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

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NCERT Solutions for Class 12 Aldehydes, Ketones and Carboxylic Acids Chapter 8 are provided here for students struggling with the concepts by SimplyAcad. The solutions are organised in a structured manner for the ease of students. Access to these solutions will allow them to look into the various sections of the chapter thoroughly and gain deep insights on the same.

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

Question1. {Question} What is meant by the following terms? Give an example of the reaction in each case. \begin{enumerate} \item Cyanohydrin \item Acetal \item Semicarbazone \item Aldol \item Hemiacetal \item Oxime \item Ketal \item Imine \item 2,4-DNP-derivative \item Schiff’s base \end{enumerate} \section*

{Solution} \begin{enumerate} \item \textbf{Cyanohydrin:} Cyanohydrin is a compound where both -OH group and cyano group are present on the same carbon atom. It is formed by the addition of HCN to a carbonyl group in a weakly acidic medium. \[ \chemfig{CH_3CHO} + \chemfig{HCN} \rightarrow \chemfig{CH_3CH(OH)CN} \] \item \textbf{Acetal:} An acetal contains two alkoxy groups attached to the same carbon atom. It is formed by the addition of two equivalents of a monohydric alcohol to one equivalent of an aldehyde in the presence of dry HCl gas. \[ \chemfig{CH_3CHO} + 2 \chemfig{CH_3OH} \xrightarrow{\text{dry HCl}} \chemfig{CH_3CH(OCH_3)_2} + \chemfig{H_2O} \] \item \textbf{Semicarbazone:} A semicarbazone is a derivative of an aldehyde or ketone obtained by reaction with semicarbazide in a weakly acidic medium. \[ \chemfig{CH_3CHO} + \chemfig{H_2N-NHCO-NH_2} \rightarrow \chemfig{CH_3CH=NNHCO-NH_2} + \chemfig{H_2O} \] \item \textbf{Aldol:} An aldol is a beta-hydroxy aldehyde or ketone formed by the condensation of two molecules of aldehydes or ketones with alpha hydrogen atoms in the presence of a dilute aqueous base. \[ 2 \chemfig{CH_3CHO} \xrightarrow{\text{dil. base}} \chemfig{CH_3CH(OH)CH_2CHO} \] \item \textbf{Hemiacetal:} A hemiacetal contains one alkoxy group and one hydroxyl group attached to the same carbon atom. It is formed by the addition of one molecule of a monohydric alcohol to an aldehyde in the presence of dry HCl gas. \[ \chemfig{CH_3CHO} + \chemfig{CH_3OH} \leftrightarrow \chemfig{CH_3CH(OH)OCH_3} \] \item \textbf{Oxime:} An oxime is formed by the reaction of aldehydes or ketones with hydroxylamine in a weakly acidic medium. \[ \chemfig{CH_3CHO} + \chemfig{H_2NOH} \rightarrow \chemfig{CH_3CH=NOH} \] \item \textbf{Ketal:} A ketal is a gem-dialkoxy alkane formed by the reaction of a ketone with ethylene glycol in the presence of dry HCl. It contains two alkoxy groups attached to the same carbon atom. \[ \chemfig{CH_3COCH_3} + \chemfig{HOCH_2CH_2OH} \xrightarrow{\text{dry HCl}} \chemfig{(CH_3)_2C(OC_2H_4O)_2} \] \item \textbf{Imine:} An imine contains a -CH=N- group and is obtained by the reaction between aldehydes or ketones with ammonia derivatives such as hydrazine, hydroxylamine, phenylhydrazine, or semicarbazide. \[ \chemfig{CH_3COCH_3} + \chemfig{PhNH_2} \rightarrow \chemfig{(CH_3)_2C=NHPh} \] \item \textbf{2,4-DNP-derivative:} These are derivatives obtained by the reaction of aldehydes or ketones with 2,4-dinitrophenylhydrazine (2,4-DNP) in a weakly acidic medium. \[ \chemfig{CH_3COCH_3} + \chemfig{2,4-DNP} \rightarrow \chemfig{(CH_3)_2C=NNH-C_6H_3(NO_2)_2} + \chemfig{H_2O} \] \item \textbf{Schiff’s Base:} Schiff’s base is an azomethine obtained by the reaction of aldehydes or ketones with primary amines (aliphatic or aromatic). \[ \chemfig{C_6H_5CHO} + \chemfig{C_6H_5NH_2} \rightarrow \chemfig{C_6H_5CH=NH-C_6H_5} + \chemfig{H_2O} \]

