Class 12 Chemistry Chapter 8 – Aldehydes, Ketones and Carboxylic Acids NCERT Solutions

Intext Questions – Class 12 Chemistry Chapter 8 – Aldehydes, Ketones and Carboxylic Acids

8.1

Ans –

8.2

Ans –

(ii) (C₆H₅CH₂)₂Cd + 2CH₃COCl → 2CH₃COCH₂C₆H₅ + CdCl₂

8.3

Ans – The sequence of the elements based on their boiling levels will be represented as: CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO <CH₃CH₂OH
The molecular masses of each of these substances are similar. The greatest boiling point is exhibited by CH₃CH₂OH, which experiences significant intermolecular Il-bonding. Dipole-dipole relations are powerful in CH₃CHO compared to CH₃OCH₃ because CH₃CHO is considerably more polarized than CH₃OCH₃. Consequently, CH₃CHO’s boiling level > CH₃OCH₃. CH₃CH₂CH₃ exhibits the lowest boiling point & only fragile Van Der Waals interactions occur amongst its molecular compounds.

8.4

Ans – (i) Butanone < propanone < propanal < ethanal represents the ascending sequence of responsiveness among carbonyl groups over nucleophilic addition processes. Two elements determine the sensitivity such as electronic & steric variables.

(ii) Acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde is the ascending pattern of reactive elements.

Justification: Acetophenone is the most poorly reactive ketone when it comes to nucleophilic interaction. Every other substance is an aldehyde. Considering the group composed of CH3 near the para position concerning the -CHO group is going to boost the electron density over the carbonyl carbon atom resulting in the hyperconjugation impact, p-tolualdehyde remains less reactive compared to benzaldehyde. That’s why in comparison to benzaldehyde; the nucleophile assault is a bit less severe.

The nitro compound in p-nitrobenzaldehyde consists of the opposite consequence. Because of the -I and -R effects, this is an electron-withdrawing substance. The nucleophile assault is favoured when the electron density is present around the atom of carbonyl carbon drops.

8.5

Ans –

8.6

Ans – (i) 3 – Phenylpropanoic acid
(ii) 3 – Methylbut-2-enoic acid
(iii) 2-Methylcyclohexanecarboxylic acid
(iv) 2,4,6 – Trinitrobenzoic acid

8.7

Ans –

8.8

Ans – Justification: The reason for this is that the CH₃ category’s +I action elevates the electron density around the oxygen atom that is commonly found in the carboxyl category’s O-H bond, making bond breakage even more challenging. As a result, the acidic strength is significantly reduced. The -I power, or the overall impact of electron-withdrawing atoms is extremely strong for the existing atomic particles that go by the name F.

This robust atomic structure will transform the connected bond breakage process much more efficiently thereby reducing the oxygen atom’s combined electron density. As a result, the acidic nature gets highly solidified.

We must also consider the fact that how many F atoms are readily available within a specific molecule. Consequently, the F atom’s approximate location inside the carbon atom ring. Based on the earlier conversation,

(i) The stronger acid will be CH₂FCOOH.
(ii) The stronger acid will be CH₂FCOOH.
(iii) The stronger acid will be CH₃CHFCH₂COOH.

Exercise Questions – Class 12 Chemistry Chapter 8 – Aldehydes, Ketones and Carboxylic Acids

8.1

Ans – (i) The organic compounds called cyanohydrins consist of RR′(OH)CN, wherein R & R′s might represent a chemical combination of aryl or alkyl substances. Aldehydes & ketones interact via hydrogen cyanide (HCN) in the form of cyanohydrin when excessive sodium cyanide (NaCN) is present as a catalyst in the mixture. We refer to this as cyanohydrin interactions. The following is the response:

(ii) Acetals can be gem-dialkoxy alkanes with 2 alcoholic categories attached to the final carbon atom. One link is connected to an alkyl category while the additional link is connected to an atom of hydrogen.

