ALDEHYDES AND KETONES

Classification:

$R \& R^{1}-$ can be aliphatic or aromatic

$\qquad\quad -$ can be identical or different

Important carbonyl compounds :

All have very pleasant fragrances and are used in food products and pharmaceuticals.

Some other carbonyl compounds like acetone are used in manufacture of adhesives, perfumes, plastics, paints etc.

PREPARATION

1. From alcohols

a) by oxidation

Aldehydes formed are oxidised to carboxylic acids if they remain in the reaction mixture.

Mild oxidising agents like PDC or PCC stop the oxidation at the aldehyde stage.

$\underset{\text { allyl alcohol }}{\mathrm{CH}_2=\mathrm{CH}}-\mathrm{CH}_2 \mathrm{OH}\xrightarrow[\text { Jones reagent }]{\mathrm{CrO}_3-\mathrm{H}_2 \mathrm{SO}_4} \underset{\text { acrolein }}{\mathrm{CH}_2=\mathrm{CH}-\mathrm{CHO}}$

b) by catalytic dehydrogenation of alcohols (primary and secondary)

• Further oxidation of aldehydes and ketones to acids does not take place in this method.

2. From carboxylic acids

3. By hydration of alkynes

4. By ozonolysis of alkenes

5. By Wacker process

$\begin{aligned} & \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{PdCl}_2 \xrightarrow[\text { air or } \mathrm{O}_2]{\mathrm{CuCl}_2} \mathrm{CH}_3 \mathrm{CHO}+\mathrm{Pd}+2 \mathrm{HCl} \\ & \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{PdCl}_2 \xrightarrow[\text { air or } \mathrm{O}_2]{\mathrm{CuCl}_2} \mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{Pd}+2 \mathrm{HCl} \\ & \end{aligned}$

6. Acid chlorides ketones

$2 \mathrm{CH} _{3} \mathrm{COCl}+\left(\mathrm{CH} _{3} \mathrm{CH} _{2}\right) _{2} \mathrm{Cd} \xrightarrow[\text { ether }]{\text { dry }} 2 \mathrm{CH} _{3} \mathrm{COCH} _{2} \mathrm{CH} _{3}+\mathrm{CdCl} _{2}$

$2 \mathrm{R}^{\prime} \mathrm{COCl}+\mathrm{R} _{2} \mathrm{Cd} \longrightarrow 2 \mathrm{R}^{\prime}-\mathrm{CO}-\mathrm{R}+\mathrm{CdCl} _{2}$

7. Acid chlorides aldehydes (Rosenmund reaction)

$\begin{aligned} & 2 \mathrm{CH}_3 \mathrm{COCl}+\mathrm{H}_2 \dfrac{\mathrm{Pd} / \mathrm{BaSO}_4}{\text { boiling xylene }} \mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCl} \\ & \\ & \mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl}+\mathrm{H}_2 \xrightarrow{\mathrm{Pd} / \mathrm{BaSO}_4} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO} \\ & \end{aligned}$

8. From nitriles (Stephen reduction)

$\mathrm{CH} _{3} \mathrm{CN}+\mathrm{SnCl} _{2}+\mathrm{HCl} \xrightarrow{\text { dry ether }} \mathrm{CH} _{3} \mathrm{CH}=\mathrm{NH} \xrightarrow{\mathrm{H} _{2}\mathrm{O}/ \mathrm{HCl}} \mathrm{CH} _{3}-\mathrm{CHO}+\mathrm{NH} _{4} \mathrm{Cl}$

This reaction can also be carried out with DIBAL-H (di isobutyl aluminium hydride)

$ \mathrm{CH} _{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CN} \xrightarrow{\mathrm{DIBAL}-\mathrm{H}} \mathrm{CH} _{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CHO} $

• DIBAL-H does not reduce ethylenic double bonds.

• DIBAL-H also reduces esters to aldehydes

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{COOC}_2 \mathrm{H}_5 \xrightarrow[2. \mathrm{H}_2 \mathrm{O} ]{\text { 1. DIBAL-H }} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO} $

9. From nitriles

10. From hydrocarbons

a) By side chain oxidation

Etard reaction

• In both the reactions oxidation of aldehyde to the acid is prevented by trapping the aldehyde in the form of a complex which on hydrolysis gives aldehyde.

b) By side chain oxidation followed by hydrolysis

c) By Gattermann-Koch reaction

d) Friedel Crafts acylation

PHYSICAL PROPERTIES

1. Boiling points

Aldehydes and ketones have higher boiling points than corresponding hydrocarbons due to the presence of dipole-dipole interactions.

Aldehydes and ketones cannot form any intermolecular hydrogen bonds like alcohols because there is no hydrogen attached to oxygen atom in these compounds.

Thus alcohols have higher boiling points than aldehydes and ketones since intermolecular hydrogen bonding is stronger than dipole-dipole interactions.

2. Solubility

Lower aldehydes and ketones are soluble in water due to $\mathrm{H}$ bonding between the polar carbonyl group and water molecules. As the size of alkyl group increases, solubility in water decreases. But carbonyl compounds are easily soluble in organic solvents.

3. Dipole moments

Aldehydes and ketones have large dipole moments as compared to alcohols or ethers. This is because of the shifting of $\pi$-electrons towards oxygen in the $\mathrm{C}=0$ bond due to the higher electronegativity of oxygen.

CHEMICAL PROPERTIES

I. NUCLEOPHILICADDITION REACTIONS

Due to greater electronegativity of oxygen as compared to carbon, the carbon atom of carbonyl group acquires positive charge and behaves as electrophile. Thus they react with nucleophiles at the carbonyl carbon.

Relative reactivities of aldehydes and ketones:

i. Inductive effect- Greater the number of alkyl groups attached to carbonyl group, greater is the electron density on the carbonyl carbon, hence lower is its reactivity towards nucleophilic addition.

Thus aldehydes are more reactive than ketones due to electronic factors.

ii. Steric effect - As the number and size of alkyl groups increases, attack of nucleophile on the carbonyl group becomes more and more difficult due to steric hinderance.

Thus aldehydes are more reactive than ketones due to steric factors.

1. Addition of Water

2. Addition of hydrogen cyanide [HCN]

Cyanohydrins are useful synthetic intermediates

3. Addition of sodium bisulphite

These bisulphite adducts can be easily decomposed to give back the carbonyl compound on treatment with a base or acid. Thus this reaction can be used for the purification and separation of carbonyl compounds.

4. Addition of Grignard reagents

Grignard reagents add to formaldehyde, aldehydes and ketones to give primary, secondary and tertiary alcohols respectively.

5. Addition of alcohol - Acetal and Ketal formation

If instead of using two molecules of a monohydric alcohol, one molecule of a dihydric alcohol is used, a cyclic acetal or cyclic ketal is formed.

Acetals and ketals can be hydrolysed with aqueous mineral acids to give back respective aldehydes and ketones. Thus this reaction can be used to protect a carbonyl compound in a reaction.

