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[Eduwaves360]OC Oxidation Reduction of Organic Compounds E

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JEE (MAIN+ADVANCED)
OXIDATION & REDUCTION OF
ORGANIC COMPOUNDS
CONTENT
S.No
Pages
1.
Oxidation of alkenes, Alcohols
& Carbonyl compounds
02 – 07
2.
Exercise-1
08 – 11
3.
Reducing agents and their role
12 – 15
4.
Exercise-2
16 – 18
5.
Exercise-3
19 – 21
6.
Answer Key
22 – 27
OXIDATION & REDUCTION OF ORGANIC COMPOUND
OXIDATION OF ALKENES, ALCOHOLS & CARBONYL COMPOUNDS
(I)
OXIDATION OF ALKENES
OsO4
R–CH=CR2
R–CH—CR2
H2O
OH OH
Cold dil.
alkaline
KMnO4
R–CH—CR2
OH OH
*
Cold dil. alkaline KMnO4 is called as Bayer’s reagent.
*
Overall syn addition
*
Given by alkenes & alkynes
*
Benzene & Cyclopropane can not give this reaction.
If we use acidic KMnO4 or warm KMnO4 or too concentrated KMnO4 the oxidative cleavage of
Glycol occurs resulting in mixture of Carboxylic acids & Ketones.

H ,KMnO
R–CH = CR2   4  RCOOH + R2C = O

Hot acidic KMnO4, Hot acidic K2Cr2O7 & hot acidic NaIO4 gives same result with alkene. The effect
is similar to that of oxidative ozonolysis on alkenes.
Preilschaive reaction :
Epoxidation of alkenes is reaction of alkenes with peroxyacids.


O
O
||
||
CH2=CH2 + CH3  C  O  O  H  CH2–CH2 + CH 3  C  OH
O
With the decrease in nucleophilicity of double bond, rate of reaction decreases.
With the decrease in e withdrawing substituents in leaving group, rate decreases.
R
CH
CH2
H
O
+
R
dil.H2SO4
CH
R
H2O–CH
+
CH2
+
O–H
R
+
–H HO–CH
CH2–OH
CH2–OH
H2O
RCO3H
RCO3H
H 3O
HCO3H
O
OH
R
CH +
OH
CH2
OH
1 O Ag
2 2 
R
CH
CH2
O
OH
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Page # 2
OXIDATION & REDUCTION OF ORGANIC COMPOUND
(II)
OXIDATION OFALCOHOLS
Oxidising agents
Different oxidisingagentsare used tooxidisealcohols in correspondingcarbonyl compounds and carboxylic
acids.
e.g.
(I)
(II)
(III)
(IV)
mild oxidising
 R  C  H (Aldehyde)
R  CH 2  OH 
agent
||
1alcohol
O
O
OH
||
|
mild oxidising
R  CH  R '   R  C  R ' (Ketone)
2alcohol
agent
strong oxidising

 R  C  OH
R  CH 2  OH  
agent
||
1alcohol
O
CH 3
|
Cu 300C
CH 3  C  OH  
|
CH3
CH 3
CH2
C
Dehydration takes place.
CH3
(1)
Cu / 300°C (or Red hot Cu tube)
It oxidises only 1° alcohol in aldehyde and 2° alcohol into ketones. It does not oxidises 3°alcohol.
Dehydration of 3°alcohol takes place and alkene is formed.
(2)
H /KMnO4,  (Strong oxidising agent)

(i) It oxidises 1° alcohol into carboxylic acid
(ii) It oxidises 2° alcohol into ketone
(iii) It does not affect 3° alcohol
(iv) It affects carbon-carbon multiple bond and gives carboxylic acid
(3)
H/K2Cr2O7,  (Strong oxidising agent)
(i) It oxidises 1° alcohol into carboxylic acid
(ii) It oxidises 2° alcohol into ketone
(iii) It does not affect 3° alcohol
(iv) It affects carbon-carbon multiple bond and give carboxylic acid
(4)
PCC (Pyridinium chloro chromate)

