NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids 2021

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

Edited By Shivani Poonia | Updated on Mar 26, 2025 12:01 AM IST | #CBSE Class 12th

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Class 12 Chemistry talks about Aldehydes, Ketones, and Carboxylic Acids. This chapter tell us about how that compound form, how they are made, and how they react. These carbonyl compounds are really important in both our day-to-day life and industries. For instance, aldehydes and ketones are common in perfumes, medicines, and food preservatives. Carboxylic acids, such as acetic acid and citric acid, are common organic acids we come across in daily life. This chapter covers essential reactions like nucleophilic addition, oxidation, and reduction, which are crucial for grasping organic chemistry. The NCERT solutions offer straightforward explanations and step-by-step answers to help students understand the material and prepare for tests with confidence .

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

LiveCBSE Board Exams 2025 Live: Class 12 biology paper 'moderate'; analysis, answer key updatesMar 25, 2025 | 10:35 PM IST

The Central Board of Secondary Education (CBSE) will conduct the Class 12 board for accountancy subject tomorrow, March 26. The exam will be conducted in a single shift from 10:30 am to 1:30 pm.

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NCERT Exemplar Solutions For Aldehydes Ketones and Carboxylic Acid

Aldehydes Ketones and Carboxylic Acid NCERT Chapter Notes

NCERT Solutions Class 12 Aldehyde Ketone and Carboxylic Acids (Intext Question from 8.1 to 8.8)

Page no. 231

Q 8.1(i) Write the structures of the following compounds.

$α$ -Methoxypropionaldehyde

Answer :

The structure of the compound $α$ -methoxy propionaldehyde is shown here-

necrt_18898

Q8.1(ii) Write the structures of the following compounds.

3-Hydroxy butanal

Answer :

The structure of the compound 3-Hydroxy butanal is shown here-

necrt_18901

Q8.1(iii) Write the structures of the following compounds.

2-Hydroxy cyclopentane carbaldehyde

Answer :

The structure of the compound 2-Hydroxycyclopentane carbaldehyde is shown here-

necrt_18903

Q 8.1(iv) Write the structures of the following compounds.

4-Oxopentanal

Answer :

The structure of the compound 4-oxopentanal is shown here-

necrt_18912

Q8.1(v) Write the structures of the following compounds.

Di-sec. butyl ketone

Answer :

The structure of the compound Di-sec. butyl ketone is shown here-

necrt_18916

Q8.1(vi) Write the structures of the following compounds.

4-Fluoro acetophenone

Answer :

The structure of the compound 4-Fluoro acetophenone is shown here-

necrt_18918

Page no.234

Q 8.2(i) Write the structures of products of the following reactions:

(i) 1649968310983 $\to_{C S_{2}}^{A n h y d . A l C l_{3}}$

Answer :

When benzene is treated with acid chloride in the presence of anhydrous aluminum chloride $(- C O C H_{3})$ group attached to the benzene ring. this reaction is known as friedel craft acylation reaction.

necrt_18924

Q8.2(ii) Write the structure of products of the following reactions;

(ii) $(C_{6} H_{5} C H_{2})_{2} C d + 2 C H_{3} C O C l \to$

Answer :

Reaction of acyl chloride with dialkylcadmium ( $(C_{6} H_{5} C H_{2})_{2} C d$ ), prepared from reaction of cadmium chloride and grignard reagents ,gives ketone

necrt_189301

Q8.2(iii) Write the structures of products of the following reactions:

(iii) $H_{3} C - C \equiv C - H \overset{H g^{2 +}, H_{2} S O_{4}}{\to}$

Answer :

When propyne reacts with $H g^{2 +}$ in presence of dil. sulphuric acid, water molecules as a nucleophile attack on suitable postion[ $C H_{3} - C^{+} = C H^{-}$ , primary anion are stable] and then tautomerisation occurs to get the final product

necrt_18930

Q8.2(iv) Write the structures of products of the following reactions:

(iv) 1649968366695 $\to_{2. H_{3} O^{+}}^{1. C r O_{2} C l_{2}}$

Answer :

Chromyl chloride oxidise the methyl group into a chromium complex, which on hydrolysis give aldehyde group.

necrt_18938

Page no. 236

Q8.3 Arrange the following compounds in increasing order of their boiling points.

$C H_{3} C H O, C H_{3} C H_{2} O H, C H_{3} O C H_{3}, C H_{3} C H_{2} C H_{3}$

Answer :

Increasing order in their boiling points-

$C H_{3} C H_{2} C H_{3} < C H_{3} O C H_{3} < C H_{3} C H O < C H_{3} C H_{2} O H$

Alcohol has the highest boiling point due to more extensive intermolecular H-bonding. Aldehyde is more polar than ether so, ethanal has high BP than ethyl ether. And alkane has the lowest BP.

Page no. 243

Q8.4(i) Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

Ethanal, Propanal, Propanone, Butanone.

Answer :

necrt_189451

By the above structure, we can see that, due to +I effect of alkyl group te electron density at Carbonyl carbon increase from ethanal to bytanone. And so the tendency of attacking nucleophile is decreased.
Thus the increasig order (reactivity towards nucleophile)-

Butanone < propanone < propanal < ethanal

Q8.4(ii) Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

Benzaldehyde, p-Tolualdehyde, p-nitrobenzaldehyde, Acetophenone.