Hemiacetals happen whenever two monohydric alcohol substitutes are exposed to dry HCl gases; these hemiacetals then combine with a different alcohol molecule to make acetal. The following is the response:

(iii) When ketone or aldehyde & semicarbazide condense, aldehydes and ketones are produced as semicarbazones. The following is a prime instance of how semicarbazone is formed:

(iv) It is known that an aldol corresponds to a β-hydroxy aldehyde or ketone. Two separate molecules comprising one or two different aldehydes or ketones produce a reaction called condensation when a fundamental substance is present. Here is an illustration of how aldol gets formed:

(v) α-alkoxyalcohols have been referred to as hemiacetals. Aldehyde interacts with an individual monohydric alcohol molecular structure when dry HCl gas is present. The following is the response:

(vi) The typical equation for oxides is RR′CNOH, wherein R is an organic secondary chain while R′ is either an organic side chord or hydrogen. Organic compounds are oxides. If R′ is an organic terminal cord, it is commonly referred to as ketoxime. When its chemical symbol is H, then it’s classified as aldoxime.

During a moderately acid medium, hydroxylamine is used for treating aldehydes or ketones produced by oximes. The following constitutes the response:

(vii) Ketals were gemodioxyalcanes that have 2 alcoholic compounds upon the same atom of carbon throughout the chain of molecules. Two distinct sets of alkyls are connected to each other via 2 carbon linkages. Ketones and ethylene glycol combine to form ethylene glycol ketals, a cycle-like molecule when HCl dry gas is present. Here is an illustration of how a ketal gets formed:

(viii) Chemical compounds with a dual carbon-nitrogen link are known as imines. Whenever ketones & aldehydes interact with ammonia & their byproducts, imines are produced. The following constitutes the response:

(ix) Dinitrophenylhydrazones 2, 4 under the scenario of aldehydes or ketones, 2, 4-dinitrophenylhydrazine is used to create 2, 4-DNP− compounds under weakly acidic media. The following constitutes the response:

(x) Schiff’s bases are comprised of aldehyde or ketone substances in which an imine or azomethine category has taken over the role of the carbonyl group in their structure. Schiff’s base, also known as azomethine, is a chemical substance that contains an aryl or alkyl unit with no atoms of hydrogen. This is composed of carbon and nitrogen elements which is two times of nitrogen. Their equation is the standard R1R2C=NR3 since an imine. It bears the work of famous scientist Hugo Schiff. The following is the response:

8.2

Ans – (i) 4-Methyl pentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-en-l-al
(iv) Pentane-2,4-dione
(v) 3,3,5-Trimethyl-hexan-2-one
(vi) 3,3-Dimethyl butanoic acid
(vii) Benzene-1,4-dicarbaldehyde

8.3

Ans –

8.4

Ans – (i) IUPAC name – Heptan-2-one
Common name – Methyl n-pentyl ketone

(ii) IUPAC name – 4-Bromo-2-methylhexanal
Common name – γ -Bromo- α -methyl caproaldehyde

(iii) IUPAC name – Heptanal
Common name – NA

(iv) IUPAC name – 3-Phenylprop-2-enal
Common name – β-phenylacrolein

(v) IUPAC name – Cyclopentanecarbaldehyde
Common name – NA

(vi) IUPAC name – Diphenylmetthanone
Common name – Benzophenone

8.5

Ans –

8.6

Ans –

8.7

Ans – The Cannizzaro reaction was given its name in honour of a renowned scientist Stanislao Cannizzaro. It’s a type of chemical procedure in which two distinct molecular substances that form non-enolizable aromatic compounds are disproportionated by a substance called base to produce a primary alcohol alongside a compound known as carboxylic acid.

The Cannizzaro Reaction Process describes how to convert two separate molecules concerning a certain aldehyde into one alcohol molecular structure alongside a single carboxylic acid compound. During the year 1853, chemist Stanislao Cannizzaro was able to separate potassium benzoate & benzyl alcohol using benzaldehyde.

Aldol condensation occurs in ketones & aldehydes that contain a minimum of single α-hydrogen. Several α-hydrogen atoms can be found in the substances (ii) 2-methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone & (vi) phenylacetaldehyde. Consequently, aldol condensation occurs in them.