N CrO3 Cl or
N
CrO3 + HCl
H
PCC oxidises 1° alcohol into aldehyde, further oxidation is not possible.
PCC oxidises 2° alcohol into ketone. It does not oxidises multiple bond and 3° alcohol. It oxidises allylic
primary alcohol to corresponding aldehyde further oxidation is not possible.
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Page # 3
OXIDATION & REDUCTION OF ORGANIC COMPOUND
(5)
Collin’s reagent
(
(2 mol) + CrO3 + CH2Cl2)
N
(i) It oxidises 1° alcohol into aldehyde
(ii) It oxidises 2° alcohol into ketone
(iii) Multiple bond is not affected
(iv) 3° alcohol is not affected
(6)
Jones reagent (H2CrO4 in Anhydrous acetone) or CrO3 + H2SO4 in acetone.
Sufficiently mild so that it oxidises alcohols without oxidising or rearranging double bonds (8 or 9)
(i) It oxidised 1° alcohol into carboxylic acid (ii) It oxidised 2° alcohol into ketone
(iii) Multiple bond is not affected
(iv) 3° alcohol is not affected
(7)
TsCl + DMSO + NaHCO3
TsCl
 RCH2OTs

 RCHO
RCH2OH
s DMSO
NaHCO3
TsCl
 R2CH–OTs

 R2CO
R2CHOH
s DMSO
NaHCO3
TsCl
DMSO

 R3C–OTs
R3COH
s 
NaHCO3
(i)
(ii)
(iii)
(iv)
(8)
×
It oxidised 1° alcohol into aldehyde
It oxidised 2° alcohol into ketone
Multiple bond is not affected
3° alcohol is not affected
MnO2-Oxidises onlyallylic or benzylic–OH.i.e.
1°Allylic or benzylic OH MnO
2  Aldehyde
2°Allylic or benzylic OH MnO
2  Ketone
No effect on 3° ROH and on Carbon-carbon multiple bond.
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Page # 4
OXIDATION & REDUCTION OF ORGANIC COMPOUND
(9)
Periodic cleavage
Periodic cleavage : It is done by periodic acid (H5IO6 or HIO4. 2H2O). Characteristics property of
periodic acid is the oxidative cleavage of the bonds with adjacent oxidizable groups such as 1,2-diols,
a-hydroxy carbonyl compounds, 1,2-diketones.
R – CHOH
R – CHOH
HO
OH
OH
OH
I
HO
O

O
R – CH – O
R – CH – O
OH
I
O

O
2R – CHO + IO 3¯ + H2O
(10)
Openaur oxidation
CH3
O
Al O–C–CH3
CH3 3
R – CH – R
OH
O
R–C–R
CH3 –C – CH3

R2CH–OH + Al(OCMe3)3Me3COH + Al(OCHR2 )3
(R2CH–O)2 Al–O
CR2
O
H
3Me2C
3R2C=O + (Me2C–O)3 Al
Oxidation of alcohol with aluminium tertiary
butoxide is Openaur oxidation.
OH
O
Al (OCMe3)3
acetone
(11)
NBS
It oxidises 1° alcohol into aldehyde.
It oxidises 2° alcohol into ketone.
3° alcohol is not affected
In case of alkene allylic position is brominated.
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Page # 5
OXIDATION & REDUCTION OF ORGANIC COMPOUND
(III)
OXIDATION OF CARBONYL COMPOUNDS.
1.

RCHO + [Ag(NH3)2]OH 
RCOOH + NH3 + Ag Ї

Aldehyde acts as reducing agent, they can reduce mild oxidizing agents like Tollen’s Reagent. Tollen’s
test Gentle Heating for 20 to 25 mins.
2.
Fehling’s Solutions
Fehling’sA
aq. CuSO4
COONa
H
OH
Fehling’s B
OH
H
Alk. solution of Roschelye
salt (sodium potassium tartrate)
COOK
It acts a carrier for Cu2+ as it make reversible complex with Cu2+
This test is also used is Blood and Urine test.
H 2O
 RCOOH  Cu 
RCHO + Cu2+ 



RCOO – Cu 2 O ( red ppt.)
3.
Benedict’s solution
Sodium Citrate + NaOH + NaHCO3 + CuSO4
H 2O
 RCOOH  Cu 
RCHO + Cu2+ 



