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

 

NCERT solutions for class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids- This chapter is very important for you because this chapter carries 6 out of 70 marks in the CBSE Board examination. It is recommended to go through solutions of NCERT for class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids to maxmise your score. This chapter 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 CBSE NCERT solutions for class 12 chemistry chapter 12 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 the previous chapter, you have studied organic compounds containing carbon-oxygen single bond functional groups. In this chapter along with NCERT solutions for class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids, you will learn about organic compounds containing carbon-oxygen double bond (>C=O) that is known as carbonyl group. In the organic chemistry, carbonyl group (>C=O) is one of the most important functional group. After completing the class 12 chemistry unit 12 aldehydes, ketones and carboxylic acids, students will be able to write the IUPAC names of ketones, aldehydes and carboxylic acid, their respective structures and describe important reactions of these three class of compounds and their important methods of preparation and correlate their structures with their physical and chemical properties.The NCERT solutions provided here are completely free and you can also download them for offline use also if you want to prepare or any other subject or any other class.

In aldehydes, the carbonyl functional group(>C=O) is bonded to a carbon and hydrogen atom, while in the ketones, the carbonyl group is bounded to two carbon atoms. In the carboxylic acid, carbon of carbonyl group(>C=O) is bounded to hydrogen or carbon and oxygen of hydroxyl moiety(-OH), while in amides carbon is attached to hydrogen or carbon and nitrogen of moiety-NH_{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.

                                Aldehydes, Ketones and Carboxylic Acids

Intext Questions

Exercise Questions

Find NCERT solutions for class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic acids 

Solutions to In-Text Questions Ex 12.1  to 12.8

 

Question  12.1(i)     Write the structures of the following compounds.

   \alpha -Methoxypropionaldehyde

Answer:

The structure of the compound \alpha -methoxy propionaldehyde is shown here-

 

Question 12.1(ii)     Write the structures of the following compounds.

3-Hydroxy butanal

Answer:

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

 

Question 12.1(iii)     Write the structures of the following compounds. 

2-Hydroxy cyclopentane carbaldehyde 

Answer:

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

 

 

Question 12.1(iv)     Write the structures of the following compounds.

    4-Oxopentanal

Answer:

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

 

Question 12.1(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-

 

 

Question 12.1(vi)     Write the structures of the following compounds.

4-Fluoro acetophenone

Answer:

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

 

 

Question 12.2(i)     Write the structures of products of the following reactions:

         (i)      \xrightarrow[CS_{2}]{Anhyd. AlCl_{3}} 

 

Answer:

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

 

 

Question 12.2(ii)     Write the structure of products of the following reactions;

       (ii)      (C_{6}H_{5}CH_{2})_{2}Cd + 2 CH _{3}COCl\rightarrow

Answer:

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

 

 

Question 12.2(iii)     Write the structures of products of the following reactions:

        (iii)     H_{3}C - C \equiv C - H \overset{Hg^{2+}, H_{2}SO_{4}}{\rightarrow}

Answer:

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

 

   

 

Question 12.2(iv)     Write the structures of products of the following reactions:

        (iv)     \xrightarrow[2.H_{3}O^{+} ]{1.CrO_{2}Cl_{2}}

 

Answer:

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

 

 

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

                CH_{3}CHO, CH_{3}CH_{2}OH,CH_{3}OCH_{3}, CH_{3}CH_{2}CH_{3}

Answer:

Increasing order in their boiling points-

CH_{3}CH_{2}CH_{3}<CH_{3}OCH_{3}<CH_{3}CHO<CH_{3}CH_{2}OH

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.

 

Question 12.4(i)     Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

 Ethanal, Propanal, Propanone, Butanone.

Answer:


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  

 

Question 12.4(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.

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

Acetophenone < p-tolualdehyde < benzaldehyde <p-nitrobenzaldehyde

 

Question 12.5(i)     Predict the products of the following reactions:

       (i)     + HO - NH_{2}\overset{H^{+}}{\rightarrow}

 

Answer:

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

 

 

Question 12.5(ii)     Predict the products of the following reactions:

        (ii)     

 

Answer:

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

 

 

Question 12.5(iii)     Predict the products of the following reactions:

  R-CH=CH-CHO +  \overset{H^{+}}{\rightarrow}

 

Answer:

The product of the above reaction is -

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

 

 

Question 12.5(iv)     Predict the products of the following reactions:

        (iv)     + CH_{3}CH_{2}NH_{2}\overset{H^{+}}{\rightarrow}

Answer:

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

 

 

