Aldehydes, Ketones and Carboxylic Acids

ABYSS Learning

ALDEHYDES AND KETONES

 Introduction :

Organic Compounds having C O group are called carbonyl compounds and C O group is known

as carbonyl or oxo group. It's general formula is Cn H2nO (n = 1, 2, 3......) Carbonyl compounds are
grouped into two categories.

( a ) Aldehydes : Aldehyde group is O
C H (also known as formyl group). It is a monovalent group

( b ) Ketones : The carbonyl group ( C O) is a Ketonic group when its both the valencies are satisfied

by alkyl group. It is a bivalent group.

 Ketones are further classified as : R
(i) Simple or Symmetrical ketones : Having two similar alkyl groups. CO

R

R
(ii) Mixed or unsymmetrical ketones : Having two different alkyl groups. C O

R'

Example : (Ketones) : Symmetrical Unsymmetrical

CH3 C O CH3CH2 C O
CH3 CH3

(Acetone or Dimethyl ketone) (Ethyl methyl ketone)

2–Propanone 2–Butanone

O OO O

Special Point : .. .. .. ..

C OH, C X, C NH2, C OR,

In all the compounds given above, lone pair of electrons and double bond are conjugate.

OO

.. so resonance occurs. These compounds have C group still they are not carbonyl compounds

CZ

because carbonyl group takes parts in resonance with the lone pair of electrons.

 Structure :

In C O compounds C-atom is sp2 hybridised which forms two  bonds with C and H-atom respectively
and one  bond with oxygen atom. The unhybridised atomic orbital of C-atom and the parallel 2p orbital

of oxygen atom give the  bond in C O group.

C sp2 
CO
C The C—C—O and H—C—O bond angles are of 120°.


Due to electro-negativity difference in C & O atoms, the C O group is polar.

 

C O Hence aldehydes and Ketones posses dipole moment.

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 General Methods of Preparation :

( A ) For both Aldehydes and Ketones :

( 1 ) By Oxidation of Alcohols :

( a ) By K2Cr2O7 / H2SO4 :
Oxidation of primary alcohols gives aldehyde and oxidation of secondary alcohols gives Ketones.
Here, (K2Cr2O7 / H2SO4) is a strong oxidising agent.

RCH2OH K2Cr2O7 /[OH]2SO4 (dil.) RCHO (Aldehyde)

R CH R K2Cr2O[7O/]H2SO4 RCR (Ketone)

OH O

Aldehydes are quite susceptible to further oxidation to acids -

R C H 2O H [O] R—CHO [O] R—COOH

Thus oxidation of primary alcohols is made at the temperature much above the boiling point of

aldehyde and thus aldehydes are vapourised out and prevented from being oxidised.

 Note : Aldehydes can be prepared from 1° alcohol,secondary alcohols can be oxidized to ketones, by oxidation
with pyridinium chlorchromate (PCC) in CH2Cl2 solvent, pyridinium dichromate (PDC) and with
Jones reagent(CrO3+H2SO4) in acetone.

( b ) Oppenaur Oxidation :

The oxidation of secondary alcohols to ketones by heating them with specific reagent : [(CH3)3CO]3Al
(Aluminium-t-butoxide) in presence of acetone. Primary alcohols may also be oxidized to aldehydes
if ketones is replaced by a better hydrogen acceptor, e.g. p-benzoquinone. The equilibrium can be
controlled by the amount of acetone, an excess of which favours the oxidation of the alcohol.

R + CH3 C O[(CH3)3 CO]3Al R C O+ CH3 CH OH
CHOH CH3 R CH3

R Acetone Ketone Isopropyl alcohol

2O Alcohol

RCH2OH + O O [(CH3)3 CO]3Al R—CHO + HO— —OH

10 Alcohol Quinone Aldehyde Quinol

 Note : The reaction is the reverse of Meerwein-Ponndorf -verley reduction.

( c ) Mild Oxidising Agent :

1o alcohols will get oxidised with CrO3 / Pyridine, collin's reagent Ag / O2 at 250oC

RCH2OH + [O]  RCHO + H2O

By this reaction, good yield of aldehyde is possible.

( 2 ) Dehydrogenation of alcohols :

CH3CH2OH 30C0uC CH3CHO (Acetaldehyde)

CH3 CHCH3 30C0uC O
OH CH3 C CH3 (Acetone)

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CH3 30C0uC CH2
CH3 C OH CH3 C + H2O (Isobutylene)

CH3 CH3

( 3 ) By Hydrolysis of gem dihalides :

Terminal gem-dihalides on hydrolysis give aldehydes while the non-terminal dihalides give ketone.

Cl KOH(aq ) OH H2O  CH3CHO
CH3CH Cl CH3CH OH Acetaldehyde
[unstable]
Terminal gem-dihalide

Cl OH O
CH3 C CH3 CH3 C CH3
KOH(aq ) CH3 C CH3 H2O 
Cl Acetone
OH
unstable

( 4 ) By Oxidation of diols :
With periodic acid (HIO4) & lead tetra acetate (CH3COO)4 Pb vicinal diols gets oxidised to form carbonyl
compounds

R CH CH R' + HIO4  RCHO + R'CHO + HIO3 + H2O
OH OH

RR OO
R C C R' + HIO4  R C R + R' C R + HIO3 + H2O

OH OH

( 5 ) By Ozonolysis of alkenes :
This reaction is used to determine the position of double bond in alkene.

O O

RCH CH2 + O3  RCH CH2 –H2O RCHO + HCHO

Zn

Ozone O

Ozonide

OO

R C CH2 + O3  R C CH2 –H2O R C O + HCHO

Zn

R RO R

Un b ra n ch e d a lk e n e  a ld e h yd e
B r a n ch e d a l k en e  k e t o n e

( 6 ) From Alkyne : O
( a ) Hydration : With dil H2SO4 & 1% HgSO4 at 60-800C.

CH CH + H2O  [CH2 CHOH] CH3 C H

( Ta u to m e r i s a t i o n )

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Other alkynes give ketone :

CH3C CH + H2O  CH3 C CH2 CH3 C CH3

OH O
(enol)

( b ) Hydroboration : Reaction with B2H6, 2BH3 or R2BH give dialkyl borane.

1 – a lk y n e g iv es  a ld e h yd e
o th e r a lk yn e  k et o n e

RC CH + R2BH R CH CHBR2 H2O2  RCH CHOH RTautomerism CH2 C H + R2BOH

OH–

O

CH3—C C—CH3 + R2BH    CH3 CH C CH3 HO2HO–2 

BR2

R2BOH+CH3 CH C CH3  CH3 CH2 C CH3 +R2BOH
OH O

( 7 ) By Nef's reaction :

Nitro alkanes are used in this reaction. The  of nitro alkane shows acidic nature.