Answer :

If +I effect is more its reactivity towards nucleophilic addition is less. and -I is more, more reactive toward addition.

necrt_18947

So, according to this concept, increasing order of their reactivity in nucleophilic addition reactions-

Acetophenone < p-tolualdehyde < benzaldehyde <p-nitrobenzaldehyde

Q8.5(i) Predict the products of the following reactions:

(i) $+ H O - N H_{2} \overset{H^{+}}{\to}$

Answer :

The product of the reaction is-
Water molecule is removed as a by-product in this reaction.

necrt_18948

Q 8.5(ii) Predict the products of the following reactions:

(ii) 1649968429167 →

Answer :

The product of the reaction is a hydrazone, which is formed when ketone and 2, 4-dinitrophenyl hydrazine derivative reacts with each other.

necrt_18949

Q 8.5(iii) Predict the products of the following reactions:

$R - C H = C H - C H O +$ 1649968459039 $\overset{H^{+}}{\to}$

Answer :

The product of the above reaction is -

Water molecule is eleminated as a by-product in this reaction

necrt_18950

Q8.5(iv) Predict the products of the following reactions:

(iv) $+ C H_{3} C H_{2} N H_{2} \overset{H^{+}}{\to}$

Answer :

The product of the above reaction is -
water is also obtained in this reaction as a by-product

necrt_18951

Q 8.6(i) Give the IUPAC names of the following compounds:

$P h C H_{2} C H_{2} C O O H$

Answer :

The IUPAC name of the compound $P h C H_{2} C H_{2} C O O H$ is-

3-phenyl propanoic acid

Q8.6(ii) Give the IUPAC names of the following compounds:

$(C H_{3})_{2} C = C H C O O H$

Answer :

The IUPAC name of the compound $(C H_{3})_{2} C = C H C O O H$ is -

3-methyl but-2-en-1-oic acid

Q 8.6(iii) Give the IUPAC names of the following compounds:

1649968532540

Answer :

The IUPAC name of the compound is-

2-methyl cyclopentane carboxylic acid

Q8.6(Iv) Give the IUPAC names of the following compounds:

(iv) 1649968580406

Answer :

The IUPAC name of the compound is-

2, 4, 6-trinitrobenzoic acid

Page no. 248

Q8.7(i) Show how each of the following compounds can be converted to benzoic acid?

Ethyl benzene

Answer :

Strong oxidising agents like potassium permanganate in the presence of KOH, followed by acidic hydrolysis give benzoic acid.

necrt_19039

Q 8.7(ii) Show how each of the following compounds can be converted to benzoic acid.

Acetophenone

Answer :

On strong oxidation with potassium permanganate in presenc of KOH followed by acidic hydrolysis, give benzoic acid
1594981355927

Q 8.7(iii) Show how each of the following compounds can be converted to benzoic acid?

Bromobenzene

Answer :

Make bromobenzene first Grignard reagent, react it with dry ice (carbon dioxide) followed by acidic hydrolysis gives benzoic acid.

necrt_19051

Q 8.7(iv) Show how each of the following compounds can be converted to benzoic acid?

Phenylethene (Styrene)

Answer :

On strong oxidation with potassium permanganate ( $K M n O_{4}$ ) in the presence of strong alkali (KOH) followed by acidic hydrolysis gives benzoic acid.

necrt_19055

Page no. 254

Q8.8(i) Which acid of each pair shown here would you expect to be stronger?

$C H_{3} C O_{2} H$ or $C H_{2} F C O_{2} H$

Answer :

$C H_{2} F C O_{2} H$ is stronger than $C H_{3} C O O H$ due to -I effect of Fluorine decreases the electron density at OH bond, which makes it easier to lose proton ( $H^{+}$ ). The conjugate base of $C H_{2} F C O_{2}^{-}$ is more stable than $C H_{3} C O O^{-}$ .

Q8.8(ii) Which acid of each pair shown here would you expect to be stronger?

$C H_{2} F C O_{2} H$ or $C H_{2} C l C O_{2} H$

Answer :

$C H_{2} F C O_{2} H$ is a stronger acid.

Fluorine has more -I effect than chlorine. So, $C H_{2} F C O_{2} H$ can release proton easily than $C H_{2} C l C O_{2} H$ .

Q8.8(iii) Which acid of each pair shown here would you expect to be stronger?

(iii) $C H_{2} F C H_{2} C H_{2} C O_{2} H$ or $C H_{3} C H F C H_{2} C O_{2} H$

Answer :

Since we know that inductive effect depends on distance. Greater is the distance lesser is the effect. So, -I effect in $C H_{3} C H F C H_{2} C O_{2} H$ is more.

$C H_{3} C H F C H_{2} C O_{2} H$ is more acidic than $C H_{2} F C H_{2} C H_{2} C O_{2} H$ .

Q 8.8(iv) Which acid of each pair shown here would you expect to be stronger?

1649968633265

Answer :

Due to -I effect of Fluorine in A, it is easy to release proton ( $H^{+}$ ) but in compound B due to +I effect of methyl group it becomes difficlt te release protons. Therefore, compound A is more acidic than compound B.

necrt_19082

NCERT Back Exercise (QUESTIONS AND SOLUTIONS)

Q 8.1 (i). What is meant by the following terms ? Give an example of the reaction in each case.

Cyanohydrin

Answer:

Cyanohydrin -
When ketone and aldehyde react with the hydrogen cyanide ( $H C N$ ) to yield cyanohydrin. The Reaction is very slow with the pure HCN, so it can be catalyzed by using a base.

necrt_19086

Q 8.1 (ii). What is meant by the following terms? Give an example of the reaction in each case.

Acetal

Answer:

Acetal -
When aldehyde reacts with one molecule of monohydric alcohol in presence of dry HCl, it gives intermediate compound known as hemiacetals, which on further reaction with one more molecule of alcohol gives a product (gem-dialkoxy compound), known as acetal.

For example:

necrt_19088

Q 8.1(iii). What is meant by the following terms? Give an example of the reaction in each case.

Semicarbazone

Answer:

Semicarbazone -
This is the derivative of the aldehyde and ketone and it is derived from the condensation reaction between aldehyde and ketone. For example:

necrt_19095

Q 8.1(iv). What is meant by the following terms? Give an example of the reaction in each case.

Aldol

Answer:

Aldol -
$β$ -hydroxy aldehyde is known as aldol and it can be prepared from condensation of aldehyde and ketone, having atleast one alpha-hydrogen atom in presence of dil. alkali as a catalyst.

For example:

necrt_190971

Q 8.1(v) What is meant by the following terms? Give an example of the reaction in each case.

Hemiacetal

Answer:

Hemiacetal -
Aldehyde reacts with one equivalent of a monohydric alcohol, to yield an alkoxy alcohol intermediate compound, in presence of dry hydrochloric acid. The intermediate is called hemiacetal.

For example:

necrt_19099

Q 8.1 (vi) What is meant by the following terms? Give an example of the reaction in each case.