Question 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

Solution :
(i) 4-methylpentanal

(ii) 6-Chloro-4-ethylhexan-3-one

(iii) But-2-en-1-al

(iv) Pentane-2,4-dione

(v) 3,3,5-Trimethylhexan-2-one

(vi) 3,3-Dimethylbutanoic acid

(vii) Benzene-1,4-dicarbaldehyde

Question3. 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

Solution :
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Question 4. {Question} Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names. \begin{enumerate} \item \textbf{CH$_3$CO(CH$_2$)$_4$CH$_3$} \item \textbf{CH$_3$CH$_2$CH$_2$CHBrCH$_2$CH(CH$_3$)CH$_2$CHO} \item \textbf{CH$_3$(CH$_2$)$_5$CHO} \item \textbf{Ph-CH=CH-CHO} \item \textbf{C$_5$H$_8$O} \item \textbf{Ph$_2$CO} \end{enumerate} \section*

{Solution} \begin{enumerate} \item \textbf{CH$_3$CO(CH$_2$)$_4$CH$_3$} \begin{itemize} \item \textbf{IUPAC Name:} Heptan-2-one \item \textbf{Common Name:} Methyl n-pentyl ketone \end{itemize} \item \textbf{CH$_3$CH$_2$CH$_2$CHBrCH$_2$CH(CH$_3$)CH$_2$CHO} \begin{itemize} \item \textbf{IUPAC Name:} 4-Bromo-2-methylhexanal \item \textbf{Common Name:} $\gamma$-Bromo-$\alpha$-methylhexanal \end{itemize} \item \textbf{CH$_3$(CH$_2$)$_5$CHO} \begin{itemize} \item \textbf{IUPAC Name:} Heptanal \item \textbf{Common Name:} None (Heptanal is commonly used without a specific common name) \end{itemize} \item \textbf{Ph-CH=CH-CHO} \begin{itemize} \item \textbf{IUPAC Name:} 3-Phenylprop-2-enal \item \textbf{Common Name:} Cinnamaldehyde \end{itemize} \item \textbf{C$_5$H$_8$O} \begin{itemize} \item \textbf{IUPAC Name:} Cyclopentacarbaldehyde \item \textbf{Common Name:} None (Cyclopentacarbaldehyde is commonly used without a specific common name) \end{itemize} \item \textbf{Ph$_2$CO} \begin{itemize} \item \textbf{IUPAC Name:} Diphenylmethanone \item \textbf{Common Name:} Benzophenone \end{itemize} \end{enumerate}

Question5. 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

Solution :
(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question 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

Solution :
(i)

(ii)

(iii)

(iv)

(v)

Question 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

Solution :
Aldehydes and ketones having at least one α-hydrogen undergo aldol condensation. The compounds (ii) 2−methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone, and (vii) phenylacetaldehyde contain one or more α-hydrogen atoms. Therefore, these undergo aldol condensation.

Aldehydes having no α-hydrogen atoms undergo Cannizzaro reactions. The compounds (i) Methanal, (iii) Benzaldehyde, and (ix) 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 reactions.

Aldol condensation

(ii)

(v)

(vi)

(vii)

Cannizzaro reaction

(i)

(iii)

(ix)

Question 8. How will you convert ethanal into the following compounds?

(i) Butane-1, 3-diol (ii) But-2-enal (iii) But-2-enoic acid

Solution :
(i) On treatment with dilute alkali, ethanal produces 3-hydroxybutanal gives butane-1, 3-diol on reduction.

(ii) On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal.

(iii) When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid .

Question 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.

: Aldehydes, Ketones and Carboxylic Acids Chapter 8

Solution :
(i) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.

(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.

(iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.

(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

Question 10. An organic compound with the molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.

: Aldehydes, Ketones and Carboxylic Acids Chapter 8

Solution :
It is given that the compound (with molecular formula C9H10O) forms 2, 4-DNP derivative and reduces Tollen’s reagent. Therefore, the given compound must be an aldehyde.

Again, the compound undergoes cannizzaro reaction and on oxidation gives 1, 2-benzenedicarboxylic acid. Therefore, the −CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted. Hence, the compound is 2-ethylbenzaldehyde.

The given reactions can be explained by the following equations.

