# 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$\dpi{100} -NH_{2}$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.

Intext Questions

Exercise Questions

## Solutions to In-Text Questions Ex 12.1  to 12.8

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

$\alpha$ -Methoxypropionaldehyde

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

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

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

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

4-Oxopentanal

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

Question

Di-sec. butyl ketone

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

Question

4-Fluoro acetophenone

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

Question

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

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

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

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

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

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

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

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

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

Increasing order in their boiling points-

$CH_{3}CH_{2}CH_{3}

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.

Ethanal, Propanal, Propanone, Butanone.

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

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

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

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

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

Question

(ii)

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

Question

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

The product of the above reaction is -

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

Question

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

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

Question

$Ph CH_{2}CH_{2}COOH$

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

3-phenyl propanoic acid

Question

$(CH_{3})_{2}C=CH COOH$

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

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

Question

The IUPAC name of the compound is-

2-methyl cyclopentane carboxylic acid

Question

(iv)

The IUPAC name of the compound is-

2, 4, 6-trinitrobenzoic acid

Question

Ethyl benzene

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

Question

Acetophenone

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

Question

Bromobenzene

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

Question

Phenylethene (Styrene)

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

Question

$CH_{3}CO_{2}H$ or $CH_{2}FCO_{2}H$

$CH_{2}FCO_{2}H$ 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

$CH_{2}FCO_{2}H$ or $CH_{2}Cl CO_{2}H$

$CH_{2}FCO_{2}H$ 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

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

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

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.

Cyanohydrin

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.

Acetal

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:

Semicarbazone

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

Aldol

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:

Hemiacetal

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:

Oxime

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

For example:

Ketal

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:

Imine

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:

2,4-DNP-derivative

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:

Schiff’s base

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:

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

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

IUPAC name of this compound is 4-methyl pentanal

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

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

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

$CH_{3}CH= CH CHO$

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

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

$CH_{3}COCH_{2}CO CH_{3}$

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

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

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

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

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

$(CH_{3})_{3}CCH_{2}COOH$

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

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

$OHCC_{6}H_{4}CHO -p$

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

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

3-Methyl butanal

The structure of the 3-methylbutanal is shown here-

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

p-Nitropropiophenone

The structure of p-nitro propiophenone is shown here-

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

p-methyl benzaldehyde

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

4-Methylpent-3-en-2-one

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

4-Chloropentan-2-one

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

3-Bromo-4-phenylpentanoic acid

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

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

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

$CH_{3}CO(CH_{2})_{4}CH_{3}$

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

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

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

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

$CH_{3}(CH_{2})_{5}CHO$

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

The IUPAC name of the compound is heptanal

$Ph - CH = CH - CHO$

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

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

Common name is $\beta$-phenylacrolein

The IUPAC name of the compound is cyclopentane carbaldehyde.

$Ph CO Ph$

$Ph CO Ph$
structure of the compound is

The IUPAC name of the compound is Diphenyl mentanone.

The common name is - benzophenone

The 2,4-dinitro phenyl hydrazone of benzaldehyde

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

Cyclopropanone oxime

The structure of the Cyclopropanone oxime

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

Acetaldehyde dimethyl acetal

The structure of Acetaldehydedimethylacetal

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

The semicarbazone of cyclobutanone

The structure of the semicarbazone of cyclobutanone

The ethylene ketal of hexan-3-one

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

The structure of the compound methyl hemiacetal of formaldehyde

PhMgBr and then H3O+

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

Tollens'  reagent

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-

Semicarbazide and weak acid

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.

Excess ethanol and acid

Structure of cyclohexanecarbaldehyde

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

Zinc amalgam and dilute hydrochloric acid

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

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-

2 methylpentanal

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

Benzaldehyde

Benzaldehyde Structure

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

Benzophenone

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 -

Cyclohexanone

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.

1-Phenylpropanone

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. $NaOH$

Phenylacetaldehyde

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

Butan-1-ol

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

2,2-Dimethylbutanal

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.

Butane-1,3-diol

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

But-2-enal

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

But-2-enoic acid

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

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)

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-

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

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.

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

increasing order of their acidic strength-

$\\(CH_{3})_{2}CHCOOH

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
$+I\propto 1/ acidic$

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

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.

Propanal and Propanone

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.

Acetophenone and Benzophenone

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.

Phenol and Benzoic acid

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

Benzoic acid and Ethyl benzoate

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.

Pentan-2-one and Pentan-3-one

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.

Benzaldehyde and Acetophenone

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.

Ethanal and Propanal

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.

Methyl benzoate

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.

m-nitrobenzoic acid

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).

(iii)    p-nitrobenzoic acid

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.

Phenyl acetic acid

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.

p-nitrobenzaldehyde

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.

Propanone to Propene

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.

Benzoic acid to benzaldehyde

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.

Ethanol to 3-hydroxy butanal

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

Benzene to m-nitroacetophenone

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.

Benzaldehyde to benzophenone

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.

Bromobenzene to 1-phenylethanol

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

Benzaldehyde to 3-Phenylpropan-1-ol

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

Benzaldehyde to α-hydroxy phenyl acetic acid

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

Benzoic acid to m-nitrobenzyl alcohol

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

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

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-
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

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.

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

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

$\xrightarrow[heat]{SoCl_{2}}$

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

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

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-

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

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

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

$\overset{NaCN / HCl}{\rightarrow}$

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

So the reaction gives

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

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

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

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

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

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

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

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.

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

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.

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

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.

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.

$RCOOH+R'OH\rightleftharpoons RCOOR'+H_{2}O$
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.

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 - $CH_{3}-CO-(CH_{2})_{2}-CH_{3}$

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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