RCOO – Cu 2 O ( red ppt.)
4.
RCHO + HgCl2 + H2O  RCOOH + 2HCl + Hg2Cl2

RCHO + Hg2Cl2 + H2O ® RCOOH + 2HCl + 2Hg  grey ppt.
5.
Schiff’s Reagent
Schiff’s Reagent is aq. solution of following base decolourised by passing SO2.
Aldehyde restore pink colour of Schiff’s reagent.
NH2
NH 2
SO2
C
+ NH
Colourless
RCHO
solution

(Schiff’s Reagent)
Pink colour
Cl
2
p-Rosaniline Hydrochloride
Magenta colour (Fuschin)
Ketons are not easy to oxidize so they do not give these 5 tests. These five tests can be used to distinguish
aldehyde and ketones. Both gives 2,4 DNP test
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Page # 6
OXIDATION & REDUCTION OF ORGANIC COMPOUND
KETONES ARE DIFFICULT TO OXIDIZE
Ketones can be oxidized from their enolic form at high temperature with very strong oxidizing agent.
Oxidation of ketones is sometimes governed by Popoff’s rule. According to this rule carbonyl group
remains with the smaller alkyl group. More electron rich alkene will be easy to oxidized.
O
Me – C – Me
[O]

MeCOOH + CO 2 + H 2O
Oxidation by using SeO2
SeO2 is a selective oxidizing agent with converts –CH2– group adjacent to carbonyl group into carbonyl
group. The reagent, in general, oxidises active methylene and methyl groups to ketonic and aldehydic
groups respectively.
O
O
O O
O
|
|
|
|
|
|
||
||
SeO 2
SeO2
  C  CHO
 CH 2  C    C  C  ;  C  CH 3 
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Page # 7
OXIDATION & REDUCTION OF ORGANIC COMPOUND
EXERCISE-1
Q.1

H / KMnO
CH2 = CH2   4 
(i)


/ KMnO 4
H 

(iii)


/ KMnO4
H 

(v)

/ KMnO 4
H 

(iv)
/ KMnO 4
H 

(viii)


/ KMnO 4
H 

(ix)
H / KMnO
CH3–CH= CH2   4 




/ KMnO4
H 

(vi)

(vii)

(ii)


/ KMnO 4
H 



/ KMnO 4
H 

(x)



/ KMnO4
H 

(xi)


H / KMnO
C10H10   4  HOOC  CH 2  CH 2  CH 2  COOH

|
CH 2  COOH
(xii)
Q.2
A to F alkenes with minimum possible carbon.
(i)
H / KMnO
A   4  MeCOOH as the only product
(ii)
H / KMnO
B   4 
(iii)
H / KMnO
C   4  MeCH2COOH as the only organic product



O



O

H / KMnO
D   4 
(iv)

O
(v)
O
||
H / KMnO
E   4  HOOC  C  C  C  C  C  C  C  C  COOH

||
O
(vi)
H / KMnO
F   4  acetone + ethanoic acid
Q.3
(i)



1% alkaline
KMnO4
(A)
(ii)
mCPBA
mCPBA\hydrolysis
(B)
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Page # 8
OXIDATION & REDUCTION OF ORGANIC COMPOUND
Me
(1) mCPBA
(2) hydrolysis
(iii)
C=C
(iv)
H
H
Me
H
Me
Me
H
(1) mCPBA
C=C
(v)
Me (2) hydrolysis
H
Q.4
(i)
Me
C=C
(vi)
H
Ph
(1) mCPBA
(2) hydrolysis
1
Ag2O or 2Ag + O2
2

KMnO / OH ¯, 
 ?
CH3– CH2 – CH2 – OH  4 

or KMnO 4 / H ,

K2Cr2O 7 / H , 
?
or conc. HNO 3, 
(ii)
OH
|
KMnO , H 
?
CH3  CH  CH 2  CH 3  4

or K 2 Cr2O 7 , H
OH
HO
Cu/300°C
+
P
H /KMnO 4/ 
(iii)
OH
HO
Q.5
Q
PCC or Collin's reagent
MnO2/
R
S
(i)
CH2 = CH – (CH2)3 – CH2 – OH PCC