Question 11. {Question} An organic compound (A) (molecular formula \textbf{C$_8$H$_{16}$O$_2$}) was hydrolyzed with dilute sulfuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved. \section*

{Solution} Organic compound A is an ester because its hydrolysis yields a carboxylic acid (B) and an alcohol (C). Given that the oxidation of alcohol (C) yields the carboxylic acid (B), both (B) and (C) must have the same number of carbon atoms. Since compound A has 8 carbon atoms, and both (B) and (C) each have 4 carbon atoms, we deduce that: \begin{itemize} \item The ester must be \textbf{butyl butanoate} (C$_4$H$_9$COO-C$_4$H$_9$). \item The carboxylic acid (B) is \textbf{butanoic acid} (C$_4$H$_9$COOH). \item The alcohol (C) is \textbf{butan-1-ol} (C$_4$H$_9$OH). \end{itemize} The reactions are as follows: 1. \textbf{Hydrolysis of ester A:} \[ \text{C}_4\text{H}_9\text{COO-C}_4\text{H}_9 + \text{H}_2\text{SO}_4 \xrightarrow{\text{dil. H}_2\text{SO}_4} \text{C}_4\text{H}_9\text{COOH} + \text{C}_4\text{H}_9\text{OH} \]2. \textbf{Dehydrationofbutan-1-ol:}\[ \text{C}_4\text{H}_9\text{OH}\xrightarrow{\text{Dehydration}}\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 \] 3. \textbf{Oxidation of butan-1-ol:} \[ \text{C}_4\text{H}_9\text{OH} xrightarrow{\text{CrO}_3} \text{C}_4\text{H}_9\text{COOH} \]

Question12. Arrange the following compounds in increasing order of their property as indicated:

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

(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)

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

Solution :
(i) When HCN reacts with a compound, the attacking species is a nucleophile, CN−. Therefore, as the negative charge on the compound increases, its reactivity with HCN decreases. In the given compounds, the +I effect increases as shown below. It can be observed that steric hindrance also increases in the same

Hence, the given compounds can be arranged according to their increasing reactivities toward HCN as:

Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

(ii) After losing a proton, carboxylic acids gain a negative charge as shown:

Now, any group that will help stabilise the negative charge will increase the stability of the carboxyl ion and as a result, will increase the strength of the acid. Thus, groups having +I effect will decrease the strength of the acids and groups having −I effect will increase the strength of the acids. In the given compounds, −CH3 group has +I effect and Br− group has −I effect. Thus, acids containing Br− are stronger.

Now, the +I effect of isopropyl group is more than that of n-propyl group. Hence, (CH3)2CHCOOH is a weaker acid than CH3CH2CH2COOH.

Also, the −I effect grows weaker as distance increases. Hence, CH3CH(Br)CH2COOH is a weaker acid than CH3CH2CH(Br)COOH.

Hence, the strengths of the given acids increase as:

(CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH

(iii) As we have seen in the previous case, electron-donating groups decrease the strengths of acids, while electron-withdrawing groups increase the strengths of acids. As methoxy group is an electron-donating group, 4-methoxybenzoic acid is a weaker acid than benzoic acid. Nitro group is an electron-withdrawing group and will increase the strengths of acids. As 3,4-dinitrobenzoic acid contains two nitro groups, it is a slightly stronger acid than 4-nitrobenzoic acid. Hence, the strengths of the given acids increase as:

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

Question13. Give simple chemical tests to distinguish between the following pairs of compounds: \begin{enumerate} \item Propanal and Propanone \item Acetophenone and Benzophenone \item Phenol and Benzoic acid \item Benzoic acid and Ethyl benzoate \item Pentan-2-one and Pentan-3-one \item Benzaldehyde and Acetophenone \item Ethanal and Propanal \end{enumerate} \section*

