Download aldehydes and ketones

CAK – 1
ALDEHYDES AND KETONES
C1A Physical Properties :
It has high dipole moment
Boiling point are lower than alcohols due to their inability to form intermolecular H-bonding.
B.Pt. are higher than corresponding alkanes due to dipole-dipole interaction.
Carbonyl group can form H-bond with H2O hence they are soluble in water to varying extent.
C1B
Method of Preparation :
1.
Oxidation :
(a)
(b)
(c)
This oxidation is called as oppenauer oxidation.
2.
Rossenmund Reduction :
3.
[LiAlH (O-t-C4H9)3 or Lithium tri-t-butoxy Aluminium hydride]
4.
5.
Hydrolysis of Gem-Dihalide
(a)
(b)
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 2
(c)
6.
Cl , heat
H O
2
ArCH 3 2  ArCHCl 2 

 ArCHO
Hydration of Alkynes : Hydration of alkynes gives ketones (except CH  CH which gives CH3CHO)
(a)

H O, H
CH  CH 2  CH 3 CHO (b)
HgSO 4
7.
H O, H 
CH 3 C  CH 2  CH 3 COCH 3
HgSO 4
Hydroboration Oxidation :
Hydroboration of a non-terminal alkyne followed by oxidation of the intermediate yields a ketone
but terminal alkyne yield aldehyde.
8.
9.
Use of Grignard Reagent :
(a)
With HCN aldehyde is formed.
(b)
With RCN a ketone is formed.
With Esters
(a)
(b)
10.
Decarboxylation of calcium salts of carboxylic acids :
11.
Oxo Process :
2 CH3CH = CH2 + CO + H2
CH3CH2CH2CHO +
[COH(CO)4] 
 Cobal + Carbonyl hydride
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 3
12.
Reduction of acid chloride with organocopper compounds :
13.
Friedel-Crafts Acylation :
(a)
(b)
14.
From Cadmium and Lithium Salts :
1.
2RMgX + CdCl2 
 R – Cd – R + MgCl2 + MgX2
 RCOR’ + RCdCl
R – Cd – R + R’COCl 
R 
 should be 10 alkyl or aryl (–C6H5)
15.
 Aldehyde/Ketone
Ozonylysis of Alkene (Zn/H+) 
C2
Chemical Properties :
1.
Nucleophilic Addition to the Carbon-Oxygen Double bond :
The most characteristic reaction of aldehyde and ketone is nucleophilic addition to the
carbon-oxygen double bond.
General Reaction :
Relative reactivity of Aldehydes versus Ketones :
Aldehydes are more reactive than Ketones. There are two reasons for this, they are as follows :
1.
1.
Steric Factor
2.
Electronic factor
Steric Factor : With one group being the small hydrogen atom, the central carbon of the tetrahedral
product formed from the aldehyde is less crowded and the product is more stable.
With ketones two alkyl groups at the carbonyl carbon causes greater steric crowding in the
tetrahedral product and make it less stable. Therefore small concentration is present at equilibrium.
2.
Electronic Factor : Because alkyl group are electron releasing therefore aldehydes are more reactive
on electronic grounds as well. Aldehyde have one electron releasing alkyl group to stablise the partial
positive charge on the carbon atom of the carbonyl group. Whereas ketones have two alkyl groups.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 4
(a)
Addition of cyanide :
(b)
Addition derivatives of ammonia :
(i)
(ii)
HCHO reacts with NH3 differently forming UROTROPINE [hexamethylene tetraamine].
6HCHO + 4NH3  (CH2)6N4 + 4H2O
H2N – G
(c)
Product
H2N – OH
Hydroxylamine
= N – OH
oxime
H2N – NH2
Hydrazine
= N – NH2
Hydrazone
H2N – NHC6H5
Phenylhydrazine
= N – NHC6H5
Phenylhydrazone
H2N – NHCOCH2
Semicarbazine
= NNHCOCH2
Semicarbazone
Addition of Alcohols : Acetal Formation
(i)
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 5
(ii)
Important :
(d)
Addition of Grignard reagents :
Formaldehyde with Grignard Reagent gives 10alcohol, all higher aldehydes with grignard reagent
give 20 alcohol and ketones with grignard reagent gives the 30 alcohol.
C3
Other Reactions of Aldehyde and Ketones :
(a)
Oxidation :
(i)
RCHO or ArCHO
Aldehydes (except) benzaldehyde reduce “Fehling’s Solution” (Cu2+ reduced to Cu+) which is an
alkaline solution of Cu2+ ion complexed with tetrarate ion.
(ii)
Example : Tollen’s Test :
CH 3 CHO  2Ag( NH 3 ) 2   3OH – 
 CH 3 COO   2Ag  4NH 3  2H 2 O
Colourless
Solution
Silver
mirror
Fehling Solution CH 3 CHO  2Cu 2   3OH  
 CH 3 COO   2Cu   2H 2O
Re d PPt .
Tollen’s test is cheifly given by aldehydes. Tollen’s reagent does not attack carbon-carbon double
bond. Aldehyde also reduce benedict’s solution (Cu2+ complexed with citrate ion) to Cu+
(b)
(i)
Ketones with strong oxidants and at high temperature undergo cleavage of C-C
bond on either side of carbonyl group.
Carbonyl Group after bond cleavage goes with that alkyl group which is of smaller
size.
(ii)
Ketones are also oxidised from cleavage of bond by caro’s acid (H2SO5) or
SO 5
peroxybenzoic acid (C6H5CO3H) to esters. RCOR  H
2 
 RCOOR  [Bayer’ss
[O ]
Villiger Oxidation]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 6
(c)
Haloform Test :
(i)
Methylketones :
(ii)
Hypohalite NaOX (NaOH + X2) cannot only halogenate can also oxidise alcohols
(d)
Reduction :
(a)
Reduction to alcohol
(e)
Reduction to hydrocarbons :
(f)
Reductive Amination (Discussed under Amines)
(g)
Cannizzaro reaction : In the presence of an concentrated base i.e. alkali, aldehydes
containing no -hydrogens undergo self oxidation and reduction to yield a mixture of an
alcohol.
(i)
(ii)
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 7
(h)
(i)
Crossed Cannizzaro reaction :
Tischenko Reaction :
All aldehyde in presence of aluminium ethoxide, Al(OC2H5)3 can be simultaneously oxidised
(to acid) and reduced (to alcohols) to form ester. This is called Tischenko reaction and is
thus like cannizaro reaction.
(j)
Distinction
Aldehyde and Ketones
S.No.
Test
RCHO
RCOR
1.
Schiffs
reagent
magenta
colour restored by
RCHO
no. reaction
2.
Tollen’s reagent
is reduced by RCHO
is not reduced
3.
Fehling’s solution
is reduced by RCHO
(except C6H5CHO)
is not reduced
-hydroxy, ketones
reudce Tollen’s re
agent and Fehling’s
solution
Practice Problems :
NaOH
1.
Identify (Z) in the reaction series,
(a)
2.
3.
C2H 5I
(b)
CH 2  CH 2 HBr

