NCERT Solutions for Class 12 Chemistry Chapter 12 - Aldehydes Ketones And Carboxylic Acids

NCERT Exercise

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

(i) Cyanohydrin

Ans: Cyanohydrins are organic, RR′(OH)CN chemicals, where R and R′s may be either alkyl or aryl.

In the presence of excess sodium cyanide (NaCN) as a catalyst in the field of cyanohydrin, aldehydes and ketones react with hydrogen cyanide (HCN). These are called cyanohydrins reactions. The reaction is given below:

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(ii) Acetal

Ans: Acetals are gem-dialcoxy alkanes in which the terminal carbon atom consists of two alcohol groups. There are connections between one bond and an alkyl group and the other with a hydrogen atom.

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When two monohydric alcohol equivalents are treated with dry HCl gas, the hemiacetals are generated which react further with another alcohol molecule to form acetal. The reaction is given below:

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(iii) Semicarbazone

Ans: Semicarbazones are products of the condensation process of aldehydes and ketones generated by ketone or aldehyde and semicarbazide.  An example of the formation of semicarbazone is given below:

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(iv) Aldol

Ans: An aldol is recognised to be a $\text{ }\!\!\beta\!\!\text{ -hydroxy}$ aldehyde or ketone. The condensation process in the presence of a base is generated between two molecules of one or two distinct aldehydes or ketones. An example of the formation of aldol is given below:

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(v) Hemiacetal

Ans: We can say that $\text{ }\!\!\alpha\!\!\text{ -alkoxyalcohols}$ are known as hemiacetals. In the presence of dry HCl gas, aldehyde interacts with a single molecule of monohydric alcohol. The reaction is given below:

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(vi) Oxime

Ans: Oximes have the common formula of RR′CNOH where R is an organic side chain, and R′ is hydrogen or organic side chain. Oximes are organic chemicals. If R′ is H, the aldoxime is known and R′ is known as ketoxime if it is an organic side chain.

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Aldehydes or ketones from oximes are treated with hydroxylamine in a slightly acidic media. The reaction is given below:

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(vii) Ketal

Ans: Ketals are gemodioxyalcanes in which the chain contains two alcoholic groups on the same carbon atom. The other two carbon bonds are linked to two groups of alkyles. In the presence of HCl dry gas, ketones react with ethylene glycol to a cyclic compound called ethylene glycol ketals. An example of the formation of a ketal is given below:

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(viii) Imine

Ans: Imines are chemicals that have a double bond of carbon-nitrogen. Imines are generated when ammonia and its derivatives are reacted by aldehydes and ketones. The reaction is given below:

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(ix) 2,4-DNP-derivative

Ans: 2, 4−dinitrophenylhydrazones 2, 4−DNP− derivatives that are made using 2, 4−dinitrophenylhydrazine in a low acidic medium in the case of aldehydes or ketones. The reaction is given below:

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(x) Schiff’s base

Ans: The Schiff's base (or azomethine) is a chemical molecule that has an aryl or alkyl group, but not hydrogen, carbon-nitrogen twice the atom of nitrogen. They have the usual R1R2C=NR3 formulation. As it is an imine and it is named after Hugo Schiff, a physicist. The reaction is given below:

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2. Name the following compounds according to IUPAC system of nomenclature:

(i) $\text{C}{{\text{H}}_{\text{3}}}\text{CH(C}{{\text{H}}_{\text{3}}}\text{)C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{CHO}$ 

Ans: The IUPAC name of the given compound is 4-Methylpentanal.

(ii) $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{COCH(}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{)C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl}$ 

Ans: The IUPAC name of the given compound is 6-Chloro-4-ethylhexan-3-one.

(iii) $\text{C}{{\text{H}}_{\text{3}}}\text{CH=CHCHO}$ 

Ans: The IUPAC name of the given compound is But-2-en-1-al.

(iv) $\text{C}{{\text{H}}_{\text{3}}}\text{COC}{{\text{H}}_{\text{2}}}\text{COC}{{\text{H}}_{\text{3}}}$ 

Ans: The IUPAC name of the given compound is Pentan-2,4-dione.

(v) $\text{C}{{\text{H}}_{\text{3}}}\text{CH(C}{{\text{H}}_{\text{3}}}\text{)C}{{\text{H}}_{\text{2}}}\text{C(C}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{COC}{{\text{H}}_{\text{3}}}$ 

Ans: The IUPAC name of the given compound is 3,3,5-Trimethylhexan-2-one.

(vi) ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{3}}}\text{CC}{{\text{H}}_{\text{2}}}\text{COOH}$ 

Ans: The IUPAC name of the given compound is 3, 3-Dimethylbutanoic acid.

(vii) $\text{OHC}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{4}}}\text{CHO-p}$ 

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

3. Draw the structures of the following compounds:

(i) 3-Methylbutanal

Ans: The structure of 3-Methylbutanal is given below:

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(ii) p-Nitropropiophenone

Ans: The structure of p-Nitropropiophenone is given below:

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(iii) p-Methylbenzaldehyde

Ans: The structure of p-Methylbenzaldehyde is given below:

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(iv) 4-Methylpent-3-en-2-one

Ans: The structure of 4-Methykpent-3-en-2-one is given below:

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(v) 4-Chloropentan-2-one

Ans: The structure of 4-Chloropentan-2-one is given below:

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(vi) 3-Bromo-4-phenylpentanoic acid

Ans: The structure of 3-Bromo-4-phenylpentanoic acid is given below:

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(vii) p,p’-Dihydroxybenzophenone

Ans: The structure of p,p’-Dihydroxybenzophenone is given below:

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(viii) Hex-2-en-4-ynoic acid

Ans: The structure of Hex-2-en-4-ynoic acid is given below:

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4. Write the IUPAC names of the following ketones and aldehydes. Whenever possible, give also common names.