Oxime

Answer:

Oxime -
Aldehyde and ketone on reacting with the hydroxylamine in a weakly basic medium give oximes.
The general form of oxime-

necrt_19100_11

For example:

necrt_19100_2

Q8.1 (vii) What is meant by the following terms? Give an example of the reaction in each case.

Ketal

Answer:

Ketal-

It is a cyclic compound, which is formed when a ketone reacts with the ethylene glycol in the presence of dry hydrochloric acid.

For example:

necrt_19103_1

Q 8.1(vii) What is meant by the following terms? Give an example of the reaction in each case.

Imine

Answer:

Imine -
These are the chemical compounds having carbon-nitrogen double bonds. It is formed when aldehyde and ketone react with ammonia and its derivatives.

For example:

necrt_19106

Q 8.1(ix) What is meant by the following terms? Give an example of the reaction in each case.

2,4-DNP-derivative

Answer:

2,4-DNP-derivative-

These are produced when aldehyde and ketone are reacted with the 2, 4-dinitrophenylhydrazine in a weak basic medium. 2, 4DNP test is used for distinguishing between aldehyde and ketone.

For example:

necrt_19107

Q 8.1 (x) What is meant by the following terms? Give an example of the reaction in each case.

Schiff’s base

Answer:

Schiff’s base-
When aldehyde and ketone are treated with the primary aliphatic or aromatic amines in the presence of acid produces Schiff's base.

For example:

necrt_19109

Q 8.2(i) Name the following compounds according to IUPAC system of nomenclature:

$C H_{3} C H (C H_{3}) C H_{2} C H_{2} C H O$

Answer:

$C H_{3} C H (C H_{3}) C H_{2} C H_{2} C H O$
1649968694133 IUPAC name of this compound is 4-methyl pentanal

Q 8.2(ii) Name the following compounds according to IUPAC system of nomenclature:

$C H_{3} C H_{2} C O C H (C_{2} H_{5}) C H_{2} C H_{2} C l$

Answer:

$C H_{3} C H_{2} C O C H (C_{2} H_{5}) C H_{2} C H_{2} C l$

1649968723336 The IUPAC name of the compound is 6-chloro-4ethyl hexane-3-one

Q8.2(iii) Name the following compounds according to IUPAC system of nomenclature:

$C H_{3} C H = C H C H O$

Answer:

$C H_{3} C H = C H C H O$
The structure of the compound is
1649968756828

The IUPAC name of the compound is But -2-en-1-al

Q8.2(iv) Name the following compounds according to IUPAC system of nomenclature:

$C H_{3} C O C H_{2} C O C H_{3}$

Answer:

$C H_{3} C O C H_{2} C O C H_{3}$
The structure of the given compound is-
1649968789852

The IUPAC name of the compound is - pentan-2, 4-dione.

Q8.2 (v) Name the following compounds according to IUPAC system of nomenclature:

$C H_{3} C H (C H_{3}) C H_{2} C (C H_{3})_{2} C O C H_{3}$

Answer:

$C H_{3} C H (C H_{3}) C H_{2} C (C H_{3})_{2} C O C H_{3}$
The structure of the compound is -
1649968823649

The IUPAC name of the given compound is 3, 3, 5-trimethyl hexan-2-one .

Q8.2 (vi) Name the following compounds according to IUPAC system of nomenclature:

$(C H_{3})_{3} C C H_{2} C O O H$

Answer:

$(C H_{3})_{3} C C H_{2} C O O H$
The structure of the compound is -
1649968858754

The IUPAC name of the compound is 3, 3-dimethyl butanoic acid .

Q8.2 (vii) Name the following compounds according to IUPAC system of nomenclature:

$O H C C_{6} H_{4} C H O - p$

Answer:

$O H C C_{6} H_{4} C H O - p$
The structure of the compound is-
1649968900656

The IUPAC name of the compound is Benzene-1, 4-dicarbaldehyde .

Q8.3(i) Draw the structures of the following compounds.

3-Methyl butanal

Answer:

The structure of the 3-methylbutanal is shown here-

necrt_19186

Q8.3 (ii) Draw the structures of the following compounds.

p-Nitropropiophenone

Answer:

The structure of p-nitro propiophenone is shown here-

necrt_19187

Q8.3 (iii) Draw the structures of the following compounds.

p-methyl benzaldehyde

Answer:

The strcuture of the $p$ -methyl benzaldehyde is shown here-

necrt_19189

Q8.3(iv) Draw the structures of the following compounds.

4-Methylpent-3-en-2-one

Answer:

The structure of the compound 4-methylpent-3-en-2-one is shown here-

necrt_19193

Q8.3(v) Draw the structures of the following compounds.

4-Chloropentan-2-one

Answer:

The structure of the compound 4-chloro-pentan-2-one is shown here-

necrt_19195

Q8.3 (vi) Draw the structures of the following compounds.

3-Bromo-4-phenylpentanoic acid

Answer:

The structure of the compound 3-Bromo-4-phenyl pentanoic acid is shown here-

necrt_19197

Q8.3 (vii) Draw the structures of the following compounds.

p,p’-dihydroxy benzophenone

Answer:

The structure of the compound p,p’-dihydroxy benzophenone acid is shown here-

necrt_19199

Q8.3 (viii) Draw the structures of the following compounds.

Hex-2-en-4-ynoic acid

Answer:

The structure of the compound Hex-2-en-4-ynoic acid acid is shown here-

necrt_19200

Q8.4 (i) Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

$C H_{3} C O (C H_{2})_{4} C H_{3}$

Answer:

$C H_{3} C O (C H_{2})_{4} C H_{3}$
The structure of the compound is
1649968936614 The IUPAC name of the compound is hept-2-one Common name is methyl-n-propyl ketone

Q8.4 (ii) Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

$C H_{3} C H_{2} C H B r C H_{2} C H (C H_{3}) C H O$

Answer:

$C H_{3} C H_{2} C H B r C H_{2} C H (C H_{3}) C H O$
The structure of the compound is

1649968969701 The IUPAC name of the compound is 4-bromo-2-methylhexanal
The common name is - $γ$ -bromo- $α$ -methyl-caproaldehyde

Q8.4 (iii) Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

$C H_{3} (C H_{2})_{5} C H O$

Answer:

$C H_{3} (C H_{2})_{5} C H O$
structure os the compound is

1649969010911 The IUPAC name of the compound is heptanal

Q8.4 (iv) Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

$P h - C H = C H - C H O$

Answer:

$P h - C H = C H - C H O$
structur eof the compound is

1649969068080 The IUPAC name of the sompound is 3-phenylprop-2-ene-al

Common name is $β$ -phenylacrolein

Q8.4 (v). Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

1649969125909

Answer:

1649969150883 The IUPAC name of the compound is cyclopentane carbaldehyde.

Q 8.4 (vi) Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

$P h C O P h$

Answer:

$P h C O P h$
structure of the compound is

1649969191854 The IUPAC name of the compound is Diphenyl mentanone.

The common name is - benzophenone

Q8.5 (i) Draw structures of the following derivatives.