(ii)
C6H5 – CH = CH – CH2 –OH PCC


(iii)
OH
|
CH 3  CH  CH 2  CH 2  CH 2  OH PCC

 (A)
OH
(iv)
PCC


(v)
CH2 = CH – CH2–OH MnO
2  ?
CH2OH
OH
(vi)
CH3O
MnO
CH–CH2–CH2–OH 2  ?
Acetone
CH3O
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Page # 9
OXIDATION & REDUCTION OF ORGANIC COMPOUND
(vii)
Q.6
Q.7
OH
|
(viii) C6 H 5  CH  CH 3 MnO
2  ?
CH 3
|
MnO 2
 ?
CH  C  CH  C  CH 2  OH Acetone
CCl 4
(ix)
TsCl  NaHCO3
C 6 H 5  CH  CH  CH 2  CH  CH 2  OH DMSO

   ?
|
CH 3
(i)
HO
(ii)
OH
|
Aluminium tert butoxide
  
CH 2  CH  CH 2  CH  CH 3   
Acetone
tert butoxide
Aluminium

  ?
Acetone
Which one of the following alcohols are oxidised by MnO2?
(A) C6H5 – CH2 – CH2–OH
OH
|
(B) CH 2  CH  CH 2  CH  CH 3
OH
|
(C) CH3  CH  CH  CH  CH 3
(D) CH3–CH2 – CH2 –OH
HIO
Me  CH  CH 2  OH 4 

|
OH
HIO
(ii) Me 2C — CH — Et 4 

|
|
OH OH
Q.8
(i)
OH
HIO
4 
(iii)
OH

(v)
CH 2 — CH  CH 2  CH 3 HIO
4 

|
|
OH
OH
(vii)
CH 2 — CH  CH  CH 2 HIO
4 

|
|
|
|
OH
OH OH OH
(ix)
Me  C  C  Me HIO
4 

|| ||
O O
Q.9
Which will give the Tollen test.
O
(i)
OH
H
O
(ii)
OMe
 CH 2  OH HIO 4
(iv) HO  CH 2  CH 2  CH

|

OH
(vi)
CH 2 — CH  CH  CH 3 HIO
4 

|
|
|
OH
OH OH
(viii)
Me  C  CH  Me HIO
4 

|| |
O OH
O OH
|| |
(iii) R  C  CH 2
HO
(iv)
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HO
Page # 10
OXIDATION & REDUCTION OF ORGANIC COMPOUND
O
Q.10
(a)
O
(b)
(d)
Q.11
[O]

Me2CH–C–Me
O
(c)
[O]
H3C – CH 2 – C – CH 3 
[O]

Me3C–C–Me
O [O]

(a)
CH3–CHO
(b)
Me2CO
SeO2

SeO2

Me – C – C – H
O
(c)
(d)
H3C – CH2 – C –C H3
SeO2
P1 mCPBA

 P2 LAH

 P3

O SeO
2

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Page # 11
OXIDATION & REDUCTION OF ORGANIC COMPOUND
Reducing agents and their role
Group
Product
–CHO
–CH2OH
+
>C=O
>CH–OH
–CO2H
LiAlH4
+AlCl3
–
+
+
+
+
+
–
+
+
+
+
–CH2OH
+
–
–
+
+
+
–CO2R
–CH2OH
+
–
–
+
+
+
–COCl
–CH2OH
+
+*
+
+
–
+
–CONH2
–CH2NH2
+
–
–
+
+
+
(RCO)2O
RCH2OH
+
–
–
+
+
+
–CN
–CH2NH2
+
–
–
+
+
+
>C=NOH –CH2NH 2
+
–
–
+
–
+
>C=C<
>CH–CH<
–
–
–
–
+
+
–CC–
–CH=CH–
–
–
–
+
+
+
+
–
–
+
–
+
RH
LiAlH(OCMe3)3
in THF
B2H6
in THF
H2+
** catalyst
NaBH4
in EtOH
1° RX
LAH in
ether
* Product is RCHO
** Catalyst : Ni / Pd / Pt / Ru
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Page # 12
OXIDATION & REDUCTION OF ORGANIC COMPOUND
LiAlH4 as a reducing agent :
From LAH
H
(i)
(1) LiAlH
C = O   4 
( 2 ) H 2O
C–O
H
From Solvent
Mechanism
H
Li
C=O
(ii)
+
H–Al–H 
H
H
H