{Solution} \begin{enumerate} \item \textbf{Propanal and Propanone:} \begin{itemize} \item \textbf{Tollen’s Test:} Propanal gives a silver mirror with Tollen’s reagent, while propanone does not. \item \textbf{Fehling’s Test:} Propanal gives a red precipitate with Fehling’s solution, but propanone does not. \end{itemize} \item \textbf{Acetophenone and Benzophenone:} \begin{itemize} \item \textbf{Iodoform Test:} Acetophenone gives a yellow precipitate (iodoform) with an alkaline solution of iodine, whereas benzophenone does not. \end{itemize} \item \textbf{Phenol and Benzoic acid:} \begin{itemize} \item \textbf{Reaction with Sodium Hydrogen Carbonate:} Phenol does not react with sodium hydrogen carbonate, while benzoic acid reacts, producing effervescence of carbon dioxide gas. \item \textbf{Ferric Chloride Test:} Phenol gives a violet color with ferric chloride solution, whereas benzoic acid does not. \end{itemize} \item \textbf{Benzoic acid and Ethyl benzoate:} \begin{itemize} \item \textbf{Reaction with Sodium Hydrogen Carbonate:} Benzoic acid reacts with sodium hydrogen carbonate, producing effervescence of carbon dioxide gas, while ethyl benzoate does not. \item \textbf{Saponification Test:} When ethyl benzoate is boiled with excess NaOH, it gives ethanol. Ethanol on heating with iodine gives a yellow precipitate of iodoform. This test is not given by benzoic acid. \end{itemize} \item \textbf{Pentan-2-one and Pentan-3-one:} \begin{itemize} \item \textbf{Iodoform Test:} Pentan-2-one gives a yellow precipitate (iodoform) with an alkaline solution of iodine, while pentan-3-one does not. \item \textbf{Sodium Bisulphite Test:} Pentan-2-one gives a white precipitate with sodium bisulphite, whereas pentan-3-one does not. \end{itemize} \item \textbf{Benzaldehyde and Acetophenone:} \begin{itemize} \item \textbf{Tollen’s Test:} Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent), whereas acetophenone does not. \item \textbf{Iodoform Test:} Acetophenone gives a yellow precipitate (iodoform), while benzaldehyde does not. \end{itemize} \item \textbf{Ethanal and Propanal:} \begin{itemize} \item \textbf{Iodoform Test:} Ethanal gives a yellow precipitate (iodoform) with an alkaline solution of iodine, while propanal does not. \end{itemize}

Question14. 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

(i) Methyl benzoate (ii) m-Nitrobenzoic acid

(iii) p-Nitrobenzoic acid (iv) Phenylacetic acid

(v) p-Nitrobenzaldehyde.

Solution :
(i)

(ii)

(iii)

(iv)

(v)

Question15. How will you bring about the following conversions in not more than two steps?

(i) Propanone to Propene

(ii) Benzoic acid to Benzaldehyde

(iii) Ethanol to 3-Hydroxybutanal

(iv) Benzene to m-Nitroacetophenone

(v) Benzaldehyde to Benzophenone

(vi) Bromobenzene to 1-Phenylethanol

(vii) Benzaldehyde to 3-Phenylpropan-1-ol

(viii) Benazaldehyde to α-Hydroxyphenylacetic acid

(ix) Benzoic acid to m- Nitrobenzyl alcohol

Solution :
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Question16. Describe the following:

(i) Acetylation

(ii) Cannizzaro reaction

(iii) Cross aldol condensation

(iv) Decarboxylation

Solution :
(i) Acetylation :

The introduction of an acetyl functional group into an organic compound is known as acetylation. It is usually carried out in the presence of a base such as pyridine, dirnethylaniline, etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetylating agents.

For example, acetylation of ethanol produces ethyl acetate.

(ii) Cannizzaro reaction:

The self oxidation-reduction (disproportionation) reaction of aldehydes having no α-hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.

For example, when ethanol is treated with concentrated potassium hydroxide, ethanol and potassium ethanoate are produced.

(iii) Cross-aldol condensation:

When aldol condensation is carried out between two different aldehydes, or two different ketones, or an aldehyde and a ketone, then the reaction is called a cross-aldol condensation. If both the reactants contain α-hydrogens, four compounds are obtained as products.

For example, ethanal and propanal react to give four products.

(iv) Decarboxylation:

Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda-lime.

Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolyzed. This electrolytic process is known as Kolbe’s electrolysis.

Question17. Complete each synthesis by giving missing starting material, reagent or products

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

Solution :
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

Question18. Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6 trimethylcyclohexanone does not.

(ii) There are two −NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(iii) 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.

Solution :
(i) Cyclohexanones form cyanohydrins according to the following equation.

In this case, the nucleophile CN− can easily attack without any steric hindrance. However, in the case of 2, 2, 6 trimethylcydohexanone, methyl groups at α-positions offer steric hindrances and as a result, CN− cannot attack effectively.

For this reason, it does not form a cyanohydrin.

(ii) Semicarbazide undergoes resonance involving only one of the two −NH2 groups, which is attached directly to the carbonyl-carbon atom.

Therefore, the electron density on −NH2 group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other −NH2 group is not involved in resonance; it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.

(iii) Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.

If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed.