Question 12.6(i)     Give the IUPAC names of the following compounds:

Ph CH_{2}CH_{2}COOH 

Answer:

The IUPAC name of the compound Ph CH_{2}CH_{2}COOH is-

3-phenyl propanoic acid

 

Question 12.6(ii)     Give the IUPAC names of the following compounds:

   (CH_{3})_{2}C=CH COOH

 

Answer:

The IUPAC name of the compound (CH_{3})_{2}C = CH COOH is -

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

 

Question 12.6(iii)     Give the IUPAC names of the following compounds:

   

Answer:

The IUPAC name of the compound is-

          
2-methyl cyclopentane carboxylic acid

 

Question 12.6(Iv)     Give the IUPAC names of the following compounds:

        (iv)         

 

Answer:

The IUPAC name of the compound is-


2, 4, 6-trinitrobenzoic acid

 

Question 12.7(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.

 

Question 12.7(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 

 

Question 12.7(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.

 

 

Question 12.7(iv)     Show how each of the following compounds can be converted to benzoic acid?

Phenylethene (Styrene)

Answer:

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

 

Question 12.8(i)     Which acid of each pair shown here would you expect to be stronger?

 CH_{3}CO_{2}H or CH_{2}FCO_{2}H

Answer:

CH_{2}FCO_{2}H is stronger than CH_{3}COOH  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 CH_{2}FCO_{2}^{-} is more stable than CH_{3}COO^{-}.

 

Question 12.8(ii)     Which acid of each pair shown here would you expect to be stronger?

   CH_{2}FCO_{2}H or CH_{2}Cl CO_{2}H

Answer:

CH_{2}FCO_{2}H is a stronger acid.

Fluorine has more -I effect than chlorine. So, CH_{2}FCO_{2}H can release proton easily than CH_{2}ClCO_{2}H.

 

Question 12.8(iii)     Which acid of each pair shown here would you expect to be stronger?

      (iii)     CH_{2}FCH_{2}CH_{2}CO_{2}H or CH_{3}CH FCH_{2}CO_{2}H

Answer:

Since we know that inductive effect depends on distance. Greater is the distance lesser is the effect. So, -I effect in CH_{3}CH FCH_{2}CO_{2}H is more.

CH_{3}CH FCH_{2}CO_{2}H is more acidic than CH_{2}FCH_{2}CH_{2}CO_{2}H.

 

Question 12.8(iv)     Which acid of each pair shown here would you expect to be stronger?

    

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.

 

 

Question: 12.1 (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 (HCN) to yield cyanohydrin. The Reaction is very slow with the pure HCN, so it can be catalyzed by using a base.

 

 

Q 12.1 (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:


 

Q 12.1(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:

 

 

Q 12.1(iv). What is meant by the following terms? Give an example of the reaction in each case.

             Aldol

Answer:

Aldol-
\beta-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:

 

Q 12.1(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:
 

Q 12.1 (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-

For example:

Q 12.1 (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:

                                                                         

Q 12.1(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:

Q 12.1(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:

 

Q 12.1 (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:

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

            CH_{3}CH(CH_{3})CH_{2}CH_{2}CHO    

 

Answer:

CH_{3}CH(CH_{3})CH_{2}CH_{2}CHO

IUPAC name of this compound is 4-methyl pentanal

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

            CH_{3}CH_{2}COCH(C_{2}H_{5})CH_{2}CH_{2}Cl

Answer:

CH_{3}CH_{2}COCH(C_{2}H_{5})CH_{2}CH_{2}Cl


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

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

          CH_{3}CH= CH CHO

Answer:

CH_{3}CH= CH CHO
the structure of the compound is 

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

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

           CH_{3}COCH_{2}CO CH_{3}

Answer:

CH_{3}COCH_{2}CO CH_{3}
The structure of the given compound is-

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

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

       CH_{3}CH(CH_{3})CH_{2}C(CH_{3})_{2}COCH_{3}

 

Answer:

CH_{3}CH(CH_{3})CH_{2}C(CH_{3})_{2}COCH_{3}
The structure of the compound is -

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

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

         (CH_{3})_{3}CCH_{2}COOH  

Answer:

(CH_{3})_{3}CCH_{2}COOH
The structure of the compound is -

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

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

        OHCC_{6}H_{4}CHO -p

Answer:

OHCC_{6}H_{4}CHO -p
The structure of the compound is-

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

Q 12.3(i)       Draw the structures of the following compounds.

