NCERT Solutions for Class 12 Science Chemistry Chapter 12 - Aldehydes, Ketones And Carboxylic Acids

Answer:

Addition of water to but -1- yne in the presence of Hg²⁺ ions as a catalyst produces butan-2-one.


Hence, the correct answer is option (ii).

Answer:

The reactivity of carbonyl compounds can be decided by two factors.
(i) Steric factor: The reactivity of carbonyl compounds is inversely proportional to steric factor. Lesser the steric factor, greater will be its reactivity.
(ii) Electronic factor: Greater the number of alkyl groups, lesser will be its electrophilicity. Therefore, CH₃ – CHO will be most reactive towards nucleophilic addition reaction.
Hence, the correct answer is option (i).

Answer:

Phenol is more acidic than ethanol because in phenol, the phenoxide ion obtained after deprotonation is stabalised by resonance which is not possible in the case of ethanol.
Carboxylic acid is more acidic than phenol due to the formation of a more stable conjugate base after removal of H⁺ as compared to phenol.
Chloroacetic acid is more stable than acetic acid due to the presence of an electron-withdrawing chlorine atom which stabalises the carboxylate anion.
Therefore, the correct order of increasing acidic strength is as follows:

Hence, the correct answer is option (iii).

Answer:

The compound  can be prepared by the reaction of phenol and benzoyl chloride in the presence of pyridine.

This is an example of Schotten-Baumann's reaction.

Hence, the correct answer is option (ii).
 

Answer:

Aromatic aldehydes and ketones do not give Fehling's solution test. Acetone do not have hydrogen (attached to carbon-oxygen double bond which is present in aldehydes) that will undergo oxidation. Benzaldehyde (aromatic aldehyde) does not have α- hydrogen, therefore, does not react with Fehling's solution.
Hence, the correct answer is option (iii).

Answer:

The necessary condition for Cannizzaro reaction is the absence of α- hydrogen atom. CH₃CHO has 3 α- hydrogen atoms. So, it will not give Cannizzaro reaction. The remaining three compounds do not contain α- hydrogen atom and therefore, give Cannizzaro reaction.
Hence, the correct answer is option (iv).

Answer:

Benzaldehyde is an aromatic aldehyde with no α-hydrogen. On reaction with aqueous KOH solution, it undergoes Cannizzaro reaction to produce benzyl alcohol and potassium benzoate.

Hence, the correct answer is option (ii).

Answer:

Therefore, A is Prop-l-en-2-ol which undergoes tautomerism to form the product acetone.
Hence, the correct answer is option (iv).

Answer:

The chemical reaction can be shown as :

Therefore, compounds  and  are position isomers.
Hence, the correct answer is option (ii).

Answer:

Iodoform test is used to test the presence of –COCH₃ group. The –COCH₃ group gets converted into –COOH group.
The reaction can be shown as :

Hence, the correct answer is option (iii). 

Answer:

But an-2-ol on oxidation with alkaline KMnO₄ solution produces butan-2- one or butanone.
The reaction is as follows:

Hence, the correct answer is option (ii).
 

Answer:

Clemmensen reduction is used to convert carbonyl group to CH₂ group. Zinc amalgam and Hcl act as reagent in this reaction.

Hence, the correct answer is option (i).

Answer:

The necessary condition for aldol condensation is the presence of atleast one α H-atom.
Therefore, and do not undergo aldol condensation as both do not contain any α- hydrogen atom.
Hence, the correct answers are options (ii) and (iv).

Answer:

on treatment with NaOH yields sodium phenoxide and sodium benzoate by nucleophillic substitution reaction.

Hence, the correct answer are options (ii) and (iii).
 

Answer:

Clemmensen reduction is used to convert cyclohexanone into cyclohexane and benzophenone into diphenyl methane.
The reactions can be written as :


Zinc amalgam in hydrochloric acid, Zn(Hg) in Hcl is used as reagent in clemmensen reduction.
Hence, the correct answer are options (ii) and (iv).
 

Answer:

Both Grignard reaction and aldol condensation is used to increase the number of carbon atoms in the chain.