(i) $\text{C}{{\text{H}}_{\text{3}}}\text{CO(C}{{\text{H}}_{\text{2}}}{{\text{)}}_{\text{4}}}\text{C}{{\text{H}}_{\text{3}}}$ 

Ans: The IUPAC name of the given compound is Heptan-2-one and its common name is Methyl n-pentyl ketone.

(ii) $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CHBrC}{{\text{H}}_{\text{2}}}\text{CH(C}{{\text{H}}_{\text{3}}}\text{)CHO}$ 

Ans: The IUPAC name of the given compound is 4-Bromo-2-methylhexanal and its common name is $\text{ }\!\!\gamma\!\!\text{ -Bromo- }\!\!\alpha\!\!\text{ -methylcaproaldehyde}$

(iii) $\text{C}{{\text{H}}_{\text{3}}}{{\text{(C}{{\text{H}}_{\text{2}}}\text{)}}_{\text{5}}}\text{CHO}$ 

Ans: The IUPAC name of the given compound is Heptanal.

(iv) $\text{Ph-CH=CH-CHO}$ 

Ans: The IUPAC name of the given compound is 3-Phenylprop-2-enal and its common name is beta-phenylacrolein.

(v)                                       (Image will be uploaded soon)

Ans: The IUPAC name of the given compound is Cyclopentanecarbaldehyde and its common name is also Cyclopentanecarbaldehyde.

(vi) PhCOPh

Ans: The IUPAC name of the given compound is Diphenylmetthanone and its common name is Benzophenone.

5. Draw structures of the following derivatives.

(i) The 2,4-dinitrophenylhydrazone of benzaldehyde

Ans: The structure of derivative of 2,4-dinitrophenylhydrazone of benzaldehyde will be:

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(ii) Cyclopropanone oxime

Ans: The structure of derivative of Cyclopropanone oxime is given below:

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(iii) Acetaldehyde Dimethyl Acetal

Ans: The structure of derivative of Acetaldehyde Dimethyl Acetal is given below:

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(iv) The semicarbazone of cyclobutanone

Ans: The structure of derivative of the semicarbazone of cyclobutanone is given below:

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(v) The ethylene ketal of hexan-3-one

Ans: The structure of derivative of the ethylene ketal of hexan-3-one is given below:

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(vi) The methyl hemiacetal of formaldehyde

Ans: The structure of derivative of the methyl hemiacetal of formaldehyde is given below:

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6. Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.

(i) PhMgBr and then ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}$ 

Ans: The product in this reaction will be Cyclohexyl Phenyl Carbinol. The reaction is given below:

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(ii) Tollens’ reagent

Ans: The product formed in this reaction is Cyclohexane-carboxylate ion. The reaction is given below:

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(iii) Semicarbazide and weak acid

Ans: The product formed in this reaction is Cyclohexanecarbaldehyde semicarbazone. The reaction is given below:

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(iv) Excess ethanol and acid

Ans: The product formed in this reaction is Cyclohexanecarbaldehyde diethyl acetal. The reaction is given below:

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(v) Zinc amalgam and dilute hydrochloric acid.

Ans: The product formed in this reaction will be Methylcyclohexane. The reaction is given below:

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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.

1. Methanal

2. 2-Methylpentanal

3. Benzaldehyde

4. Benzophenone

5. Cyclohexanone

6. 1-Phenylpropanone

7. Phenylacetaldehyde

8. Butan-1-ol

9. 2, 2-Dimethylbutanal

Ans: Aldehydes and ketones having at least one $\text{ }\!\!\alpha\!\!\text{ -hydrogen}$ undergo aldol condensation. The compounds (ii) 2−methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone, and (vi) phenylacetaldehyde contain one or more $\text{ }\!\!\alpha\!\!\text{ -hydrogen}$atoms. Therefore, these undergo aldol condensation.

The product formed by the aldol condensation of 2-Methylpentanal is 3-Hydroxy-2,4-dimethyl-2-propylheptanal. The reaction is given below:

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The product formed by the aldol condensation of cyclohexanone is 2-(1-Hydroxy-1-cyclohexyl)cyclohexane-1-one. The reaction is given below:

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The product formed by the aldol condensation of 1-Phenylpropanone is 3-Hydroxy-2-methyl-1,3-diphenylpentan-1-one. The reaction is given below:

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The product formed by the aldol condensation of Phenylacetyladehyde is 3-Hydroxy-2,4-diphenylbutanal. The reaction is given below:

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Aldehydes having no $\text{ }\!\!\alpha\!\!\text{ -hydrogen}$ atoms undergo Cannizzaro reactions. The compounds (i) Methanal, (iii) Benzaldehyde, and (ix) 2, 2-dimethylbutanal do not have any $\text{ }\!\!\alpha\!\!\text{ -hydrogen}$.Therefore, these undergo cannizzaro reactions.

The products formed by the Cannizaro reaction of methanal are methanol and sodium methanoate. The reaction is given below:

\[\text{2 HCHO }\xrightarrow{\text{conc}\text{. NaOH}}\text{ C}{{\text{H}}_{\text{3}}}\text{OH + HCOONa}\] 

The products formed by the Cannizaro reaction of benzaldehyde are benzyl alcohol and sodium benzoate. The reaction is given below:

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The products formed by the Cannizaro reaction of 2,2-Dimethylbutanol are benzyl alcohol and sodium benzoate. The reaction is given below:

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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.

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

1. Butane-1, 3-diol

Ans: When ethanal reacts with dilute sodium hydroxide to form 3-Hydroxybutanal. Now, 3-Hydroxybutanal reacts with $\text{NaB}{{\text{H}}_{\text{4}}}$ to form butane-1, 3-diol. The reaction is given below:

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2. But-2-enal

Ans: When ethanal reacts with dilute sodium hydroxide to form 3-Hydroxybutanal. Now, 3-Hydroxybutanal reacts with acid to form But-2-enal. The reaction is given below:

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3. But-2-enoic acid

Ans: When ethanal reacts with dilute sodium hydroxide to form 3-Hydroxybutanal. Now, 3-Hydroxybutanal reacts with acid to form But-2-enal. Now, but-2-enal reacts with Tollen’s reagent to But-2-enoic acid. The reaction is given below:

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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.