           3-Methyl butanal

Answer:

The structure of the 3-methylbutanal is shown here-

Q 12.3  (ii)   Draw the structures of the following compounds.

            p-Nitropropiophenone

Answer:

The structure of p-nitro propiophenone is shown here-

 

Q 12.3 (iii)      Draw the structures of the following compounds.

          p-methyl benzaldehyde   

Answer:

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

Q 12.3(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-

Q 12.3(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-

 

 

 

Q 12.3 (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-

 

Q 12.3  (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-

 

 

Q 12.3 (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-

 

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

         CH_{3}CO(CH_{2})_{4}CH_{3}

Answer:

CH_{3}CO(CH_{2})_{4}CH_{3}
The structure of the compound is

The IUPAC name of the compound is hept-2-one
Common name is methyl-n-propyl ketone

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

         CH_{3}CH_{2}CHBrCH_{2}CH(CH_{3})CHO

Answer:

CH_{3}CH_{2}CHBrCH_{2}CH(CH_{3})CHO
The structure of the compound is 


The IUPAC name of the compound is 4-bromo-2-methylhexanal
The common name is -\gamma-bromo-\alpha-methyl-caproaldehyde

 

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

      CH_{3}(CH_{2})_{5}CHO

 

Answer:

CH_{3}(CH_{2})_{5}CHO
structure os the compound is


The IUPAC name of the compound is heptanal

 

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

    Ph - CH = CH - CHO

Answer:

Ph - CH = CH - CHO
structur eof the compound is


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

Common name is \beta-phenylacrolein

 

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

 

             

Answer:


The IUPAC name of the compound is cyclopentane carbaldehyde.

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

           Ph CO Ph

 

Answer:

Ph CO Ph
structure of the compound is


The IUPAC name of the compound is Diphenyl mentanone.

The common name is - benzophenone

 

Q 12.5  (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

 

 

12.5  (ii)      Draw structures of the following derivatives.

        Cyclopropanone oxime

Answer:

The structure of the Cyclopropanone oxime

 

 

Q 12.5  (iii)      Draw structures of the following derivatives.

        Acetaldehyde dimethyl acetal  

 

Answer:

The structure of Acetaldehydedimethylacetal  

 

Q 12.5  (iv)     Draw structures of the following derivatives.

         The semicarbazone of cyclobutanone

Answer:

The structure of the semicarbazone of cyclobutanone

 

 

Q 12.5  (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

Q 12.5  (vi)       Draw structures of the following derivatives.

       The methyl hemiacetal of formaldehyde 

Answer:

The structure of the compound methyl hemiacetal of formaldehyde

 

Q 12.6 (i)     Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.

             PhMgBr and then H3O+

Answer:

When cyclohexane carbaldehyde reacts with PhMgBr in presence of dry ether and then H_{3}O^{+}(hydrolysis)

                                                                                                                                                           

Q 12.6 (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 (Ag) and oxidises itself to cyclohexane carboxylate ion
So the reaction is-


 

Q 12.6 (iii)    Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.

          Semicarbazide and weak acid

Answer:

Structure of cyclohexanecarbaldehyde

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

 

 

Q 12.6   (iv)     Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.

       Excess ethanol and acid

Answer:

Structure of cyclohexanecarbaldehyde

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

 

   

Q 12.6  (v)   Predict the products formed when cyclohexane carbaldehyde reacts with following reagents.

           Zinc amalgam and dilute hydrochloric acid

 

Answer:

Structure of cyclohexane carbaldehyde

When cyclohexane carbaldehyde reacts with zinc amalgam(Zn-Hg) and dil. HCl, the carbonyl group of reactant is reduced to CH_{2}, it is also known as clemmensen reduction.

 

Q 12.7  (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 \alpha- 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-

                     

Q 12.7  (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-

 

 

 

Q 12.7 (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 

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

 

 

Q 12.7  (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 -

 

Q 12.7  (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.

 

 

Q 12.7 (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-hyrdrogen atom, So it gives aldol condensation reaction. It react with an another molecule of 1-Phenylpropanone in presence of dil. NaOH


 

Q 12.7(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;

 

 

Q 12.7(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 (NaOH). One molecule of it is oxidised and the other molecule is reduced.

 

Q 12.8 (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 NaBH_{4} gives the product Butane-1,3-diol

 

Q 12.8 (ii)     How will you convert ethanal into the following compounds?

           But-2-enal 

Answer:

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

 

 

Q 12.8 (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. 