Hence, the correct answer are options (i) and (iii).

Answer:

Benzophenone can be obtained by Friedel-Craft acylation reaction. Benzoylation of benzene takes place in the presence of catalyst anhydrons aluminium chloride.

Benzophenone can also be prepared by the reaction between benzoyl chloride and diphenyl cadium.


Hence, the correct answer are options (i) and (ii).

Answer:

Carbonyl compound is planar molecule in which the nucleophile can attack from two sides the front side or the rear side.
The two orientations obtained are:

The two products formed are enantiomers. 

Hence, the correct answers are options (i) and (ii).

Answer:

Butan-1-ol contains intermolecular H-bonding to presence of O-H bond which is not present in case of butanal. Instead, butanal contains only dipole- dipole interactions.

Thus, the boiling point of butan-1-ol is higher than that of butanal.

Answer:

Pental-2-one and Pentan-3-one can be distinguished with the help of iodoform test. The organic compounds that contains  group (methyl ketone) responds positively to this test. Thus, Pentan-2-one gives positive iodoform test as it contains −COCH₃ group whereas Pentan-3-one does not.

Answer:

IUPAC names of the compounds are:
(i) 

(ii) 

(iii) 

(iv) 

Answer:

The structure of the compounds are as follows:
(i) 4-Nitropropiophenone

(ii) 2-Hydroxycyclopentanecarbaldehyde

(iii) Phenyl acetaldehyde


 

Answer:

(i) 

IUPAC name: Ethan-1, 2-dial
(ii) 

IUPAC name: Benzene-1, 4-dicarbaldehyde
(iii) 

IUPAC name: 3-Bromobenzaldehyde

Answer:

Benzal chloride can be obtained from chlorination of toluene in presence of sunlight. Benzal Chloride on heating with boiling water produces benzaldehyde.
The reaction involved is as follows:

Answer:

Benzene on reaction with benzoyl choride in the presence of anhydrous AlCl₃ leads to the formation of benzophenone, through the intermediate benzoylinium cation. C₆H₅CO⁺ (benzoylium cation) is the electrophile generated.

This is an example of Friedel-craft acylation reaction.

Answer:

Oxidation of 2, 5-dimethylhexan-3-one can be shown as:

Answer:

(i) The acidic strength increases with the increase in electron-withdrawing groups or substituents.
(ii) Alcohols are weakly acidic than carboxylic acids.
(iii) Presence of phenyl group (C₆H₅) increases the acidity due to presence of sp² hybridisation and resonance.
Therefore, the correct decreasing order of acidic strength is:
FCH₂COOH > ClCH₂COOH > C₆H₅CH₂COOH > CH₃COOH > CH₃CH₂OH

Answer:

The above reaction is an example of cross aldol condensation.

Answer:

The given information suggests that Compound 'A' is an acid. Compound 'B' is an alcohol and it produces acid 'A' on oxidation with alkaline KMnO₄. The acid 'A' is reduced with LiAlH₄ to produce alcohol 'B'.
The two Compounds 'A' and 'B' react in the presence of H₂SO₄ to form ester 'C'. This is called esterification reaction.
The reactions involved are as follows:




 

Answer:

The acidic strength increases as the number of electron withdrawing groups or substituents linked to ∝-carbon or carboxylic group of carboxylic acid increases. Also, the electron withdrawing ability of NO₂, F and C₆H₅ are as follows:
$-{\mathrm{NO}}_2>-\mathrm{F}>-{\mathrm{C}}_6{\mathrm{H}}_5$
Therefore, the decreasing order of acidic strength is:

NO₂CH₂COOH > FCH₂COOH > C₆H₅COOH

Answer:

The nature of bond between and  is different.


                               (Due to electronegativity difference between carbon and oxygen)

In , Carbon attains partial positive charges and oxygen attains partial negative charge, as a result of which nucleophilic addition reaction occurs to the electrophilic Carbonyl Carbon.

 on the other hand undergoes electrophilic addition reaction due to the nucleophilic nature of  as it contains π bond.