 (X) Hydrolysis
 (Y) I
( Z)
2 ( excess )
C2H5OH
(c)
CHI3
(d)
CH3CHO
A compound (X) of the formula C3H8O yields a compound C3H6O on oxidation. To which of the
following class of compounds could (X) belong
(a)
aldehyde
(b)
secondard alcohol
(c)
alkene
(d)
tert. alcohol
Which statement is incorrect in the case of acetaldehyde and acetone
(a)
both react with hydroxylamine
(b)
both react with NaHSO3
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 8
4.
(c)
both react with hydrazine
(d)
both reduce ammonical silver nitrate
Which of the following undergoes Cannizzaro’s reaction
(a)
5.
7.
8.
(c)
(CH3)2CHCHO (d)
HCHO
acetaldehyde
(b)
formaldehyde
(c)
acetone
phenol
(d)
(a)
Cannizzaro’s reaction
(b)
Clemmensen’s reduction
(c)
Rosenmund’s reaction
(d)
Tischenko reaction
When acetaldehyde is treated with aluminium ethoxide, it forms
(a)
ethyl acetate
(b)
ethyl alcohol
(c)
acetic acid
(d)
methyl propionate
(c)
acetone (d)
Chloretone is formed when chloroform reacts with
formaldehyde
(b)
acetaldehyde
benzaldehyde
Which of the following reagent reacts differently with HCHO, CH3CHO and CH3COCH3
(a)
10.
CH3CH2CHO
Hydrocarbons are formed when aldehydes and ketones are reacted with amalgamated zinc and
conc. HCl. The reaction is called
(a)
9.
(b)
Urotropine is formed by the action of ammonia on
(a)
6.
CH3CHO
HCN
(b)
NH2OH
(c)
C6H5NHNH2
(d)
NH3
In the following sequence of reactions, the end product is
2
/ H 2SO 4
O]
)2
2O
CaC2 H

