(i) Tollens Reagent Test: Tollens reagent is an ammoniacal silver nitrate solution. When an aldehyde is warmed with Tollens reagent a bright silver mirror is produced due to the formation of silver metal. This reaction is also called silver mirror test for aldehydes.
\({ CH }_{ 3 }CHO+2\left[ Ag\left( { NH }_{ 3 } \right) _{ 2 } \right] ^{ + }+3{ OH }^{ - }\longrightarrow { CH }_{ 3 }CO{ O }^{ - }+\underset { Silver\ mirror }{ 4{ NH }_{ 3 }+2Ag+2{ H }_{ 2 }O } \) 
(ii) Fehlings solution Test:
(a) Fehling's solution is prepared by mixing equal volumes of Fehlings solution A containing aqueous copper sulphate and Fehlings solution 'B' containing alkaline solution of sodium potassium tartarate (Rochelle salt).
(b) When aldehyde is warmed with Fehlings solution deep blue colour solution is changed to red precipitate of cuprous oxide.
\(\\ \\ { CH }_{ 3 }CHO+\underset { blue }{ 2{ Cu }^{ 2 } } +5{ OH }^{ - }\longrightarrow { CH }_{ 3 }{ COO }^{ - }+\underset { red }{ { Cu }_{ 2 }O\downarrow } +3{ H }_{ 2 }O\)
(iii) Schiffs' reagent Test: Dilute solution of aldehydes when added to Schiff's reagent (Rosaniline hydrochloride dissolved in water and its red colour decolourised by passing SO₂) yields its red colour. This is known as Schiff's test for aldehydes. Ketones do not give this test. Acetone however gives a positive test but slowly.

(i) Hydrolysis: Acid anhydride are slowly hydrolysed, by water to form corresponding carboxylic acids.

(ii) Reaction with alcohol:

Some common electrophilic substitution reactions of benzoic acid are given below
(i) Halogenation:

(ii) Nitration:

(iii) Sulphonation:

(iv) Benzoic acid does not undergo friedal craft's reaction. This is due to the strong deactivating nature of the carboxyl group.

(i) CCl₃COOH > CHCl₂COOH > CH₂CICOOH > CH₃COOH
(ii) p-nitrophenol > m-nitrophenol > phenol> cresol.
(i) CCl₃COOH > CHCl₂COOH > CH₂CICOOH > CH₃COOH :
(a) Any factor that weakens the -COOH bond will facilitate the cleavage to release H⁺ more easily. The degree of ionisation increases and acid becomes relatively a stronger acid.
(b) The acid strength increases with increase in electronegativity of substituents.
(c) In monochloro acetic acid, the -I effect of chlorine increases the strength of the acid.
(d) Thus monochloro acetic acid is stronger than acetic acid. The strength of dichloro acetic acid is greater than monochloro acetic acid.

(ii) p-nitrophenol> m-nitrophenol > phenol > cresol:
(a) In aromatic acid, the nitro group (electron withdrawing group) especially at ortho or para position increases its strength due to -I effect.
(b) So p-nitro phenol is more acidic than m-nitro phenol and it is more acidic than phenol.
(c) But cresol contains CH₃ group which has +I effect (electron repelling group) and due to this strength of the acid decreases. So phenol is more acidic than cresol.

(i) Formic acid (HCOOH) is unique because it contains both an aldehyde group and carboxyl group also.

(ii) Hence it can act as a reducing agent. It reduces Fehling's solution Tollen's reagent and decolourises pink coloured KMnO₄ solution.
(iii) Whereas in acetic acid, there is no aldehyde group and it cannot act as reducing agent.
(iv) Formic acid reduces ammoniacal silver nitrate solution (Tollen's reagent) to metallic silver.
HCOOH + Ag₂O⟶H₂O + CO₂ + 2Ag↓ (metallic silver)
(v) Formic acid reduces Fehling's solution. It reduces blue coloured cupric ions to red coloured cuprous ions.
\(\mathrm{HCOO}^{-}+2 \mathrm{Cu}^{2+}+5 \mathrm{OH}^{-} \longrightarrow \mathrm{CO}_{3}^{2-}+\mathrm{Cu}_{2} \mathrm{O}+3 \mathrm{H}_{2} \mathrm{O}\\ \quad \quad \quad \quad (blue) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (red)\)

Acidic nature of acids:
(i) Acid molecule ionises to produce H⁺ ions in solution.