EtOH
C — OLi
C — OH
(1) LiAlH
R  C  NH 2   4  R – CH2 – NH2
( 2 ) H 2O
||
O
Mechanism
H
+
Li
R–C–NH2
O
H–Al–H
H
–H2O, –AlH3

R–C–NH Li
O

R–C= NH
O¯Li
+
H–Al–H
H
H
R–C–O–AlH2
+
NH Li
R–CH=NH
H
+
Li

R–C–O – Al–H Li
NH
H

H–Al–H 
H

R–CH2–NH Li
(iii)
H
 H–OH
(1) LAH
 R–CH2–OH + R'–OH
R  C  OR '  
||
( 2) H3O
O
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Page # 13
OXIDATION & REDUCTION OF ORGANIC COMPOUND
Mechanism
H
Li
+
H–Al–H 
H
H
–AlH3
R–C–OR'
R–C–OR'
O
O–Li
R–C–H + R'–OLi
O
+
H–OH
R'–OH
LAH
H2O
R–CH2–OH
(iv)
1) LAH
R – C  N (

 R–CH2–NH2
( 2 ) H 2O
Mechanism
H
R – C N +
+
H
H–Al–H 
Li


R–C=N Li + AlH2
H
H
R–CH2–NH2

R–CH2–N–Li
AlH2
(v)
1) LAH
R  C  OH (

 R–CH2–OH
( 2 ) H 2O
||
O
Mechanism
H
Li
R–C–OH
+
H
H–Al–H
H
–H2
O
R–C–O¯Li
+
AlH3

R–C–O–Al–H Li
O

H
O
H
R–CH2–OH
(1) LiAlH4
(2) H2 O
R–C–H
O
+
Li O¯AlH 2
R–C–OAlH2
O Li

BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
Page # 14
OXIDATION & REDUCTION OF ORGANIC COMPOUND
Wolf Kischner reduction :
 NH 2 / OH ¯
C = O NH
 2
  
CH2 + N2
Mechansim
–H2O
C = O + NH2 – NH2
C = N–NH2

–H2O OH
H
C – N=N–H
HO–H

C = N–NH

–H2O OH
H

C–N=N
H–OH
CH2 + N2
Clemension reduction
Zn ( Hg )


C=O 
HCl( conc.)
CH2
Mechanism
C=O
H

C = O–H
Zn Zn + 2e
2+

C – OH
H
–H2O
CH
CH–OH2
H


CH – OH

2e
CH
H

CH2
BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
Page # 15
OXIDATION & REDUCTION OF ORGANIC COMPOUND
EXERCISE-2
Q.1
How many alkene on catalytic reduction give normal butane as product.
(i)
(iii)
(v)
Q.2
2 / Pt
(A) H
 n-butane
2 / Pt
(C) H
 Neo-pentane
(iv)
2 / Pt
(B) H
 Iso-pentane
2 / Pt
(D) H
 Cyclopentane
H2
(E) 
Pt
Give the expected major product for each reaction, including stereochemistry where applicable.
(a)
H2
CH3–CH2–CH=CH2 
P1
Pt
H 3C
(c)
(b)
H
C=C
H
CH3
 P
D2
Pt
P2
Me
D2
(e)
Q.3
(ii)
Ni
3
H2/Pt
excess
P5
H2/Pt
1 eq.
P6
(d)
(i)
H / Pt
CH2 = CH–CH2 – CH = O 2  P1
(ii)
H 2 / Pt
 P2
CH2=CH–CH2–CN 
Me
Ni
/ H2  P4
excess
excess
Q.4
Identify the product?
(i)
NaBH4

  Me–CHO LiAlH

4 
(ii)
LiAlH 4
NaBH4


  Me2CO  
(iii)
NaBH4

  Me–COCl LiAlH

4 
(iv)
LiAlH 4
NaBH4


  Me–COOEt  
(v)
LiAlH 4
NaBH4


  Me–COOH  
(vi)
NaBH4

  Me–COOMe LiAlH

4 
(vii)
NaBH4

  Me–CONH2 LiAlH

4 
(viii)
LiAlH 4
NaBH4


  Me–CONH–Me  
(ix)
(x)
LiAlH 4
NaBH4


  Me–CH=NH  
(xi)
NaBH4

  Me–CONMe2 LiAlH

4 
NaBH4

  CH3–CH = CH2 LiAlH

4 
Q.5
Give product in following reactions.
(i)
NaBH4


LiAlH 4

O  
(ii)
NaBH4

  H–N
(iii)
NaBH4


LiAlH 4

O  
(iv)
NaBH4


O LiAlH

4 
(vi)
NaBH4


(v)
NaBH4


N
O
N
LiAlH 4

O  
LiAlH

4 
H
O
LiAlH 4

O  
O
BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
Page # 16
OXIDATION & REDUCTION OF ORGANIC COMPOUND
LiAlH