O R CH OH CH N ONaH3O R CHO + N2O + H2O + NaOH
N NaOH R
R CH2 N
O OO

(Nitro form) (Aci form) (Aldehyde)

1° nitro alkane

RO R N OH R C N ONaH3OR C O + N2O + H2O + NaOH
CH N C NaOH OR
OR
RO R

(Nitro form) (Aci form) (Ketone)
NaOH
2° nitro alkane No Reaction

RO
R CN

RO
(3° nitro alkane)

( 8 ) By hydrolysis of carbonyl derivatives :

R — C H N — O H H2O/ H  R—CHO + NH2—OH
(Aldoxime) (Aldehyde) (Hydroxyl amine)

R H2O/ H  R O + NH2 OH
C N OH C (Ketone)

R R
(Ketoxime)

R OR H2O/ H  R
C C O + 2ROH

H OR H
(Acetal) (Aldehyde) (Alcohol)

R OR H2O/ H  R
C C O + 2ROH

R OR R
(Ketal) (Ketone) (Alcohol)

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( 9 ) By oxidation of alkyl halides:

Oxidation takes place by (CH3)2S O dimethyl sulphoxide (DMSO).

R—CH2—X + (CH3)2S O   RCHO + (CH3)2S + HX

Alkyl halide (Aldehyde) (Dimethyl thio ether)

X O
RC R
R CH R  CH3 2 SO  Ketone

2° halide
( 1 0 ) From Grignard reagents :

( a ) By Cyanides :

RMgX + R C N R C N MgX H2O/H R C OH
R + NH3 + Mg X

RO

(Ketone)

( b ) By Esters : HCHO can't be prepared by this method.

O OMgX OH

RMgX + H C OR  H C OR H2OH C OR –ROH  H C O

(Alkyl formate) RR R

(Hemiacetal) (Aldehyde)

O OMgX OH

RMgX + R C OR  R C OR H2OR C OR  R C O

R R R
(Hemiacetal) (Ketone)
(Alkyl Alkanoate)

( c ) By acid chlorides :

R' C R' X
Cl + RMgX   R C O + Mg
O
( 1 1 ) From -keto acids : Cl

The decarboxylation reaction takes place via formation of six membered ring transition state.

( a ) HCOCH2COOH 110C CH3CHO+CO2

 OH 

(b) CH3 C CH2 C –CO2 CH3 C CH3

OO O

( B ) For Aldehydes only :

( 1 ) Reduction of acyl halides, esters and nitriles :

( a ) Acyl chlorides can be reduced to aldehydes by treating them with lithium-tri-tert-butoxyaluminium

hydride, LiAIH[OC(CH3)3], at – 78°C. O
O

R (i) LAH( t(ii)BHuO2O)3 ,78C R
Cl H

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( b ) Both esters and nitriles can be reduced to aldehydes by DIBAL-H. Reduction must be carried out
at low temperatures. Hydrolysis of the intermediates gives the aldehyde.

OO

R (i) DIBAL (iHi ),Hhe2xOane,78C R
OR' H

R—C N (i) DIBAL (iHi ),Hhe2xOane,78C R O
( 2 ) Rosenmund's reduction : H

Quinoline or sulphur act as a poisoned catalyst, controls the further reduction of aldehyde to alcohols.

RCOCl + H2 QuPindol/inBe oarSsOulp4hur  RCHO + HCl

RCOCl + H2 Pd R C H O  R CH 2OH

Formaldehyde can not be prepared by this method.

E x am p l e : C2H5COCl + H2 QuPindol/inBe oarSsOulp4hur  C2H5CHO + HCl
Propionyl Chloride Propanal

( 3 ) Stephen's reduction :

Alkyl cyanides are reduced by SnCl2 and HCl.

R — C N SnCl2 /HCl  R — C H N H H3O  RCHO+NH3

C2H5—C N SnCl2 /HCl  C 2H 5C H N H H3O  C2H5CHO + NH3

( 4 ) Oxo reaction or hydroformylation :

In this reaction symmetrical alkene gives 10 aldehyde while unsymmetrical alkene gives isomeric aldehyde
(Chain isomers).

CH2 CH2 + COH2 15C0oC CH3CH2CHO

Water gas

CH3—CH CH2 + CO + H2 15C0oC CH3CH2CH2CHO + CH3 CH CH3
CHO
( C ) For Ketones only :
( 1 ) From Grignard's reagent :

RMgX + R—C N  R C NMgX 2H2O R C X
O + NH3 + Mg
OH

RR

O OMgX OH O

RMgX + R C Cl  R C Cl H2OR C Cl  R C + HCl

RR R
O

RMgX + R C NH2  RH+ (RCONH)Mg(X) ; Ketone doe s not forms.

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O OMgX OH O

RMgX + R C OR  R C OR H2OR C OR  R C + ROH

RR R

( 2 ) From dialkyl Cadmium :

RCdR' (dialkyl Cadmium) is a organometallic compound.

RCOCl + RCdR'  RCOR' + RCdCl

This reaction is superior than Grignard Reaction because the ketones formed, further reacts with Grignard

reagent to form 3° alcohols.

C2H5 CH3 C2H5
O + Cd
Example : CH3COCl + Cd  C
Cl
C2H5 C2H5

( 3 ) From R2CuLi :

R2CuLi + R'COCl  R'COR + RCu + LiCl

( 4 ) By hydrolysis of Aceto Acetic Ester (AAE) :

CH3 C CH2 C OC2H5 H3O CH3   OH  CH3 C CH3
C CH2 C
–C2H5OH –CO2

OO OO O

( - keto acid) (Acetone)

Ex : CH3 C CH C OC2H5 H3O  CH3   OH  CH3 C CH2 CH3
C CH C
–C2H5OH –CO2

O CH3 O O CH3 O O

(–keto acid) (Butanone)

 Other methods for aldehyde and ketone :

( 1 ) By dry distillation of Ca-salts of carboxylic acid :

O

R COO OCH
R COO Ca + Ca O C H  
2RCHO + 2CaCO3

O

Calcium formate (Also R C R and HCHO formed)

O

H COO   HCHO + CaCO3
Ca

HCO O

R COO   R
Ca C O + CaCO3

RCO O R

Calcium-alkanoate Ketone

Calcium salts of acids other then formic acid on heating together gives ketone

O O
  2R C R' + 2CaCO3
R COO Ca + Ca O C R'
Ketone
R COO O C R'

O

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To prepare ethyl methyl ketone Calcium acetate and Calcium propionate are used :

O

CH3 OCO O C CH2CH3   2 CH3 C O + 2CaCO3
Ca + Ca C2H5
CH3 OCO O C CH2CH3

O Ethyl methyl ketone
Calcium Acetate Calcium propionate

( 2 ) By Thermal decomposition of carboxylic acids :

Vapour of carboxylic acids when passed over MnO/3000C give carbonyl compounds

2 HCOOH 3M00nOC HCHO + H2O + CO2

2 CH3COOH 3M00nOC CH3 C O + CO2 + H2O
CH3

RCOOH + HCOOH 3M00nOC RCHO + CO2 + H2O

RCOOH + R'COOH 3M00nOC RCOR' + CO2 + H2O

( 3 ) Wacker process :

In this reaction double bond is not cleaved so same C-atom aldehyde and ketones are formed.

CH2 CH2 + PdCl2 + H2O CuCl2  CH3CHO + Pd + HCl
All other alkenes gives ketone.