The outcome of 2-Methylpentanal’s aldol condensation is defined as, 3-Hydroxy-2,4-dimethyl-2-propylheptanal. The following constitutes the response:

2-(1-Hydroxy-1-cyclohexyl) cyclohexane-1-one will be the final substance of cyclohexanone’s aldol condensation. The following constitutes the response:

The outcome of 1-phenylpropanone’s aldol condensation becomes 3-Hydroxy-2-methyl-1,3-diphenylpentan-1-one. The following applies to the response:

The product created by the aldol condensation of Phenylacetyladehyde will be, 3-Hydroxy-2,4-diphenylbutanal. Regarding the reaction, what follows is relevant:

Cannizzaro events are shown by aldehydes that lack α-hydrogen atoms. There’s no α-hydrogen present in the substances (i) methanal, (iii) benzoaldehyde & (ix) 2, 2-dimethylbutanal. Those elements experience Cannizzaro effects.

Methanol & sodium methanoate were the end byproducts of the Cannizaro process that produces methanal. The following constitutes the response:

Benzyl alcohol & sodium benzoate constitute the end byproducts of the Cannizaro response in reacting with benzaldehyde. The following represents the explanation:

When 2,2-Dimethylbutanol undergoes the Cannizaro response, sodium benzoate, and benzyl alcohol become the end compounds. The following serves as the response:

Mixture (iv) (viii) Butan-1-ol constitutes an alcohol, whereas benzophenone serves as a ketone without having an α-hydrogen element. As a result, neither Cannizzaro interactions nor aldol condensation occurs when using these substances.

8.8

Ans –

8.9

Ans –

8.10

Ans – The reason for this is due to the organic compound (C₉H₁₀O) decreases the Tollen reagent & produces a chemical product of 2, 4-DNP. Therefore, the substance has to be an aldehyde. The result undergoes Cannizzaro’s process again, whereas oxidation yields 1,2-benzenedicarboxylic alcohol. As a result, the CHO unit is ortho-substituted in place of benzaldehyde & instantly joined by a benzene chain. Hence, the chemical substance involved is 2-ethyl benzaldehyde.

8.11

Ans – An equation for a molecule Alcohol (C) with diluted sulfuric acid & carboxylic acid (B) hydrolyze an element called C₈H₁₆O₂. Thus, compound A’s composition must be a type of ester. Additionally, alcohol C supplies acid B for the oxidation of chromic acid. Consequently, B & C must have the same quantity of atoms made of carbon. Compound A possesses eight carbon atoms, whereas compounds B & C apiece have four carbon atoms.

Whenever alcohol C becomes dehydrated, but-1-ene is once more produced. Because C is straight ahead, butane-1-ol is produced. Once butan-1-ol undergoes oxidation, a substance called butanoic acid is produced. Butanoic acid consequently counts as a B acid. Therefore, the chemical structure of ester C₈H₁₆O₂ represents butyl butanoate. Listed below are all the responses for the inquiries such as,

8.12

Ans – (i) The process that occurs when the nucleophile (CN– ion) assaults the carbon atom as part of the carbonyl compound ends in the formation of cyanohydrin derivatives. When the alkyl group’s +I impact increases, the sequence of reactivity diminishes. It also drops when the steric barrier rises because of the alkyl groups’ length and quantity. Given the facts provided, the reactivity’s descending sequence is as follows:

(ii) We are aware because the strength of acid is reduced by alkyl groups through the +I action. Compared to n-propyl groups, isopropyl groups have a greater +I impact. Comparably the acidic strength is increased by bromine (Br) due to the -I-event. The more powerful the acid, the more closely it is in proximity to the carboxyl (COOH) unit in the corresponding carbon atom cord, and the greater its -I-effect. Given the above, the acidic strength increases in the following sequence: (CH₃)₂CHCOOH< CH₃CH₂CH₂COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br) COOH.