 

Answer:

Carboxylic acid contain carbonyl group but don't not show the nucleophilic addition reaction like aldehyde and ketone. This is due to resonance present in carboxylic acid, as a result of which, the partial positive charge on the carbonyl carbon atom is reduced.

In case of aldehydes and ketones, the carbonyl group undergoes resonance as:

The Carbonyl carbon of carboxylic acid is less electrophilic compared to the carbonyl carbon in aldehydes and ketones. Thus, nucleophilic addition reaction is not shown in the case of the carbonyl group of carboxylic acid.
 

Answer:

The chemical reaction can be represented as:

Therefore compounds A, B, C are as follows:

$$\begin{gathered} \mathrm{A}={\mathrm{CH}}_3\mathrm{MgBr} \\ \mathrm{B}={\mathrm{CH}}_3\mathrm{COOH} \\ \mathrm{C}={\mathrm{CH}}_3{\mathrm{COOCH}}_3 \end{gathered}$$
 

Answer:

Carboxylic acids are more acidic than alcohol or phenol. This can be explained on the basis of stability of conjugate base obtained after the removal of H⁺ ion from acid, alcohol or phenol. The Carboxylate ion is more stable compared to the phenoxide ion. This is due to the fact that the electron charge on the carboxylate ion is more dispersed in comparison to that of phenoxide ion. Therefore, the dissociation of O–H bond in case of the carboxylic acid is easier than phenol. Hence carboxylic acids are more acidic than alcohols or phenols.

Answer:

The products that complete the given reaction sequence are shown below.

Answer:

Ethyl benzene is generally prepared by acetylation followed by reduced. 
The reaction involved in as follows:

The direct alkylation produces polysubstituted product because of which direct alkylation is not preferred for the preparation of alkylbenzenes. For this reason only, Friedel-Craft's acylation is used.

Answer:

In Gattermann-Koch reaction, benzene or its derivative on reaction with CO and HCl in the presence of anhydrous ${\mathrm{AlCl}}_3$ to produce benzaldehyde.
The reaction is as follows:

In Friedel-Craft acylation, acyl group is transferred to carbon of benzene and acyl benzene is formed. It involves the reaction between benzene with acyl chloride in the presence of anhydrous ${\mathrm{AlCl}}_3$.


Yes, Gatterman-Koch reaction can be considered similar to Friedel Craft's acylation reaction.

Answer:

Common names	Structure	IUPAC names
(i) Cinnamaldehyde		3-Phenylprop-2-enal
(ii) Acetophenone		1-Phenylethanone
(iii) Valeraldehyde		Pentanal
(iv) Acrolein		Prop-2-enal
(v) Mesityl oxide		4-Methylpent-3-en-2-one

Answer:

Acids	Structure	IUPAC names
(i) Phthalic acid		Benzene-1,2-dicarboxylic acid
(ii) Oxalic acid		Ethane-1,2-dioic acid
(iii) Succinic acid		Butane-1,4-dioic acid
(iv) Adipic acid		Hexane-1,6-dioic acid
(v) Glutaric acid		Pentane-1,5-dioic acid

Answer:

Answer:

Column I (Example)	Column II (Reaction)
(i)	(e) Rosemund's reduction
(ii)	(d) Cannizzaro's reaction
(iii)	(a) Friedel-Crafts acylation
(iv)	(b) HUZ reaction
(v)	(f) Stephen's reaction
(vi)	(c) Aldol Condensation

Answer:

The Structure of formaldehyde is as follows:

The Carbon atom is ${\mathrm{sp}}^2$ hybridised.
It is a planar molecule due to ${\mathrm{sp}}^2$ hybridised carbon atom.
Hence, the correct answer is option (i).

Answer:

The compounds containing –CHO group are easily oxidised to corresponding carboxylic acids due to the electron-withdrawing nature of $\mathrm{C}=\mathrm{O}$ group. The C–H bond in aldehydes is weak and easily oxidised to the corresponding carboxylic acids with the help of mild oxidising agent like Fehling's solution and Tollen's reagents.
Hence, the correct answer is option (ii).