RCH CH2 + PdCl2 + H2O CuCl2  O
R C CH3 + Pd + HCl

 Physical Properties :

 State :

Only formaldehyde is gas, all other carbonyl compounds upto C11 are liquids and C12 & onwards solid.
 Odour :

Lower aldehydes give unpleasant smell, higher aldehydes and all ketones have pleasant smell.

 Solubility :

C1 to C3 (formaldehyde, acetaldehyde and propionaldehyde) and acetone are freely soluble in water

 

due to polarity of C O bond and can form H—bond with water molecule. C5 onwards are insoluble

in water.

  H So lub ility  1
H O
CO Molecular weight

H-bonding

 Boiling point : Boiling point  Molecular weight

Boiling point order is - Alcohol > Carbonyl compounds > Alkane

This is because in alcohols intermolecular H-bonding is present but in carbonyl compounds H-bonding
doesn't exist, instead dipole-dipole vander waal force of at traction is present. Alkanes are non polar.

   

CO CO

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 Density : Density of carbonyl compounds is lower than water.

 Chemical Properties :

Reactions of both aldehydes and ketones :

Due to strong electronegativity of oxygen, the mobile  electrons pulled strongly towards oxygen,

leaving the carbon atom deficient of electrons.

CO 

   C O



Carbon is thus readily attacked by N u . The negatively charged oxygen is attacked by electron deficient
(electrophile) E+.

C O bond in carbonyl group is stronger than C=C bond in alkanes.

C O Bond energy is 84.0 K Cals C C Bond energy is 83.1 K Cals
C O Bond energy is 178 K Cals C C Bond energy is 146 K Cals

Reactivity of carbonyl group   Magnitude of +ve charge  – I group    1
Ex : Why carbonyl compound gives nucleophilic addition reaction (NAR) ? + I group

  
E
S o l . C O  C O C OE  No Reaction
.. Carbocation (Less stable due to incomplete octet)
Nu
C   C OE

O E 

Nu Nu
Anion (More stable due to complete octet)

E x . Arrange the following for reactivity in decreasing order

H (ii) CH3 C O (iii) CH3 C O
(I) (i) C O H CH3

H

(II) (i) ClCH2CHO (ii) NO2CH2CHO (iii) CH3CHO (iv) CH3CH2CHO
(III) (i) CH3CHO (ii) ClCH2CHO
(iii) HCCl2CHO (iv) CCl3CHO

CH3 CH3CH2 (CH3)2CH CCl3
(ii) C O
(IV) (i) CO (iii) C O (iv) CO
CH3
CH3 CH3 CH3

S o l . (A) I > II > III (B) II > I > III > IV (C) IV > III > II > I (D) IV > I > II > III

[Hint : CH3— is +I group, decreases the intensity of +ve charge on C-atom of C O group.

Cl – is –I group increases the intensity of +ve charge on C-atom of C O group.]

Cl
(-I)
Ex. In Cl C C O CH3(+I) O , which one is more reactive ?
and C

Cl H H

Chloral

S o l . I is more reactive than II.

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 Chemical Reactions :
Carbonyl compounds in general under goes neucleophilic addition reaction :

( A ) Nucleophilic addition reactions :
( 1 ) Addition of HCN :

C O + HCN  C OH H2O C OH
Partial hydrolysis CONH2
CN -Hydroxy amide
(Cyanohydrin) H3O
Complete C OH
hydrolysis COOH
-Hydroxy acid
Ni/H2
C OH
CH2NH2
-Amino alcohol

H2O H
Partial hydrolysis C OH

H H H3O H
C O + HCN  C OH CONH2
Complete
H H hydrolysis Glyconamide
CN
H
(Formaldehyde cyanohydrin) C OH

Ni/H2 H
COOH

Glycolic acid

H
C OH

H
CH2NH2

2-Amino ethanol

H2O CH3 C OH

Partial hydrolysis H

CONH2

Lactamide (2-Hydroxy propanamide)

CH3 C O + HCN CH3 C* OH H3O CH3 C OH
H H Complete H
CN hydrolysis COOH

Acetaldehyde Cyanohydrin Ni/H2 Lactic acid (2- Hydroxy propanoic acid)
*(Racemic mixture)
CH3 C OH
H
CH2NH2

1-Amino – 2-Propanol

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( 2 ) Addition of NaHSO3 :
This reaction is utilized for the separation of carbonyl compounds from non - carbonyl compounds.

C O + NaHSO3    C OH H3O     Carbonyl compounds. *(Regain)
SO3Na

Sodium bi sulphite Bisulphite compound
(Crystalline)

Mechanism : NaHSO3  Na  HSO3

HSO3  H   S O 2
3

C O + SO–32 Slow O H+  OH
C SO3 C SO3
Fast
Na+

OH
C

SO3Na

( 3 ) Reaction with ammonia derivatives :

These are condensation or addition elimination reaction. These proceeds well in weakly acidic medium.

NH3  NH2Z (Ammonia derivative)

C O + H2 N Z HZ C N Z + H2O

Addition - elimination (Condensation)
Mechanism :

 .. IMPE C  –H2O C  –H C NZ
C OH + NH2Z
C OH OH2 NHZ
NH2Z
NHZ

 Ammonia derivatives (NH2Z) :
Z = OH  NH2OH (Hydroxyl amine)
Z = NH2  NH2NH2 (hydrazine)
Z = NHC6H5  NH2NHC6H5 (Phenyl hydrazine)

NO2 NO2

Z= NO2  NH2 NH NO2
NH

2, 4–Dinitro phenyl hydrazine (DNP) Brady's reagent.

Z = NHCONH2  NH2NHCONH2
Semi Carbazide.

RR (Aldoxime)
C O + H2NOH C NOH

HH

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RR (Hydrazone)
C O + H2 NNH2 C NNH2 (Phenyl hydrazone)

HH
RR

C O + H2NNHC6H5  C NNHC6H5
HH

R NO2 NO2 R C NO2 NO2
C O + H2NNH H NNH

H

2,4-DNP (Brady's reagent) (2, 4 - dinitro phenyl hydrazone)

RR
C O + H2NNHCONH2  C NNHCONH2 (Semi Carbazone)

HH
( 4 ) With alcohol and thioalcohol :

H H OH HRCOl(Hg ) H OR
C O + ROH HCl(g) C C

R R OR R OR

Hemi-acetal Acetal

R R OR
C O + 2ROH HCl(g) C
R OR + H2O
R

Ketal

Tri ethoxy methane [HC(OC2H5)3] remove the water formed during the react ion and so the react ion
proceeds in forward direction.

R O + 2RSH HHCl2(Og ) R SR [O] R SO2R
C Thioalcohol C C

H H SR H SO2R

Mercaptal

R O + 2RSH HHCl2(Og ) R SR [O] R SO2R
C C C

R R SR R SO2R

Mercaptal (Thio Ketal) Sulphones Compound

All sulphones compounds are hypnotic compounds.