6. Addition of ammonia derivatives

This is a nucleophilic addition reaction followed by elimination of a molecule of water

a) Reaction with ammonia

b) Reaction with amine

c) Reaction with hydroxylamine

d) Reaction with hydrazine

e) Reaction with phenylhydrazine

f) Reaction with 2,4-dinitrophenylhydrazine [Brady’s reagent]

g) Reaction with semicarbazide

Control of $\mathrm{pH}$ during addition of ammonia derivatives to aldehydes and ketones

In case, the medium is too acidic, the ammonia derivatives being basic in nature will form their respective ammonium salts. Due to the absence of a lone pair of electrons on the nitrogen atom, these ammonium salts will no longer be nucleophilic and hence the reaction will not occur. If the medium is only very slightly acidic, the protonation of the carbonyl group will occur. This will increase the electron deficiency [or + ve charge] on the carbon atom of the carbonyl group and hence weak nucleophiles like ammonia derivatives will be able to react.

• These important derivatives of aldehydes and ketones can be used to characterise and identify them.

II Oxidation Reactions

1. Reaction with Tollen’s reagent-

Tollen’s reagent is an ammoniacal solution of silver nitrate

$$ \mathrm{RCHO}+2\left[\mathrm{Ag}\left(\mathrm{NH} _{3}\right) _{2}\right]+3 \mathrm{OH}^{-} \xrightarrow{\Delta} \mathrm{RCOO}^{-}+2 \mathrm{Ag} \downarrow+4 \mathrm{NH} _{3}+2 \mathrm{H} _{2} \mathrm{O} $$

Both aliphatic and aromatic aldehydes give this silver mirror test.

2. Reaction with Fehling’s solution

Aromatic aldehydes do not reduce Fehling’s solution.

Same reaction with cupric ions takes place with Benedict’s solution (citric acid instead of tartaric acid).

3. lodoform Reaction

Compounds having form yellow precipitate with $\mathrm{NaOH} / \mathrm{I} _{2}$ or $\mathrm{NaOI}$.

$$ \begin{aligned} \mathrm{CH} _{3} \mathrm{CHO}+3 \mathrm{NaOI} \rightarrow & \mathrm{CHI} _{3} \downarrow+\mathrm{HCOONa}+2 \mathrm{NaOH} \\ & \text { yellow (iodoform) } \\ \mathrm{CH} _{3} \mathrm{COCH} _{3}+3 \mathrm{NaOI} \rightarrow & \mathrm{CHI} _{3} \downarrow+\mathrm{CH} _{3} \mathrm{COONa}+2 \mathrm{NaOH} \end{aligned} $$

Other halogens can also be used in place of iodine and the general reaction is known as haloform reaction.

III. Reduction Reactions

1. Reduction to alcohol

$\underset{\text { Aldehyde }}{\mathrm{RCHO}} \xrightarrow[\text { or LiAlH } \mathrm{H} _{4} \text { or } \mathrm{NaBH} _{4}]{\mathrm{H} _{2}+\mathrm{Ni} / \mathrm{Pt} \text { or } \mathrm{Pd}} \underset{1^{\circ} \mathrm{Al} _{2} \mathrm{O} \text { hol }}{\mathrm{RCH} _{2} \mathrm{OH}}$

2. Reduction to hydrocarbon

i. Clemmensen reduction

$\underset{\text{Aldehyde}}{\mathrm{RCHO}}+4[\mathrm{H}] \xrightarrow{\mathrm{Zn}-\mathrm{Hg}, \mathrm{HCl}} \underset{\text{alkane}}{\mathrm{R}}-\mathrm{CH} _{3}+\mathrm{H} _{2} \mathrm{O}$

ii. Wolff-Kishner reduction

$\underset{\text{Aldehyde}}{\mathrm{R}-\mathrm{CH}}=0 \xrightarrow[{-\mathrm{H} _{2} \mathrm{O}}]{\mathrm{NH} _{2}-\mathrm{NH} _{2}} \mathrm{R}-\mathrm{CH}=\mathrm{NNH} _{2} \dfrac{\mathrm{KOH}, \text { alycol }}{453-473 \mathrm{~K}} \underset{\text{Alkane}}{\mathrm{R}-\mathrm{CH} _{3}}+\mathrm{N} _{2}$

iii. With $H 1+P($ Red)

$\underset{\text{Acetaldehyde}}{\mathrm{CH} _{3} \mathrm{CHO}}+4 \mathrm{HI} \xrightarrow{\text { RedP, } 423 \mathrm{~K}} \underset{\text{Ethane}}{\mathrm{CH} _{3}-\mathrm{CH} _{3}}+\mathrm{H} _{2} \mathrm{O}+2 \mathrm{I} _{2}$

$\underset{\text{Acetone}}{\mathrm{CH} _{3} \mathrm{COCH} _{3}}+4 \mathrm{HI} \xrightarrow{\text { RedP, 423K }} \underset{\text{Propane}}{\mathrm{CH} _{3} \mathrm{CH} _{2}} \mathrm{CH} _{3}+\mathrm{H} _{2} \mathrm{O}+2 \mathrm{I} _{2}$

IV. Other Reactions

1. Aldol Condensation

Aldehydes or ketones having $\alpha$ hydrogen show this reaction. In this reaction, two molecules of aldehyde or ketone condense in presence of dilute alkali to form $\beta$ hydroxyaldehyde or $\beta$ hydroxyketone.

Mechanism

Step 1. In the first step, the base, i.e. $\mathrm{OH}^{-}$ion abstracts one of the acidic $\alpha$-hydrogens to form the enolate ion which is stabilized by resonance.

Step 2. The enolate ion being a strong nucleophile attacks the carbonyl group of the second molecule of acetaldehyde (which acts as electrophile) to form the anion (I).

Step 3. The anion (I) finally abstracts a proton from water to form aldol.

Dehydration of aldols: The products of aldol condensation when heated with dilute acids undergo dehydration leading to the formation of $\alpha, \beta$ - unsaturated aldehydes or ketones. For example

Formaldehyde, benzaldehyde and benzophenone do not undergo aldol condensation since they do not contain $\alpha$-hydrogen atoms.

Cross Aldol Condensation

It takes place between two different aldehydes or between aldehyde and ketone or between two different ketones.

(a) If both have $\alpha$ hydrogens it gives a mixture of 4 products.

(b) When only one reactant has $\alpha \mathrm{H}$

Intramolecular Aldol Condensation

If a compound contains two aldehyde / ketone groups or one aldehyde and one ketone group at 1, 6 or 1,7-positions w.r.t. each other, then the enolate ion of one carbonyl group can add to the carbonyl group of the other. This reaction is called intramolecular aldol condensation. In this reaction, first an aldol is formed which subsequently loses a molecule of $\mathrm{H} _{2} \mathrm{O}$ to form an $\alpha$, $\beta$-unsaturated aldehyde / ketone. The driving force for elimination of $\mathrm{H} _{2} \mathrm{O}$ is the stability achieved due to conjugation of the double bond with the carbonyl group. For example,

• Reaction can not take place between $\mathrm{C}-2$ and $\mathrm{C}-8$ because it will give 7 membered cyclic compound which is unstable.

2. Cannizzaro Reaction :

Aldehydes which do not have $\alpha$ hydrogen atom on treatment with concentrated alkali solution, undergo disproportionation i.e. self oxidation-reduction. As a result, one of the molecule of aldehyde is reduced to alcohol and other is oxidised to corresponding carboxylic acid.

$ \underset{\text { benzaldehyde }}{2 \mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CHO}}+\mathrm{NaOH} \longrightarrow \underset{\text { benzyl alcohol }}{\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH} _{2} \mathrm{OH}}+\underset{\text { sod. benzoate }}{\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{COONa}} $

It is a hydride transfer reaction.

Cross Cannizzaro Reaction: Takes place between 2 different aldehydes.

$ \underset{\text { Benzaldehyde }}{\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CHO}}+\underset{\text { Formaldehyde }}{\mathrm{HCHO}}+\underset{(50 \%)}{\mathrm{NaOH}} \longrightarrow \underset{\text { Benzyl alcohol }}{\mathrm{N} _{6} \mathrm{H} _{5} \mathrm{CH} _{2} \mathrm{OH}}+\underset{\text { Sod. formate }}{\mathrm{HCOONa}} $

If one of the aldehydes is formaldehyde, it always undergoes oxidation to form sodium formate.

Intramolecular Cannizzaro reaction

3. Reaction with Aluminium ethoxide (Tischenko reaction)

All aldehydes (with or without $\propto-$ H’s) can undergo Cannizzaro reaction on treatment with aluminium ethoxide. Under these conditions, alcohol and acid produced as a result of Cannizzaro reaction, combine together to form esters.

$ \underset{\text { propionaldehyde }}{2 \mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{CHO}} \xrightarrow{\mathrm{Al}\left(\mathrm{OC} _{2} \mathrm{H} _{5}\right) _{3}} \underset{\substack{\text { n-propyl propionate }}}{\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{COOCH} _{2}} \mathrm{CH} _{2} \mathrm{CH} _{3} $

4. Reaction with alcoholic KCN (Benzoin Condensation)

5. Reaction with Schiff’s reagent

aldehyde + $\underset{\text{(colourless)}}{\text{Schiff’s reagent}} \longrightarrow$ pink colour

6. Reaction with Chloroform $\left(\mathrm{CHCl} _{3}\right)$

7. Reaction with $\mathrm{PCl} _{5}$

8. Reaction with primary amines

$ \underset{\text { aldehyde }}{\mathrm{CH} _{3} \mathrm{CHO}}+\underset{1^{\circ} \text { amine }}{\mathrm{CH} _{3} \mathrm{NH} _{2}} \xrightarrow{\mathrm{H}^{+}, \Delta} \underset{\text { Schiff’s base }}{\mathrm{CH} _{3} \mathrm{CH}=\mathrm{NCH} _{3}}+\mathrm{H} _{2} \mathrm{O} $

ELECTROPHILIC SUBSTITUTION IN AROMATIC ALDEHYDES

Since aldehydic and ketonic groups are electron withdrawing, they are deactivating and m-directing

1. Halogenation

2. Nitration

3. Sulphonation

Acetophenone does not undergo Friedel Crafts reaction, due to formation of complex between $>\mathrm{C}=0$ group and $\mathrm{AICl} _{3}$ and electron withdrawing nature of $>\mathrm{C}=0$ group.

CARBOXYLIC ACIDS AND THEIR DERIVATIVES

Carboxylic acids are also called as fatty acids because some members of aliphatic series of acids (from $\mathrm{C} _{12}-\mathrm{C} _{18}$ ) are obtained by hydrolysis of natural fats or oils.

The number of carboxyl groups are indicated by adding the prefix di, tri etc. to the term-oic acid.

When $-\mathrm{OH}$ group of a carboxyl acid is replaced by some other group, then they are called acid derivatives. Some of the common acid derivatives are,

PREPARATION

1. From primary alcohols and aldehydes

Alcohols are oxidised to carboxylic acids using strong oxidising agents like $\mathrm{KMnO} _{4}$ (in neutral, acidic or alkaline medium), $\mathrm{K} _{2} \mathrm{Cr} _{2} \mathrm{O} _{7} / \mathrm{H} _{2} \mathrm{SO} _{4}$ or $\mathrm{CrO} _{3}$ in acidic media

$\underset{1^\circ \text{ alcohol}}{\mathrm{RCH}_2 \mathrm{ OH}} \xrightarrow[ 2) \mathrm{H_3} \mathrm{O}^+]{ 1) \mathrm{KMnO_4/KOH}, \Delta} \underset{\text{carboxylic acid}}{\mathrm{RCOOH}}$

Aldehydes are easily oxidised to the corresponding carboxylic acids even with mild oxidising agents such as Tollen’s reagent.

RCHO $\xrightarrow[\text { Reagent }]{\text { Tolen’s }} \mathrm{RCOOH}$

2. From alkylbenzenes

Aromatic acids are obtained by vigorous oxidation of alkylbenzenes with a number of oxidising agents such as acidic or alkaline potassium permanganate, acidified potassium dichromate [chromic acid] or even dil. $\mathrm{HNO} _{3}$. During these oxidations, the aromatic nucleus remains intact but the entire side chain is oxidised to $\mathrm{COOH}$ group irrespective of the length of the carbon chain.

3. From nitriles and amides

4. From Grignard reagents

5. From acyl or acid halides, anhydrides and esters

$ \underset{\text { Ethanoyl chloride }}{\mathrm{CH} _{3} \mathrm{COCl}+\mathrm{H} _{2} \mathrm{O} \xrightarrow{-\mathrm{OH}}} \underset{\text { Ethanoic acid }}{\mathrm{CH} _{3} \mathrm{COOH}}+\mathrm{HCl} $

$ \underset{\text { Benzoic anhydride }}{\left[\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CO}\right] _{2} \mathrm{O}}+\mathrm{H} _{2} \mathrm{O} \xrightarrow{\mathrm{OH}} \underset{\text { Benzoic acid }}{2 \mathrm{C} _{6} \mathrm{H} _{5} \mathrm{COOH}} $

$\underset{\text {Ethyl benzoate}}{\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{COOC} _{2} \mathrm{H} _{5}}+\mathrm{H} _{2} \mathrm{O} \xrightarrow{\mathrm{OH}^{-}} \underset{\text {Benzoic acid}}{\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{COOH}}+\mathrm{C} _{2} \mathrm{H} _{5} \mathrm{OH}$

PHYSICAL PROPERTIES

1. Solubility

Carboxylic acids are miscible in water due to formation of hydrogen bonds. Solubility decreases as the number of carbon atoms increases due to increased hydrophobic interactions of the hydrocarbon part.

Benzoic acid is insoluble in water since the non polar hydrocarbon part outweighs the effect of polar-COOH part.

Carboxylic acids are soluble in less polar organic solvents like benzene, ether, chloroform etc.

2. Boiling points

Boiling points of carboxylic acids are higher than hydrocarbons. Carboxylic acids form stronger hydrogen bonds than alcohols. In vapour phase and in aprotic solvents, most of the carboxylic acids exist as cyclic dimers in which two molecules of acid are held together by strong hydrogen bonds.

3. Melting points

They show oscillation / alternation effect. Melting points of an acid containing even number of carbon atoms is higher than the next higher homologue containing odd number of carbon atoms.

CHEMICAL PROPERTIES

I. Acidity - carboxylic acids are the most acidic among organic compounds but less acidic than mineral acids. They interact with water molecules as,

The ionisation (or dissociation) equilibrium of a carboxylic acid is,

The dissociation or acidity constant $\mathrm{K} _{\mathrm{a}}$ for this reaction is

$\mathrm{K} _{\mathrm{a}}=\xrightarrow[\mathrm{RCOOH}]{[\mathrm{RCOO}^-]\left[\mathrm{H}^{+}\right]}$

The value of $\mathrm{K} _{\mathrm{a}}$ is directly proportional to the concentration of hydrogen ions and is a measure of the acidity of the individual acid. Larger value of $\mathrm{K} _{\mathrm{a}}$ means the acid is stronger.

$\mathrm{pK} _{\mathrm{a}}=-\log \mathrm{K} _{\mathrm{a}}\left(\mathrm{pK} _{\mathrm{a}}\right.$ is the more common unit of acidity)

Causes of Acidity - (i) Resonance effect - Delocalization of charge through resonance is the major factor responsible for large acidity of carboxylic acids when compared to alcohols.

Both acid and carboxylate ion can exist in two canonical forms

In II, there is separation of opposite charges and therefore it has more energy and is less stable than I.

III $\&$ IV have equivalent structures of equal stability. Thus stabilization is more for carboxylate ion than for acid and therefore more ionization and stronger acid.

ii) Effect of substituents : a) chloroacetic acid is 80 times stronger than acetic acid. In acetic acid the electron donating effect of methyl group destabilizes the anion. Thus acetic acid has a lower $K _{a}$ than formic acid. In chloroacetic acid, the electron withdrawing effect of chlorine stabilizes the anion due to dispersal of charge thus favoring more ionisation. Thus, in general, electron donating substituents weaken the acidity whereas electron withdrawing substituents make the acid stronger. Thus trichloroacetic acid is stronger than dichloroacetic acid and monochloro acetic acid. Fluoroacetic acid is stronger than chloroacetic acid which is stronger than bromo and iodoacetic acid.

This influence also decreases with distance of halogen from the carboxylic group and becomes ineffective when there are four or more carbon atoms in a chain.

Benzoic acid $\left(\mathrm{pK} _{\mathrm{a}}=4.2\right)$ is slightly stronger than simple aliphatic acids because the carboxylate group is attached to a more electronegative carbon ( $\mathrm{sp}^{2}$ hybridised). Also the acidic character increases from saturated to unsaturated acids because carbon becomes more electron with drawing as its s-character increases. In case of substituted benzoic acid, electron withdrawing groups make the acid stronger because they stabilize the conjugate base.

e.g.

Carbonyl carbon in carboxyl group is less electrophilic than the carbonyl carbon in aldehydes and ketones due to resonance. The less electrophilic nature of the carbonyl carbon puts a partial positive charge on the hydroxyl 0-atom, thereby making the hydroxyl group of carboxylic acids more acidic than the hydroxyl group of alcohols and phenols.

1. Reaction with metals

$\underset{\text { Acetic acid }}{2 \mathrm{CH} _{3} \mathrm{COOH}}+2 \mathrm{Na} \longrightarrow \underset{\text { Sod. acetate }}{2 \mathrm{CH} _{3} \mathrm{COONa}}+\mathrm{H} _{2}$

$\underset{\text { Acetic acid }}{2 \mathrm{CH} _{3} \mathrm{COOH}}+\mathrm{Zn} \longrightarrow \underset{\text { Zinc acetate}}{\left[\mathrm{CH} _{3} \mathrm{COO}\right] _{2}} \mathrm{Zn}+\mathrm{H} _{2}$

2. Reaction with carbonates and bicarbonates

$\underset{\text { Acetic acid }}{2 \mathrm{CH} _{3} \mathrm{COOH}}+\mathrm{Na} _{2} \mathrm{CO} _{3} \longrightarrow \underset{\text { Sod. acetate }}{2 \mathrm{CH} _{3} \mathrm{COONa}}+\mathrm{CO} _{2}+\mathrm{H} _{2} \mathrm{O}$

$\mathrm{CH} _{3} \mathrm{COOH}+\mathrm{NaHCO} _{3} \longrightarrow \mathrm{CH} _{3} \mathrm{COONa}+\mathrm{CO} _{2}+\mathrm{H} _{2} \mathrm{O}$

This reaction of carboxylic acid with $\mathrm{NaHCO} _{3}$ is used to detect the presence of a carboxyl group in an organic compound.

3. Reaction with alkalies

$\underset{\text{Acetic acid}}{\mathrm{RCOOH}}+\mathrm{NaOH} \longrightarrow \underset{\text { Sod. Carboxylate }}{\mathrm{RCOONa}}+\mathrm{H}_2 \mathrm{O}$

$\mathrm{RCOOH}+\mathrm{NH}_4 \mathrm{OH} \longrightarrow \underset{\text{Amm. Carboxylate}}{\mathrm{RCOONH}_4}+\mathrm{H}_2 \mathrm{O}$

II. FORMATION OF FUNCTIONAL DERIVATIVES

1. Formation of anhydride (Action of heat in presence of $\mathrm{P} _{2} \mathrm{O} _{5}$ )

$\underset{\text { Ethanoic acid }}{2 \mathrm{CH} _{3} \mathrm{COOH}} \xrightarrow[\text { or conc } \mathrm{H} _{2} \mathrm{SO} _{4}]{\mathrm{P} _{2} \mathrm{O} _{4} \Delta} \underset{\text { ethanoic anhydride }}{\left(\mathrm{CH} _{3} \mathrm{CO}\right) _{2} \mathrm{O}}+\mathrm{H} _{2} \mathrm{O}$

2. Formation of esters (Reaction with alcohols or phenols)

3. Formation of acid chloride (Reaction with $\mathrm{SOCl} _{2}, \mathrm{PCl} _{5}, \mathrm{Or} \mathrm{PCl} _{3}$ )

$\underset{\text { Acetic acid }}{\mathrm{CH}_3 \mathrm{COOH}}+\mathrm{SOCl}_2 \longrightarrow \underset{\text { Acetyl chloride }}{\mathrm{CH}_3 \mathrm{COCl}}+\mathrm{SO}_2 \uparrow+\mathrm{HCl} \uparrow$

$\underset{\text { Acetic acid }}{\mathrm{CH}_3 \mathrm{COOH}}+\mathrm{PCl}_5 \longrightarrow \underset{\text { Acetyl chloride }}{\mathrm{CH}_3 \mathrm{COCl}}+\mathrm{POCl}_3+\mathrm{HCl}$

$3 \mathrm{CH} _{3} \mathrm{COOH}+\mathrm{PCl} _{3} \longrightarrow 3 \mathrm{CH} _{3} \mathrm{COCl}+\mathrm{H} _{3} \mathrm{PO} _{3}$

Thionyl chloride $\left[\mathrm{SOCl} _{2}\right]$ is the best because its by products $\mathrm{SO} _{2}$ and $\mathrm{HCl}$ are gases.

4. Formation of amides (Reaction with $\mathrm{NH} _{3}$ )

$\underset{\text { Acetic acid }}{\mathrm{CH}_3 \mathrm{COOH}}+\mathrm{NH}_3 \longrightarrow \mathrm{CH}_3 \mathrm{COO}^{-} \mathrm{NH}_4^{+} \xrightarrow{\text { Heat }} \underset{\text { Acetamide }}{\mathrm{CH}_3 \mathrm{CONH}_2}+\mathrm{H}_2 \mathrm{O} $

III. REACTIONS INVOLVING CARBOXYL [COOH] GROUP AS A WHOLE

1. Reduction

a) Reduction to alkanes

$\underset{\text { Acetic acid }}{\mathrm{CH} _{3} \mathrm{COOH}}+6 \mathrm{HI} \xrightarrow{\text { Red P, 473 K }} \underset{\text { Ethane }}{\mathrm{CH} _{3}-\mathrm{CH} _{3}}+2 \mathrm{H} _{2} \mathrm{O}+3 \mathrm{I} _{2}$

b) Reduction to alcohols

$\underset{\text { Acetic acid }}{\mathrm{CH}_3 \mathrm{COOH}} \xrightarrow[\text { ii) } \mathrm{H}_3 \mathrm{O}^{+}] {\text {i) } \mathrm{LiAlH}_4 \text {/ether }} \underset{\text { Ethanol }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}}$

2. Decarboxylation

Using soda-lime

$\mathrm{CH} _{3} \mathrm{COONa}+\mathrm{NaOH} \xrightarrow{\mathrm{CaO} / 630 \mathrm{~K}} \underset{\text{Methane}}{\mathrm{CH} _{4}}+\mathrm{Na} _{2} \mathrm{CO} _{3}$

3. Electrolytic decarboxylation

Electrolysis of aqueous solution of sodium or potassium salts of fatty acids gives alkanes having twice the number of carbon atoms present in the alkyl group of the acid. This process is called Kolbe’s Electrolytic reaction.

$\begin{aligned}2 \mathrm{RCOONa} \longrightarrow 2 \mathrm{RCOO}^{-}+2 \mathrm{Na}^{+} \text {(ionisation) } \\ 2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 2 \mathrm{OH}^{-} +2 \mathrm{H}^{+} \text { (ionisation) } \end{aligned}$

$\text { At anode : } \quad 2 \mathrm{RCOO}^{-}-2 \mathrm{e}^{-} \longrightarrow \underset{\text { (unstable) }}{2 \mathrm{RCOO}{ }^\bullet} \longrightarrow \mathrm{CO}_2+\underset{\text { Alkane }}{\mathrm{R}-\mathrm{R}}$

At cathode : Both $\mathrm{Na}^{+}$and $\mathrm{H}^{+}\left(\right.$from $\left.\mathrm{H} _{2} \mathrm{O}\right)$ are present, but $\mathrm{H}^{+}$ions are preferentially discharged due to their lower discharge potential w.r.t. $\mathrm{Na}^{+}$ions and thus hydrogen gas is evolved at the cathode.

4. Decarboxylation of silver salt of carboxylic acids in presence of bromine

$ \underset{\text{Silver acetate}}{\mathrm{CH}_3 \mathrm{COOAg}}+\mathrm{Br}_2 \xrightarrow{\mathrm{CCl}_4 ,\text { Reflux }} \underset{\text{Methyl bromide}}{\mathrm{CH}_3-\mathrm{Br}+\mathrm{CO}_2}+\mathrm{AgBr} $

This reaction is called Hunsdiecker reaction.

IV. Reactions of the Alkyl group Halogenation

[Hell Volhard Zelinsky (HVZ) reaction]

Carboxylic acids (except formic acid which does not contain an alkyl group) react with chlorine or bromine in presence of small quantities of red phosphorus to give exclusively $\alpha$-chloro or $\alpha$-bromo acids.

$\underset{\text { Acetic acid }}{\mathrm{CH}_3 \mathrm{COOH}} \xrightarrow[-\mathrm{HCl}]{\mathrm{Cl}_2 \text {, red } \mathrm{P}} \underset{\text { chloroacetic acid }}{\mathrm{CICH}_2 \mathrm{COOH}} \xrightarrow[-\mathrm{HCl}]{\mathrm{Cl}_2, \text { red } \mathrm{P}} \underset{\text { Dichloroacetic acid }}{\mathrm{Cl}_2 \mathrm{CHCOOH}} \xrightarrow[{-\mathrm{HCl}}]{\mathrm{Cl}_2 \text { red } \mathrm{P}} \underset{\text { Trichloracectic acid }}{\mathrm{Cl}_3 \mathrm{CCOOH}}$

Halogenation exclusively occurs at the $\alpha$-position and the reaction stops when all the $\alpha$-hydrogens have been replaced by the halogen atoms;

V. ELECTROPHILIC SUBSTITUTION IN AROMATIC ACIDS

Since the $-\mathrm{COOH}$ group is electron withdrawing, therefore, it is meta-directing. Thus electrophilic substitution reactions in benzoic acid occur at the meta-position. Since the $\mathrm{COOH}$ group is deactivating reactions take place only under drastic condition.

Carboxylic acids, however do not undergo Fridel-Crafts reactions because the carboxyl group is strongly deactivating and the catalyst $\mathrm{AlCl} _{3}$ (Lewis acid) gets bonded to the carboxyl group strongly.

EXCEPTIONAL BEHAVIOUR OF FORMIC ACID

1. Formic acid is unique since it contains a hydrogen atom instead of an alkyl group.

2. It is regarded both as an aldehyde as well as a carboxylic acid.

3. Formic acid acts as reducing agent.

$ \mathrm{HCOOH}+[0] \longrightarrow \mathrm{CO} _{2}+\mathrm{H} _{2} \mathrm{O} $

Like aldehydes, it reduces :-

a) Fehling’s solution to red precipitate

b) Tollen’s reagent to silver mirror

c) Decolourizes acidified $\mathrm{KMnO} _{4}$ solution

d) Turns $\mathrm{K} _{2} \mathrm{Cr} _{2} \mathrm{O} _{7}$ solution green

e) Reduces mercuric salts to mercurous salts

$\underset{\begin{array}{l} \text { mercuric } \\ \text { chloride } \end{array}}{2 \mathrm{HgCl}_2}+\underset{\begin{array}{c} \text { formic } \\ \text { acid } \end{array}}{\mathrm{HCOOH}} \longrightarrow \underset{\begin{array}{c} \text { mercurous chloride } \\ \text { (white ppt) } \end{array}}{\mathrm{Hg}_2 \mathrm{Cl}_2}+\mathrm{HCl}+\mathrm{CO}_2$

Solved Example

Question 1- Acetone is treated with excess of ethanol in the presence of hydrochloric acid. The product obtained is

Show Answer

Answer - is d

Question 2- Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound $\mathrm{E}$. compound $\mathrm{E}$ on further treatment with aqueous KOH yields compound F. Compound F is

Show Answer

Answer - a

This is ozonolysis followed by aldol condensation

Question 3- In the given reaction sequence,

Show Answer

Answer - is $\mathrm{c}$. The reaction sequence involves ketal formation followed by hydrolysis to give back the original carbonyl compound and alcohol.

Question 4- The number of aldol reaction [s] that occurs in the given transformation is

(a) 1

(b) 2

(c) 3

(d) 4

Show Answer

Answer - is $\mathrm{c}$ i.e. three aldol reactions. This is because after three aldol reactions. one cannizzaro reaction takes place to give the product

Question 5- Which of the following combination of aldehydes gives cross Cannizzaro reaction?

a) $\mathrm{CH} _{3} \mathrm{CHO}, \mathrm{HCHO}$

b) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CHO}, \mathrm{CH} _{3} \mathrm{CHO}$

c) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CHO}, \mathrm{HCHO}$

d) all of these

Show Answer

Answer - c

This is because in both a and b, $\mathrm{CH} _{3} \mathrm{CHO}$ is having $\propto-\mathrm{H}$ ’s and will undergo aldol condensation

Question 6- Compound ’ $A$ ’ (molecular formula $\mathrm{C} _{3} \mathrm{H} _{8} \mathrm{O}$ ) is treated with acidified potassium dichromate to form a product ’ $\mathrm{B}$ ’ (molecular formula $\mathrm{C} _{3} \mathrm{H} _{6} \mathrm{O}$ ). ‘B’ forms a shining silver mirror on warming with ammoniacal silver nitrate. ‘B’ when treatred with an aqueous solution of $\mathrm{H} _{2} \mathrm{NCONHNH} _{2} \cdot \mathrm{HCl}$ and sodium acetate gives a product ’ $C$ ‘. Identify the structure of ’ $C$ ‘.

(a) $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CH}=\mathrm{NNHCONH} _{2}$

(b) $\mathrm{CH} _{3}-\mathrm{C}=\mathrm{NNHCONH} _{2}$

(c) $\mathrm{CH} _{3}-\mathrm{C}=\mathrm{NCONHNH} _{2}$

(d) $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CH}=$ NCONHNH

Show Answer

Answer - a

$ \begin{aligned} & \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{OH} \xrightarrow{\mathrm{K} _{2} \mathrm{Cr} _{2} \mathrm{O} _{7} / \mathrm{H}^{+}} \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CHO} \xrightarrow{\mathrm{NH} _{2} \mathrm{CONHNH} _{2} \cdot \mathrm{CH} _{3} \mathrm{COONa}} \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CH}=\mathrm{NNHCONH} _{2} \\ & \left(\mathrm{C} _{3} \mathrm{H} _{8} \mathrm{O}\right)(\mathrm{A}) \quad \qquad \qquad \qquad\left(\mathrm{C} _{3} \mathrm{H} _{6} \mathrm{O}\right)(\mathrm{B}) \end{aligned} $

Since (B) gives silver mirror test it is an aldehyde which reacts with semicarbazide and sodium acetate.

Question 7- A carbonyl compound with molecular weight 86 , does not reduce Fehling’s solution but forms crystalline bisulphite derivative and gives iodoform test. The possible compound can be

a) 2-pentanone and 3-pentanone

b) 2-pentanone and 3-methyl-2-butanone

c) 2-pentanone and pentanal

d) 3-pentanone and 3-methyl-2-butanone

Show Answer

Answer - b

Since the compound does not reduce Fehlings solution, it is a ketone and since it gives iodoform test it is a methyl ketone.

Question 8-

The compound B is

Show Answer

Answer - a

$\beta$-keto acids undergo decarboxylation easily on heating

Question 9- o-Toluic acid on reaction with $\mathrm{Br} _{2}+\mathrm{Fe}$ gives

Show Answer

Answer - C

Since $-\mathrm{CH} _{3}$ group is para directing and $-\mathrm{CO} _{2} \mathrm{H}$ group is m-directing .

Question 10-. An organic compound $A$ upon reacting with $\mathrm{NH} _{3}$ gives $B$. On heating $B$ gives $C$. C in the presence of $\mathrm{KOH}$ reacts with $\mathrm{Br} _{2}$ to give $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{NH} _{2}$. A is

(a) $\mathrm{CH} _{3} \mathrm{COOH}$

(b) $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{COOH}$

(c)

(d) $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{COOH}$

Show Answer

Answer - d

Question 11-. In a set of reactions, ethylbenzene yielded a product $D$.

‘D’ should be

Show Answer

Answer - a

PRACTICE QUESTIONS

Question 1- Identify the product $Y$ in the following reaction sequence

(a) Pentane

(b) Cyclobutane

(c) Cyclopentane

(d) Cyclopentanone

Show Answer

Answer:- a

Question 2- Ozonolysis of an organic compound ’ $A$ ’ produces acetone and propionaldehyde in equimolar mixture. Identify ’ $A$ ’ from the following compounds

(a) 2-methyl-1-pentene

(b) 1-Pentene

(c) 2-pentene

(d) 2-methylpent-2-ene

Show Answer

Answer:- d

Question 3- Identify the product of the reaction

$\mathrm{PhC} \equiv \mathrm{CMe} \xrightarrow{\mathrm{H} _{3} \mathrm{O}^{+}, \mathrm{Mg}^{2+}}$ ?

(a) $\mathrm{PhCH} _{2} \mathrm{CH} _{2} \mathrm{CHO}$

(b) $\mathrm{PhCOCH} _{2} \mathrm{CH} _{3}$

(c) $\mathrm{PhCH} _{2} \mathrm{COCH} _{3}$

(d) $\mathrm{PhCOCOMe}$

Show Answer

Answer:- b

Question 4- $\mathrm{CH} _{3} \mathrm{CH} _{2}-\mathrm{C} \equiv \mathrm{N} \xrightarrow{x} \mathrm{CH} _{3} \mathrm{CH} _{2}-\mathrm{CHO}$

The reagent/s $X$ is [are]

(a) $\mathrm{SnCl} _{2} / \mathrm{HCl}, \mathrm{H} _{2} \mathrm{O} /$ boil

(b) $\mathrm{H} _{2} / \mathrm{Pd}-\mathrm{BaSO} _{4}$

(c) $\mathrm{LiAlH} _{4} /$ ether

(d) $\mathrm{NaBH} _{4} /$ ether, $\mathrm{H} _{3} \mathrm{O}+$

Show Answer

Answer:- a

Question 5- Consider the following reaction,

The product $A$ is

(a) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CHO}$

(b) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{OH}$

(c) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{COCH} _{3}$

(d) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{Cl}$

Show Answer

Answer:- a

Question 6- In the given reaction,

$[x]$ will be :

Show Answer

Answer:- a

Question 7- Consider the structure of given alcohol :

The alcohol can be prepared from

Show Answer

Answer:- d

Question 8- In the given reaction:

$\mathrm{C} _{6} \mathrm{H} _{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO} \xrightarrow{[x]} \mathrm{C} _{6} \mathrm{H} _{5}-\mathrm{CH} _{2}-\mathrm{CH} _{2}-\mathrm{CH} _{2} \mathrm{OH}$

$[\mathrm{x}]$ will be

(a) $\mathrm{NaBH} _{4}$

(b) Aluminium isopropoxide

(c) $\mathrm{LiAlH} _{4}$

(d) All of the above

Show Answer

Answer:- c

Question 9- Which one of the combinations will give propionaldehyde on dry distillation?

(a) $\quad\left[\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{COO}\right] _{2} \mathrm{Ca}$ and $[\mathrm{HCOO}] _{2} \mathrm{Ca}$

(b) $\quad\left[\mathrm{CH} _{3} \mathrm{COO}\right] _{2} \mathrm{Ca}$ and $\left[\mathrm{CH} _{3} \mathrm{CH} _{2}-\mathrm{COO}\right] _{2} \mathrm{Ca}$

(c) $\quad\left[\mathrm{CH} _{2} \mathrm{CH} _{2}-\mathrm{COO}\right] _{2} \mathrm{Ca}$ and $[\mathrm{HCOO}] _{2} \mathrm{Ca}$

(d) $\quad\left[\mathrm{CH} _{3} \mathrm{COO}\right] _{2} \mathrm{Ca}$ and $\left[\mathrm{CH} _{3} \mathrm{COO}\right] _{2} \mathrm{Ca}$

Show Answer

Answer:- c

Question 10- The reagent with which both acetaldehyde and acetone react easily is

(a) Tollen’s reagent

(b) Schiff’s reagent

(c) Grignard reagent

(d) Fehling reagent

Show Answer

Answer:- c

Question 11- The compound that will not give iodoform on treatment with alkali and iodine is

(a) acetone

(b) ethanol

(c) diethyl ketone

(d) isopropyl alcohol

Show Answer

Answer:- c

Question 12- Which of the following has the most acidic hydrogen?

(a) 3-hexanone

(b) 2,4-hexanedione

(c) 2,5-hexanedione

(d) 2,3-hexanedione

Show Answer

Answer:- b

Question 13- A mixture of benzaldehyde and formaldehyde on heating with aqueous $\mathrm{NaOH}$ solution gives

(a) benzyl alcohol and sodium formate

(b) sodium benzoate and methyl alcohol

(c) sodium benzoate and sodium formate

(d) benzyl alcohol and methyl alcohol

Show Answer

Answer:- a

Question 14- Butan $-2-$ one can be converted to propanoic acid by which of the following?

(a) $\mathrm{NaOH}, \mathrm{Nal} / \mathrm{H}^{+}$

(b) Fehling solution

(c) $\mathrm{NaOH}, \mathrm{I} _{2} / \mathrm{H}^{+}$

(d) Tollen’s reagent

Show Answer

Answer:- c

Question 15- In the given reaction:

$[A]$ and $[B]$ will respectively be :

(a) $\mathrm{H} _{3} \mathrm{C}-\mathrm{CH} _{2}-\mathrm{CHO}$ and $\mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{CHO}$

(b) $\mathrm{H} _{3} \mathrm{C}-\mathrm{CHO}$ and $\mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{CHO}$

(c) $\mathrm{H} _{3} \mathrm{C}-\mathrm{CHO}$ and $\mathrm{CH} _{3}-\mathrm{CHO}$

(d)

Show Answer

Answer:- b

Question 16- In the given reaction:

$[A]+[B] \xrightarrow{\mathrm{NaOH} / \triangle} \mathrm{C} _{6} \mathrm{H} _{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$

$[A]$ and $[B]$ will be :

(a) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CHO}$ and $\mathrm{HCHO}$

(b) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CHO}$ and $\mathrm{CH} _{3} \mathrm{CH} _{2}-\mathrm{CHO}$

(c) $\mathrm{C} _{6} \mathrm{H} _{5}-\mathrm{CHO}$ and $\mathrm{CH} _{3}-\mathrm{CHO}$

(d) $\mathrm{C} _{6} \mathrm{H} _{5}-\mathrm{CH} _{2}-\mathrm{CHO}$ and $\mathrm{CH} _{3}-\mathrm{CHO}$

Show Answer

Answer:- c

Question 17- Base catalysed aldol condensation occurs with

(a) Propionaldehyde

(b) Benzaldehyde

(c) 2,2-dimethyl butanaldehyde

(d) 2,2-dimethyl propionaldehyde

Show Answer

Answer:- a

Question 18- Which of the following compounds will give a yellow precipitate with iodine and alkali?

(a) 2-hydroxypropane

(b) Benzophenone

(c) Methylacetate

(d) acetamide

Show Answer

Answer:- a

Question 19- Which of the following is an example of aldol condensation?

(c) $2 \mathrm{HCHO} \xrightarrow{\text { dil. } \mathrm{NaOH}} \mathrm{CH} _{3} \mathrm{OH}+\mathrm{HCOONa}$

(d) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CHO}+\mathrm{HCHO} \xrightarrow{\text { dil. } \mathrm{NaOH}} \mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH} _{2} \mathrm{OH}+\mathrm{HCOONa}$

Show Answer

Answer:- b

Question 20- The product of the reaction

will be

Show Answer

Answer:- a

Question 21- Aldol condensation between acetaldehyde and propionaldehyde will give how many products?

(a) Two

(b) Four

(c) Three

(d) One

Show Answer

Answer:- b

Question 22- In the reaction:

$[x]$ will be

Show Answer

Answer:- d

Question 23- In the scheme given below, the total number of intramolecular aldol condensation products formed from ’ $y$ ’ is

(a) 1

(b) 2

(c) 3

(d) 4

Show Answer

Answer:- a

Question 24- Predict the product in the given reaction

Show Answer

Answer:- c

Question 25- In the cannizzaro reaction given below $2 \mathrm{Ph}-\mathrm{CHO} \xrightarrow{\mathrm{OH}^{-}} \mathrm{Ph}-\mathrm{CH} _{2} \mathrm{OH}+\mathrm{PhCO} _{2}^{-}$, the slowest step is

(a) the attack of ${ }^{-} \mathrm{OH}$ at the carbonyl group

(b) the transfer of hydride ion at the carbonyl group

(c) the abstraction of a proton from the carboxylic acid

(d) the deprotonation of $\mathrm{Ph}-\mathrm{CH} _{2} \mathrm{OH}$

Show Answer

Answer:- b

Question 26- Trichloroacetaldehyde was subjected to Cannizzaro’s reaction by using $\mathrm{NaOH}$. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is