The 2,4-dinitro phenyl hydrazone of benzaldehyde

Answer:

The structure of 2,4-dinitro phenyl hydrazone of benzaldehyde

necrt_19290

Q8.5 (ii) Draw structures of the following derivatives.

Cyclopropanone oxime

Answer:

The structure of the Cyclopropanone oxime

necrt_19291

Q 8.5 (iii) Draw structures of the following derivatives.

Acetaldehyde dimethyl acetal

Answer:

The structure of Acetaldehydedimethylacetal

necrt_19292

Q 8.5 (iv) Draw structures of the following derivatives.

The semicarbazone of cyclobutanone

Answer:

The structure of the semicarbazone of cyclobutanone

necrt_19293

Q8.5 (v) Draw structures of the following derivatives.

The ethylene ketal of hexan-3-one

Answer:

The structure of the ethylene ketal of hexan-3-one

necrt_19294

Q8.5 (vi) Draw structures of the following derivatives.

The methyl hemiacetal of formaldehyde

Answer:

The structure of the compound methyl hemiacetal of formaldehyde

necrt_19302

Q 8.6 (i) Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.

PhMgBr and then H ₃ O ⁺

Answer:

When cyclohexane carbaldehyde reacts with $P h M g B r$ in presence of dry ether and then $H_{3} O^{+}$ (hydrolysis)

1649969233356

Q 8.6 (ii) Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.

Tollens' reagent

Answer:

Structure of cyclohexanecarbaldehyde

Cyclohexanecarbaldehyde reacts with tollen's reagent and reduce it to silver ( $A g$ ) and oxidises itself to cyclohexane carboxylate ion
So the reaction is-

1594981421573

Q 8.6 (iii) Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.

Semicarbazide and weak acid

Answer:

Structure of cyclohexanecarbaldehyde
necrt_19307
The reaction of cyclohexane carbaldehyde with semicarbazide and weak acid is-
The $- N H_{2}$ group which is not involving in resonance with carbonyl group act as a nucleophile and attack on -CHO group to form product.

necrt_19307_2

Q 8.6 (iv) Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.

Excess ethanol and acid

Answer:

Structure of cyclohexanecarbaldehyde
necrt_19307
reaction of cyclohexanecarbaldehyde with the excess of ethanol and acid to form cyclohexanecarbaldehyde diethyl acetal.

necrt_19309

Q 8.6 (v) Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.

Zinc amalgam and dilute hydrochloric acid

Answer:

Structure of cyclohexane carbaldehyde
necrt_19307
When cyclohexane carbaldehyde reacts with zinc amalgam( $Z n - H g$ ) and dil. $H C l$ , the carbonyl group of reactant is reduced to $C H_{2}$ , it is also known as clemmensen reduction.

necrt_19312

Q 8.7 (i) 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.

Methanal

Answer:

The compounds of ketones or aldehyde having at least one $α$ - hydrogen atom give aldol condensation reaction. Here, methanal has not any alpha-hydrogen atom. So, it gives a cannizzaro reaction.

The product of the reactions are, in which in the product is reduced and another is oxidised to carboxylic acid salt-

necrt_19318

Q 8.7 (ii) 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.

2 methylpentanal

Answer:

In 2-methylpentanal, there is one alpha-hydrogen atom so it gives aldol condensation reaction. Hence aldol product is-

necrt_19321

Q 8.7 (iii) 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.

Benzaldehyde

Answer:

Benzaldehyde Structure
necrt_19326
In this structure we can clearly see that there is no alpha-hydrogen atom. So, it gives cannizzaro reaction.

necrt_19326_2

Q8.7 (iv) 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.

Benzophenone

Answer:

It does not give any of the reaction because it is a keto compound and having no alpha -hydrogen atom. Hence it does not give cannizzaro as well as aldol reaction.

Structure -
necrt_19330

Q 8.7 (v) 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.

Cyclohexanone

Answer:

Yes, this compound will give aldol condensation reaction due to the presence of alpha-hydrogen atom. Two moles of cyclohexanone react with each other in which one molecule act as nucleophile and other act as an electrophile.

necrt_19333

Q 8.7 (vi) 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.

1-Phenylpropanone

Answer:

1-Phenylpropanone has two alpha-hydrogen atom, So it gives aldol condensation reaction. It reacts with another molecule of 1-Phenylpropanone in the presence of dil. $N a O H$ .

necrt_19336

Q8.7(vii) 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.

Phenylacetaldehyde

Answer:

Phenylacetaldehyde, it has two alpha-hydrogen, which makes it possible to perform aldol condensation reaction. So, the reaction is shown here;

necrt_19339

Q8.7 (viii) 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.

Butan-1-ol

Answer:

It does not give any of the reaction, either aldol reaction or cannizzaro reaction because it is a alcohol.

Q 8.7(ix) 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.

2,2-Dimethylbutanal

Answer:

In this compound, there is no alpha hydrogen, which means it gives cannizzaro reaction in the presence of conc. sodium hydroxide ( $N a O H$ ). One molecule of it is oxidised and the other molecule is reduced.

necrt_19343

Q 8.8 (i) How will you convert ethanal into the following compounds?

Butane-1,3-diol

Answer:

On treating ethanal with dil. alkali it gives 3-hydroxybutanal, which on further reduction with $N a B H_{4}$ gives the product Butane-1,3-diol

necrt_19344

Q8.8 (ii) How will you convert ethanal into the following compounds?