CH3 C SO2R CH3 C SO2R C2H5 C SO2R

CH3 SO2R C2H5 SO2R C2H5 SO2R

Sulphonal Tr i o n a l Tetronal

( 5 ) Reaction with glycol (group protection) :

H O CH2 HHCl2(Og ) C O CH2
C O+ O CH2

H O CH2 Cyclic acetal / ketal

(neutral)

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( 6 ) Reaction with sodium alkynide :

C O + HC   C  Acid C OH

CNa ONa C CH
Acetylinic alcohol
Sodium Alkynide C CH

( 7 ) Reaction with Grignard reagent :

H H OMgl H2OCH3CH2OH + Mg I
C O + CH3 Mgl  C

H H OH

CH3

Ethanol (1° alcohol)

CH3CHO + CH3Mgl   CH3 OMgl OH
C HH2O CH3 OH
CH3
C H + Mg
I

CH3

2–Propanol (2° alcohol)

CH3 C O + CH3+ Mgl  CH3 OMgl OH I
CH3 C CH3 H2O CH3 C CH3 + Mg OH
CH3 CH3

2–Methyl–2–propanol (3° alcohol)

( 8 ) Reaction with H2O : It is a reversible reaction.

C O + H2O WeaHk2aOcid OH
C

OH

(neutral) unstable hydrate

Ex : Which compound form more stable hydrate with H2O?

(A) HCHO (B) CH3CHO (C) CH3COCH3 (D) CH3COC2H5

[Hint : HCHO since it is more reactive towards this reaction.]

Ans. (A)

Ex : Which carbonyl compound not gives reversible reaction with water ?

S o l . Chloral hydrate.

Cl Cl HO

Cl C CH + H2O  Cl C CH

Cl O Cl HO

(Chloral) (Chloral hydrate)
Stable by intra molecular hydrogen bonding.

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( B ) Other Reactions :
( 1 ) Wittig Reaction :

Wittig reaction affords an important and useful method for the synthesis of alkenes by the treatment
of aldehydes or kentones with alkylidenetriphenylphosphorane (Ph3P = CR2) or simply known as phosphorane

Ph Ph Ph Ph Ph Ph

P+ CH2 + P

Ph CH2 Ph O Ph Ph O

(1-phenylvinyl) benzene triphenylphosphine oxide

The wittig reagent, alkylidenetriphenylphosphorane (ylide), is prepared by treating trialkyl or triarylphosphine
usually the latter with an alkyl halide in either soultion. The resulting phosphonium salt is treated with
strong base (such as C6H5Li, BuLi, NaNH2, NaH, C2H5ONa, etc.)

Ph Ph Ph  Ph  Ph
P + H3C—Br Ph P CH2 + C6H6+ LiBr
Ph P CH3 Br C6H5 Li Ph  CH2
Ph Ph
P (II)

Ph Ph
(I)

( 2 ) Cannizaro's reaction :

Those aldehydes which do not contain  -H atom give this reaction, with conc. NaOH or KOH; Products

are Salt of carboxylic acid + alcohol.
In this reaction one molecule of carbonyl compounds is oxidised to acid, while other is reduced to
alcohol, such type of reactions are called disproportionation reaction. (Redox reaction)

HCHO + HCHO Conc. HCOONa + CH3OH

NaOH

Mechanism : (Cannizaro reaction)

(a) Rapid reversible addition of OH to one molecule of HCHO.

HC H OH OH
HC H

O O



(b) Transfer of hydride ion H to second molecule of HCHO

OH O O H O
H C H + C H  H C+ H C H

OH OO



Formic Methoxide
acid ion

Proton exchange

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(c) Proton exchange

H C OH + CH3O  HCOO  + CH3OH
O  HCOONa
HCOO + Na

When molecules are same  Simple cannizaro reaction
Two different molecules  Mixed cannizaro reaction.

In mixed or crossed cannizaro reaction more reactive aldehyde is oxidised and less reactive aldelyde
is reduced.

HCHO + C6H5CHO NaOH HCOONa + C6H5CH2OH

Oxidized Reduced

(Sodium formate) (Benzyl alcohol)

Ex : CH3CHO + HCHO Ca(OH)2  C(CH2OH)4 + (HCOO)2 Ca, explain mechanism ?
2, 2-Dihydroxy methyl –1, 3–propane diol.(Penta erythritol)

( 3 ) Tischenko reaction :

It is a modified cannizaro reaction. All aldehydes undergo this reaction in presence of (C2H5O)3Al, to
form ester.

2RCHO (R 'O)3Al O
RCH2 O C R

Ester

Example : CH3CHO  CH3CHO (C2H5 O)3 Al  CH3 COOH CH3 CH2 OH
Esterification 

CH3—COOCH2CH3 (Ethyl acetate)

( 4 ) Reaction With Halogen :

( a ) Replacement of -H atoms :

This reaction is not shown by formaldehyde (HCHO), since  -H atoms are absent, as enolisation

does not takes place in HCHO.

Cl

 C H + Cl2 –HCl  CH2 C HCl2  CH C H–CHl2Cl CCl3 C H

CH2 –HCl

HO Cl O Cl O O

Chloral

Example : CH3 C CH3 + 3Cl2 –3HCl  CH3 C CCl3
O O

Tri chloro acetone

Example : CH3 CH2 C CH3 + 3Cl2 –3HCl  CH3 CH2 C CCl3
OO

Example : CCl3 C R + NaOH CHCl3+ NaO C
OO

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Example : CH3 CH2 C CH2 CH3 + 2Cl2 –2HCl  CH3 CH2 C CCl2 CH3
O O

( b ) Replacement of O-atom of C O group : It takes place by PCl5 or SOCl2.

C O + PCl5  Cl
C Cl + POCl3

Phosphorus penta chloride

C O+ SOCl2  Cl
C Cl + SO2

Thionyl chloride
( c ) Haloform reactions :

Chlorine or bromine replaces one or more -hydrogen atoms in aldehydes and ketones, e.g.,
acetone may be brominated in glacial acetic acid to give monobromoacetone :

CH3COCH3 + Br2  CH3COCH2Br + HBr (43–44%)
The halogenation of carbonyl compounds is catalysed by acids and bases. Let us consider the
case of acetone. In alkaline solution, tribromoacetone and bromoform are isolated. Thus, the
introduction of a second and a third bromine atom is more rapid than the first. In aqueous sodium
hydroxide, the rate has been shown to be independent of the bromine concentration, but first
order with respect to both acetone and base i.e.,

Rate = k [acetone] [OH]

O O
CH3COCH3+ OH Slow H2O + CH3—CCH2
CH3C=CH2 Br2 CH3COCH2Br + Br

fast

( 5 ) Aldol Condensation :

Two molecules of an aldehyde or a ketone undergo condensation in the presence of a base to yield
a -hydroxyaldehyde or a -hydroxyketone. This reaction is called the aldol condensation. In general
Carbonyl compounds which contain -H atoms undergo aldol condensation with dil. NaOH. Aldol contains
both alcoholic and carbonyl group.
 Mechanism of aldol condensation : It takes place in the following two stages :
(a) Formation of Carbanion
(b) Combination of carbanion with other aldehyde molecule.
( a ) Formation of Carbanion :

-H atom of C O group are quite acidic which can be removed easily as proton, by a base.