Question19. An organic compound contains 69.77\% carbon, 11.63\% hydrogen and the rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound. \section*

{Solution} \textbf{Step I: Determine the Molecular Formula of the Compound} Given: \begin{itemize} \item Carbon: 69.77\% \item Hydrogen: 11.63\% \item Oxygen: 100\% – 69.77\% – 11.63\% = 18.60\% \item Molecular Mass: 86 \end{itemize} \textbf{Calculate the Empirical Formula:} \begin{itemize} \item Moles of Carbon: \[ \text{Moles of Carbon} = \frac{69.77}{12} \approx 5.81 \] \item Moles of Hydrogen: \[ \text{Moles of Hydrogen} = \frac{11.63}{1} \approx 11.63 \] \item Moles of Oxygen: \[ \text{Moles of Oxygen} = \frac{18.60}{16} \approx 1.16 \] \end{itemize} \textbf{Mole Ratios:} \begin{itemize} \item Divide by the smallest number (1.16): \[ \text{Carbon} = \frac{5.81}{1.16} \approx 5 \] \[ \text{Hydrogen} = \frac{11.63}{1.16} \approx 10 \] \[ \text{Oxygen} = \frac{1.16}{1.16} \approx 1 \] \end{itemize} \textbf{Empirical Formula:} \[ \text{C}_5\text{H}_{10}\text{O} \] \textbf{Molecular Formula:} \[ \text{Molecular Formula} = \text{Empirical Formula} \times n \] \[ \text{Where } n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{86}{86} = 1 \] \[ \text{Molecular Formula} = 1 \times \text{C}_5\text{H}_{10}\text{O} = \text{C}_5\text{H}_{10}\text{O} \] \textbf{Step II: Predict the Structure of the Compound} \begin{itemize} \item Formation of an addition compound with sodium hydrogen sulphite indicates the presence of an aldehyde or ketone. \item Since the compound does not reduce Tollen’s reagent, it is not an aldehyde but a ketone. \item A positive iodoform test indicates a methyl ketone. \item On oxidation, the compound forms ethanoic and propanoic acids, suggesting the structure is such that oxidation results in these products. \end{itemize} \textbf{Possible Structure:} The structure of the compound, given it is a methyl ketone and oxidizes to give a mixture of ethanoic and propanoic acids, is: \[ \text{CH}_3\text{CO}\text{C}_3\text{H}_7 \] Thus, the compound is \textbf{3-Pentanone}. \textbf{Reactions:} 1. **Formation of Addition Compound with Sodium Hydrogen Sulphite:** \[ \text{CH}_3\text{CO}\text{C}_3\text{H}_7 + \text{NaHSO}_3 \rightarrow \text{Addition Compound} \] 2. **Iodoform Test:** \[ \text{CH}_3\text{CO}\text{C}_3\text{H}_7 + \text{I}_2 + \text{NaOH} \rightarrow \text{CH}_3\text{COI} + \text{CH}_3\text{I} + \text{NaI} + \text{NaOH} + \text{H}_2\text{O} \] 3. **Oxidation:** \[ \text{CH}_3\text{CO}\text{C}_3\text{H}_7 \xrightarrow{\text{O}}\text{CH}_3\text{COOH}+ \text{CH}_3\text{CH}_2\text{COOH} \]

Question20. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Solution :
Resonance structures of phenoxide ion are:

It can be observed from the resonance structures of phenoxide ion that in II, III and IV, less electronegative carbon atoms carry a negative charge. Therefore, these three structures contribute negligibly towards the resonance stability of the phenoxide ion. Hence, these structures can be eliminated. Only structures I and V carry a negative charge on the more electronegative oxygen atom.

Resonance structures of carboxylate ion are:

In the case of carboxylate ion, resonating structures I′ and II′ contain a charge carried by a more electronegative oxygen atom.

Further, in resonating structures I′ and II′, the negative charge is delocalized over two oxygen atoms. But in resonating structures I and V of the phexoxide ion, the negative charge is localized on the same oxygen atom. Therefore, the resonating structures of carboxylate ion contribute more towards its stability than those of phenoxide ion. As a result, carboxylate ion is more resonance-stabilized than phenoxide ion. Hence, carboxylic acid is a stronger acid than phenol.

Question21. 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

Solution :
(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question22. Write the structures of products of the following reactions;

(i)

(ii)

(iii)

(iv)

Solution :
(i)

(ii)

(iii)

(iv)

Question23. Arrange the following compounds in increasing order of their boiling points.

CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3

Solution :
The molecular masses of the given compounds are in the range 44 to 46. CH3CH2OH undergoes extensive intermolecular H-bonding, resulting in the association of molecules. Therefore, it has the highest boiling point. CH3CHO is more polar than CH3OCH3 and so CH3CHO has stronger intermolecular dipole − dipole attraction than CH3OCH3⋅ CH3CH2CH3 has only weak van der Waals force. Thus, the arrangement of the given compounds in the increasing order of their boiling points is given by:

CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH

Question24. 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.