But-2-enal

Answer:

On treating with he presence of dil. alkali ( $N a O H$ ), it gives 3-hydroxybuttanal which on further dehydration gives but -2-ene-al.

necrt_19345

Q 8.8 (iii) How will you convert ethanal into the following compounds?

But-2-enoic acid

Answer:

When the given substrate is reacting with the tollens reagent it produces but-2-enoic acid.

necrt_19349

Q 8.9 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?

Answer:

Propanal - $C H_{3} C H_{2} C H O$

butanal - $C H_{3} C H_{2} C H_{2} C H O$

In cross-aldol condensation, there are total four cases of reaction-

When two molecules of propanal react with each other
When two molecules of butanal reacts with eath other
When propanal act as a nucleophile and attacking on butanal(as an electrophile)
When butanal act as a nucleophile and attacking on propanal(as an electrophile)

Now the reactions are shown here:
(i)

necrt_19352_1
(ii)

necrt_19352_2
(iii)

necrt_19352_3
(iv)

necrt_19352_4

Q 8.10 An organic compound with the molecular formula $C_{9} H_{10} O$ forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid. Identify the compound.

Answer:

According to given information, the molecular formula is $C_{9} H_{10} O$ and it reduces tollens reagent. So, there must be an aldehyde group.And also it gives cannizaro reaction, it means the given copound has no $α$ -hydrogen atom. On oxidation it gives 1,2-benzenedicarboxylic acid. Thus, the aldehyde group ( $- C H O$ ) is directly attached with the benzene ring and also it should be para-substituted.

Hence, the structure of the compound (2-ethyl benzaldehyde) is-

necrt_19356_01

By following reaction we can undrstand this prblem-

necrt_19356_2

Q 8.12 (i) Arrange the following compounds in increasing order of their property as indicated:

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

Answer:

Increasing order (reactivity towards HCN)-
Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

necrt_19444
The attacking power of $C N$ nucleophile can be affected by two ways-

hindrance near carbonyl group
negative charge on $- C O$ group (through +I effect of methyl)

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By observing the structure of all compound we can arrange them in order to reactivity towards HCN.

Q 8.12 (ii) Arrange the following compounds in increasing order of their property as indicated:

$C H_{3} C H_{2} C H (B r) C O O H, C H_{3} C H (B r) C H_{2} C O O H, (C H_{3})_{2} C H C O O H$ $, C H_{3} C H_{2} C H_{2} C O O H$ (acid strength)

Answer:

increasing order of their acidic strength-

$(C H_{3})_{2} C H C O O H < C H_{3} C H_{2} C H_{2} C O O H < C H_{3} C H (B r) C H_{2} C O O H < C H_{3} C H_{2} C H (B r) C O O H$

Due to presence of Bromine, which shows -I effect and we know that -I grows weaker as distance increases. And in case of n-propyl and isopropyl, the +I effect is more in isopropyl, so it is weak in acidic strength.
$- I \propto a c i d i c$ nature
$+ I \propto 1 / a c i d i c$

Q 8.12 (iii) Arrange the following compounds in increasing order of their property as indicated:

Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

Answer:

Increasing order of acidic strength -

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

As we know, electron withdrawing groups increase the strength of the acid and electron donating group decreases the strength of acids. Therefore, 4-methoxybenzoic acid is the weakest acid among them and 3,4-Dinitrobenzoic acid is the strongest due to the presence of 2 electron withdrawing groups and then 4-Nitrobenzoic acid.

Q 8.13 (i) Give simple chemical tests to distinguish between the following pairs of compounds.

Propanal and Propanone

Answer:

Both the copound can be distinguished by following tests-

tollens test
iodoform test
fehlings test

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Propanal reduces the tollens reagent but poropanone does not.
Akdehyde responds to fehlings test but ketone does not. So, Propanal gives positive fehlings test but not by propanone.

Q 8.13 (ii) Give simple chemical tests to distinguish between the following pairs of compounds.

Acetophenone and Benzophenone

Answer:

Acetophenone gives ppositive iodoform test, it react with $N a O I$ to give yellow ppt. but benzophenone does not give this test.

$C_{6} H_{5} C O C H_{3} + N a O I \to C_{6} H_{5} C O O N a + C H I_{3} + N a O H$

benzophenone + $N a O I$ $\to$ no yellow ppt.

Q 8.13 (iii) Give simple chemical tests to distinguish between the following pairs of compounds.

Phenol and Benzoic acid

Answer:

Phenol gives ferric chloride test ( $F e C l_{3}$ ) , on reaction with ferric chloride phenol gives violet coloured solution but benzoic acid does not.

necrt_19450

Q 8.13 (iv) Give simple chemical tests to distinguish between the following pairs of compounds.

Benzoic acid and Ethyl benzoate

Answer:

Both can be distinguish by sodium bicarbonate test, In this test, benzoic acid react with $N a H C O_{3}$ to give brisk effervescence due to evoltion of carbon dioxide ( $C O_{2}$ ) but Ethyl benzoate does not.

necrt_19451

Q8.13 (v) Give simple chemical tests to distinguish between the following pairs of compounds.

Pentan-2-one and Pentan-3-one

Answer:

They can be distinguished by iodoform test,
This test is given by compounds having methyl ketone group. So, pentan-2-one give this test and pentan-3-one does not.

necrt_19452

Q 8.13 (vi) Give simple chemical tests to distinguish between the following pairs of compounds.

Benzaldehyde and Acetophenone

Answer:

Both can be distinguished by-

Iodoform test
In acetophenone, methyl ketone group present.So it give positive iodoform test by forming a yellow ppt.
Tollen's test
Aldehyde give positive tollens test. Benzaldehyde reduces the tollens reagent and give red brown ppt.but acetophenone does not give this test.

Q 8.13 (vii) Give simple chemical tests to distinguish between the following pairs of compounds.