(A) Hg


(B) [
((C) Ca
( OH


(D) heat
(E)
(a)
11.
acetaldehyde
(b)
formaldehyde
(c)
acetic acid
(d)
acetone
In the following sequence of reactions, the end product is
2
/ H 2SO 4
[O]
HC  CH Hg


(A) CH
3MgX


((B) 
(C)
[ H 2O ]
12.
13.
(a)
acetaldehyde
(b)
isopropyl alcohol
(c)
acetone
(d)
ethyl alcohol
HCN
(a)
CH3COOH
(b)
CH3CHOHCOOH
(c)
CH3CH2NH2
(d)
CH3CONH2
A compound, C5H10O, forms a phenyl hydrazone and gives negative Tollen’s and iodoform tests. The
compound on reduction gives n-pentane. The compound A is
(a)
14.
H O
2
In the following sequence of reactions, the end product is CH 3CHO 
( A ) 

 ( B)
pentanal
(b)
pentanone-2
(c)
pentanone-3
(d)
amyl alcohol
CHI3
(d)
CH3CHO
ethyl acetate
(d)
acetylene
The product Z in the series is
Na 2 CO 3
CH 2  CH 2 HBr

 X Hydrolysis
  Y I
 Z
2 ( excess )
(a)
15.
(b)
C2H5OH
(c)
If formaldehyde and KOH are treated together, we get
(a)
16.
C2H 5I
methane
(b)
methanol
(c)
The correct order of reactivity in nucleophilic addition reaction CH3CHO, CH3COC2H5 and
CH3COCH3 is
(a)
CH3CHO > CH3COCH3 > CH3COC2H5
(b)
C2H5COCH3 > CH3COCH3 > CH3CHO
(c)
CH3COCH3 > CH3CHO > C2H5COCH3
(d)
CH3COCH3 > C2H5COCH3 > CH3CHO
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 9
17.
18.
19.
20.
21.
To distinguish between 2-pentanone and 3-pentanone which reagent can be used
(a)
NaOH/I2
(c)
K 2Cr2O7/H
+
Tollen’s reagent
(d)
Zn–Hg, HCl
CH3CH = CHCHO is oxidised to CH3 – CH = CHCOOH, using oxidising agent as
(a)
alkaline KMnO4
(b)
K2Cr2O7/conc. H2SO4
(c)
ammonical AgNO3
(d)
dilute HNO3
m-chloro benzaldehyde on reaction with conc. KOH at room temperature gives
(a)
potassium m-chloro benzoate and m-hydroxy benzaldehyde
(b)
m-chloro benzyl alcohol and m-hydroxy benzaldehyde
(c)
m-chloro benzyl alcohol and m-hydroxy benzyl alcohol
(d)
m-chloro benzyl alcohol and potassium m-chloro benzoate
The reagent which can be used to distinguish acetophenone from benzophenone is :
(a)
2, 4-dinitrophenyl hydrazine
(b)
benedict reagent
(c)
I2 and Na2CO3
(d)
aqueous solution of NaHSO3
Best starting material to synthesize 2-methyl-2-butenoic acid is
(a)
(b)
(c)
(d)
22.
CH3CH2CH2CHO
X and Y can be distinguished by :
(a)
23.
(b)
silver-mirror test(b)
iodoform test
(c)
both
(d)
CH3  C CCl3
(b)
CH 3  C  CH 2 I
CH 3  C CH 2Cl
(d)
none
Which of the following will give haloform test
O
||
(a)
(c)
|
OH
all
||
O
[Answers : (1) c (2) b (3) d (4) d (5) b (6) b (7) a (8) c (9) d (10) d (11) c (12) b (13) c (14) c (15) b
(16) a (17) a (18) a (19) d (20) c (21) c (22) c (23) d]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 10
SINGLE CORRECT CHOICE TYPE
1.
2.
Which of the following reagent reacts differently
with HCHO, CH3CHO and CH3COCH3
HCN
(b)
NH2OH
(a)
Benedict’s solution
(c)
C6H5NHNH2
(d)
NH3
(b)
Tollen’s reagent
(c)
Fehling’s solution
(d)
NaHCO3
In the following sequence of reactions, the end
product is
Condensati on
Dehydratin g agent
mild alkali
heat
(a)
aldol
(b)
crotonaldehyde
(c)
paraldehyde
(d)
4.
5.
9.
metaldehyde
Which of the following reaction is a condensation
reaction
10.
(a)
Cannizzaro’s reaction
(b)
Aldol condensation
(c)
Claisen’s reaction
(d)
Tischenko reaction
The reaction involving condensation of acetic anhydride with an aromatic aldehyde by a carboxylate ion is an example of
HCHO  Paraformaldehyde
(b)
CH3CHO  Paraldehyde
(a)
aldol condensation
(c)
CH3COCH3  Mesityl oxide
(b)
benzoin condensation
(d)
CH2 = CH2  Polyethylene
(c)
Perkin reaction
(d)
Wurtz reaction
A compound A has molecular formula C2Cl3OH. It
reduces Fehling’s solution and on oxidation gives a
monocarboxylic acid. A is obtained by the action of
Cl2 on ethyl alcohol. The compound A is
11.
Benzaldehyde when refluxed with aqueous alcoholic
KCN forms
(a)
chloroform
(a)
benzoin
(b)
chloral
(b)
benzene
(c)
trichloro ethanol
(c)
phenyl cyanide
(d)
trichloro acetic acid
(d)
phenyl isocyanide
12.
The product Z in the series is
Hydrolysis
CH 2  CH 2 
 X   Y
Na 2 CO 3
I
 Z
2 ( excess )
7.
Al ( OC H )
5 3