Ans: 

1. Taking propanal as nucleophile as well as electrophile. The product will be 3-Hydroxy-2-methylpentanal. The reaction is given below:

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2. Taking propanal as electrophile and butanal as nucleophile. The product will be 2-Ethyl-3-hydroxypentanal. The reaction is given below:

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3. Taking butanal as electrophile and propanal as nucleophile. The product will be 3-Hydroxy-2-methylhexanal. The reaction is given below:

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4. Taking butanal as nucleophile as well as electrophile. The product will be 2-Ethyl-3-hydroxyhexanal. The reaction is given below:

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10. An organic compound with the molecular formula ${{\text{C}}_{\text{9}}}{{\text{H}}_{\text{10}}}\text{O}$ forms 2, 4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.

Ans: It is because the chemical (${{\text{C}}_{\text{9}}}{{\text{H}}_{\text{10}}}\text{O}$) generates a derivative of 2, 4-dnp and reduces the reagent of Tollen. The chemical must thus be an aldehyde. The product is again subjected to cannizzaro's reaction and 1,2-benzenedicarboxylic acid is provided by oxidation. The −CHO group is therefore immediately connected to a benzene ring and ortho-substituted for this benzaldehyde. Thus, 2-ethylbenzaldehyde is the chemical.

The reactions given in the questions are given below:

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11. An organic compound (A) (molecular formula ${{\text{C}}_{\text{8}}}{{\text{H}}_{\text{16}}}{{\text{O}}_{\text{2}}}$) was hydrolyzed with dilute sulphuric 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.

Ans: A molecular molecular formula A molecule ${{\text{C}}_{\text{8}}}{{\text{H}}_{\text{16}}}{{\text{O}}_{\text{2}}}$ provides hydrolysis with dilute sulphuric acid carboxylic acid (B) and alcohol (C) Compound A therefore has to be an ester. Alcohol C also provides acid B for chromic acid oxidation. Therefore, the number of carbon atoms must be identical in B and C. Compound A has 8 carbon atoms, while 4 carbon atoms are included in each of B and C. Alcohol C again produces but-1-ene when it is dehydrated. C is thus straight and hence butane-1-ol. Butan-1-ol produces butanoic acid when oxidised. Therefore, butanoic acid is B acid. Thus, butylbutanoate is the molecular ester ${{\text{C}}_{\text{8}}}{{\text{H}}_{\text{16}}}{{\text{O}}_{\text{2}}}$.

All the reactions in questions are given below:

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

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

Ans: When HCN interacts with a molecule, the attacking species is nucleophile, CN- Consequently, its reactivity to HCN diminishes as the negative charge of the compound rises. The +I effect rises in the given compounds as indicated below. The steric impediment also rises in this. It may be noticed.

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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.

2. $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CH(Br)COOH, C}{{\text{H}}_{\text{3}}}\text{CH(Br)C}{{\text{H}}_{\text{2}}}\text{COOH, (C}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{CHCOOH, }$ 

$\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{COOH (acid strength)}$

Ans: Carboxylic acids are negatively charged after losing a proton, as illustrated:

\[\text{R-COOH }\to \text{ R-CO}{{\text{O}}^{\text{-}}}\text{+}{{\text{H}}^{\text{+}}}\] 

Now any group which helps stabilise the negative load will enhance the carboxylic ion's stability and hence improve the acid strength. This will reduce the strength of acids in groups with +I effect and raise the strength of acids in groups with −I effect. The group $\text{-C}{{\text{H}}_{\text{3}}}$ exhibits +I and $\text{B}{{\text{r}}^{\text{-}}}$ action in the following compounds. $\text{B}{{\text{r}}^{\text{-}}}$ containing acids are hence stronger. Now, +I is more than the n-propyl isopropyl group effect.

Therefore, ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{2}}}\text{CHCOOH}$ is a weaker acid than $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{COOH}$.

Also, the –I effect grows weaker as distance increases. Hence, $\text{C}{{\text{H}}_{\text{3}}}\text{CH(Br)C}{{\text{H}}_{\text{2}}}\text{COOH}$is a weaker acid than $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CH(Br)COOH}$.

Hence, the order the strength is given below:

$\text{C}{{\text{H}}_{\text{3}}}\text{CH(C}{{\text{H}}_{\text{3}}}\text{)COOH}$ < $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{COOH}$ < $\text{C}{{\text{H}}_{\text{3}}}\text{CH(Br)C}{{\text{H}}_{\text{2}}}\text{COOH}$ < $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CH(Br)COOH}$ 

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

Ans: As we saw in the past case, electron-donor groups reduce acid strengths whereas electron-donor groups enhance acid strengths. Benefits of 4-methoxybenzoic acid is a lesser acid than benzoic acid as a group of electron-donor. Nitro group is a retracting electron group and increases acid strength. As it includes 2 nitrogen groups of 3,4-dinitrobenzoic acid, it is a little stronger 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    

13. Give simple chemical tests to distinguish between the following pairs of compounds.

1. Propanal and Propanone

Ans: 

1. Tollen’s test

An aldehyde is Propanal. It reduces the reagent of Tollen. But, propanone being a ketone does not reduces Tollen’s reagent.

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CHO + 2 }\!\![\!\!\text{ Ag(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}{{\text{ }\!\!]\!\!\text{ }}^{\text{+}}}\text{+3O}{{\text{H}}^{\text{-}}}\text{ }\to \text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CO}{{\text{O}}^{\text{-}}}\text{+Ag}\downarrow \text{+4N}{{\text{H}}_{\text{3}}}\text{+2}{{\text{H}}_{\text{2}}}\text{O}\] 

2. Fehling’s test

Aldehydes are answering the test of Fehling, while ketones are not. Aldehyde-propanal converts the solution to a red-brown Cu2O precipitate, although propanone does not.