(iii) We now know the fact that benzoic acid’s capability of acidic nature is reduced by the group that donates electrons (OCH₃). The electron-withdrawing component (NO₂) also rises at the identical period. In light of the aforementioned, the pH level increases in the following order of importance:

8.13

Ans – (i) a) Tollen’s test

Tollens’ test, often referred to as the silver mirror assessment, is a qualitative scientific procedure used to differentiate between ketone & aldehyde. The process makes use of the fact that ketones aren’t easily oxidized, but the compounds of aldehydes do. Propanal acts as a type of aldehyde which decreases the sensitivity of the Tollen reagent. However, Tollen’s reagent cannot be reduced by propanone as it is part of ketone.
CH₃CH₂CHO + 2 [Ag(NH₃)₂]⁺ + 3OH^– → CH₃CH₂COO^– + Ag↓ + 4NH₃ + 2H₂O

b) Fehling’s test

Fehling’s test is an analytical procedure used to identify reductive sugars & distinguish between functional units of ketone & water-soluble polysaccharides. Ketones don’t succeed in the Fehling examination. Meanwhile, the compound aldehydes have cracked it. On the other hand, the substance propanone failed to modify the above compounds into a reddish-brown Cu2O precipitate, but the combination of aldehyde-propanal has been able to do that.
CH₃CH₂CHO + 2Cu²⁺ + 5OH^– → CH₃CH₂COO^– + Cu₂O + 3H₂O

c) Iodoform test

Aldehydes & ketones are produced when one of the methyl groups connected to the atomic structure of carbonyl carbon responds to the assessment performed using iodoform. Sodium hypoiodite (NaOI) can oxidize them & produce iodoform. Propanal isn’t a methyl ketone, but propanone is,
CH₃COCH₃ + 3NaOI → CH₃COONa + CHI₃ + 2NaOH

(ii) Iodoform test
While benzophenone is unlikely to produce the Iodoform examination, acetophenone would.
C₆H₅COCH₃ + 3NaOI → C₆H₅COONa + CHI₃ + 2NaOH

(iii) Ferric chloride test
When utilized alongside neutral ferric chloride, phenol gives off a violet hue.
6C₆H₅OH + FeCl₃ → [Fe(OC₆H₅)₆]^3- (Violet colour) + 3H⁺ + 3Cl^–
When mixed together with neutral ferric chloride, benzoic acid can generate the buff coloring.
3C₆H₅COOH + FeCl₃ → [Fe(OOC₆H₅)₃]^3- (buff colour) + 3H⁺ + 3Cl^–

(iv) Sodium bicarbonate test
Carbon dioxide is produced when benzoic acid & sodium bicarbonate interact.
C₆H₅COOH + NaHCO₃ → C₆H₅COONa + CO₂ + H₂O

(v) Iodoform test
In the iodoform examination, pentan-2-one is going to respond meanwhile pentan-3-one won’t.
CH₃CH₂CH₂COCH₃ + 3NaOI → CH₃CH₂CH₂COONa + CHI₃ + 2NaOH

(vi) Iodoform test
Benzaldehyde is unlikely to pass the Iodoform assessment, while acetophenone can.
C₆H₅COCH₃ + 3NaOI → C₆H₅COONa + CHI₃ + 2NaOH

(vii) Iodoform test
Aldehydes & ketones having at least a single methyl group connected to the carbonyl carbon atomic structure are detected by the iodoform test. Ethanol containing a single methyl group joined by an additional carbonyl atom passes this examination. Propanal, however, is unresponsive to this situation because it lacks a group of methyl compounds associated with carbonyl carbon dioxide.
CH₃CHO + 3NaOI → HCOONa + CHI₃ + 2NaOH

8.14

Ans – (i) Bromobenzene is created when benzoene and bromine combine. Benzoic acid will eventually be produced from the threshold of bromobenzene. Methyl benzoate is currently being produced using benzoic acid. The one that follows constitutes the response:

(ii) Bromobenzene is created when benzoene and bromine interact. Benzoic acid will be produced from these bromobenzene. m-nitro benzoic acid is currently being produced from the above benzoic acid. The following represents the answer:

(iii) Toluene is going to be produced from benzoene. P-nitrotoluene has now been generated from toluene. During the oxidation procedure, this p-nitrotoluene gets completely transformed into p-nitrobenzoic acid. The following serves as the response:

(iv) Toluene is tend to be modified from benzoene. The chemical Tolune is currently undergoing a conversion to become benzyl bromide, which will first go through another conversion to produce benzyl cyanide. Phenylacetic acid is supposed to be produced by hydrolyzing benzoyl cyanide. The one that follows constitutes the response:

(v) Toluene is going to be produced from benzoene. P-nitrotoluene will shortly be produced from toluene. Upon oxidative reactions, this p-nitrotoluene will turn into p-nitrobenzaldehyde. The one that follows represents the response:

8.15

Ans –

8.16

Ans – (i) An ester is considered a substance that gets the end molecule of an acetylation process in which the hydrogen atom of the group that contains alcohol is swapped in exchange for an acetyl compound. Freestanding hydroxyl compounds are available in the substrate element for these types of peculiar interactions.