 CH3COOCH2CH3. This
2CH3CHO  2 
reaction is called
(a)
HBr
6.
Formic acid and formaldehyde can be distinguished
by treating with
(a)
CH 3CHO   
(A )    
(B)
3.
8.
(a)
C2H 5I
(b)
C2H5OH
(c)
CHI3
(d)
CH3CHO
13.
acetone
(b)
acetaldehyde
(c)
propanal
(d)
HCHO
(a)
Wurtz reaction
(b)
Cannizzaro’s reaction
(c)
Perkin reaction
(d)
Claisen reaction
Anhyd .AlCl
3  (X) + HCl
C6H6 + CO + HCl   
The compound (X) is
Cyanohydrin of which of the following gives lactic
acid on hydrolysis
(a)
Benzyl alcohol is obtained from benzaldehyde by
14.
(a)
C6H5CHO
(b)
C6H5COOH
(c)
C6H5CH2Cl
(d)
C6H5CH3
Which of the following is the weakest acid
(a)
benzene sulphonic acid
In the reaction CH3CHO + HCN  CH3CHOHCN,
a chiral centre is produced. The product is
(b)
benzoic acid
(c)
benzyl alcohol
(a)
meso compound
(d)
phenol
(b)
laevorotatory
(c)
dextrorotatory
(d)
racemic mixture
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 11
15.
In the Cannizzaro’s reaction given below
OH 
2Ph  CHO 
 Ph  CH 2 OH  PhCO
20.
O
||

2
(a)
the slowest step is
16.
17.
18.
(a)
attack of OH— at the carbonyl group
(b)
the transfer of hydride to the carbonyl
group
(c)
the abstraction of proton from the
carboxylic acid
(d)
the deprotonation of Ph–CH2OH
Which of the following will give haloform test
(b)
CH3  C CCl3
CH 3  C  CH 2 I
|
OH
(c)
CH 3  C CH 2Cl
||
O
(d)
all
The reagent(s) which can be used to distinguish
acetophenone from benzophenone is (are) :
(a)
2, 4-dinitrophenyl hydrazine
(b)
benedict reagent
(c)
I2 and Na2CO3
(d)
aqueous solution of NaHSO3
21.
cannizzaro
  

(Y is alcohol, D is deuterium)
X and Y
X and Y will have structure :
m-chloro benzaldehyde on reaction with conc. KOH
at room temperature gives
(a)
potassium m-chloro benzoate and
m-hydroxy benzaldehyde
(b)
m-chloro benzyl alcohol and
m-hydroxy benzaldehyde
(c)
m-chloro benzyl alcohol and m-hydroxy
benzyl alcohol
(d)
m-chloro benzyl alcohol and potassium
m-chloro benzoate
(a)
(b)
A natural compound (X), C4H8O2, reduces Fehling’s
solution, liberates hydrogen when treated with
sodium metal and gives a positive iodoform test.
The structure of (X) is
(a)
CH3CHOHCH2CHO
(b)
HOCH2CH2CHO
(c)
CH3COCH2CHO
(d)
CH3COCH2CH2OH
(c)
(d)
none is correct
NaOH
 A

22.
A is :
(a)
19.
(b)
X and Y can be distinguished by :
(a)
silver-mirror test(b)
iodoform test
(c)
both
none
(d)
(c)
(d)
Einstein Classes,
none is correct
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 12
23.
24.
A carbonyl compound with molecular weight 86,
does not reduce Fehling’s solution but forms
crystalline bisulphite derivatives and gives
iodoform test. The possible compounds 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
In the Cannizzaro’s reaction given below,