 

Q 12.9     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 -CH_{3}CH_{2}CHO

butanal - CH_{3}CH_{2}CH_{2}CHO
 

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

  1. When two molecules of propanal react with each other

  2. When two molecules of butanal reacts with eath other

  3. When propanal act as a nucleophile and attacking on butanal(as an electrophile)

  4. When butanal act as a nucleophile and attacking on propanal(as an electrophile)

Now the reactions are shown here:
(i)


(ii)


(iii)


(iv)


 

Q 12.10     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 \alpha-hydrogen atom. On oxidation it gives 1,2-benzenedicarboxylic acid. Thus, the aldehyde group (-CHO) is directly attached with the benzene ring and also it should be para-substituted.

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

By following reaction we can undrstand this prblem-


 

Q 12.12  (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


The attacking power of CN nucleophile can be affected by two ways-

  1. hindrance near carbonyl group

  2. negative charge on -CO group (through +I effect of methyl)

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

 

Q 12.12  (ii)     Arrange the following compounds in increasing order of their property as indicated: 

           CH_{3}CH_{2}CH(Br)COOH, CH_{3}CH(Br)CH_{2}COOH, (CH_{3})_{2}CHCOOH ,CH_{3}CH_{2}CH_{2}COOH (acid strength)

Answer:

increasing order of their acidic strength-

\\(CH_{3})_{2}CHCOOH<CH_{3}CH_{2}CH_{2}COOH<CH_{3}CH(Br)CH_{2}COOH<CH_{3}CH_{2}CH(Br)COOH

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 acidic nature
+I\propto 1/ acidic

Q 12.12  (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 12.13  (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

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 12.13 (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 NaOI to give yellow ppt. but benzophenone does not give this test.

C_{6}H_{5}COCH_{3}+NaOI\rightarrow C_{6}H_{5}COONa+CHI_{3}+NaOH

benzophenone + NaOI\rightarrow no yellow ppt.

Q 12.13 (iii)    Give simple chemical tests to distinguish between the following pairs of compounds.

            Phenol and Benzoic acid

Answer:

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

 

Q 12.13 (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 NaHCO_{3} to give brisk effervescence due to evoltion of carbon dioxide (CO_{2}) but Ethyl benzoate does not.

 

Q 12.13    (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.

 

Q 12.13  (vi)     Give simple chemical tests to distinguish between the following pairs of compounds.

           Benzaldehyde and Acetophenone

Answer:

Both can be distinguished by-

  1. Iodoform test
    In acetophenone, methyl ketone group present.So it give positive iodoform test by forming a yellow ppt.

  2. 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 12.13 (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.

 

Q 12.14(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 Mg in presence of ether it becomes Grignard reagents (R-MgX). 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.

Q 12.14 (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 Mg in presence of ether it becomes grignard reagents (R-MgX). 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).

 


 

 

Q 12.14 (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(KMnO_{4}/KOH) gives salts of carboxylic acid, which on further acidic hydrolysis gives p-nitrobenzoic acid.

 

 

Q 12.14 (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 AlCl_{3} gives toluene, which on further reaction with Br_{2}/\Delta /hv 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.


 

Q 12.14(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 HNO_{3}/H_{2}SO_{4} gives p- nitrotoluene and react it with the the carbon disulphide /CrO_2Cl_{2} followed by acidic hydrolysis to form p-nitrobenzaldehyde.


 

Q 12.15(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 (LiAlH_{4}) to get propan-2-ol and then by dehydration remove water molecule to get propene.

Q 12.15(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.

alternate method-

 


First react it with SOCl_{2} which replace OH group and then hydogenation reaction.

Q 12.15 (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.

 

Q 12.15  (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(CH_{3}CO)_{2}CO, it gives acetophenone And further reacting acetophenone with nitric acid in presence of sulphuric acid give meta substituted product, m-nitroacetophenone.

 

 

Q 12.15 (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 CaCO_{3} it formed calcium benzoate. After distillation we get benzophenone.

 

 

Q 12.15 (vi)    How will you bring about the following conversions in not more than two steps? 

             Bromobenzene to 1-phenylethanol 

Answer:

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

 

12.15 (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 

 

 

Q 12.15 (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 NaCN in presence of HCl gives benzaldehyde cyanohydrine, which on acidic hydrolysis formed α-Hydroxyphenylacetic acid

 

Q 12.15(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.

 

 

Q 12.16 (i)     Describe the following.

           Acetylation 

Answer:

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

 

Q 12.16 (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.

 

 

Q 12.16 (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.


If the reactants contain alpha-hydrogen atoms then,

 

Q 12.16  (iv)    Describe the following.

            Decarboxylation

Answer:

Decarboxylation-
The reaction in which the carboxylic acid loses carbon dioxide to hydrocarbon, in the presence of sodalime(NaOH and CaO in the ratio of 3:1). This reaction is known as decarboxylation.