In the field of chemistry, the acetylation method represents an organic esterification procedure that incorporates acetic acid. A chemical-based molecular unit is offered to link with acetyl compounds. These kinds of compounds are commonly referred to as acetates or acetate esters.

In organic molecules, the terminology “acetylation” refers to the inclusion of the acetyl functional element in the compound. Usually, a base like a pyridine & dimethylaniline has been employed during the process. This involves substituting an ensemble of acetyls for a functional hydrogen atomic particle. Acetic anhydride & acetyl chloride are the two most often used acetylating reagents. For instance, ethyl acetate is produced by ethanol acetylation.

(ii) The Cannizzaro Reaction Process describes how to convert two separate molecules concerning a certain aldehyde into one alcohol molecular structure alongside a single carboxylic acid compound. During the year 1853, chemist Stanislao Cannizzaro was able to separate potassium benzoate & benzyl alcohol using benzaldehyde.

The Cannizzaro response or an individual oxidation-reduction process (disproportion) are other names for the response of aldehydes without α-hydrogen towards saturated alkaline substances. The procedure involves the reduction of a single aldehyde molecular structure to alcohol & the oxidation of another molecule towards carboxylic acid. For instance, saturated potassium hydroxide treatment results in the production of ethanol alongside potassium ethanoate.

(iii) Aldol condensation produces β-hydroxy aldehydes, often known as aldols when aldehydes containing α-hydrogen react against a diluted base. Aldol condensation is a frequently used term for this process. It is referred to as crossed aldol condensation when the process of condensation takes place amid two distinct carbonyl molecules.

It happens when two different aldehydes, 2 different ketones, or an aldehyde with a ketone conduct aldol condensation. 4 compounded byproducts are created when the two separate reacting substances containing α-hydrogen are interlinked. For instance, ethanol and propanal react to 4 different compounds.

(iv) A chemical procedure known as decarboxylation releases carbon dioxide (CO₂) when the carboxyl component is eliminated. Decarboxylation is a generic name for a reaction between carboxylic acid substances that removes a carbon atom of a carbon cord. The initial chemical step in the process of photosynthesis is carboxylation, which is the incorporation of CO₂ into an existing material.

Oxidation is a counterproductive process that occurs as the initial chemical-based stage in photosynthesis. Carbohydrate decarboxylation is triggered by enzymatic processes called decarboxylases. This refers to the process through which carboxylic acids form hydrocarbons over losing carbon dioxide once their sodium salts undergo heating using soda-lime.

8.17

Ans –

8.18

Ans – (i) The remainder of this formula describes how cyclohexanones convert into cyanohydrins.

Nucleophiles (CN^–) do readily navigate this situation while encountering a steric barrier. However, α-positioned methyl chains in 2-2, 6-trimethylcyclohexanone provide steric hurdles, which prevent CN^– from attacking properly.

The body is unable to produce cyanohydrins as a result.

(ii) The more active of each of these compounds -NH₂ that is directly connected alongside the carbonyl carbon particles makes semicarbazide resonance.

As a result, the concentration of electrons at -NH₂ correspondingly falls when the resonance energy increases. Effectively as a consequence, the compounds fail to function in the capacity of nucleophiles. The second type of compound, -NH₂, could operate as a nucleophile enzyme and target aldehydrogen alongside ketone carbonic atomic particles to produce semicarbazones because it is not engaged with a state of resonance.

(iii) Both carboxylic acid & alcohol are permanently converted into ester & water when acidic substances remain present.

The components that react respond to the reaction’s reversible change if water or ester has not been removed during formation. Consequently, in order to move the equilibrium ahead in an additional ester manner, either of those elements needs to be removed.