2Ph  CHO OH

 PhCH 2 OH  PhCOO
the slowest step is
(a)
the attack of OH— at the carbonyl group
(b)
the transfer of hydride to the carbonyl
group
(c)
the abstraction of proton from the
carboxylic acid
(d)
the deprotonation of Ph–CH2OH
Einstein Classes,
ANSWERS (SINGLE CORRECT
CHOICE TYPE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
d
b
c
b
c
b
d
d
d
c
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
a
b
a
c
b
c
d
a
c
d
21.
22.
23.
24.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
a
a
b
b
CAK – 13
EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
4.
Structure of K formed is
Comprehension-1
H CrO
(a)
Cyclohexanol 2 4  A(C 6 H10O)
acetone
( 1) CH MgI
HA
 3

 B(C 7 H 14 O ) 
 C(C 7 H 12 )

heat
( 2 ) H 3O
(b)
(1 ) O 3
(1) Ag 2O , OH 
( 2 ) Zn , HOAc
( 2 ) H 3O
 D(C 7 H 12O 2 )    
 E(C 7 H 12O 3 )

1.
(c)
The structure of compound B formed is
(a)
(b)
(d)
5.
(c)
Thus L is
(d)
(a)
2.
The D formed in the above sequence is
(b)
(a)
(c)
(b)
(d)
(c)
(d)
3.
6.
HOOC(CH2)4COOH
CH3CH = CHCO2Et
Thus M is
(a)
CH3CH2CH2CO2Et
–
D on treatment with Ag 2 O, OH and upon
acidification forms
(b)
(a)
(c)
(d)
(b)
CH3CH2COOH
Comprehension-3
(c)
(d)
HOOC(CH2)4COOH
A
B

Comprehension-2

The following reaction sequence shows how the
carbon chain of an aldehyde may be lengthened by
two carbon atoms.
(1) BrCH CO Et , Zn
( i ) NaOH
 
 Product(s) ?

HA , heat
Ethanal   2 2  K (C 6 H 12 O 3 )  
( ii ) H
( 2 ) H 3O
H , Pt
( 1) DIBAL  H
L(C 6 H 10O 2 ) 2  M(C 6 H 12O 2 )    
7.
( 2 ) H 2O
butanal
8.
Einstein Classes,
The missing reagent A is
(a)
OsO4, NaHSO3 (b)
dilute KMnO4
(c)
any one of these (d)
O3/H2O2
The missing reagent B is
(a)
HIO4
(b)
(c)
any one of these (d)
NaIO4
None
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 14
9.
The product(s) formed is/are
8.
(a)
(c)
(b)
both of these
(d)
none
MULTIPLE CORRECT CHOICE TYPE
1.
2.
3.
4.
5.
6.
7.
A new carbon-carbon bond formation is possible
in
(a)
Cannizoro reaction
(b)
Friedal-craft reaction
(c)
Clemensen reduction
(d)
Reimen-Tiemann reaction
Which of the following undergoes Aldol
Condensation
(a)
Acetaldehyde
(b)
propionaldehyde
(c)
Benzaldehyde
(d)
Trideuteroacetaldehyde
9.
10.
(c)
2-Methyl propionaldehyde
(d)
2, 2-Dimethyl-propionaldehyde
On heating which of the following get
decarboxylated ?
(a)
Methylmalonic acid
(b)
succinic acid
(c)
2, 2-Dimethyl acetoacetic acid
(d)
Maliec acid
Cannizaro reaction will be given by
(a)
(b)
(c)
(d)
Which of the following is/are correct about an
acetal ?
(a)
it is a condensation product of two
molecules of aldehyde
(b)
it is the condensation product of two
molecules of ketones
(c)
it is the condensation product of one
molecular of aldehyde two molecules of
Alcohols
it is a diether
Which of the following on reduction with LiAlH4
with give ethyl alcohol ?
(a)
(CH3CO)2O
(b)
CH3COCl
(d)
(c)
CH3CO NH2
(d)
CH3COOC2H5
Assertion-Reason Type
Acetophenone is obtained when
(a)
Benzoylchloride treated with dimethyl
cadmium
(b)
Acetylchloride is treated with dimethyl
cadmium
(c)
Acetyl chloride is treated with benzene
in presence of anhydrous AlCl3
(d)
Benzoyl chloride is reduced with H2 in
presence of Lindar’s catalyst
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
Which of the following on oxidation with alk.
KMnO4 followed by acidification with HCl gives
benzoic acid
(a)
Methyl benzene (b)
Ethyl benzene
(c)
O-xylene
p-xylene
(d)
Which of the following statements about
Benzaldehyde is/are true ?
CCl3 CHO
1.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(C)
Statement-1 is True, Statement-2 is False
(D)
Statement-1 is False, Statement-2 is True
STATEMENT-1 : H2CO is a planar molecule.
STATEMENT-2 : In H2CO, C is sp2 hybridised
2.
STATEMENT-1 : H 2 + PhCH 2 CH 2 COCl
Pd / C
(a)
Reduces Tollen’s reagent
S
  Ph CH2CH2CHO
or quinoline
(b)
Undergoes Aldol condensation
above reaction is correct.
(c)
Undergoes cannizaro’s reaction
(d)
Does not form an addition compound
with NaHSO3
STATEMENT-2 : The reduction stops at the
aldehyde stage because the catalyst is poinsoned
by adding quinoline.
Base-catalysed Aldol condensation occurs with
(a)
propionaldehyde
(b)
Benzaldehyde
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 15
8.
3.
STATEMENT-1 :
H
SO