4 
(vii)
NaBH4

  Me–CO–N
(ix)
LiAlH 4
NaBH4


  Ph–CH=CH–CHO  
Q.6
(viii)
LiAlH 4
NaBH4


  CH2=CH–CHO  
(x)
NaBH4


LiAlH

4 
O
Give product in following reactions.
O
O
P  HI
H Re
d

(a)
P  HI
Re
d

(b)
OH
CHO
COCH3
(c)
CH2OH
HO
Q.7
P  HI
Re
d

Give product in following reactions.
(i)
N
(iii)
O
(ii)
LAH


H3CCOO
LAH
CH3– CH – CH2
O
(iv)
COOCH3
LiAlH

4  (A) + (B) + (C)
LAH
AlCl3
O
(v)
Q.8
O
Give product in following reactions.
CHO
(i)
NaBH4
O
COOEt
O
(ii)
LAH
 (A)NaOI
 (B)
C—OEt
BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
Page # 17
OXIDATION & REDUCTION OF ORGANIC COMPOUND
(iii)
LiAlH4, D 2O
LiAlH4
H2 O
Q.9
(a)
Zn ( Hg )
.
 (A) NBS


 (D)

 (B) alc
(C) HCl
O=C—CH 3  
HCl
KOH
O
Zn
( Hg

) 
(b)
(c)
HCl, 
O
Zn
( Hg

) 
(d)
HCl, H 2O
O
Q.10
O
||
Zn ( Hg )


Ph  C  CH 3 
HCl
O
(e)
i ) H 2 NNH 2
(


(ii ) KOH , heat
N2H4
KOH, heat
O
(f)
R 2O 2
(excess)
O
Zn(Hg)
HCl
(excess)
Suggest appropriate reagents for following conversion.
OH
A
O
O
B
OH
C
BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
Page # 18
OXIDATION & REDUCTION OF ORGANIC COMPOUND
EXERCISE-3
OBJECTIVE
Q.1
Which of the following will decolourise alkaline KMnO4 solution?
(A) C3H8
(B) CH4
(C) CCl4
[JEE 1980]
(D) C2H4
Q.2
The reagent with which both acetaldehyde and acetone react easily is
(A) Tollen's reagent
(B) Schiff's reagent
(C) Grignard reagent
[JEE 1981]
(D) Fehling reagent
Q.3
When acetaldehyde is treated with Fehling's solution, it gives a precipitate of
[JEE 1982]
(A) Cu
(B) CuO
(C) Cu2O
(D) Cu + Cu2O + CuO
Q.4
Baeyer's reagent is
(A) alkaline permanganate solution
(C) neutral permanganate solution
[JEE 1984]
(B) acidified permanganate solution
(D) aqueous bromine solution
Q.5
Hydrogenation of benzoyl chloride in the presence of Pd on BaSO4 gives
(A) benzyl alcohol
(B) benzaldehyde
(C) benzoic acid
(D) phenol
[JEE 1992]
Q.6
The appropriate reagent for the following transformation:
[JEE 2000]
O
CH2CH3
CH3
HO
(A) Zn(Hg), HCl
HO
(B) NH2NH2, OH¯
(C) H2 / Ni
(D) NaBH4
Q.7
Statement-1: Dimethyl sulphide is commonly used for the reduction of an ozonide of an alkene to get
the carbonyl compound.
[JEE 2001]
Statement-2: It reduces theozonide giving water soluble dimethyl sulpoxide and excess of it evaporates.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Q.8
1-propanol and 2-propanol can be best distinguished by
(A) Oxidation with alkaline KMnO4 followed by reaction with Fehling solution.
(B) Oxidation with acidic dichromate followed by reaction with Fehling solution
(C) Oxidation by heating with copper followed by reaction with Fehling solution
(D) Oxidation with concentrated H2SO4 followed by reaction with Fehling solution.
[JEE 2001]
Q.9
Butan-2-one can be converted to propanoic acid by which of the following?
(A) NaOH, NaI / H+
(B) Fehling solution
(C) NaOH, I2 / H+
(D) Tollen's reagent
[JEE 2006]
BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
Page # 19
OXIDATION & REDUCTION OF ORGANIC COMPOUND
COMPREHENSION :
In the following sequence, product I, J and L are formed. K represents a reagent.
( i ) Mg / Ether
Cl
( ii ) CO
K
i ) NaBH 4
 J 
Hex-3-ynal (
  I  2
H3C