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CHO + 2C}{{\text{u}}^{\text{2+}}}\text{+5O}{{\text{H}}^{\text{-}}}\text{ }\to \text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CO}{{\text{O}}^{\text{-}}}\text{+C}{{\text{u}}_{\text{2}}}\text{O+3}{{\text{H}}_{\text{2}}}\text{O}\] 

3. Iodoform test

As one methyl group associated with the carbonyl carbon atom reacts to the iodoform test, with aldehydes and ketones. They are oxidised to yield iodoform using sodium hypoiodite (NaOI). Propanone is a methyl ketone, while propanal is not.

\[\text{C}{{\text{H}}_{\text{3}}}\text{COC}{{\text{H}}_{\text{3}}}\text{+3NaOI }\to \text{ C}{{\text{H}}_{\text{3}}}\text{COONa+CH}{{\text{I}}_{\text{3}}}\text{+2NaOH}\] 

2. Acetophenone and Benzophenone

Ans: Iodoform test

Acetophenone will give the Iodoform test but benzophenone will not.

\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COC}{{\text{H}}_{\text{3}}}\text{+3NaOI }\to \text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COONa+CH}{{\text{I}}_{\text{3}}}\text{+2NaOH}\] 

3. Phenol and Benzoic acid

Ans: Ferric chloride test

Phenol will give the violet coloration with neutral ferric chloride.

\[\text{6}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{OH+FeC}{{\text{l}}_{\text{3}}}\text{ }\to \text{ }\underset{\text{Violet color}}{\mathop{{{\text{ }\!\![\!\!\text{ Fe(O}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3-}}}}}\,\text{+3}{{\text{H}}^{\text{+}}}\text{+3C}{{\text{l}}^{\text{-}}}\] 

Benzoic acid will give the buff coloration with neutral ferric chloride.

\[\text{3}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH+FeC}{{\text{l}}_{\text{3}}}\text{ }\to \text{ }\underset{\text{buff color}}{\mathop{{{\text{ }\!\![\!\!\text{ Fe(OO}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{)}}_{3}}\text{ }\!\!]\!\!\text{ }}^{\text{3-}}}}}\,\text{+3}{{\text{H}}^{\text{+}}}\text{+3C}{{\text{l}}^{\text{-}}}\] 

4. Benzoic acid and Ethyl benzoate

Ans: Sodium bicarbonate test

Benzoic acid will react with sodium bicarbonate and will evolve carbon dioxide.

\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH+NaHC}{{\text{O}}_{\text{3}}}\text{ }\to \text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COONa+C}{{\text{O}}_{\text{ 2}}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\]

5. Pentan-2-one and Pentan-3-one

Ans: Iodoform test

Pentan-2-one will react to the Iodoform test but Pentan-3-one will not react.\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{COC}{{\text{H}}_{\text{3}}}\text{+3NaOI }\to \text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{COONa+CH}{{\text{I}}_{\text{3}}}\text{+2NaOH}\] 

6. Benzaldehyde and Acetophenone

Ans: Iodoform test

Acetophenone will give the Iodoform test but benzaldehyde will not.

\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COC}{{\text{H}}_{\text{3}}}\text{+3NaOI }\to \text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COONa+CH}{{\text{I}}_{\text{3}}}\text{+2NaOH}\]

7. Ethanal and Propanal

Ans: Iodoform test

The iodoform test responds to aldehydes and ketones with a minimum of a methyl group associated with the carbonyl carbon atom. This test is followed by ethanal with one methyl group attached to the carbonyl atom. But propanal doesn't contain a carbonyl carbon dioxide-related methyl group and does not thus respond to this condition.

\[\text{C}{{\text{H}}_{\text{3}}}\text{CHO+3NaOI }\to \text{ HCOONa+CH}{{\text{I}}_{\text{3}}}\text{+2NaOH}\] 

14. 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.

1. Methyl benzoate

Ans: Benzene will react with bromine to form Bromobenzene. This bromobenzene will be converted into benzoic acid. Now, this benzoic acid will be converted into Methyl benzoate. The reaction is given below:

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2. m-Nitrobenzoic acid.

Ans: Benzene will react with bromine to form Bromobenzene. This bromobenzene will be converted into benzoic acid. Now, this benzoic acid will be converted into m-Nitro benzoic acid. The reaction is given below:

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3. p-Nitrobenzoic acid

Ans: Benzene will be converted into toluene. Now, Toluene will be converted into p-nitrotoluene. Now, this p-nitrotoluene on oxidation will form p-nitrobenzoic acid. The reaction is given below:

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4. Phenylacetic acid

Ans: Benzene will be converted into toluene. Now, Toluene will be converted into Benzyl bromide; this will be further converted into benzyl cyanide. Benzyl cyanide will be hydrolyzed to Phenylacetic acid. The reaction is given below:

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5. p-Nitrobenzaldehyde

Ans: Benzene will be converted into toluene. Now, Toluene will be converted into p-nitrotoluene. Now, this p-nitrotoluene on oxidation will form p-nitrobenzaldehyde. The reaction is given below:

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

1. Propanone to Propene

Ans: Propanone will react with $\text{NaB}{{\text{H}}_{\text{3}}}$ to form 2-Propanol. When 2-propanol will react with concentrated sulfuric acid to form propene. The reaction is given below:

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2. Benzoic acid to Benzaldehyde

Ans: Benzoic acid will be converted into Benzoyl chloride which will react with hydrogen and barium sulphate to give Benzaldehyde. The reaction is given below:

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3. Ethanol to 3-Hydroxybutanal

Ans: Ethanol will be converted into ethanal which will react with dilute NaOH to form 3-Hydroxybutanal. The reaction is given below:

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4. Benzene to m-Nitroacetophenone

Ans: Benzene will be converted into acetophenone which will react with concentrated nitric acid and sulfuric acid to form m-Nitroacetophenone. The reaction is given below:

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5. Benzaldehyde to Benzophenone

Ans: Benzaldehyde will be converted into benzoic acid. The benzoic acid will be converted into benzoyl chloride which will react with benzene to form Benzophenone. The reaction is given below:

(Image will be uploaded soon

6. Bromobenzene to 1-Phenylethanol

Ans: Bromobenzene will be converted into Phenyl magnesium bromide which will react with ethanal to form 1-Phenylethanol. The reaction is given below:

(Image will be uploaded soon)

7. Benzaldehyde to 3-Phenylpropan-1-ol

Ans: Benzaldehyde will be converted into 3-Phenylprop-2-enal which will react with hydrogen to form 3-Phenylpropan-1-ol. The reaction is given below:

(Image will be uploaded soon)

8. Benzaldehyde to $\text{ }\!\!\alpha\!\!\text{ }$-Hydroxyphenylacetic acid

Ans: Benzaldehyde will be converted into Benzaldehyde cyanohydrin which will react with acidic water to form $\text{ }\!\!\alpha\!\!\text{ -Hydroxyphenylacetic acid}$. The reaction is given below:

(Image will be uploaded soon)

9. Benzoic acid to m- Nitrobenzyl alcohol

Ans: Benzoic acid will be converted into m-nitrobenzoic acid which will be further converted into m-Nitrobenzoyl chloride. m-Nitrobenzoyl chloride will react with $\text{NaB}{{\text{H}}_{\text{4}}}$ to form m-Nitrobenzyl alcohol. The reaction is given below:

(Image will be uploaded soon)

16. Describe the following:

1. Acetylation

Ans: Acetyl functional group is referred to as acetylation in the organic molecule. It is generally done with a basis such as pyridine, dimethylaniline, etc. This includes replacing an active hydrogen atom with a group of acetyls. The most popular acetylating agents include acetyl chloride and acetic anhydride. For example, ethanol acetylation generates ethyl acetate.

\[\underset{\text{Ethanol}}{\mathop{\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}}}\,\text{ + }\underset{\text{Acetyl Chloride}}{\mathop{\text{C}{{\text{H}}_{\text{3}}}\text{COCl}}}\,\xrightarrow{\text{pyridine}}\underset{\text{Ethyl acetate}}{\mathop{\text{ C}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}}}\,\text{+HCl}\]  

2. Cannizzaro reaction

Ans: The reaction of aldehydes that have no $\text{ }\!\!\alpha\!\!\text{ }$-hydrogen to concentrated alkalis is sometimes called the Cannizzaro Reaction or Self-oxidation-reduction (disproportion). Two aldehyde molecules are involved in this process, one reduced to alcohol and the other to carboxylic acid oxidised.

Ethanol and potassium ethanoate are, for example, produced when treated with concentrated potassium hydroxide.

(Image will be uploaded soon)

3. Cross aldol condensation

Ans: The reaction is called a cross-aldol condensation if aldol condensation is carried out by two distinct aldehydes or two separate ketones or an aldehyde and a ketone. If the $\text{ }\!\!\alpha\!\!\text{ }$-hydrogen is present in the two reactants, four compound products are produced.

Ethanal and propanal, for example, respond to four products.

(Image will be uploaded soon)

4. Decarboxylation

Ans: Decarboxylation is the reaction in which carboxylic acids, when their salts are heated by soda-lime, lose carbon dioxide in order to produce hydrocarbons.

\[\text{C}{{\text{H}}_{\text{3}}}\text{-COONa }\xrightarrow{\text{Soda lime}}\text{ C}{{\text{H}}_{\text{4}}}\text{ + N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\]

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

1.                                 (Image will be uploaded soon)

Ans: The products formed in this reaction are potassium benzoate and then benzoic acid. The reaction is given below:

(Image will be uploaded soon)

2.                                (Image will be uploaded soon)

Ans: The product formed in this reaction is Phthaloyl chloride. The reaction is given below:

(Image will be uploaded soon)

3. ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{CHO}\xrightarrow{{{\text{H}}_{\text{2}}}\text{NCONHN}{{\text{H}}_{\text{2}}}}$ 

Ans: The product formed in this reaction will be Benzaldehyde semicarbazone. The reaction is given below:

\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{CHO}\xrightarrow{{{\text{H}}_{\text{2}}}\text{NCONHN}{{\text{H}}_{\text{2}}}}\text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{CH=NNHCON}{{\text{H}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O}\] 

4.                                (Image will be uploaded soon).

Ans: In this reaction, another reactant is benzoyl chloride and the catalyst is aluminium chloride. The reaction is given below:

(Image will be uploaded soon)

5.                                (Image will be uploaded soon)

Ans: The tollen’s reagent will only reduce the aldehyde part. The reaction is given below:

(Image will be uploaded soon)

6.                                (Image will be uploaded soon)

Ans: The product formed in this reaction will be 2-(1-Hydroxy Cyanomethyl) benzoic acid. The reaction is given below:

7. ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{CHO + C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CHO }\xrightarrow{\text{dil}\text{.NaOH}}$ 

Ans: The product formed in this reaction will be 2-Methyl-3-phenylprop-2-enal. The reaction is given below:

(Image will be uploaded soon)

8. $\text{C}{{\text{H}}_{\text{3}}}\text{COC}{{\text{H}}_{\text{2}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\xrightarrow[\text{(ii)}{{\text{H}}^{\text{+}}}]{\text{(i)NaB}{{\text{H}}_{\text{4}}}}$ 

Ans: The product formed in this reaction is Ethyl-3-hydroxybutanoate. The reaction is given below:

(Image will be uploaded soon)

9.                               (Image will be uploaded soon)

Ans: The product formed in this reaction will be Cyclohexanone. The reaction is given below:

(Image will be uploaded soon)

10.                                (Image will be uploaded soon)

Ans: The complete reaction is given below:

(Image will be uploaded soon)

11.                                (Image will be uploaded soon)

Ans: The reactant for this reaction will be Cyclohexylidene Cyclohexanone. The reaction is given below:

(Image will be uploaded soon)

18. Give plausible explanation for each of the following:

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

Ans: Cyclohexanones form cyanohydrins according to the following equation.