(ii) The carboxylate anion is stabilised by resonance.

Hence the cleavage of -COOH bond to release H⁺ is favoured, thereby making it to behave as acids.
(iii) Any factor that weakens the -COOH bond will facilitate the cleavage to release H⁺ more easily. The degree of ionisation is increased and acid becomes relatively a stronger acid.
(iv) When the following acids are compared
Their strength varies as the above order.
(v) Methyl group is a + I group (electron pair repelling group). Acid strength decreases: with increasing number of electron repelling substituent attached to the α- carbon atom.

The +1 effect is felt in increasing the strength of the -OH bond and the cleavage to form H+ becomes difficult. Hence they are weaker acid than formic acid (HCOOH) which does not have + I methyl group.
(vi) Acid strength increases with increase in electronegativity of substituents

This effect weakness the -O- H bond there by facilitating the removal of H⁺ from -OH group. This acid becomes stronger than formic acid.

(i) NH₃ with HCHO:
Formaldehyde forms hexamethylene tetramine with NH₃.
6HCHO + 4NH₃ ➝ (CH₂)₆N₄ + 6Hp
(ii) NH₃ with CH₃CHO:
Acetaldehyde reacts with NH₃ to form aldimine.
\(CH_{ 3 }CHO+{ NH }_{ 3 }\longrightarrow { H }_{ 3 }C-\overset { \underset { | }{ { NH }_{ 2 } } }{ \underset { \overset { | }{ H } }{ C } } -OH\overset { -{ H }_{ 2 }O }{ \longrightarrow } \underset { aldimine }{ CH_{ 3 }-CH=NH } \) 
(iii) NH₃ with C₆H₅CHO:
Benzaldehyde undergoes condensation with ammonia to form hydrobenzamide

(iv) NH₃ with CH₃CO - CH₃:
Acetone with ammonia forms acetone ammonia initially at room temperature. On heating, it forms diacetone amine 
\({ H }_{ 3 }C-\overset { \underset { | }{ { CH }_{ 3 } } }{ C } =O+{ HNH }_{ 2 }\longrightarrow \underset { acetone \ ammonia }{ CH_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { NH }_{ 2 } } }{ C } } -{ OH }_{ 2 } } \) 
\({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { NH }_{ 2 } } }{ C } } -OH+{ CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 } \overset { high \ temp }{ \underset { -{ H }_{ 2 }O }{ \longrightarrow } } \underset {Diacetone \ amine}{{ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } } { \underset { \overset { | }{ { NH }_{ 3 } } }{ C } } -{ CH }_{ 2 }}-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }\)

Cannizaro reaction involves three steps.
Step 1:  Attack of OH- on the carbonyl carbons

Step 2: Hydride ion transfer

Step 3: Acid - base reaction
\({ C }_{ 6 }{ H }_{ 5 }-\overset { \underset { || }{ O } }{ C } -OH+{ C }_{ 6 }{ H }_{ 5 }H_{ 2 }{ O }^{ - }+{ Na }^{ + }\overset { Proton }{ \underset { exchange }{ \longrightarrow } } \underset { Sodium\quad benzoate }{ { C }_{ 6 }{ H }_{ 5 }-\overset { \underset { || }{ O } }{ C } -ONa } +\underset { Benzyl \ alcohol }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }OH } \)
Cannizaro reaction is a characteristic of aldehyde having no a - hydrogen

Aldol condensation can also takes place between two different aldehydes or ketones or between one aldehyde and one ketone such an aldol condensation is called crossed or mixed aldol condensation. This reaction is not very useful as the product is usually a mixture of all possible condensation products and cannot be separated easily.
\(\underset { formldehyde }{ HCHO } +\underset { acataldehyde }{ { CH }_{ 3 }CHO } \overset { dil.NaOH }{ \longrightarrow } \underset { 3-hydroxy \ propanol }{ HO-{ CH }_{ 2 }-{ CH }_{ 2 }-CHO } \) 
\(\underset { formaldehyde }{ { HCHO } } +\underset { acetone }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 } } \overset { dil.NaOH }{ \longrightarrow } \underset { 4-hydroxy \ butan-2-one }{ HO-{ CH }_{ 2 }-{ CH }_{ 2 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 } } \)

