.

• Among B and C former is an aldehyde(responds to Felling’s test) and latter is obviously a ketone as both forms iodoforms on treatment with I₂ and NaOH. And from this clue, we can identify the alkene ‘A’ (Mol. formula C₅H₁₀) which is 2 Methyl but-2-ene also as ozonolysis does not disrupt the symmetry of the molecule.

• The equation for ozonolysis of alkene
A :

•

‘A’ ‘ C’ ‘B’

• Now, for compound ‘B’ (acetaldehyde) the iodoform reaction is as follows :

• It also gives Felling’s test which is the way to identify it as an aldehyde :

• Compound ‘C’ which is a ketone (acetone) the iodoform reaction is as follows:

But it is unable to give Fehling’s test.

compound A is obviously is a ketone because it gives positive 2,4- DNP and positive iodoform test but does not respond to Tollen’s or Felling’s solution ( not aldehyde).

• With the molecular formula of C₈H₈O and D.B.E. calculation of : 8 (carbon no.s) –(8/2) (hydrogens no./2) +0 (no nitrogen) +1 = 5, therefore it has 5 unsaturations in forms of the ring or double bonds or both.

• Hence the structure of A should be :

i.e. acetophenone, justified by the 4 double bonds, 1 ring, and one carbonyl carbon.

• It gives positive 2,4-DNP test :

• It gives a positive iodoform test:

‘B’ (yellow ppt)

Hence the yellow compound formed on iodoform reaction is CHI_3.

• On drastic oxidation with potassium permanganate, it forms a carboxylic acid ‘C’ which is actually benzoic acid which is formed from acetophenone.

structure of ‘C’

The reaction is as follows :

If we want to write the structural isomers of the molecular formula C₃H₆O, we have to calculate the double bond equivalent (D.B.E.) i.e. the degree or no. Of unsaturation this molecular formula has.

• D.B.E. = C – () + () + 1 where, C=the no. Of carbon atoms, H=the no. Of hydrogen atoms, N= the no. Of nitrogen atoms.

• For C₃H₆O , D.B.E. will be = 3 – () + 0 + 1= 3 -3+ 1 = 1 ; hence it will have only one unsaturation in all of its isomers.( Unsaturation stands for double or ripple bonds and cyclic rings also.)

• Therefore the structures are,

, , , , ,

propanal Propanone (acetone) allyl alcohol 1-propenol

, , ,

Oxacyclobutane cyclopropanol propylene oxide methyl vinyl ether.

• Among all these isomers only propanal (aldehyde) and acetone will react with HCN; which is a type of nucleophilic addition reaction. (HCN being a nucleophile i.e. love for nucleus or positive charge) .

• But the corresponding aldehyde will give a faster and more productive reaction as it is more active than its ketonic counterpart due to-

(I)Electronic factor: the presence of an α-H which leads to resonance stabilisation (by hyperconjugation) on the carbonyl carbon atom which makes it more electrophillic and more prone towards nucleophilic addition and

(II). Steric factor: The steric hindrance in acetone due to two alkyl group on each side of the carbonyl carbon which makes it more difficult for the nucleophile to attack.

Mechanism of the reaction as follows :

HCN has no lone pair of electrons thus cannot act as a nucleophile alone. However if the reaction is carried out in presence of a base (OH ^- ); it yields cyanide ions CN^- which acts as a nuclephile and the reaction proceeds and cyanohydrins, a new compound is formed.

• At these reaction conditions, the nucleophilic addition reaction is reversible hence , the reaction will not lead to the completion of the reaction with the conversion of the whole reactant into the product at reaction conditions, that will be partial.

• If a strong acid will be added to this reaction mixture , the cyanohydrins will undergo hydrolysis which will lead to formation of the reaction lead to the completion with the conversion of the whole reactant into product at reaction conditions α hydroxy butanoic acid (CH₃CH₂CH(OH)COOH) which will be produced in more concentrated state as the reaction is irreversible as more stable product is formed.

• Obviously the liquid ‘A’ is the aldehyde among the two because it gives a positive silver mirror test also reacts with sodium hydrogen sulphite to give a white crystalline solid (the bisulphite addition compound of the corresponding aldehyde A)

‘A’ Silver mirror

• Ketones also respond to this reaction with sodium hydrogen sulphite and form the same type crystalline addition product.

• But when it comes to tollen’s reagents ketone do not give this test because ketones do not get oxidised by mild oxidising agents for the absence of an α-H atom.

• Hence tollen’s reagent test is the clear distinction between aldehydes and ketones. Therefore in the given problem liquid, ‘A’ must be an aldehyde because it gives a positive tollen’s test and the other liquid ‘B’ must be a ketone as though it does not give tollen’s test but it reacts with NaHSO₃ and forms the addition crystalline product.