( ii ) PBr3
Q.10
( iii ) H 3O
[JEE 2008]
2
 H

 L
Pb / BaSO 4 quinoline
O
The structure of the product I is
(A) H C
3
(B) H3C
Br
Br
(C) H C
3
Q.11
(D)
Br
H3C
Br
The structures of compound J and K, respectively, are
(A)
(C)
H3C
COOH and SOCl2
H3 C
(B) H C
3
COOH
& SOCl2 (D) H3C
and SO2Cl2
COOH and CH3SO2Cl
COOH
Q.12
The structure of product L is
(A)
H3 C
CHO
(B) H C
3
CHO
(D) H C
3
CHO
H3 C
(C)
CHO
Q.13
The most suitable reagent for the conversion of R–CH2 – OH  R – CHO is :[JEE Main 2014]
(A) K2Cr2O7
(B) CrO3
(C) PCC (Pyridinium Chlorochromate)
(D) KMnO4
Q.14
Reagent(s) which can be used to bring about the following transformation is(are) [JEEAdvance 2016]
O
C
O
O
O
C
OH
H
COOH
(A) LiAlH4 in (C2H5)2O
(C) NaBH4 in C2H5OH
O
COOH
(B) BH3 in THF
(D) Raney Ni/H2 in THF
BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
Page # 20
OXIDATION & REDUCTION OF ORGANIC COMPOUND
SUBJECTIVE
Q.1
One mole of a hydrocarbon A reacts with 1 mol of bromine giving a dibromo compound, C5H10Br2.
CompoundAon treatment with cold dilute alkaline potassium permanganate solution forms a compound,
C5H12O2. On ozonolysis,Agives equimolar quantities of propanone and ethanal. Deduce the structural
formula ofA.
[JEE 1981]
Q.2
Awhite precipitate was formed slowlywhen silver nitrate was added to compoundAwith the molecular
formula C6H13Cl. CompoundAon treatment with hot alcoholic potassium hydroxide gave a mixture of
two isomeric alkenes B and C, having the formula C6H12. The mixture of b and C, on ozonolysis,
furnished four compounds.
[JEE 1986]
(a) CH3CHO
(b) C2H5CHO
(c) CH3COCH3
(d) CH 3  CH  CHO
|
CH 3
What are the structures of A, B and C?
Q.3
Write the balanced chemical equations for the following:
[JEE 1991]
"Ethylene glycol is obtained by the reaction of ethylene with potassium permanganate".
Q.4
When gas A is passed through dry KOH at low temperature, a deep-red coloured compound B and a
gas C are obtained. Gas A, on reaction with but-2-ene, followed by treatment with Zn/H2O yields
acetaldehyde. IdentifyA, B and C.
[JEE 1994]
Q.5
An organic compound E(C5H8) on hydrogenation gives compound F(C5H12). Compound E on ozonolysis
gives formaldehyde and 2-ketopropanal. Deduce the structure of compound E.
[JEE 1995]
Q.6
HydrocarbonAadds 1 mol of hydrogen in the presence of a platnium catalyst to form n-hexane. When
A is oxidised vigorouslywith KMnO4, a single carboxylic acid, containing three carbon atoms, is isolated.
Give the structure ofAand explain.
[JEE 1997]
Q.7
Monomer Aof a polymer on ozonolysis yields 2 mol of HCHO and 1 mol of CH3COCHO.
(a)
Deduce the structure ofA.
[JEE 2005]
(b)
Write the structure of all cis forms of polymer of compoundA.
BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
Page # 21
OXIDATION & REDUCTION OF ORGANIC COMPOUND
ANSWER KEY
EXERCISE - 1
Q.1
(i)
CO 2
(ii)
O + CO2 + HOOC – CH2 – CH2 – CH2–COOH
(iii)
(iv)
O + CH3–COOH
COOH
COOH
(v)
(vi)
O
(vii)
(xi)
Q.2
O
COOH
(viii)
O
(ix)
CH3–COOH + CO2
COOH CH 2  COOH
|
+ |
COOH CH 2  COOH
O
COOH
HOOC
2CH3–COOH
CH 2  COOH
+ |
O CH 2  COOH
O
(x)
O
(xii)
CH  C  CH 2  CH  CH 2  C  CH
|
CH 2
|
C  CH
(i)
Me–CH=CH–CH3
(ii)
(iii)
Me–CH2–CH=CH2
(iv)
(v)
(vi)
OH
Q.3
OH
OH
(i) (A)
O
(ii)
(B)
OH
OH
Me
OH
(iii)
Me
OH
Me
(iv)
(v)
OH
OH
Me
OH HO
H
H
H
HO
Me
OH
Me
H
H
Me
H
OH
Me
Me
(vi)
H
C—C
Ph
H O
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Page # 22
OXIDATION & REDUCTION OF ORGANIC COMPOUND
Q.4
–COO
(i)
CH3–CH2
, CH3–CH2–COOH
(ii)
O
||
CH 3  C  CH 2  CH 3
O
(iii)
OH
O
(P)
(Q)
O
OH
Q.5
O
(S)
O
O
O
OH
HO
(R)
CHO
OH
O
O
O
COOH
CHO OH
O