(Image will be uploaded soon)

In this scenario, nucleophiles $\text{C}{{\text{N}}^{\text{-}}}$  can approach freely without steric obstruction. But for 2-2, 6-trimethylcyclohexanone, $\text{ }\!\!\alpha\!\!\text{ }$-positioned methyl groups present steric barriers and as a result $\text{C}{{\text{N}}^{\text{-}}}$ cannot successfully attack.

(Image will be uploaded soon)

That’s why it cannot form cyanohydrins.

2. There are two $\text{-N}{{\text{H}}_{\text{2}}}$ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

Ans: Semicarbazide is resonant with just one of two groups $\text{-N}{{\text{H}}_{\text{2}}}$ which is immediately linked to the carbonyl carbon atom.

(Image will be uploaded soon

Consequently, the electron density at $\text{-N}{{\text{H}}_{\text{2}}}$ likewise decreases with the resonance. It cannot therefore operate as a nucleophile. Since the other group $\text{-N}{{\text{H}}_{\text{2}}}$ does not engage itself in resonance, it may function as a nucleophile and can attack aldehydrogen and ketone carbonic atoms to generate semicarbazones.

3. 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.

Ans: In presence of acid, ester along with water is reversibly made of carboxylic acid and alcohol.

\[\text{RCOOH+R }\!\!'\!\!\text{ OH}\xrightarrow{{{\text{H}}^{\text{+}}}}\text{RCOOR }\!\!'\!\!\text{ +}{{\text{H}}_{\text{2}}}\text{O}\] 

If water or ester is not eliminated as it forms, then the reactants react to the reversibility of the reaction. Therefore, one of the two should be eliminated to push the balance forward in a more ester way.

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 additional 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.

Ans: The percentage of carbon atom = 69.77%

The percentage of hydrogen atom = 11.63%

The percentage of oxygen atom = 100 – (69.77 + 11.63) = 18.6%

Thus, the ratio of the number of carbon, hydrogen, and oxygen atoms in the organic compound can be given as:

\[\text{C : H : O = }\frac{\text{69}\text{.77}}{\text{12}}\text{ : }\frac{\text{11}\text{.63}}{\text{1}}\text{ : }\frac{\text{18}\text{.6}}{\text{16}}\]

= 5.81 : 11.63 : 1.16

= 5 : 10 : 1

Hence the empirical formula of the compound is ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{10}}}\text{O}$. So, the mass of the compound is = 5 $\text{ }\!\!\times\!\!\text{ }$ 12 + 10 $\text{ }\!\!\times\!\!\text{ }$ 1 + 1 $\text{ }\!\!\times\!\!\text{ }$ 16 = 86

Since the given molecular mass is also 86. The molecular formula is also ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{10}}}\text{O}$.

As Tollen's reagent is not reduced by that molecule, it is not an aldehyde. The chemical once again provides a positive Iodoform test to form sodium hydrogen sulphate supplements. The molecule must be a methyl ketone, because it is not an aldehyde. A combination of ethanoic acid and propanoic acid also results in the given chemical. The chemical is thus pentan−2−ol.

The reactions in the question are given below:

(Image will be uploaded soon)

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

Ans: Resonance structures of phenoxide ion are:

(Image will be uploaded soon)

Resonance structures of phenoxide ion may be shown to have a negative charge for fewer electronegative carbon atoms in II, III and IV. Those three configurations therefore make a negligible contribution to the phenoxide ion's resonance stability. Such constructions can thus be removed. The most electronegative oxygen atom carries a negative load only in structures I and V.

The resonating structures of carboxylate ion are given below:

(Image will be uploaded soon)

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 phenoxide 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.

INTEXT SOLUTIONS

1. Write the structures of the following compounds.

1. α-Methoxypropionaldehyde

Ans: The structure of α-Methoxypropionaldehyde is given below:

(Image will be uploaded soon)

2. 3-Hydroxybutanal

Ans: The structure of 3-Hydroxybutanal is given below:

(Image will be uploaded soon)

3. 2-Hydroxycyclopentane carbaldehyde

Ans: The structure of 2-Hydroxycyclopentane carbaldehyde is given below:

(Image will be uploaded soon)

4. 4-Oxopentanal

Ans: The structure of 4-Oxopentanal is given below:

(Image will be uploaded soon)

5. Di-sec-butyl ketone

Ans: The structure of Di-sec-butyl ketone is given below:

(Image will be uploaded soon)

6. 4-Fluoroacetophenone

Ans: The structure of 4-Fluoroacetophenone is given below:

(Image will be uploaded soon)

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

1.                              (Image will be uploaded soon)

Ans: The product formed in this reaction is Propiophenone. The reaction is given below:

(Image will be uploaded soon)

2. ${{\text{(}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{)}}_{\text{2}}}\text{Cd + 2C}{{\text{H}}_{\text{3}}}\text{COCl }\to $ 

Ans: The product formed in this reaction is 1-Phenylpropan-2-one. The reaction is given below: \[{{\text{(}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{C}{{\text{H}}_{\text{2}}}\text{)}}_{\text{2}}}\text{Cd + 2C}{{\text{H}}_{\text{3}}}\text{COCl }\to \text{ 2C}{{\text{H}}_{\text{3}}}\text{COC}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{ + CdC}{{\text{l}}_{\text{2}}}\]

3. $\text{C}{{\text{H}}_{\text{3}}}\text{-C}\equiv \text{CH }\xrightarrow{\text{H}{{\text{g}}^{\text{2+}}}\text{,}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}}$

Ans: The product formed in this reaction is Propanone. The reaction is given below:

(Image will be uploaded soon)

4.                               (Image will be uploaded soon)

Ans: The product formed in this reaction is p-Nitrobenzaldehyde. The reaction is given below:

(Image will be uploaded soon)

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

\[\text{C}{{\text{H}}_{\text{3}}}\text{CHO, C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH, C}{{\text{H}}_{\text{3}}}\text{OC}{{\text{H}}_{\text{3}}}\text{, C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}\] 

Ans: The molecular masses are in the range 44 to 46 of the listed compounds. $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}$ has the greatest boiling point is consequently subjected to substantial, intermolecular H-bonding, culminating in the combination between molecules.

$\text{C}{{\text{H}}_{\text{3}}}\text{CHO}$ is more polar than $\text{C}{{\text{H}}_{\text{3}}}\text{OC}{{\text{H}}_{\text{3}}}$, thus the dipole is greater than $\text{C}{{\text{H}}_{\text{3}}}\text{OC}{{\text{H}}_{\text{3}}}$, and $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}$ is weaker than the van der Waals.

The order will be:

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}\text{  C}{{\text{H}}_{\text{3}}}\text{OC}{{\text{H}}_{\text{3}}}\text{  C}{{\text{H}}_{\text{3}}}\text{CHO  C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}\] 

4. Arrange the following compounds in increasing order of their reactivity in Nucleophilic addition reactions.

1. Ethanal, Propanal, Propanone, Butanone.

Ans: The structures of all the compounds are given below:

(Image will be uploaded soon)

The alkyl group +I impact increases:

Ethanal < Propanal < Propanone < Butanone

With the +I effect increasing, the electron density of carbonyl carbon rises. As a consequence, the possibilities of a nucleophile attack are reduced. Thus, in nucleophilic addition reactions, the increasing order of reactivities of given carbonyl compounds is Butanone < Propanone < Propanal < Ethanal.

2. Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone

Ans: The structure of the following compounds is given below:

(Image will be uploaded soon)

The impact of +I in ketone is higher than aldehyde. Acetophenone is therefore the least reactive in nuclear processes. The +I effect in p-tolualdehyde is greatest among aldehydes, due to the electron-donating group $\text{-C}{{\text{H}}_{\text{3}}}$ and the presence of the electron-donating group $\text{-N}{{\text{O}}_{\text{2}}}$ and the existence of the p-nitrobenzaldehyde.

Therefore, the increasing order of the compound reactivities is:

Benzaldehyde < p-Nitrobenzaldehyde < Acetophenone < p-tolualdehyde.

5. Predict the products of the following reactions:

1.                              (Image will be uploaded soon)

Ans: The complete reaction is given below:

(Image will be uploaded soon)

2.                               (Image will be uploaded soon)

Ans: The complete reaction is given below:

                                           (Image will be uploaded soon)

3.                              (Image will be uploaded soon)

Ans: The complete reaction is given below:

(Image will be uploaded soon)

4.                               (Image will be uploaded soon)

Ans: The complete reaction is given below:

(Image will be uploaded soon)

6. Give the IUPAC names of the following compounds:

1. $\text{PhC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{COOH}$

   Ans: The IUPAC name of the compound is 3-Phenylpropanoic acid.

2. ${{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{2}}}\text{C=CHCOOH}$ 

    Ans: The IUPAC name of the compound is 3-Methylbut-2-enoic acid.

3.                       (Image will be uploaded soon)

   Ans: The IUPAC name of the compound is 2-Methylcyclopentanecarboxylic  acid.

4.                        (Image will be uploaded soon)

Ans: The IUPAC name of the compound is 2, 4, 6-Trinitrobenzoic acid.

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

1. Ethylbenzene

Ans: Ethylbenzene will convert into potassium benzoate. Now, this potassium benzoate will be hydrolyzed to form benzoic acid. The reaction is given below:

(Image will be uploaded soon)

2. Acetophenone

Ans: Acetophenone will convert into potassium benzoate. Now, this potassium benzoate will be hydrolyzed to form benzoic acid. The reaction is given below:

(Image will be uploaded soon)

3. Bromobenzene

Ans: Bromobenzene will react with magnesium to form Phenyl magnesium bromide. This will be converted into benzoic acid. The reaction is given below:

(Image will be uploaded soon)

4. Phenylethene (Styrene)

Ans: Phenylethene will convert into potassium benzoate. Now, this potassium benzoate will be hydrolyzed to form benzoic acid. The reaction is given below:

(Image will be uploaded soon)

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

1. $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{O}}_{\text{2}}}\text{H o rC}{{\text{H}}_{\text{2}}}\text{FC}{{\text{O}}_{\text{2}}}\text{H}$ 

Ans: The structures of the compounds are given below:

(Image will be uploaded soon)

The $\text{-C}{{\text{H}}_{\text{3}}}$  Group's +I action increases the electron density on the O-H connection. Thus, it is tough to release the proton. The −I impact of F on the O-H bond, on the other hand, reduces the electron density. Proton may therefore readily be released. $\text{C}{{\text{H}}_{\text{2}}}\text{FC}{{\text{O}}_{\text{2}}}\text{H}$ is hence stronger than $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{O}}_{\text{2}}}\text{H}$.

2. $\text{C}{{\text{H}}_{\text{2}}}\text{FC}{{\text{O}}_{\text{2}}}\text{H or C}{{\text{H}}_{\text{2}}}\text{ClC}{{\text{O}}_{\text{2}}}\text{H}$ 

Ans: F has stronger −I effect than Cl. Therefore, $\text{C}{{\text{H}}_{\text{2}}}\text{FC}{{\text{O}}_{\text{2}}}\text{H}$ can release proton more easily than $\text{C}{{\text{H}}_{\text{2}}}\text{ClC}{{\text{O}}_{\text{2}}}\text{H}$. Hence, $\text{C}{{\text{H}}_{\text{2}}}\text{FC}{{\text{O}}_{\text{2}}}\text{H}$ is stronger acid than $\text{C}{{\text{H}}_{\text{2}}}\text{ClC}{{\text{O}}_{\text{2}}}\text{H}$.

3. $\text{C}{{\text{H}}_{\text{2}}}\text{FC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{2}}}\text{H or C}{{\text{H}}_{\text{3}}}\text{CHFC}{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{2}}}\text{H}$ 

Ans: The structures of the compounds are given below:

(Image will be uploaded soon)

Inductive effect decreases with increase in distance. Hence, the +I effect of F in $\text{C}{{\text{H}}_{\text{3}}}\text{CHFC}{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{2}}}\text{H}$ is more than it is in $\text{C}{{\text{H}}_{\text{2}}}\text{FC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{2}}}\text{H}$. Hence, $\text{C}{{\text{H}}_{\text{3}}}\text{CHFC}{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{2}}}\text{H}$ is stronger acid than $\text{C}{{\text{H}}_{\text{2}}}\text{FC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{O}}_{\text{2}}}\text{H}$.

4.                             (Image will be uploaded soon)

Ans: Due to the −I action of F, the release of proton for the chemical is simpler (A). However, proton release is problematic in the case of compound (B), because of the +I action of a group of $\text{-C}{{\text{H}}_{\text{3}}}$. (A) is hence stronger than acid (B).

An Overview: NCERT Solutions For Class 12 Chemistry Chapter 12 PDF

Organic Chemistry is a scoring part of the syllabus, given that the students understand the topics well. Students, while studying it should focus on qualitative preparation rather than quantitative preparation. To do this, students can take the help of solutions which are readily available on various websites in PDF formats.

Several reliable e-learning platforms offer Aldehydes Ketones and Carboxylic Acids Class 12 NCERT Books Solutions PDF at free of cost. Students can easily download and study these as per their convenience. The answers are elaborately explained along with several illustrations to help students understand the concepts better. Moreover, these solutions are curated by an experienced faculty, which ensures high accuracy of these answers.

Important Topics Covered in NCERT Solutions Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids PDF

The important topics covered in Chapter 12 of Class 12 Chemistry are listed below. The NCERT Solutions for Class 12 Ch-12 Aldehydes, Ketones, and Carboxylic Acids will help students to develop a strong foundation for the concepts of nomenclature, structure, preparation, physical properties, chemical reactions, and other aspects of aldehydes, ketones, and carboxylic acids.

• Nomenclature of Carbonyl Group

• Structure of Carbonyl Group

• Nomenclature of Carboxyl Group

• Structure of Carboxyl Group

• Preparation of Aldehydes and Ketones

• Physical Properties 

• Chemical Reactions

• Methods of Preparation of Carboxylic Acids

• Physical Properties

• Chemical Reactions

• Uses of Aldehydes and Ketones

• Uses of Carboxylic Acids

Chapter 12 Chemistry Class 12

Students, while studying the twelfth chapter of Class 12 NCERT Chemistry, will learn about the compounds Aldehydes, Ketones, and Carboxylic Acids in detail. Moreover, there are approximately ten subtopics through which students will learn about their structure, preparation, properties, and reactions.

However, to evaluate their level of understanding in each of these subtopics, they must solve the exercises. To solve these, they can refer to NCERT Solutions for Class 12 Chemistry Chapter 12. These solutions will help them to solve the following types of exercises:

1. Intext Questions:

The intext questions are the question that assesses the level of understanding regarding a particular section. There are approximately eight intext questions. When students can solve these questions with ease, they can move on to the next section. Moreover, students can refer to ch 12 Chemistry Class 12 NCERT solutions, ch 12 Chemistry Class 12 NCERT solutions; this will help them to overcome their doubts, if any, they will also be able to understand the pattern of writing an answer.

2. Exercise Question:

There are approximately twenty exercise questions given at the end of the chapter. To solve these questions, students must have comprehensive knowledge about the entire chapter. They can also take the help of Class 12 Aldehydes Ketones and Carboxylic Acids NCERT solutions to solve these questions accurately. Moreover, students studying from solutions will get to learn the following topics quickly:

• The structures of Aldehydes, Ketones, and Carboxylic Acids (like the one shown in the figure)

(image will be uploaded soon)

• Determinants of Carbolic Acids acidity.

• The physical properties of these compounds.

• Chemical reactions of Aldehydes, Ketones and Carboxylic Acids.

• Preparation of Carboxylic Acids and the other two compounds.

Marks Distribution

Chapter 12 of Class 12 NCERT Chemistry comprises six marks. The chapter has a share of around 8.5% in the exam. Therefore, students should study this chapter appropriately. Even if they are preparing by themselves, they should solve the questions and check the answers from Class 12 Chemistry Chapter 12 NCERT solutions. This will help them to identify their mistakes and rectify them.

Related Study Materials for Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids

• Aldehydes, Ketones, and Carboxylic Acids

• Class 12 Chemistry Revision Notes for Chapter 12 - Aldehydes, Ketones and Carboxylic Acids

• Important Questions for CBSE Class 12 Chemistry Chapter 12 - Aldehydes, Ketones and Carboxylic Acids

• NCERT Exemplar for Class 12 Chemistry - Aldehydes, Ketones and Carboxylic Acids - Free PDF Download

• CBSE Class 12 Chemistry Sample Papers with Solutions

• Class 12 Chemistry Previous Year Question Papers

Advantages of NCERT Solutions For Class 12 Chemistry Chapter 12

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• It is easy to understand.

• Elaborate explanations along with illustrations.

• Accurate solutions.

• Helps to self-evaluate.

Considering these advantages, students should refer to NCERT Solutions for Class 12 Chemistry Chapter 12 for scoring well in exams. The detailed knowledge offered by these solutions is helpful for other competitive exams as well.

Each answer covered in the NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids are prepared as per the latest guidelines for the CBSE Term II Class 12 Chemistry examination. Students can download all the study resources on this chapter from Vedantu at absolute zero cost and prepare for their exams. Going through the NCERT Exemplar book solutions, revision notes, important questions, and other relevant study materials will definitely help students to boost their preparation and secure good scores in the examination.