Ethanal and Propanal

Answer:

Ethanal and Propanal can be distinguished by Iodoform test.
Ethanal has one methyl group attached with Carbonyl group, so it gives a positive iodoform test but propanal does not give this test.
necrt_19455

Q 8.14(i) 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

Answer:

Benzene on bromination formed bromobenzene and after that we have to react with $M g$ in presence of ether it becomes Grignard reagents ( $R - M g X$ ). Grignard reagents react with carbon dioxide (dry ice) to form salts of carboxylic acids, which on acidic hydrolysis gives benzoic acid. On esterification, it gives methyl benzoate.

necrt_19457

Q 8.14 (ii) 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.

m-nitrobenzoic acid

Answer:

Benzene on bromination formed bromobenzene and after that we have to react with $M g$ in presence of ether it becomes grignard reagents ( $R - M g X$ ). Grignard reagents reacts with carbon dioxide (dry ice) to form salts of carboxylic acids, which on acidic hydrolysis gives benzoic acid. Finally after doing nitration we get our desired product (m-Nitrobenzoic acid).

necrt_19458

Q 8.14 (iii) 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

(iii) p-nitrobenzoic acid

Answer:

By friedel craft alkylation of benzene give methyl benzene, which on nitration give meta and para substituted methyl benzene, para substituted is major product. Oxidation of para substituted methyl benzene( $K M n O_{4} / K O H$ ) gives salts of carboxylic acid, which on further acidic hydrolysis gives p-nitrobenzoic acid.

necrt_19460

Q 8.14 (iv) 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

Phenyl acetic acid

Answer:

Alkylation of benzene in presence of anhydrous $A l C l_{3}$ gives toluene, which on further reaction with $B r_{2} / Δ / h v$ gives benzyl bromide and after that reacts with alc. KCN, CN replace the Br and formed benzyle cyanide which on acidic hydrolysis gives phenyl acetic acid.

necrt_19461

Q 8.14(v) 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

p-nitrobenzaldehyde

Answer:

Alkylationof benzene gives toluene, which on further reaction with $H N O_{3} / H_{2} S O_{4}$ gives p- nitrotoluene and react it with the the carbon disulphide / $C r O_{2} C l_{2}$ followed by acidic hydrolysis to form $p$ -nitrobenzaldehyde.

necrt_19463

Q 8.15(i) How will you bring about the following conversions in not more than two steps?

Propanone to Propene

Answer:

First, reduce propanone with the help of lithium aluminum hydride ( $L i A l H_{4}$ ) to get propan-2-ol and then by dehydration remove water molecule to get propene.

1650361743870

Q 8.15(ii) How will you bring about the following conversions in not more than two steps?

Benzoic acid to benzaldehyde

Answer:

With the help of mild reducing agent we can reduce benzoic acids into benzaldehyde.
necrt_19465
alternate method-

necrt_19465_1

First react it with $S O C l_{2}$ which replace OH group and then hydogenation reaction.

Q 8.15 (iii) How will you bring about the following conversions in not more than two steps?

Ethanol to 3-hydroxy butanal

Answer:

Ethanol on heating at 573K in presence of copper it convert into ethanal and then by aldol condensation we get 3-hydroxybutanal.

necrt_19470

Q 8.15 (iv) How will you bring about the following conversions in not more than two steps?

Benzene to m-nitroacetophenone

Answer:

Benzene on reacting with acetic anhydride $(C H_{3} C O)_{2} C O$ , it gives acetophenone And further reacting acetophenone with nitric acid in presence of sulphuric acid give meta substituted product, m-nitroacetophenone.

necrt_19471_1

Q 8.15 (v) How will you bring about the following conversions in not more than two steps?

Benzaldehyde to benzophenone

Answer:

On oxidation of benzaldehyde, it converted into benzoic acid, which on further reacting with $C a C O_{3}$ it formed calcium benzoate. After distillation we get benzophenone.

necrt_19473

Q 8.15 (vi) How will you bring about the following conversions in not more than two steps?

Bromobenzene to 1-phenylethanol

Answer:

Bromobenzene on reacting with $M g$ in presence of dry ether it formed grignard reagents ( $R - M g - X$ ), which on further reacting with ethanal followe by acidic hydrolysis gives 1-phenyl ethanol

necrt_19474

Q 8.15 (vii) How will you bring about the following conversions in not more than two steps?

Benzaldehyde to 3-Phenylpropan-1-ol

Answer:

BY cross aldol-condesation of benzaldehyde and acetaldehyde gives 3-phenylprop-2-ene-al, whixh on catalytic hydrogenation gives 3-Phenylpropan-1-ol

necrt_19478

Q 8.15 (viii) How will you bring about the following conversions in not more than two steps?

Benzaldehyde to α-hydroxy phenyl acetic acid

Answer:

Nucleophilic reaction of benzaldehyde with $N a C N$ in presence of HCl gives benzaldehyde cyanohydrine, which on acidic hydrolysis formed α-Hydroxyphenylacetic acid

necrt_19481

Q 8.15(ix) How will you bring about the following conversions in not more than two steps?

Benzoic acid to m-nitrobenzyl alcohol

Answer:

On nitration of benzoic acid gives meta substituted nitro benzoic acid which on further reacting with reducing agent and followed by acidic hydrolysis gives nitrobenzyl alcohol.

necrt_19484

Q 8.16 (i) Describe the following.

Acetylation

Answer:

Acetylation-
The addition of acetyl functional group ( $C H_{3} C O -$ ) in an organic compound is called acetylation. Acetic anhydride( $(C H_{3} C O)_{2} C O$ ) and acetyl chloride ( $C H_{3} C O C l$ ) are mostly used as acetylating agents. This reaction is happens in presence of a base such as pyridine etc.

necrt_19486

Q 8.16 (ii) Describe the following.

Cannizzaro reaction

Answer:

Cannizzaro reaction-
Aldehyde which do not have any alpha- hydrogen, undergo self oxidation as well as reduction reaction on treating with concentrated alkalies. In this reaction, one molecule gets reduced and other is oxidised to carboxylic acid salt.

necrt_19488

Q 8.16 (iii) Describe the following.

Cross aldol condensation

Answer:

Cross aldol condensation -
In aldol condensation, if the reactants are two different aldehyde or ketone, then it is called cross-aldol condensation. If both of the reactants contains alpha-hydrogen atoms then it gives a mixture of four products.

necrt_19489_1
If the reactants contain alpha-hydrogen atoms then,

necrt_19489_2

Q 8.16 (iv) Describe the following.