8.19

Ans – As we all know,
69.77% is the proportion of atoms made up of carbon.
11.63% is the proportion of atoms that are hydrogen.
The oxygen atomic particles proportion will be,
100 – (69.77 + 11.63) = 18.6%.
Consequently, the organic element’s carbon, hydrogen & oxygen atomic particles proportion may be expressed in the following manner:

Therefore, C₅H₁₀O constitutes the compound’s empirical structure. Thus, the compound’s composition is equal to 5 × 12 + 10 × 1 + 1 × 16 = 86. Given that the specified molecular weightage mass is 86 as well. C₅H₁₀O represents the chemical formula in the above equation.

The above-mentioned compound doesn’t constitute a substance called aldehyde since it does not eliminate Tollen’s reagent. Another instance in which the synthetic substance produces sodium hydrogen sulfate which gets supplemented with an affirmative Iodoform analysis.

Since it does not belong to an aldehyde, the molecular structure is bound to be a form of methyl ketone. The defined chemical substance is also produced by interlinking propanoic acid with ethanoic acid. As a result, pentan−2−ol will be serving here as the chemical compound. Given below are the reactions in the question:

8.20

Ans – The phenoxide ion’s resonance configurations consist of the following situations:

For a smaller electronegative atomic particle of carbon surrounding II, III & IV, the configuration could have demonstrated that the resonance arrangements of ions formed due to phenoxide possess a negative electrical charge. As a result, the resonance equilibrium of the phenoxide ion is little affected by all three of these combinations. Consequently, these kinds of structures could potentially be eliminated. Particularly in formations I & V the majority of electronegative in nature atoms of oxygen carry the most adverse burden. The carboxylate ion’s resonant networks have been located similar to the ones that follow:

An electrical charge that is generated from an extremely high electronegative element like an oxygen atom is available within resonating patterns I′ & II′ to create the carboxylate ion. Additionally, the negative energy becomes dispersed across a pair of oxygen molecules in resonant formations I′ & II′. However, the negative electrical charge is concentrated on the exact identical oxygen atom that exists within the phenoxide ion’s resonant configurations I alongside V.

As a result, the carboxylate ion’s resonant properties influence a greater percentage of its ability to remain stable as opposed to the phenoxide ion features. Consequently, compared to phenoxide ions, carboxylate ions are significantly more resonance-stabilized. Because of this, carboxylic acid-based substances appear to be more potent as opposed to phenol.

Related Study Resources of Chapter 8 – Aldehydes, Ketones and Carboxylic Acids

Students can use the links below to get extra study materials for Class 12 Chemistry Chapter 5: Coordination Compounds

Sl No.	Related Links
1	Class 12 Chemistry Chapter 8 Aldehydes, Ketones and Carboxylic Acids – Important Questions
2	NCERT Official Textbook (PDF download)

Chapter Overview: Coordination Compounds – Class 12 NCERT Chemistry

Focused on three significant class of organic compounds including the carbonyl group, Aldehydes, Ketones and Carboxylic Acids, Chapter 8 of Class 12 Chemistry describes physical characteristics, nomenclature (IUPAC and common names), physical properties, and techniques of aldehyde, ketone, and carboxylic acid production. Students also learn about the characteristics of the carbonyl group, its reactivity, and significant reactions such oxidation-reduction systems and nucleophilic addition.

NCERT textbook questions in this chapter test a student’s grasp of reaction mechanisms, chemical conversions, and functional group differentiating techniques. Students can use the aldehydes ketones and carboxylic acids NCERT solutions, which provide thorough, step-by-step responses to all textbook activities, to learn these ideas. These answers also serve to clarify common doubts about reagents utilized, reaction conditions, and product development.

Since advanced concepts in organic chemistry build the basis for CBSE test preparation, this chapter is absolutely vital. In-depth explanations of topics including aldol condensation, Cannizzaro reaction, and the acidity of carboxylic acids together with their uses abound. Students who practice the NCERT questions and go over the NCERT answers for aldehydes ketones and carboxylic acids would be more confident in handling both objective and subjective questions on the board exams.

By means of a range of chemical reactions and conversions, Chapter 8 not only develops fundamental ideas in organic chemistry but also improves problem-solving ability. For students hoping for highest marks on the CBSE board exams, the Class 12 Chemistry Chapter 8 solutions are a great tool for revision, homework help, and idea reinforcement.

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