(i) CH₃COO^-

(ii) CH₃CONH₂

(iii) C₆H_sCHO

(iv) C₆H₅COO^-

(i) \(\underset { Isobutylalcohol }{ { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ CH } -CHO } \overset { Zn/Hg }{ \longrightarrow } \underset { 2-methylpropane }{ { CH }_{ 3 }-\underset { \overset { | }{ { CH }_{ 3 } } }{ CH } -CH_{ 3 } } \) 
(ii) \(\underset { Acetone }{ { C }_{ 6 }{ H }_{ 5 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 } } \overset { Zn/Hg }{ \underset { HCI[H] }{ \longrightarrow } } \underset { Ethylbenzene }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ CH }_{ 3 } } \) 
(iii) \(\underset { Acetone }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 } } \overset { Zn/Hg }{ \underset { HCI[H] }{ \longrightarrow } } \underset { Propane }{ { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 3 } } \) 
(iv) \(\underset { Benzophenone }{ { C }_{ 6 }{ H }_{ 5 }-\underset { \overset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 } } \overset { Zn/Hg }{ \underset { HCI[H] }{ \longrightarrow } } \underset { diphyenyl\quad metahne }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ C }_{ 6 }{ H }_{ 5 } } \)

This reaction is catalysed by base. The carbanion generated is nucleophilic in nature. Hence it can bring about nucleophilic attack on carbonyl group
Step 1: The carbanion is formed as the a-hydrogen atom is removed as a proton by the base

Step 2: The carbanion attacks the carbonyl carbon of another unionised aldehyde molecule

Step 3: The alkoxide ion formed is protonated by water to give 'aldol'.

\(\underset { Acetaldehyde }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -H+4(H) } \overset { { NH }_{ 2 }{ NH }_{ 2 } }{ \underset { { C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } \underset { Ethane }{ { CH }_{ 3 }-{ CH }_{ 3 }+{ H }_{ 2 }O+{ N }_{ 2 } } \)
(i) Clemmensen reduction: Aldehydes and Ketones when heated with zinc amalgam and concentrated hydrochloric acid gives hydrocarbons. 
Example:
\(\underset { Acetald4ehyde }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ H }^{ + }4(H) } \overset { Zn^{ - }Hg }{ \underset { Conc.HCl }{ \longrightarrow } } \underset {Ethane}{{ CH }_{ 3 }-{ CH }_{ 3 }\pm { H }_{ 2 }O}\) 
\(\underset { Acetone }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }+4\left( H \right) } \overset { { Zn }^{ - }Hg }{ \underset { Conc.HCl }{ \longrightarrow } } { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 3 }+{ H }_{ 2 }O\)
(ii) Wolff-Kishner Reduction: Aldehydes and Ketones when heated with hydrazine (NH₂NH₂) and sodium ethoxide, hydrocarbons are formed Hydrazine acts as a reducing agent and sodium ethoxide as a catalyst.
Example :
\(\underset { Acetaldehyde }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -H+4(H) } \overset { { NH }_{ 2 }{ NH }_{ 2 } }{ \underset { { C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } \underset { Ethane }{ { CH }_{ 3 }-{ CH }_{ 3 }+{ H }_{ 2 }O+{ N }_{ 2 } } \) 
\(\underset { Acetone }{ { CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }+4(H) } \overset { { NH }_{ 2 }{ NH }_{ 2 } }{ \underset { { C }_{ 2 }{ H }_{ 5 }ONa }{ \longrightarrow } } { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 3 }+{ H }_{ 2 }O+{ N }_{ 2 }\)
Aldehyde (or) ketones is first converted to its hydrazone which on heating with strong base gives hydrocarbons.
(iii) Reduction to pinacols: Ketones, on reduction with magnesium amalgam and water, are reduced to symmetrical diols known as pinacol.

(i) With NH₃ :
Formaldehyde forms hexa m ethylenetetramine with NH₃
\(\underset { Formaldehyde }{ 6{ CH }_{ 2 }O+4N{ H }_{ 3 } } \longrightarrow \underset { Hexamethylene\quad tetramine }{ \left( { CH }_{ 2 } \right) _{ 6 }{ N }_{ 4 }+6{ H }_{ 2 }O } \) 

(ii) With CH₃MgI
Formaldehyde gives primary alcohol

(iii) With NaOH :
Cannizaro Reaction
\(\underset { Formaldehyde }{ HCHO } +HCHO\overset { NaOH }{ \longrightarrow } \underset { Sodiumformate }{ HCOONa } +\underset { Methnol }{ { CH }_{ 3 }OH } \)