Decarboxylation

Answer:

Decarboxylation -
The reaction in which the carboxylic acid loses carbon dioxide to hydrocarbon, in the presence of sodalime( $N a O H$ and $C a O$ in the ratio of 3:1). This reaction is known as decarboxylation.

necrt_19496

Q 8.17(i) Complete each synthesis by giving missing starting material, reagent or products.

$\to_{K O H, h e a t}^{K M n O_{4}}$

Answer:

Potassium permanganate oxidise the substituted ethyl group to $- C O O H$ group and due to presence of KOH, formed carboxylic acid salts (potassium benzoate)
necrt_19497

Q 8.17(ii) Complete each synthesis by giving missing starting material, reagent or products.

1649969333694 $\to_{h e a t}^{S o C l_{2}}$

Answer:

The -OH groups of both carboxylic acid is replaced by the chlorine
So, the overall reaction is

necrt_19498

Q 8.17 (iii) Complete each synthesis by giving missing starting material, reagent or products

$C_{6} H_{5} C H O \overset{H_{2} N C O N H N H_{2}}{\to}$

Answer:

In this reaction, the $- N H_{2}$ group which not directly attached with the carbonyl group act as a nucleophile and attack on -CO group of benzaldehyde and water molecule is also produced as a by-product.

So, the final product is-

necrt_19499

Q8.17(iv) Complete each synthesis by giving missing starting material, reagent or products.

1649969382126

Answer:

Benzene reacts with benzoyl chloride in presence of anhydrous aluminium chloride ( $A l C l_{3}$ ) to give benzophenone and also HCl as a by-product.

necrt_19500

Q 8.17(v) Complete each synthesis by giving missing starting material, reagent or products.

$\overset{{[A g {(N H_{3})}_{2}]}^{+}}{\to}$

Answer:

SInce, the give reactant is a aldehyde compound, So it give positive tollen's test by reducing the tollens reagent.

necrt_19501

Q 8.17 (vi) Complete each synthesis by giving missing starting material, reagent or products.

1649969442329 $\overset{N a C N / H C l}{\to}$

Answer:

In this reaction, the nucleophile CN attacks on -CHO group because it has more reactivity towards $C N^{-}$ .

So the reaction gives

necrt_19503

Q8.17 (vii) Complete each synthesis by giving missing starting material, reagent or products.

$C_{6} H_{3} C H O + C H_{3} C H_{2} C H O \to_{Δ}^{d i l N a O H}$

Answer:

The given eaction is cross aldol reaction between benzaldehyde and propanal in presence of dil. sodium hydroxide

necrt_19504

Q 8.17 (viii) Complete each synthesis by giving missing starting material, reagent or products.

$C H_{2} C O C H_{2} C O O C_{2} H_{5} \to_{i i . H^{+}}^{i . N a B H_{4}}$

Answer:

In this reaction $N a B H_{4}$ is a reducing agent, which reduces the CO group of the compound.

necrt_19505

Q8.17(ix) Complete each synthesis by giving missing starting material, reagent or products.

$\overset{C r O_{2}}{\to}$

Answer:

$C r O_{2}$ oxidises the OH(alcohol) group into keto-group (-CO) and finally the product produced is cyclohexanone

necrt_19506

Q 8.17(x) Complete each synthesis by giving missing starting material, reagent or products.

necrt_19507_1

Answer:

Diborane reacts with an alkene to give trialkyl boranes (addition product) and this is oxidised to alcohol by using hydrogen peroxide.

After then pyridinium chlorochromate (PCC) is reacted with alcohol to convert it into Aldehyde group.

necrt_19507

Q 8.17 (xi) Complete each synthesis by giving missing starting material, reagent or products.

$\to_{i i . Z n - H_{2} O}^{i . O_{3}}$ 1649969486840

Answer:

The above reaction is ozonolysis reaction in which ozone molecule breaks the double bond of the alkene and formed a ketone or aldehyde as a product.

necrt_19509

Q 8.18(i) Give plausible explanation for each of the following:

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

Answer:

Structure of 2,2,6-trimethylcyclo hexanone
necrt_19511
Cyclohexanone forms cyanohydrin when the nucleophile ( $H C N / C N^{-}$ ) attack on cyclohexanone.
necrt_19511_1
In this case, there is less hindrance around the CO group so the nucleophile can attack easily. But in case of 2,2,6-trimethylcyclo hexanone at the alpha ( $α$ ) position, there is a hindrance because of the methyl groups. Thus the nucleophile cannot attack effectively.

Q 8.18(ii) Give plausible explanation for each of the following:

There are two –NH ₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

Answer:

In Semicarbazide,

necrt_19512

One of the $- N H_{2}$ group is involved in resonance with the carbonyl group, which is directly attached. Therefore, it cannot act as a nucleophile (its electrons are less available or less electron density at that $- N H_{2}$ group). But the other $- N H_{2}$ group is not participating in resonance, so it can act as nucleophile and attack on the carbonyl carbon of ketone and aldehyde.

Q 8.18(iii) Give plausible explanation for each of the following:

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.

Answer:

$R C O O H + R^{'} O H ⇌ R C O O R^{'} + H_{2} O$
If either water or the ester is not removed immediately then ester can react with water and again and give back the reactants.

Thus to shift the reaction in forward direction we need to produce more ester or remove either of the two.