Hint:Consider steric effect and electronic effect.

Solution :
(i)

The +I effect of the alkyl group increases in the order:

Ethanal < Propanal < Propanone < Butanone

The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is:

Butanone < Propanone < Propanal < Ethanal

(ii)

The +I effect is more in ketone than in aldehyde. Hence, acetophenone 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 and the lowest in p-nitrobezaldehyde because of the presence of the electron-withdrawing −NO2 group. Hence, the increasing order of the reactivities of the given compounds is:

Acetophenone < p-tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde

Question 25. Predict the products of the following reactions:

(i)

(ii)

(iii)

(iv)

Solution :
(i)

(ii)

(iii)

(iv)

Question26. Give the IUPAC names of the following compounds:

(i) PhCH2CH2COOH (ii) (CH3)2C=CHCOOH

(iii)(iv)

Solution :
(i) 3-Phenylpropanoic acid

(ii) 3-Methylbut-2-enoic acid

(iii) 2-Methylcyclopentanecarboxylic acid

(iv)2,4,6-Trinitrobenzoic acid

Question27. Show how each of the following compounds can be converted to benzoic acid.

(i) Ethylbenzene (ii) Acetophenone

(iii) Bromobenzene (iv) Phenylethene (Styrene)

Solution :
(i)

(ii)

(iii)

(iv)

Question28. {Question} Which acid of each pair shown here would you expect to be stronger? \begin{enumerate} \item $\text{CH}_3\text{COOH}$ \text{ or } $\text{CH}_2\text{FCOOH}$ \item $\text{CH}_2\text{FCOOH}$ \text{ or } $\text{CH}_2\text{ClCOOH}$ \item $\text{CH}_2\text{FCH}_2\text{CH}_2\text{COOH}$ \text{ or } $\text{CH}_3\text{CHFCH}_2\text{COOH}$ \end{enumerate} \section*

{Solution} \textbf{A:} The acidity of carboxylic acids is influenced by the inductive effect. An electron-withdrawing group ($-I$ effect) increases the acidity of carboxylic acids by stabilizing the conjugate base, while an electron-releasing group ($+I$ effect) decreases the acidity. \[ \text{CH}_2\text{FCOOH} \text{ is stronger acid than } \text{CH}_3\text{COOH} \] This is because fluorine, with its high electronegativity, has a stronger $-I$ effect compared to the $+I$ effect of the methyl group in $\text{CH}_3\text{COOH}$. The conjugate base $\text{CH}_2\text{FCOO}^-$ is more stabilized compared to $\text{CH}_3\text{COO}^-$, making $\text{CH}_2\text{FCOOH}$ a stronger acid. \textbf{B:} Fluorine is more electronegative than chlorine, so it has a stronger $-I$ effect. Therefore: \[ \text{CH}_2\text{FCOOH} \text{ is a stronger acid than } \text{CH}_2\text{ClCOOH} \] The $-I$ effect of fluorine stabilizes the conjugate base more effectively than chlorine, making $\text{CH}_2\text{FCOOH}$ a stronger acid. \textbf{C:} The strength of the inductive effect decreases with increasing distance from the carboxyl group. In $\text{CH}_3\text{CHFCH}_2\text{COOH}$, the $-I$ effect of fluorine is closer to the carboxyl group than in $\text{CH}_2\text{FCH}_2\text{CH}_2\text{COOH}$, making the former a stronger acid. \[ \text{CH}_3\text{CHFCH}_2\text{COOH} \text{ is more acidic than } \text{CH}_2\text{FCH}_2\text{CH}_2\text{COOH} \] \textbf{D:} $\text{CF}_3$ group has a strong $-I$ effect compared to $\text{CH}_3$ group, which has a $+I$ effect. Therefore: \[ \text{CF}_3\text{COOH} \text{ is more acidic than } \text{CH}_3\text{COOH} \] The $-I$ effect of $\text{CF}_3$ stabilizes the conjugate base more than the $+I$ effect of $\text{CH}_3$, increasing the acidity of $\text{CF}_3\text{COOH}$.

These are the entire questions prescribed in your NCERT Chemistry Textbook Aldehydes, Ketones and Carboxylic Acids. Practise them regularly to score well and visit us to get help on any related matter at SimplyAcad.