Answer:

The $\alpha$-hydrogen atom in carbonyl compounds is acidic in nature due to the presence of electron-withdrawing carbonyl  group. The anion formed after the loss of $\alpha$-hydrogen atom is resonance stabalised. 
Hence, the correct answer is option (iv).

Answer:

The aromatic aldehydes and formaldehyde do not contain $\alpha$-hydrogen, therefore, undergo cannizzaro reaction.
Formaldehyde is more reactive than corresponding aromatic aldehydes. This is due to the fact that in formaldehyde, HCHO, there is no electron donating group to reduce the density of C atom of CHO group. 
Hence, the correct answer is option (iii).

Answer:

Both aldehydes and ketones contain carbonyl group, but only aldehydes react with Tollen's reagent to form silver mirror. This is because an aldehyde molecule has H-atom (R–CHO), due to which they can be easily oxides. On the other hand, ketones (RCOR') do not contain H-atom, therefore, they do not get oxidised by Tollen's reagents. 
Hence, the correct answer is option (iv).

Answer:

Given molecular formula of compound A = ${\mathrm{C}}_5{\mathrm{H}}_{10}$
Degree of unsaturation = $\left({\mathrm{C}}_{\mathrm{n}}+1\right)-\frac{{\mathrm{H}}_{\mathrm{n}}}{2}$
Where, ${\mathrm{C}}_{\mathrm{n}}$ = number of carbon atoms
             ${\mathrm{H}}_{\mathrm{n}}$ = number of hydrogen atoms
$\therefore$Degree of unsaturation = $\left(5+1\right)–\frac{10}{2}=1$
Possible Structures of alkene 'A' are:
(i) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{CH}={\mathrm{CH}}_2$
(ii) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3$
(iii)
 
(iv)

Ozonolysis of Structure (i), (ii), (iii) and (iv) is as follows:

(i) 

(ii)

(iii)

(iv)



The Ozonolysis of structure (i), (ii), (iii) produces only aldehydes whereas structure (iv) produces both aldehyde and ketone.
Since on ozonolysis of alkene 'A' gives a mixture of two compounds 'B' and 'C' Whereas Compound 'B' gives positive Fehling's and Iodoform test and Compound 'C' does not give Fehling's test but gives Iodoform test suggests that out of all the possible structure of alkenes, structure (iv) is correct one.

Formation of Iodide from 'B' and 'C'


 

Answer:

Given molecular formula of compound 'A' = ${\mathrm{C}}_8{\mathrm{H}}_8\mathrm{O}$
Degree of unsaturation = $\left({\mathrm{C}}_{\mathrm{n}}+1\right)-\frac{{\mathrm{H}}_{\mathrm{n}}}{2}$
                                      = $\left(8+1\right)-\frac{8}{2}=9-4=5$
Since, degree of unsaturation is greater than 5, it means the structure contains benzene ring.
Since the aromatic compound 'A' does not give Tollen's or Fehling's test, it must contain ketone. Also, the compound 'A' gives positive 2,4-DNP and Iodoform test, it means it is methyl ketone.
Complete reaction can be represented as:

 

Answer:

Functional isomers of ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$ with a carbonyl group are as follows:

Compound I, which is propanal will react faster with HCN. This is because there is less steric hinderance and electronic factors, whihc increases the electrophilicity of carbon atom of carbonyl group. 
Also, there are two methyl groups in structure (II), that are increasing the electron density on carbonyl carbon and thus, decreasing the rate of nucleophilic attack.
The reaction involved is as follows:




The above reaction does not complete as it is reversible in nature, therefore, equilibrium is established.
If strong acid is added to the reaction mixture, the reaction is inhibited because the production of ${\mathrm{CN}}^{-}$ ions is prevented.
 

Answer:

Since, liquid 'A' reduces ammonical silver nitrate (Tollen's reagent), therefore, A is an aldehyde and liquid 'B' does not give test with ammonical silver nitrate, thus, 'B' is a ketone.
Further, liquid 'B' form a white crystalline solid with sodium hydrogen sulphite, therefore liquid 'B' must be methyl ketone.
The chemical reactions involved are: