NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids: Free PDF

Ncert Solutions Chemistry Class 12th 2023 ( Chemistry Ncert Solutions Class 12th )

Updated on Sep 27, 2023 10:52 IST

NCERT Class 12 Chemistry Aldehydes, Ketones and Carboxylic Acids Solutions: The solutions of NCERT Chemistry Class 12 for Aldehyde, Ketone and Carboxylic Acid is provided here by highly experienced subject matter expert in easy to understand language. It will help students understand the concepts easily which would be beneficial in scoring good marks in the exams.

In aldehydes, the carbonyl group is bonded to a carbon and hydrogen while in the ketones, it is bonded to two carbon atoms. The carbonyl compounds in which carbonyl group is bonded to oxygen are known as carboxylic acids.

Topics Covered in NCERT Chemistry Class 12 Aldehyde, Ketones and Carboxylic Acids Chapter

Students can find below the list of all topics covered in Aldehyde, Ketone and Carboxylic Acid chapter of NCERT Chemistry Class 12.

  1. Nomenclature and Structure of Carbonyl Group
  2. Preparation of Aldehydes and Ketones
  3. Physical Properties
  4. Chemical Reactions
  5. Uses of Aldehydes and Ketones
  6. Nomenclature and Structure of Carboxyl Group
  7. Methods of Preparation of Carboxylic Acids
  8. Physical Properties
  9. Chemical Reactions
  10. Uses of Carboxylic Acids

NCERT Chemistry Class 12th Solution PDF - Aldehyde, Ketones and Carboxylic Acids Chapter Download

Candidates can check here NCERT Class 12 Chemistry Aldehyde, Ketone and Carboxylic Acids solutions provided by subject experts in easy to understand language. It will help students to score good marks in the examination. 

Download Here:NCERT Solution for Class XII Chemistry Aldehyde, Ketone and Carboxylic Acids PDF

 

Aldehyde, Ketones and Carboxylic Acids Solutions and FAQs

Get here all the questions of NCERT Class 12th Aldehyde, Ketones and Carboxylic Acids Chapter.

Q12.1 Write the structures of the following compounds. (i) α-Methoxypropionaldehyde (ii) 3-Hydroxybutanal (iii) 2-Hydroxycyclopentane carbaldehyde (iv) 4-Oxopentanal (v) Di-sec. butyl ketone (vi) 4-Fluoroacetophenone

A 12.1 (i) The compound contains ether group and an aldehydic functional group with the longest chain having 3 carbon atoms. α means the position of the carbon atom attached to attached to the carbon atom of the functional group, in this case, it is an aldehyde functional group.

  • (ii) The compound contains an alcoholic group and an aldehydic functional group with the longest chain having 4 carbon The numbering of the chain starts from a carbon atom of aldehyde group.
  • (iii) The compound contains alcoholic group and an aldehydic functional group with the longest chain having 5 carbon atoms which are cyclic in The numbering of the chain starts from a carbon atom of aldehyde group attached to the cyclopentane ring.
  • (iv) The compound contains ketone group and an aldehydic functional group with the longest chain having 5 carbon The numbering of the chain starts from a carbon atom of aldehyde group.
  • (v) The compound contains ketone functional group with the longest chain having 8(butyl is used because there is two similar butyl group attached to the carbonyl carbon of ketone group) carbon
  • (vi) The compound contains a fluorine atom and a ketone functional group with the longest chain which is a derivative of a phenyl group and an acetyl substituent.

Q 12.2 Write the structures of products of the following reactions

A 12.2 dia

Q 12.3 Arrange the following compounds in increasing order of their boiling points.

  • CH3 CHO, CH3 CH2OH, CH3OCH3 , CH3 CH2 CH3

A 12.3 The molecular masses of the given compounds are in the range of 44-66 g/mol. The second molecule, CH3CH2OH contains OH alcoholic group due to which it undergoes extensive H bonding with each other, leading to the association of the molecules. Therefore it has a highest boiling point. In the molecule CH3CHO , there is strong intermolecular dipole-dipole attraction due to presence of - CHO aldehydic group( as H will have partial positive charge and oxygen will have partial negative charge, so there will attraction between molecules),which is weak in case of CH3OCH3 as both CH3 groups have +I effect which results in decrease in electron affinity for the oxygen attached to it( with no partially positive H atom). And in CH3CH2CH3 there are only weak van der Waals forces of attraction. So the compounds in increasing order of their boiling point are:

CH3CH2CH3< CH3OCH3< CH3CHO< CH3CH2OH.

Q 12.4 Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i) Ethanal, Propanal, Propanone, Butanone. (ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

A 12.4 (i) Ethanal, Propanal, Propanone, Butanone

The +I effect increases in the order as:

Ethanal

Therefore the electron density of the carbonyl carbon increases as the +I effect due to the alkyl group increases. As a result, the chances of attack by a nucleophile (which are rich in electron and carries a negative charge, not necessary usually have a lone pair of an electron) decreases because the carbonyl carbon has electron density on it. Hence the increasing order of the reactivity of given compounds in nucleophilic addition reaction is:

Ethanal > Propanal> Propanone > Butanone.

(ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone:

The +I effect is more in ketone than in aldehyde because in ketone there are 2 alkyl groups contributing in the +I effect whereas in aldehydes there is only one alkyl group. Hence, acetophenone(being ketone group attached to it) is the least reactive in nucleophilic addition reactions.

Among aldehydes, the +I effect is the highest in p-tolualdehyde because of the presence of the electron- donating –CH3 group(which increase the electron density on the carbonyl carbon via resonance through the benzene ring) and the lowest in p-nitrobenzaldehyde because of the presence of the electron- withdrawing –NO2 group(which decreases the electron density on the carbonyl carbon via resonance through the benzene ring). Hence, the increasing order of the reactivity of the given compounds is:

Acetophenone<p-tolualdehyde p-Nitrobenzaldehyde

Q 12.5 Predict the products of the following reactions:

A 12.5 Cyclopentanone reacts with hydroxylamine to give cyclopentanone oxime. As you can see the cyclopentanone has ketone functional group with oxygen attached to carbonyl carbon and the hydroxylamine has two H atoms attached to N. And hence in the product this O and two H atoms are removed and form a water molecule.

Remember if any aldehyde or ketone bearing compound reacts with primary or secondary amine then for finding the product formed simply remove the oxygen atom from the aldehyde or ketone functional group and two H atoms attached to N in amines and add a double bond between C and N.

 

 

Q 12.6 Give the IUPAC names of the following compounds:

A 12.6 (i) Ph CH2 CH2 COOH The compound contains carboxylic acid as the functional group and Ph means a phenyl group. The longest chain contains 3 carbon atom with phenyl group of carbon 3 so the IUPAC name of the compound is 3-phenylpropanoic

  • (ii) (CH3 ) 2 C=CHCOOH

The longest chain contains 4 carbon atom with a methyl group as a substituent and a carboxylic acid as a functional group. The no. Of the chain starts from the carbon atom of the –COOH group with double bond on carbon 2 so the IUPAC name of the compound is 3-Methylbut-2-enoic acid.

  • (iii) The longest chain has 5 carbon atom(all are saturated) which are cyclic with the carboxylic group attached and a methyl group as a substituent. The no. Of the chain starts from the carbon atom of the – COOH group so the IUPAC name of the compound is 2-Methylcyclopentane carboxylic acid.
  • (iv) The longest chain is derivative of benzene with the carboxylic group attached and 3 nitro groups. The number of the chain starts from the carbon atom of the –COOH group so the IUPAC name of the compound is 2,4,6-Trinitrobenzoic acid.

Q 12.7 12.7 Show how each of the following compounds can be converted to benzoic acid. (i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene)

A 12.7 (i) Ethylbenzene

In the presence of nascent oxygen i.e. [O] along with KMnO4/ KOH(it is a very strong oxidizing agent which oxidizes the whole alkyl group to carboxylic salt) followed by hydrolysis leads to the formation of benzoic acid. Carbon dioxide and water are formed as byproducts.