2
4
STATEMENT-2 : Carbonyl compounds cannot
form hydrogen bonding with H2O.
9.
STATEMENT-2 : In Pinacol-pinacolone
rearrangement formation of the cyclopbtyl C+ is
unlikely because the ring strain decreases.
STATEMENT-1 : Boiling points of carbonyl
compounds are higher than those of alkanes.
STATEMENT-2 : Boiling points of carbonyl
compound are higher because of hydrogenbonding.
The above reaction is pinacol-pinacolone
rearrangement.
4.
STATEMENT-1 : Aldehydes are more reactive
than carboxyl group.
10.
STATEMENT-1 : When C 6 H 6 react with
CO + HCl in the presence of CuCl/AlCl3, it form
C6H5CHO.
STATEMENT-1 : The reaction of aldehydes and
ketones with LiAlH4 and NaBH4 is uncleophilic
addition reaction.
STATEMENT-2 : Aldol condensation reaction is
not given by C6H5CHO.
STATEMENT-2 : The electrophilic aromatic
substitution take place in above reaction.
5.
STATEMENT-1 : Under wollff-kishner reduction
RCHO reduce to CH4.
STATEMENT-2 : Under Clemmenson reduction
pHCH2CHO reduce to pHCH2CH3
6.
STATEMENT-1 : H2CO is always oxidised in the
crossed cannizzaro reaction.
STATEMENT-2 : H 2CO is the most reactive
aldehyde, it exists in aqueous OH– solution mainly
as the conjugate base of its hydrate, H2C(OH)O–.
STATEMENT-1 : PCC oxidises 1 0 alcohol to
aldehyde and 20 alcohol to ketone.
7.
STATEMENT-2 : Cu dehydrates 30 alcohol to
alkene.
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
a
2.
b
3.
a
7.
c
8.
c
9.
c
4.
c
5.
c
6.
a
5.
a, b
6.
a, c
5.
D
6.
A
MULTIPLE CORRECT CHOICE TYPE
1.
b, d
2.
a, b, d 3.
a, b, d 4.
a, c
7.
a, c
8.
a, c
9.
b, c, d
10.
c, d
ASSERTION-REASON TYPE
1.
B
2.
A
3.
C
4.
A
7.
B
8.
C
9.
C
10.
B
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 16
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
Complete the following equations and write down
the names of the products :
A ketone (A) which undergoes haloform reaction
gives compound (B) on reduction. (B) on heating
with sulphuric acid gives compound (C), which
forms monozonide (D). (D) on hydrolysis in
presence of zinc dust gives only acetaldehyde.
Identify (A), (B) and (C). Write down the reactions
involved.
8.
Show, by equations, how acetaldehyde may be
converted into :
. KCN / H 2SO 4
1
  ?
(i)
=O
(ii)
i ) BD 3 ,THF
CH 3CH 2 C  CH (
  ?
2. LiAlH 4
( ii ) H 2 O 2 , OH
2.
7.
Identify the (A), (B), (C) and (D) in the following
reactions sequence :
(i)
CH 3  CH  CH  CHO NaBH
4 (
HCl
(i) EtOH; (ii) CH3COCl; (iii) CHI3; (iv) MeCOEt;
(v) CH3CO2Et; (vi) CH3CHOHCO2H; (vii) EtNH2;
(viii) CH3CHBrCHBrCO2H.
9.
Convert ClCH2CH2CHO into HOCH2CHOHCHO.
10.
Compound (A) and (B) on reaction in ether
medium and subsequent acidification and
oxidation give 2, 5-dimethyl 3-hexanone. What are
(A) and(B) ?
11.
Identify (A) to (E) as reactant, reagent, product or
name of the reaction in following :
KCN
(A) ZnCl
2 (B) 