–   –
OH+ H CH2 C H CH2 C H + H2O

OO

Base Acetaldehyde Carbanion

Carbanion thus formed is stable because of resonance -

– H CH2 CH
CH2 C O–

O

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(b) Combination of carbanion with other aldehyde molecule :

OH O–
– CH3 C CH2 CHO

CH3 C + CH2 C O H
H
H+ H2O
Aldehyde
(other molecule) OH H

O

CH3 C CH2 C O CH3 CH CH CH

H  –Unsaturated aldehyde

Aldol

Aldol condensation is possible between :
1 . Two aldehyde (Same or different)
2 . Two ketones (Same or different)
3. One aldehyde and one ketone

Identical carbonyl compounds  S im p l e o r s e l f a l d o l co n d e n s a t io n .
Different carbonyl compounds  M ix e d o r cr o s s e d a l d o l co n d en s a t io n .

Simple or Self condensation :

CH3 CH + HCH2CHO dil  CH3 CH CH CHO   CH3 CH CH CHO

NaOH –H2O

O OH H Crotonaldehyde

CH3 O CH3

CH3 C CH3 + H CH2 C CH3  CH3 C CH2 C   CH3 C CH COCH3 + H2O

O O OH CH3

diacetone alcohol Mesityl oxide or
4–Methyl– 3–pentene– 2–one

CH3 C CH O C CH3 HCl gas  CH3 C O C CH3
CH3 C CH3 + O CH3 –H2O CH C CH CH3
CH3
Mixed or Crossed Phorone

aldol Condensation :

CH3CH + CH3 C CH3WBaeseek Total (4) products (2) simple
(2) mixed
OO

CH3  COCH3 OH CH3 CH CH2 COCH3  CH3 CH CH COCH3

CH + CH2 Week base –H2O

OH OH

(Aldol)

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CH3  CH3 CH3
C CH2 C CH
CH3  CHO OH CH3 CHO  CH3 CHO

C + CH2 Week base –H2O

OH OH

(Aldol)

Ex : CH3CHO + CH3CH2CHO (WOBH)  Total 4 product s. Write str ucture of products ?

Sol. CH3  + CH2CHO  OH (Aldol) – H2 O  CH3—CH CH—CHO

CH

OH

CH3 CH2  CHO  OH (Aldol) –H2O CH3—CH2—CH CH—CHO

CH + CH2

OH

CH3 CH3

CH3  CHO  OH (Aldol) –H2O CH3 CH C CHO

CH + CH

OH

CH3 CH3

CH3 CH2  CHO  OH (Aldol) – H2 O  CH3 CH2 CH C CHO

CH + CH

OH

 Intramolecular aldol condensation :

CH3 C CH2 CH2 C CH3 OH 

(–H2O) CH3 C CH2 CH2 C CH2
OO
OO

H3C O   H3C O H3C O
HO H  O
–H2O

( – Unsaturated Ketone) (Aldol)

Here 5 membered ring is more stable than 3 membered ring so above product is formed as a major

product.

Note :

 If in crossed aldol condensation reaction , only one carbonyl compound have –H than total two product
formed.

CH3CHO + C6H5CHO  OH  Total 2 product.

WB

CH3  CHO  OH (Aldol) –H2O CH3—CH CH—CHO
(Crotonaldehyde)
CH + CH2

OH

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C6H5  CHO  OH (Aldol) –H2O C6H5—CH CH—CHO

CH + CH2

OH

(Cinnamaldehyde)
( 6 ) Claisen condensation :

When two molecules of ester undergo a condensation reaction, the reaction is called Claisen condensation.
The product of the claisen condensation is a -keto ester.

2H3C OC2H5 CH3CH2ONa H3C CH3 + H3C OH
O O C2H5

OO

-keto ester

After nucleophilic attack, the aldol addition and the Claisen condensation differ. In the claisen condensation,
the negatively charged oxygen reforms the carbon oxygen -bond and eliminates the OR group.

 Mixed claisen condensation :

OO OO

O CH3 + H3C O CH3 (i) CH3CH2O O CH3

(ii) H

H3C
-keto ester

( 7 ) Intramolecular claisen condensation :

Dieckmann condensation : The addition of base to a 1,6-diester causes the diester to undergo

intramolecular claisen condensation, thereby forming a five membered ring -keto ester. An intramolecular
claisen condensation is called a Dieckmann condensation.

(i) CH3O + CH3OH
(ii) H H3C

OO
CH3

O CH3 O
O
(i) CH3O OCH3 + H3C—OH
(ii) H O

O

OCH3

( 8 ) Perkin reaction :

In perkin reaction, condensation has been effected between aromatic aldehydes and aliphatic acid anhydride
in the presence of sodium or potassium salt of the acid corresponding to the anhydride, to yield ,
-unsaturated aromatic acids.

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The acid anhydride should have at least two -H.
C6H5CHO + (CH3CO)2O 17A0cO18N0aC C6H5—CH=CH—COOH

(CH3CO)2O, AcONa O
 OH

OO
O

3-(-furyl) acrylic acid

O

O OH

O (CH3CO)2O, AcONa O

 O
phthalyl acetic acid
O
phthalic anhydride

( 9 ) Knoevenagel Reaction :

Condensation of aldehydes and ketones with compounds having active methylene group in the presence
of basic catalyst to form a, b-unsaturated compounds is called Knoevenagel Reaction. The basic catalyst
may be ammonia or its derivative. Thus 1°, 2°, 3° amines i.e., aniline, di-or tri - alkyl amines, pyridine
or piperidine are used.

O H5C6 O O H5C6 O
O O R
Pyridine (i) H2O
C6H5CHO + R Piperidine

O O (ii) , –CO2 OH
OR R 

Malonic Ester Cinnamic acid

( 1 0 ) Reformatsky Reaction :

A similar reaction like the addition of organometallic compounds on carbonyl compounds that involves
the addition of an organozinc reagent to the carbonyl group of an aldehyde or ketone. This reaction,
called Reformatsky reaction, extends the carbon skeleton of an aldehyde or ketone and yileds b-hydroxy
esters. It involves treating an aldehyde or ketone with an -bromo ester in the presence of zinc metal;
the solvent most often used is benzene. The initial product is a zinc alkoxide, which must be hydrolysed
to yield the -hydroxy ester.

ZnBr

O OO H3O OH O

O+ Br Zn/C6H6

Aldehyde O O—R O—R
or ketone R
-bromoester -hydroxy ester

( 1 1 ) Schmidt Reaction :

This is the reaction between a carbonyl compound and hydrazoic acid in the presence of a strong acid
concentrated sulphuric acid. Aldehydes give a mixture of cyanide and formyl derivatives of primary
amines, whereas ketones give amides :

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RCHO + HN3 H2SO4 RCN + RNHCHO + N2

RCOR + HN3 H2SO4 RCONHR + N2

Reaction with primary amine : C O + H2 NR  C NR + H2O

Schiff's Base
( 1 2 ) Benzoin condensation :

The benzoin condensation is essentially a dimerisation of two aromatic aldehydes under the catalytic
influence of cyanide ions to give benzoin (I).