Q 8.19 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 hydrogen sulphite and give a positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Answer:

In the question, it is given that,
% of carbon = 69.77
% of Hydrogen = 11.63
% of Oxygen = (100-69.77-11.63) = 18.6

Now, the no. of moles of the components are-
$n_{c} = \frac{69.77}{12} = 5.81$
$n_{H} = \frac{11.63}{1} = 11.63$
$n_{O} = \frac{18.6}{16} = 1.16$
Thus ration of C:H:O = 5.81:11.63:1.16
To get empirical formula we need to divide them by 1.16 .So,
C:H:O= 5:10:1
So, the empirical formula is $C_{5} H_{10} O$
The molecular mass of the compound = $(5 \times 12) + (10 \times 1) + (1 \times 16) = 86$

Since the compound does not give tollens test it means it is not an aldehyde. It is given that, the compound gives positive iodoform test it means the compound must be methyl ketone.
On oxidation, it gives ethanoic acid and propanoic acids. Therefore, the compound should be

Pentan-2-one
the structure of the compound is - $C H_{3} - C O - (C H_{2})_{2} - C H_{3}$

Q 8.20 Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Answer:

In carboxylate ion the electron is delocalised between oxygen, which is electronegative in nature. But in case of phenoxide ion, the electron are delocalise between less electronegative atom and also phenoxide ion has non-equivalent resonance structure. Therefore, carboxylate ion is more resonance stable than phenoxide ion and we know that more stable the conjugate base of an acid, high strong is the acidic

necrt_19518

Learn More About Aldehyde Ketone and Carboxylic Acid NCERT solutions:

The carbonyl functional group(>C=O) is bonded to a carbon and hydrogen atom, while in the ketones, the carbonyl group is bound to two carbon atoms. In the carboxylic acid, carbon of the carbonyl group(>C=O) is bonded to hydrogen or carbon and oxygen of hydroxyl moiety(-OH), while in amides carbon is attached to hydrogen or carbon and nitrogen of moiety, and in acyl halides, carbon is attached to hydrogen or carbon and to halogens. Read further to know more about these NCERT solutions for the Class 12 Chemistry PDF download.

In aldehydes, the carbonyl functional group(>C=O) is bonded to a carbon and hydrogen atom, while in ketones, the carbonyl group is bound to two carbon atoms. In the carboxylic acid, carbon of carbonyl group(>C=O) is bonded to hydrogen or carbon and oxygen of hydroxyl moiety(-OH), while in amides carbon is attached to hydrogen or carbon and nitrogen of moiety $- N H_{2}$ and in acyl halides, carbon is attached to hydrogen or carbon and to halogens. Following are the basic structures of aldehydes, ketones, carboxylic acids, acyl halide, and amide.

An Insight to Aldehydes Ketones and Carboxylic Acids NCERT Solutions:

NCERT Class 12 Chemistry solutions deals with aldehydes, ketones, and carboxylic acid compounds, their IUPAC names, physical properties, and chemical reactions. In this chapter, there are 20 questions in the exercise. The NCERT solutions for class 12 chemistry Aldehydes, Ketones and Carboxylic Acids will help you in your preparation of CBSE board exams as well as competitive exams like JEE Mains, NEET, BITSAT, etc. In Class 12 Chemistry NCERT solutions, students have studied organic compounds containing carbon-oxygen single-bond functional groups. In NCERT solutions for class 12 chemistry Aldehydes, Ketones and Carboxylic Acids, students will learn about organic compounds containing carbon-oxygen double bond (>C=O) that is known as carbonyl group. In organic chemistry, carbonyl group (>C=O) is one of the most important functional groups.

NCERT Solutions For class 12 Chemistry Aldehydes, Ketones, and Carboxylic Acids

8.1 Aldehydes, Ketones, and Carboxylic Acids( Introduction )

8.2 Nomenclature and Structure of Carbonyl Compounds

8.3 Preparation Of Aldehydes And Ketones

8.4 Physical Properties of Aldehydes and Ketones

8.5 Chemical Reactions of Aldehydes and Ketones

8.6 Uses of Aldehydes and Ketones

8.7 Carboxylic Acids : Structure and Nomenclature

8.8 Physical Properties of Carboxylic Acids

8.9 Chemical Reactions of Carboxylic Acids

8.10 Uses of Carboxylic Acids

NCERT Solutions for class 12 Chemistry

Chapter 1	NCERT solutions for class 12 chemistry chapter 1 Solutions
Chapter 2	NCERT Solutions for class 12 chemistry chapter 2 Electrochemistry
Chapter 3	NCERT solutions for class 12 chemistry chapter 3 Chemical Kinetics
Chapter 4	NCERT Solutions for class 12 chemistry chapter 4 The d and f block elements
Chapter 5	NCERT solutions for class 12 chemistry chapter 5 Coordination compounds
Chapter 6	NCERT Solutions for class 12 chemistry chapter 6 Haloalkanes and Haloarenes
Chapter 7	NCERT solutions for class 12 chemistry chapter 7 Alcohols, Phenols and Ethers
Chapter 8	NCERT solutions for class 12 chemistry chapter 8 Aldehydes, Ketones and Carboxylic Acids
Chapter 9	NCERT solutions for class 12 chemistry chapter 9 Amines
Chapter 10	NCERT solutions for class 12 chemistry chapter 10 Biomolecules

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Frequently Asked Questions (FAQs)

1. Where can I find complete solutions for NCERT class 12 Chemistry?

For complete solutions : https://school.careers360.com/ncert/ncert-solutions-class-12-chemistry

2. What are the important topics of the chapter Aldehydes, Ketones and Carboxylic Acids?

Aldehydes
Ketones
Hydration of Alkynes
Formation of Alcohols
Wolff- Kishner Reduction
Halogenation
Cannizzaro Reaction
Friedel- Crafts Acylation.
Nuclei Addition Reactions
Formation of Cyanohydrins

3. What is the weightage of NCERT Class 12 Chemistry Aldheyde ketones and carboxylic acids in NEET?

Weightage of this chapter is 3 percent. The questions of NEET can be practiced using NCERT book exercises, NCERT exemplar pronlems and NEET previous year papers.

4. What is the weightage of NCERT class 12 Chemistry Aldheyde ketones and carboxylic acids in JEE Mains ?

This chapter holds weightage of 2-3 marks in JEE mains

5. What is the weightage of NCERT class 12 Chemistry Aldheyde ketones and carboxylic acids in CBSE Board Exams?

Weightage of this chapter in CBSE Board exams is 5-6 marks.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Read Complete Answer

Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer

Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

NCERT Solutions Class 12 Aldehyde Ketone and Carboxylic Acids (Intext Question from 8.1 to 8.8)

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JEE Main Important Mathematics Formulas

Learn More About Aldehyde Ketone and Carboxylic Acid NCERT solutions:

An Insight to Aldehydes Ketones and Carboxylic Acids NCERT Solutions:

NCERT Solutions For class 12 Chemistry Aldehydes, Ketones, and Carboxylic Acids

NCERT Solutions for class 12 Subject-wise

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