(C)

H
(ii)
( alc.)
Br2
CH 3CH 2CH 2 Br KOH

(A) 
(
Hg 2
KOH ( alc.)
(i)
(B) followed
 by
NaNH
2 (C) H
(D)
2SO 4 ( dil.)
(iii)
Aldol condensation
(ii)
PCl5
Ag 2 O
CH 3CHO 
(A) 
(B)
Rosenmund’s reaction
(iii)
CH 3 MgBr
Pd
/ BaSO
4 (C)  
(D)

H 2O / H
3.
What happens when (Give equation only) Chloral
is heated with aqueous sodium hydroxide.
4.
How will you differentiate between
5.
6.
(C)
CH 3COCl  H 2 
(D)
NH 2 NH 2
H2
(i)
A)
3(CH 3 ) 2 C  O (
( B)
Acetaldehyde and Acetone.
( E ) C
 CH 3CH 2 CH 3 (F)
2 H 5 ONa
12.
Write the structural formula of the main organic
prudent formed when : the compound obtained by
the hydration of ethyne is treated with dilute
alkali.
13.
How can we convert PhCH = CHCOCH3 to
How would you bring the following conversions :
(a)
PhCH = CHCOOH
(a)
Acetone from acetaldehyde.
(b)
PhCH = CHCHOHCH3
(b)
Methanal to ethanal (not more than
3 steps)
(c)
PhCH2CH2COCH3
(d)
PhCH = CHCH2CH3
(c)
Acetone from methane.
(e)
Ph(CH2)3CH3
(d)
2-Butanone from ethyl alcohol.
(e)
3-Hexanone from n-propyl alcohol.
14.
An organic compound (A) contains 40% carbon
and 6.7% hydrogen. Its V.D. is 15. On reaction with
a conc. solution of KOH, it gives two compounds
(B) and (C). When (B) is oxidised, the original compound (A) is obtained. (C) give effervecence with
NaHCO3. Write the structures of A, B, C and D and
explain the reactions.
15.
An organic compound (A), C4H9Cl on reacting with
aqueous KOH gives (B) and on reaction alc. KOH
gives (C) which is also formed on passing the
vapours of (B) over heated copper. The compound
Arrange the following :
(i)
CH3CHO, CH3COCH3, HCHO,
C2H5COCH3 in decreasing order of
nucleophilic addition.
(ii)
CH3COCH2CHO,CH3COCH3,
CH3CHO, CH3COCH2COCH3 in
increasing order of expected enol content.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 17
(C) readily decolourizes bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to give (F) and the
compound (E) reacts with NaOH to give an alcohol
(G) and sod. salt (H) of an acid. (D) can also be
prepared from propyne on treatment with water in
presence of Hg++ and H2SO4. Identify (A) to (H) with
proper reasoning.
16.
How would you convert :
17.
An organic compound A, C8H6, on treatment with
dilute H2SO4 containing HgSO4 gives a compound
B which can also be obtained from the reaction of
benzene with an acid chloride in the presence of
anhydrous AlCl3. The compound B when treated
with iodine and aq. KOH, yields C and a yellow
compound D. Identify A, B, C and D with
justification. Show how C and D is formed.
18.
A certain hydrocarbon A was found to contain
85.7% C and 14.3 % H. This compound consumes
1 molar equivalent of hydrogen to give a saturated
hydrocarbon B. 1.0 g of hydrocarbon A just
decolourised 38.05 of a 5 percent s o l u t i o n
(by weight) of Br 2 in CCl 4 . Compound A on
oxidation with concentrated KMnO 4 gave
compound C (molecular formulae C4H 8O) and
acetic acid. Compound C could easily be prepared
by action of acidic aqueous HgSO4 on 2-Butyne.
Determine the molecular formula of A and deduce
the structure of A, B and C
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
How would you bring the following conversions :
(a)
(b)
(ii)
Ethanal to 2-hydroxy-3-butenoic acid.
5.
Compound (A), C10H12O gives off hydrogen on
treatment with sodium metal and also decolourises
Br2 in CCl4 to give (B), C10H12OBr2. (A) on treatment with I2 in NaOH gives iodoform and an acid
(C) after acidification. Give structures of (A) to (C)
and also of all the geometrical and optical isomers
of (A).
6.
A organic compound (A), C9H12O was subjected in
a series of test in the laboratory. It was found that
this compound
Propanal to
2.
Convert : (i) n-butanol into 2-ethylhexan-1-ol;
(ii) bromocyclohexane into cyclopentancarboxylic
acid.
3.
Suggest what compounds the letters stand for in
the following equations :
(i)
i ) NaNH 2
MeCOMe  CH  CH (
 B
( ii ) A
(ii)
(i)
Rotates the plane of polarised light.
(ii)
Evolves hydrogen with sodium
(iii)
Reacts with NaOH and I2 to produce a
pale yellow solid compound.
(iv)
Does not react with Br2/CCl4.
(v)
Reacts with hot KMnO4 to form
compound (B) C7H6O2 which can also be
synthesised by the reaction of benzene
and carbonyl chloride followed by
hydrolysis.
(vi)
looses optical activity as a result of
formation of compound (C) on being
heated with HI and P.
(vii)
Reacts with Lucas reagent in about 5 min.
Give structures of (A) to (C) with proper
reasoning and draw fischer projections
for (A). Give reactions for the steps
wherever possible.
NaNH 2
PhCOMe  ClCH 2 CO 2 Et  C
(iii)
 BuOK
 ClCH 2 CO 2 Et t