O

2 C6H5CHO KCN H3C OH

H5C6
(I)

The hydrogen atom attached to the carbonyl group of aldehyde is not active enough to be removed
easily but the addition of the cyanide ion to the carbonyl carbon places this hydrogen in the alpha
position of the nitrile thus rendering it relatively acidic. The carbanion, thus generated, attacks the
carbonyl carbon of the second aldehyde molecule in a rate-determining step forming an unstable cyanohydrin
of benzoin which immediately breaks down into benzoin and hydrogen cyanide.

O OH OH

O proton exchange

H5C6  H5C6 CN  CN H5C6 C
N
+ CN H5C6
H
H

OH O O OH O

OH O C6H5 Slow H3C H3C CH3 H3C 
+
H5C6 CN H CH3 OH + CN

CN H CN H H5C6

( 1 3 ) Benzilic acid rearrangement :

The addition of a strong base to a carbonyl group results in the formation of an anion. The reversal
of the anionic charge may cause expulsion of the attached group X, e.g.

X X O O
OH
HO

O

OH

However, in a 1, 2-diketone the group X may migrate to the adjacent electron-deficient carbonyl carbon
forming -hydroxy acid.
Thus, benzil on treatment with a strong base forms benzilic acid (salt), hence the name benzilic acid
rearrangement.

O O O OH
H5C6 OH O

C6H5 NaOH H5C6 H5C6
O H5C6
H5C6 ONa

benzil sodium salt of benzilic acid

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( 1 4 ) The Beckmann rearrangenment :

The acid catalysed transformation of a ketoxime to an N-substituted amide is known as the Beckmann

rearrangement.

C=O + H2N—OH –H2O C=N—OH
Oxime

H

RR O—H

N H  H2O

N –H2O R'—C R'

R' OH R' O H2 N—R N—R

–H O—H O
N—R R'
R'
NH—R

The rearrangement is catalysed by a variety of acidic reagents such as H3PO2, H2SO4, SOCl2,
PCl5, etc.
( C ) Oxidation Reactions :

( a ) By K2Cr2O7/H2SO4 :
On oxidation with K2Cr2O7/H2SO4 1° alc. gives aldehyde, which on further oxidation gives acid
with same C-atom. While, 20 alcohol on oxidation gives ketone which on further oxidation gives acid
with less C-atom.

R—CH2OH [O] R—CH O [O] R—COOH
(1° alcohol)

CH3CH2CH2 CH CH3 [O] CH3CH2CH2 C CH3 [O] CH3CH2COOH+CH3COOH

OH O
(2° alcohol)

(i) 3° alcohol is not oxidised within 2 or 3 minutes.

(ii) 1° and 2° alcohol converts orange colour of K2Cr2O7 to green in 2-3 minutes.
( b ) SeO2 (Selenium Oxide) :

Ketones or aldehydes on oxidation with SeO2 gives dicarbonyl compounds. This reaction is possible
only in compounds containing –carbon.

HCHO doesn't show this reaction.



CH3CHO + SeO2  H C C H + Se + H2O

OO
Glyoxal



CH3 C CH3 + SeO2  CH3 C C H + Se + H2O

O OO
Methyl glyoxal (Pyruvaldehyde)

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( c ) Baeyer's Villiger oxidation :
Both aldehyde and ketones are oxidized by peroxy acids. This reaction, called the Baeyer-villiger
oxidation, is especially useful with Ketones, because it converts them to carboxylic esters. For
example, treating acetophenone with a peroxy acid converts it to the ester phenyl acetate.

O

H5C6 OR O O
Mechanism : CH3 H5C6—O

OH

H3C
phenyl acetate

H :O H O
R
O A



O
:O: H—A :  O :
:
O O
:
+ :OH :H3C

H3C C6H5 H3C C6H5 R H

C6H5

 H
O
: : :O
OH A H H O H—A
OH R
OO
R + H3C  OO
O H3C O R O
C6H5 C6H5
H3C C6H5

O

H3C OC6H5

The product of this reaction show that a phenyl group has a greater tendency to migrate then a methyl

group. Had this not been the case, the product would have been C6H5COOCH3 and not CH3COOC6H5.
This tendency of a group to migrate is called is migratory aptitude. Studies of the Baeyer-villiger oxidation

and other reaction have shown that the migratory aptitude of groups H > phenyl > 3° alkyl > 2° alkyl

1° alkyl > methyl. In all cases, this order is for groups migrating with their electron pairs, that is, as anions.

E x . CH3 C C2H5 Peracid  ?
O

S o l . CH3 C OC2H5
O

CH3
E x . CH3 C C CH3 Peracid  ?

CH3 O

CH3
S o l . CH3 C O C CH3

CH3 O

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( D ) Reduction :
( a ) The wolf kishner reduction :
When a ketone or an aldehyde is heated in a basic solution of hydrazine, the carbonyl group
is converted to a methylene group this process is called Deoxygenation because an oxygen is
removed from the reactant. The reaction is known as the Wolf-kishner Reduction.

COCH3 CH3

(i) NH2–NH2
(ii) OH/HOCH2–CH2OH

( b ) Clemmensen Reduction :
The reduction of carbonyl groups of aldehydes and ketones to methylene groups with amalgamated
zinc and concentrated hydrochloric acid is known as Clemmensen reduction.

COCH3

Zn–Hg, HCl CH3
Reflux

The nature of product depends upon the reducing agent used. It can be summarized as.

(i) C O  CH2 Reducing agents are

(ii) C O  CHOH  Red P/HI at 150°C
 Zn-Hg/HCl [Clemensen's reduction]
 NH2—NH2/C2H5OH,OH [Wolff Kischner's reduction]

Reducing agents are

 LiAlH4 (Nicetron brown)
 Na/C2H5OH (Bouvalt blank)
 NaH/Benzene (Darzen reaction)

 [(CH3)2CHO]3Al (Aluminium isopropoxide)
 (CH3)2 CHOH (Isopropyl alcohol)
Reduction with aluminium isopropoxide is excess of isopropanol is called MPV (Meerwein Ponndroff
Verley) reduction. Other reducible groups are not attacked like —NO2, —CH CH2, —C C—.

Example : CH3 C O MPV Re duction CH3 CHOH
CH2 CH
CH2 CH

( E ) Reactions given by only aldehydes :
( 1 ) Polymerisation : It is a reversible process.

Formaldehyde :

(a) nHCHO Evaporation (CH2O)n , H2O Paraformaldehyde is a linear polymer
which show reducing character with
Formalin Paraformaldehyde Tollen's reagent, Fehling solution etc.