(ii) Acetophenone

The oxidation with alcoholic KMNO4 followed by hydrolysis leads to the formation of benzoic acid. The reaction is given below:

(iii) Bromobenzene is first reacted with magnesium in presence of dry ether to give an intermediate which on reaction with solid carbon dioxide followed by hydrolysis leads to the formation of benzoic acid.

(iv) Phenylethene (Styrene)

Here the oxidation of styrene with alcoholic KMNO4 is followed by hydrolysis leads to the formation of benzoic acid.

Q 12.8 Which acid of each pair shown here would you expect to be stronger? (i) CH3CO2H or CH2FCO2H (ii) CH2FCO2H or CH2ClCO2H (iii) CH2 FCH2 CH2 CO2H or CH3 CHFCH2 CO2H

(i) CH3CO2H or CH2FCO2H (ii) CH2FCO2H or CH2ClCO2H

(iii) CH2 FCH2 CH2 CO2H or CH3 CHFCH2 CO2H

A 12.8 (i) The +I effect of –CH3 group increases the electron density on the O-H bond. Therefore, the release of proton becomes difficult. On the other hand, the -I effect of F decreases the electron density on the O-H bond. Therefore, the proton can be released easily. Hence, CH2FCO2H is a stronger acid than CH3CO2H.

  • (ii) F has stronger -I effect than Cl(as fluorine is more electronegative than chlorine). Therefore, CH2FCO2H can release proton more easily than CH2ClCO2H. Hence, CH2FCO2H is a stronger acid than CH2ClCO2H.
  • (iii) Inductive effect decreases with increase in distance (the more will be the distance of the inductive group from the carboxylic group less will be its effect of polarisability). Hence, the -I effect of F in CH3CHFCH2CO2H is more than it is in CH2FCH2CH2CO2H . Hence, CH3CHFCH2CO2H is a stronger acid than CH2FCH2CH2CO2H .
  • (iv) Due to the -I effect of F (being electronegative), it is easier to release proton in the case of compound (A). However, in the case of compound (B), the release of the proton is difficult due to the +I effect of –CH3 e. methyl group. Hence, (A) is a stronger acid than (B).

Exercise Q 12.1 What is meant by the following terms? Give an example of the reaction in each case. (i) Cyanohydrin (ii) Acetal (iii) Semicarbazone (iv) Aldol (v) Hemiacetal (vi) Oxime (vii) Ketal (vii) Imine (ix) 2,4-DNP-derivative (x) Schiff’s base

A 12.1 (i) Cyanohydrin:

Cyanohydrins are those organic compounds having the formula RR“2C(OH)CN, wherever R and R“2 can be alkyl or aryl groups.

Aldehydes and ketones react with compound (KCN) within the presence of excess cyanide (NaCN) as a catalyst to field organic compound.

  • (ii) Acetal:

Acetals are gem - dialkoxy alkanes within which 2 alkoxy teams groups attached to the terminal atom. One bond is connected to associate degree alkyl whereas the opposite is connected to the hydrogen atom.

When aldehydes are treated with 2 equivalents of a monohydric alcohol within the presence of dry HCl gas, hemiacetals are produced which are further reacted with one more molecule to alcohol to yield acetal as shown below:

  • (iii) Semicarbarbazone:

Semicarbazones are the derivatives of organic compounds produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

Semicarbazones square measure helpful for identification and characterization of aldehydes and ketones.

  • (iv) Aldol:

An aldol is a β-hydroxy organic compound. It's produced by the by the condensation reaction of 2 molecules of an equivalent or one molecule every of 2 totally different aldehydes or ketones within the presence of a base.

The reaction is shown as below:

  • (v) Hemiacetals are α-alkoxyalcohols.

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

The following reaction shows the formation of hemiacetal:

  • (vi) Oxime:

Oximes are a category of organic compounds having the final formula RR“2CNOH, where R is an associate organic aspect chain associated R“2 is either H or an organic aspect chain. If R“2 is H, then it's called aldoxime associated if R“2 is an organic aspect chain, it's called ketoxime.