(a) Trichloromethanol

(b) 2,2,2-Trichloropropanol

(c) Chloroform

(d) 2,2,2-Trichloroethanol

Show Answer

Answer:- d

Question 27- If heavy water is taken as solvent instead of normal water while performing Cannizzaro reaction, the products of the reaction are

(a) $\mathrm{RCOO}^{-}+\mathrm{RCH} _{2} \mathrm{OH}$

(b) $\mathrm{RCOO}^{-}+\mathrm{RCH} _{2} \mathrm{OD}$

(c) $\mathrm{RCOOD}+\mathrm{RCD} _{2} \mathrm{OD}$

(d) $\mathrm{RCOO}^{-}+\mathrm{RCD} _{2} \mathrm{OD}$

Show Answer

Answer:- b

Question 28-

Show Answer

Answer:- b

Question 29- A compound $A$ undergoes Cannizzaro reaction and $B$ undergoes positive iodoform test. Therefore,

(a) $A=$ Acetaldehyde $\quad \quad B=1-$ Pentanal

(b) $\mathrm{A}=\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH} _{2} \mathrm{CHO}$ $\quad B=3-$ Pentanone

(c) $A=$ Formaldehyde $\quad \quad B=2-$ Pentanone

(d) $A=$ Propionaldehyde $\quad B=1,2-$ Pentanol

Show Answer

Answer:- c

Question 30- The order of reactivity of phenylmagnesium bromide ( $\mathrm{PhMgBr})$ with the following compounds; is

I. $\mathrm{CH} _{3} \mathrm{CHO}$

II. $\mathrm{CH} _{3} \mathrm{COCH} _{3}$

III. $\mathrm{PhCOPh}$

(a) III > II > I

(b) II > I > III

(c) I > III > I

(d) I > II > III

Show Answer

Answer:- d

Question 31- During reduction of aldehydes with hydrazine and potassium hydroxide, the first step is formation of

(a) $\mathrm{R}-\mathrm{C}=\mathrm{N}$

(b) $\mathrm{R}-\mathrm{CO}-\mathrm{NH} _{2}$

(c) $\mathrm{R}-\mathrm{CH}=\mathrm{NH}$

(d) $\mathrm{R}-\mathrm{CH}=\mathrm{N}-\mathrm{NH} _{2}$

Show Answer

Answer:- d

Question 32- Which of the following reagent reacts differently with $\mathrm{HCHO}, \mathrm{CH} _{3} \mathrm{CHO}$ and $\mathrm{CH} _{3} \mathrm{COCH} _{3}$ ?

(a) $\mathrm{HCN}$

(b) $\mathrm{NH} _{2} \mathrm{NH} _{2}$

(c) $\mathrm{NH} _{2} \mathrm{OH}$

(d) $\mathrm{NH} _{3}$

Show Answer

Answer:- d

Question 33- In the given transformation which of the following is the most appropriate reagent?

(a) $\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$

(b) $\mathrm{H} _{2}-\mathrm{Pd} / \mathrm{BaSO} _{4}$

(c) $\mathrm{NaBH} _{4}$

(d) $\mathrm{LiAlH} _{4}$

Show Answer

Answer:- a

Question 34- The reagent used for separation of acetaldehyde and acetophenone is

(a) $\mathrm{NaHSO} _{3}$

(b) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{NHNH} _{2}$

(c) $\mathrm{NH} _{2} \mathrm{OH}$

(d) $\mathrm{NaOH}+\mathrm{I} _{2}$

Show Answer

Answer:- a

Question 35- Consider the reaction :-

$\mathrm{RCHO}+\mathrm{NH} _{2} \mathrm{NH} _{2} \longrightarrow \mathrm{RCH}=\mathrm{N}-\mathrm{NH} _{2}$

What sort of reaction is it?

(a) Electrophilic addition-elimination reaction

(b) Free radical addition-elimination reaction

(c) Nucleophilic addition reaction

(d) Nucleophilic addition-elimination reaction

Show Answer

Answer:- d

Question 36- A compound ’ $A$ ’ $\left(\mathrm{C} _{5} \mathrm{H} _{10} \mathrm{Cl} _{2}\right)$ on hydrolysis gives $\mathrm{C} _{5} \mathrm{H} _{10} \mathrm{O}$ which reacts with $\mathrm{NH} \mathrm{H} _{2} \mathrm{OH}$, forms iodoform but does not give fehling test. A is

Show Answer

Answer:- a

Question 37- Which of the following acids on heating loses a molecule of $\mathrm{H} _{2} \mathrm{O}$ to form an $\alpha, \beta$-unsaturated acid?

(a) $\mathrm{CH} _{3} \mathrm{CHOHCOOH}$

(b) $\mathrm{HOCH} _{2} \mathrm{COOH}$

(c) $\mathrm{CH} _{3} \mathrm{CHOHCH} _{2} \mathrm{COOH}$

(d) $\mathrm{HOCH} _{2} \mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{COOH}$

Show Answer

Answer:- c

Question 38- Which of the following carboxylic acid undergoes decarboxylation easily?

Show Answer

Answer:- a

Question 39- The compound that does not liberate $\mathrm{CO} _{2}$, on treatment with aqueous sodium bicarbonate solution is.

(a) benzoic acid

(b) benzenesulphonic acid

(c) salicylic acid

(d) carbolic acid (Phenol)

Show Answer

Answer:- d

Question 40- When propionic acid is treated with aqueous sodium bicarbonate, $\mathrm{CO} _{2}$ is liberated. The $\mathrm{C}$ of $\mathrm{CO} _{2}$ comes from

(a) methyl group

(b) carboxylic acid group

(c) methylene group

(d) bicarbonate group

Show Answer

Answer:- d

Question 41- Benzoyl chloride is prepared from benzoic acid by

(a) $\mathrm{Cl} _{2}$, hv

(b) $\mathrm{SO} _{2} \mathrm{Cl} _{2}$

(c) $\mathrm{SOCl} _{2}$

(d) $\mathrm{Cl} _{2}, \mathrm{H} _{2} \mathrm{O}$

Show Answer

Answer:- c

Question 42- The compound that undergoes decarboxylation most readily under mild condition is

Show Answer

Answer:- b

Question 43- Which of the following reagents may be used to distinguish between phenol and benzoic acid?

(a) Tollen’s reagent

(b) Molisch reagent

(c) Neutral $\mathrm{FeCl} _{3}$

(d) Aqueous $\mathrm{NaOH}$

Show Answer

Answer:- c

Question 44- In the reaction

$$ \mathrm{CH} _{3} \mathrm{COOH} \xrightarrow{\mathrm{LiAlH} _{4}} \mathrm{~A} \xrightarrow{\mathrm{PCI} _{5}} \mathrm{~B} \xrightarrow{\text { Alc. } \mathrm{KOH}} \mathrm{C} $$

The product $\mathrm{C}$ is

(a) Acetaldehyde

(b) Acetylene

(c) Ethylene

(d) Acetyl chloride

Show Answer

Answer:- c

Question 45-

the product $Z$ is

(a) Benzaldehyde

(b) Benzoic acid

(c) Benzene

(d) Toluene

Show Answer

Answer:- b