(40% HCHO) n = 6–50
Hydrated white crystal

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(b) nHCHO HC2SoOnc4. (CH2O)n H2O
Poly oxy methylene
n > 100

(c) 3HCHO Allowed tos tan d (C H 2 O )3

at room temp. CH2
OO
Meta formaldehyde (Trioxane)

Cyclic polymer (Trioxy methylene) H2C

Cyclic polymer doesn't show reducing CH2

character with Tollen's reagent etc. O

(d) 6HCHO CaOH2  C6H12O6
or BaOH
2 Formose sugar

A linear polymer (  -acrose)

Acetaldehyde :

CH2

CH
OO

(a) 3CH3CHO Conc. H2SO4 (CH3CHO)3 H3C HC CH CH3

Room temp.

O

Para acetaldehyde

Paraldehyde (cyclic polymer)

Pleasent smelling liquid

Hypnotic compound

CH3 CH3

O

(b) 4 C H 3C H O conc0. HC2SO4  (CH3CHO)4 OO

O CH3
CH3

Meta aldehyde

White crystalline solid.

Cyclic polymer

Used as solid fuel or killing snails

( 4 ) Reaction with ammonia :

Except formaldehyde, all other aldehydes give addition reactions (HCHO give addition elimination i.e.
condensation reaction)

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6CH2O + 4NH3 Condensation (CH2)6N4 + 6H2O

Re action Urotropine (Hexamine)
White crystalline solid
Used in preparation of explosive
Used in treatment of urine infection diseases

CH3 CH + NH2 Addition  CH3 CH NH2 Conc. H2SO4 CH3—CH NH

Re action OH H2O
Acetaldehyde
OH Acetaldimine
ammonia
used in preparation

of cyclic polymer

3CH3—CH NH 3 H2O  CH3

Trimerisation CH
HN NH
HC CH .3H2O
H3C N CH3

H
Trimethyl hexahydro Triazine trihydrate

( 5 ) Reducing character :

Aldehydes are easily oxidised so they are strong reducing agents.

( a ) Tol len's reagent :

It oxidises aldehyde s. Tollen's reagent is ammonical si lver nitrate solut ion

( Ag NO 3+NH 4O H )    [Ag(NH3)2]OH
RCHO+[Ag(NH3)2]OH  RCOOH + Ag + H2O
Silver mirror

AgNO3 + NH4OH    A g OH  Ag 2O
RCHO + Ag2O    RCOOH + Ag  (Silver mirror)
( b ) Fehling's solution :

It is a mixture of CuSO4, NaOH and sodium potassium tartrate.
Fehling solution A– (aq.) solution of CuSO4
Fehling solution B– Roschelle salt (Sodium potassium tartrate + NaOH)

Fehling solution A + Fehlings solution B(Dark blue colour of cupric tartrate)
RCHO + Cu+2 + OH–  RCOOH + C u 2O

(Cuprous oxide–Red ppt.)

C u 2+  Cu+

(Cupric - Blue) (Cuprous - Red ppt.)

( c ) Benedict's solution :

It is a mixture of CuSO4 + sodium citrate + Na2CO3. It provides Cu+2. It is reduced by aldehyde
to give red ppt of cuprous oxide.

RCHO + Cu2+ + –  RCOOH + Cu2O

OH

(Cuprous oxide–Red ppt.)

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( d ) Mercuric chloride :
HgCl2 is a corrosive sublimate. It is reduced by aldehyde to give white ppt of mercurous chloride
(Calonal) which further react with aldehyde to give black ppt of Hg.
RCHO + HgCl2 + H2O  RCOOH + Hg2Cl2 + HCl
(calomal)
RCHO + Hg2Cl2 + H2O  RCOOH + Hg + HCl
(black ppt)

( e ) Reaction with schiff's reagent :
Schiff's reagent is dil solution of p-roseniline hydrochloride or magenta dye.
Its pink colour is discharged by passing SO2 gas and the colourless solution is called schiff's
reagent, Aldehyde reacts with this reagent to restore the pink colour.

( F ) Reaction of only ketones :

( 1 ) Reduction : Acetone is reduced by magnesium amalgam and water to give pinacol.

CH3 C O+ O C CH3 Mg Hg CH3 CH3 CH3
CH3 CH3 C C CH3
water OH OH

Pinacol

( 2 ) Reaction with chloroform :

CH3 C O+ CH3 C OH OH COOH
CH3 CHCl3  CH3 CCl3 aq.NaOHCH3 *CH

(Chloretone) (Lactic acid)
2-Hydroxy propanoic acid

( 3 ) Reaction with HNO2 :

OO
CH3 C CH3 + O N OH –H2O CH3 C CH N OH

Oximino acetone

( 4 ) Oxidation reaction : According to popoff's rule C O group stays with smaller alkyl group.
O

CH3 CH2 C CH3  CH3COOH + CH3COOH

( 5 ) Condensation reaction :
(a) In presence of dry HCl - aldol condensation takes place

CH3 O CH3 O

CH3 C O + CH3 C CH3 –H2O CH3 C CH C CH3

Mesityl oxide

CH3 CH3 CH3 O CH3
C CH
CH3 COCH3+ O C CH3–H2O CH3 C CH C CH C CH3

(Phorone) or 2,6–Dimethyl–2,5–
hepta diene–4–one

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(b) In presence of conc.H2SO4 CH3
O

3CH3 C CH3 Conc.H2SO4  Mesitylene
CH3
condensation

Polymerisation H3C

C Fe
3CH3 CH
Addition Polymerisation

( 6 ) Reaction with ammonia :

CH3 O CH3 O

CH3 C O + HCH2C CH3 –H2O CH3 C CH2 C CH3

HNH2 NH2
Diacetone amine

( 7 ) Pyrolysis :

CH2 C O  CH2 C O + CH4 HCHO   CO  H 2 
 pyrolysis 
H CH3 Ketene
Acetone CH3CHO   CH 4  CO 

TEST FOR HCHO, CH3CHO, CH3COCH3

S.No. Te st HCHO CH3CHO CH3COCH3
1. – Red Red
Legel's test :Na [Fe(NO)(CN)3]
2. sodium nitroprusite (alk.) – – –
– yellow ppt yellow ppt
Only methyl C O

compound gives this test
Iodoform test
(I2 + NaOH)

OH

3. Pyragallol OH white ppt. – –

OH

4. Orthonitro benzaldehyde – – Blue
5 . Tollen's reagent - Silver mirror Silver mirror –
Red ppt Red ppt –
Fehling's reagent - Black ppt Black ppt –
Mercuric chloride - Pink colour Pink colour –
Schiff's reagent - Orange Orange Orange
6. DNP colour colour colour

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BENZALDEHYDE (C H CHO)
65
Oil of bitter almonds

General Method of Preparation :

C6H6 CO/ HCl + AlCl3 *(Gattermann-kosch reaction)
C6H5CH3 CrO2Cl2 *(Etard reaction)
C6H5COCl Pd/BaSO4 *(Rosenmund reaction)
C6H5CN (i) SnCl2/HCl (ii) H2O *(Stephen's reaction)
(C6H5COO)2 Ca (HCOO)2Ca/
C6H5CHCl2 aq. KOH C6H5CHO
C6H5CH2OH
C6H5CH2Cl [O] (Gattermann)
C6H5MgBr (CH2)6N4
C6H6 (i) HCOOC2H5 (ii) H2O
(i) HCN/HCl (ii) H2O