 D
t  BuOH
4.
Complete the following equations and give reasons
for your answer.
(i)

H
Me 2 C(OH)CH 2OH 
?
Einstein Classes,

H
PhMeC(OH)C(OH)MePh 
?
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CAK – 18
7.
8.
A neutral organic compound (A) on analysis gave
C = 52.17%; H = 13.04%. On mild oxidation (A)
gave (B) which on treatment with dilute NaOH gave
another neutral compound (C) with double the
molecular mass. Compound (C) readily loses
water on heating to form (D) which on oxidation
gave CH3CH = CHCOOH (crotonic acid). Identify
(A), (B), (C) and (D) and explain the reactions.
(a)
11.
Identify A, B, in the following set up

O3
A 
B OH



(b)
9.
10.
An alkyne with 5 carbon atoms per
molecule when passed through dilute
sulphuric acid containing mercuric
sulphate gives a compound which forms
an oxime, but has no effect on Fehling’s
solution. The compound on oxidation
gives dimethyl acetic acid. It reacts with
sodamide to forms a hydrocarbon. What
is the structure of the alkyne ?
When 0.0088 g of compound (A) was dissolved in
0.50 g. of camphor, the melting point of camphor
was lowered by 80C. Analysis of (A) gave 68.18% C
and 13.16% H. Compound (A) showed the
following reactions : (i) It reacted with acetyl
chloride and evolved hydrogen with sodium, (ii)
When reacted with HCl + ZnCl2, a dense oily layer
separated out immediately. Compound (A) was
passed over Al2O3 at 3500C to give compound (B).
(B) on ozonolysis followed by hydrolysis gave two
neutral compounds (C) and (D) which gave
positive tests with carboyl reagents but only (C)
gave a positive test with Fehling solution. Identify
(A), (B), (C) and (D) with proper reasoning. Kf for
camphor = 40 mol–1 kg–1.
Suggest the appropriate structures for the missing
compound. (The number of carbon atoms remain
the same throughout the reactions)
Identify A to G.
12.
Wolff-Kishner reduction of the compound shown
gave compound A. Treatment of compound A with
m-chloroperoxy benzoic acid gave B which on
reduction with LiAlH 4 gave C. Oxidation of
compound C with chromic acid gave compound D
(C9H14O). Identify A, B, C and D.
13.
An optically active alkyne A contains 89.52% C and
10.48 % H. After hydrogenation over a Pd catalyst
it is converted to 1-Methyl-4-propyl cyclohexane.
When compound A reacts with CH3MgBr no gas
is liberated. Hydrogenation of A over Lindlar
catalyst, followed by ozonylysis and reaction with
KMnO4 gives product B. Product B reacts with
I2/NaOH and gives a yellow precipitate, which is
filtered off. Acidification of the filterate gives an
optically active product C. Give the structures of
A, B and C and account for all observations.
14.
Give the structure for compounds A – F :
. CH 3 MgI
/
2 CrO 4
C 6 H12 O H
 B 1

 C HA



A
acetone
2. H 3 O

.O3
,OH
D 1
 E Ag
2O

F

2. Zn / CH 3COOH
15.
2. H 3 O
C6H12O 6 can be glucose or fructose. To decide
nature, one can carry out following steps :

dil
. KMnO
4  A HIO

4  B HO

 C

/ HI
3O
C6 H12O 6 HCN

 H
 P
 ?
What can be the end product starting with glucose
and fructose ?
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111