 

 

Q 12.17(i)     Complete each synthesis by giving missing starting material, reagent or products.

               \xrightarrow[KOH, heat ]{KMnO_{4}}

 

Answer:

Potassium permanganate oxidise the substituted ethyl group to -COOHgroup and due to presence of KOH, formed carboxylic acid salts (potassium benzoate)

 

Q 12.17(ii)       Complete each synthesis by giving missing starting material, reagent or products.

               \xrightarrow[heat]{SoCl_{2}}

 

Answer:

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

 

 

Q 12.17 (iii)      Complete each synthesis by giving missing starting material, reagent or products

           C_{6}H_{5}CHO\overset{H_{2}NCONHNH_{2}}{\rightarrow}

Answer:

In this reaction, the -NH_{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-

 

Q 12.17(iv)    Complete each synthesis by giving missing starting material, reagent or products.

               

 

Answer:

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

 

 

Q 12.17(v)      Complete each synthesis by giving missing starting material, reagent or products.

             \overset{\left [ Ag\left ( NH_{3} \right )_{2} \right ]^{+}}{\rightarrow}  

Answer:

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

 

Q 12.17 (vi)     Complete each synthesis by giving missing starting material, reagent or products.

            \overset{NaCN / HCl}{\rightarrow}

 

Answer:

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

So the reaction gives

 

Q 12.17 (vii)    Complete each synthesis by giving missing starting material, reagent or products.

              C_{6}H_{3}CHO + CH_{3}CH_{2}CHO \xrightarrow[\Delta ]{dil NaOH}

Answer:

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

 

 

Q 12.17  (viii)    Complete each synthesis by giving missing starting material, reagent or products.

           CH_{2}COCH_{2}COOC_{2}H_{5}\xrightarrow[ii. H^{+}]{i. NaBH_{4}}  

Answer:

In this reaction NaBH_{4} is a reducing agent, which reduces the CO group of the compound.

Q 12.17(ix)       Complete each synthesis by giving missing starting material, reagent or products.

            \overset{CrO_{2}}{\rightarrow}

 

Answer:

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


 

Q 12.17(x)         Complete each synthesis by giving missing starting material, reagent or products.

        

 

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.

 


reacts

Q 12.17 (xi)      Complete each synthesis by giving missing starting material, reagent or products.

            \xrightarrow[ii. Zn - H_{2}O]{i. O_{3}} 

 

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.

 

 

Q 12.18(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

Cyclohexanone forms cyanohydrin when the nucleophile (HCN/CN^{-}) attack on cyclohexanone.

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 (\alpha) position, there is a hindrance because of the methyl groups. Thus the nucleophile cannot attack effectively.

Q 12.18(ii)     Give plausible explanation for each of the following:

              There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones. 

Answer:

In Semicarbazide,
 


One of the -NH_{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 -NH_{2} group). But the other -NH_{2} group is not participating in resonance, so it can act as nucleophile and attack on the carbonyl carbon of ketone and aldehyde.

Q 12.18(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:

RCOOH+R'OH\rightleftharpoons RCOOR'+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 12.19     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 - CH_{3}-CO-(CH_{2})_{2}-CH_{3}

Q 12.20     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

NCERT Solutions for class 12 chemistry

Chapter 1

CBSE NCERT solutions for class 12 chapter 1 The Solid State

Chapter 2

NCERT solutions for class 12 chemistry chapter 2 Solutions

Chapter 3

Solutions of NCERT class 12 chemistry chapter 3 Electrochemistry

Chapter 4

CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

Chapter 5

Solutions of NCERT class 12 chemistry chapter 5 Surface chemistry

Chapter 6

NCERT solutions for class 12 chemistry General Principles and Processes of isolation of elements

Chapter 7

CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements

Chapter 8

Solutions of NCERT class 12 chemistry chapter 8 The d and f block elements

Chapter 9

NCERT solutions for class 12 chemistry chapter 9 Coordination compounds

Chapter 10

Solutions of NCERT class 12 chemistry chapter 10 Haloalkanes and Haloarenes

Chapter 11

CBSE NCERT solutions for class 12 chemistry chapter 11 Alcohols, Phenols and Ethers

Chapter 12

NCERT solutions for class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids

Chapter 13

NCERT solutions for class 12 chemistry chapter 13 Amines

Chapter 14

CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules

Chapter 15

Solutions of NCERT class 12 chemistry chapter 15 Polymers

Chapter 16

NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

 

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