AlCl3

Chemical properties :

C6H5CH(OH)CN (Similar) (different) C6H5COCl
HCN 170°/Cl2
C6H5CH(OH)SO3Na
NaHSO3 2NH3 C6H5 CH N CH C6H5
C6H5CH N Z 2C6H5CHO C6H5 CH N
NH2 Z
C6H5CH3 2C6H5CHO C6H5 CH C C6H5 Benzoin
Red P + HI alc.KCN OH O
C6H5COOH
Oxidation (CH3CO)2O,CH3COONa C6H5 CH CH COOH
C6H5CHCl2 Perkin
PCl5
Silver mirror test C6H5CHO H2CH NO2 C6H5 CH CH NO2
Tollen HO– Nitrostyrene
C6H5COONa + C6H5CH2 OH reagent
Cannizaro reaction Br
C6H5CH N Ar
Schiff's base (i) Zn+CH2COOC2H5 Cinnamic acid
Pink colour
C6H5COOCH2C6H5 (ii) H2O Reformatsky reaction
Tischenko reachion
NaOH
OH
C6H5 CH C6H5 HNO3/H2SO4 m – nitrobenzaldehyde

Ar NH2 Fuming H2SO4 m–formyl benzene
sulphonic acid
Schiff's
reagent Cl2/FeCl3 Chloro benzaldehyde

AlCl3 CH3 C CH3
Al(OR)3
O C6H5CH CH C CH3
C6H5MgBr 
H2O OH

O

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SOLVED EXA MPLES

Ex 1. What is A in the following reaction ?

O Cl O
+ O ttBBuuOOHK A
H5C2

O H3C O O O
OH
O O (D)
–– – O C2H5
C2H5= C2H5
(A) (B) ––
=(C)
=
O

Ex 2. is the final product obtained when one of the following is reacted with base :
CH3

O O (B) H3C OO
(A) H3C CH3 CH3

O O O O
(C) H3C (D) H3C CH3
CH3

Sol. O O O
H3C CH3 Alkali  CH3

Ex 3. CH3 O O
CH3 CC
C – OH
H (A)

H

The product (A) in the given reaction would be :

OO O OH
CC
CO
(A) CH3 C OH (B) C–C–H

H CH3 H3C CH3

(C) CH3 O OH O
HC (D) C=C C–H
C H
CH3 CH3
CH3

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O H
C – OH
CH3 O H+ CH3 OO –H+ H3C OH
CH3 CC H –CO2 H3C H
Sol. C C–C–O–H C=C
unstable enol
CH3 H

unstable

H3C H O
H3C C–C H

Ex 4. End product of the following sequence of reactions is :
CHCH CH3MgBr CO2 /H3O HgSO4 /H2SO4 Ag2O

O O O O
O O
(A) H3C OH (B) HO (C) H3C (D) H
OH
O O OH

Sol. CHCH + CH3MgBr CHC–MgBr

CO2/H3O

–COOHis (EWG) CH C COOH

HgSO4/H2SO4

OHC–CH2–COOH

Ag2O

HOOC–CH2–COOH

 (B)
Ex 5. In which of the following substrates, rate of Benzoin condensation will be maximum ?

(A) O2N CHO (B) H3C CHO

(C) HO CHO (D) NH2 CHO

:
: ::
:
: ::

:
: ::
O :O O :O: O :O–H
 C + CN N C–H C:
N :H CN :O N
 (I) :O (II) CN
Sol. :
:O =
:
–=
:
––

(III)

Benzoin condensation is due to stability of intermediate (III) when negative charge on C extensively delocalised

in benzene ring, nitro and CN group. In all other cases, such dispersal is not extensively possible. On

the other hand, NO2– is also creating a positive charge center on carbonyl carbon, making it more susceptible
to nucleophilic attack of CN–.

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Ex 6. This intermediate is converted into product in the wittig reaction :

Ph3 P O

R–C C–R1 O=P – Ph3 + R–C = C–R1
R1 R2 R1 R2
–
Sol. Out of following which statements are correct ?–
Ex 7. (A) C–O bond is weaker as compared to P–O bond–
(B) Lone pair of oxygen atom participate in p–d bonding with phosphorous atom–
(C) C–P bond is weaker as compare to C–C bond–
(D) C–C bond is weaker as compare to C–O bond =
(A), (B), (C)
Consider the following sequence :

O CH3
H
OO O

OH  CHSt3eCpHIIO  H3C O
Step I O
H3C H2O Step III
O H3C

O O CH3
CH—CH3 H
Step IV
OH

H3C O H3C O

Which of following statements are correct for above reaction sequence ?

(A) Step I is acid-base reaction (B) Step II is nucleophilic addition reaction

(C) Step III is acid base reaction (D) Step IV is elimination reaction

Sol. (A), (B), (C), (D)
Ex 8.
Which of the following oxidation reaction can be carried out with chromic acid in aqueous acetone at
Sol.
Ex 9. 5–10°C.
Sol.
O

(A) CH3(CH2)3CC–CH–CH3  CH3(CH2)3CC–C–CH3
OH

(B) CH3(CH2)3CH=CH–CH2OH  CH3(CH2)3CH=CH–CHO

(C) C6H5CH3  C6H5COOH

(D) CH3(CH2)3CH2OH  CH3(CH2)3CHO

(A), (B), (C), (D)

O
CH3 – C – CH3 SeO2  A ; A will :

(A) Reduce Tollen's reagent (B) Give iodoform test

(C) Form oxime (D) Give Cannizaro reaction

SeO2 oxidises –  – CH2 – a w.r.t. keto group
 (A), (B), (C) and (D)

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Ex 10. 3HCHO + CH3CHO NaOH A. A found can

(A) Reduce Tollen's reagent (B) Give Cannizaro reaction
(C) React with Na (D) Give green colour with Cr2O72–/H+
S o l . A is by aldol condensation

CH2OH
HOH2C – C – CHO

CH2OH

 (A), (B), (C) and (D)
Ex 11. 2CH3 – C – CH3 MgH/Hg Product, product in the reaction is :

O

CH3 CH3 (B) CH3 – C – O – C – CH3
(A) H3C – C – C – CH3 OO

OH OH

(C) CH3 – CH – CH – CH3 (D) None of these
OH OH

Sol. CH3 CH3
(A) 2CH3 – CO – CH3 MHg2/OHg H3C – C – C – CH3

OH OH
(Pinacol)

Ex 12. Benzaldehyde on reaction with acetophenone in the presence of sodium hydroxide solution gives :

(A) C6H5CH = CHCOC6H5 (B) C6H5COCH2C6H5
(C) C6H5CH = CHC6H5 (D) C6H5CH(OH)COC6H5

Sol. O
(A) C6H5CHO + CH3COC6H5 NHaO2OH C6H5 – CH = CH – C – C6H5

Ex 13. Product in following reaction is :

CH3MgI + HCHO  Product

(A) CH3CHO (B) CH3OH (C) C2H5OH (D) CH3 – O – CH3

Sol. OH
(C) H – CHO + CH3MgI  CH3 – CH2 – OH + Mg I

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