On treatment with hydroxylamine in a very decrepit acidic medium, aldehydes or ketones kind oximes.

  • (vii) Ketal:

Ketals area gem – dialkoxyalkanes in which two which 2 alkoxy teams are attached inside the chain. The opposite 2 bonds of the atom are unit connected to 2 alkyl groups.

The general structure of ketal is shown as below:

Ketones react with glycol within the presence of dry HCl gas to grant a cyclic product referred to as glycol ketals.

(vii) Imine:

An organic compound which contains a carbon–nitrogen double bond. The structure of imine is shown as below:

Imines are produced when aldehydes and ketones react with ammonia Gas.

  • (ix) 2, 4 - DNP - derivative:

2, 4 - DNP - derivative is a substituted hydrazine,

The molecular formula for 2,4-DNP–derivative is C6H3(NO2)2NHNH2.

The other name is Dinitrophenylhydrazine is the chemical compound. Dinitrophenylhydrazine is a red to orange solid. An aldehydes or ketones react with 2, 4 - dinitrophenylhydrazine to form yellow, orange, or red coloured derivatives named as 2,4 dinitrogenphenylhydrazones. These are also known as 2,4-DNP derivative.

  • (x) Schiff's base:

Aldehydes and ketones on react with primary amines in the presence of trace of an acid yields a Schiff's base.

Q 12.2 Name the following compounds according to IUPAC system of nomenclature:

(i) CH3CH(CH3 )CH2CH2CHO                                          (ii) CH3CH2COCH(C2H5 )CH2CH2Cl

(iii) CH3CH=CHCHO                                                         (iv) CH3COCH2COCH3

(v) CH3CH(CH3 )CH2C(CH3 )2COCH3                            (vi) (CH3 )3CCH2COOH

(vii) OHCC6H4CHO-p

A 12.2 

Q 12.3 Draw the structures of the following compounds. (i) 3-Methylbutanal (ii) p-Nitropropiophenone (iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one (v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid (vii) p,p’-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid

A 12.3 

Q 12.4 Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

  • CH3CO(CH2 )4CH3
  • CH3CH2CHBrCH2CH(CH3 )CHO
  • CH3 (CH2 )5CHO
  • Ph-CH=CH-CHO

A 12.4 

Q 12.5 Draw structures of the following derivatives. (i) The 2,4-dinitrophenylhydrazone of benzaldehyde (ii) Cyclopropanone oxime (iii) Acetaldehydedimethylacetal (iv) The semicarbazone of cyclobutanone (v) The ethylene ketal of hexan-3-one (vi) The methyl hemiacetal of formaldehyde

A 12.5 

Q 12.6 Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents. (i) PhMgBr and then H3O+ (ii) Tollens’ reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid

A 12.6 

Q 12.7 Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction. (i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal

A 12.7

  • Aldehydes and ketones having at least one α-hydrogen undergo aldol The compounds

(ii) 2-methylpentanal (v)cyclohexanone

  • 1-phenylpropanone
  • phenylacetaldehyde

contain one or more α-hydrogen atoms. Therefore, these undergo aldol condensation.

  • Aldehydes having no α-hydrogen atoms undergo Cannizzaro The compounds

(i) Methanal (iii)Benzaldehyde

  • 2, 2-dimethylbutanal

do not have any α-hydrogen. Therefore, these undergo cannizzaro reactions.

  • Compound (iv) Benzophenone is a ketone having no α-hydrogen atom and compound (viii) Butan-1-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or cannizzaro

Structures of the expected products of aldol condensation and Cannizzaro reaction- Aldol condensation-

(ii) 2-methylpentanal-

  • cyclohexanone-
  • 1-phenylpropanone-
  • phenylacetaldehyde –

Cannizzaro reaction-

(i) Methanal-

(iii) Benzaldehyde –

(ix) 2, 2-dimethylbutanal –

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