O
HO
(i)
CH2 = CH – (CH2)3 – CHO
(iii)
O
||
(A) H 3C  C  CH 2  CH 2  CHO
(ii)
Ph – CH = CH – CHO
(v)
H2C = CH–CHO
(vii)
CH 3
|
CH  C  CH  C  CHO
O
(iv)
CHO
O
(vi)
MeO
C–CH2–CH2–OH
MeO
(viii)
O
||
Ph  C  CH 3
(ix)
O
Q.6
(i)
Q.8
(i)
(iii)
(ii)
Ph  CH  CH  CH 2  CH  CHO
|
CH 3
O
||
CH 2  CH  CH 2  C  CH 3
Me–CHO + HCHO
CHO
CHO
(ii)
Q.7
C
O + Et – CHO
(iv)
HO–CH2–CH2–CHO + HCHO
(v)
HCHO + CH3– CH2–CHO
(vi)
HCHO + HCOOH + CH3–CHO
(vii)
2HCHO + 2HCOOH
(viii)
Me–COOH + Me–CHO
(ix)
2 Me–COOH
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Page # 23
OXIDATION & REDUCTION OF ORGANIC COMPOUND
Q.9
(i); (iii) ; (iv)
O
Q.10
(a) 2CH3 – COOH
Me CH–C–Me
(b) Me2CO + MeCOOH
O
Me CH–C–Me
(c) Me3C–COOH + CO2 + H2O
(d)
COOH
COOH
SeO
(b) Me – C – C – H
SeO
(a) CH – CHO
Q.11
CH –CHO
O
O O
O O
O
O
|| ||
|
|
||
SeO2
mCPBA
LAH
C

C

C

C





C
–
C
–
C
–
C
 CH3–CH2–OH
C

C

O

C
 C
(c) C – C – C –C 
O
O SeO
(d)
O
O
EXERCISE - 2
Q.1
(i) cis & trans 2-butene & 1-butene; (ii) C  C  C  C C  C  C  C C  C  C  C
||
|
|
C
C
C
(iii) zero (Neo-pentane can not be prepared by catalytic hyrogenation of alkene); (iv) One
(v) Including optical = 4
, Excluding optical = 3
D
Q.2
(a)
(b)
D
CH3
(c)
H
D
CH3
D
H
CH3
D
H
CH3
H
D
(d)
H
CH3
H
CH3
*
(e)
2 G.I.
Q.3
(i)
P1
Q.4
(i)
CH3–CH2–CH2–CH2–OH
(ii)
P2
CH3–CH2–CH2–CH2–NH2
MeCH2–OH, Me–CH2OH
(ii)
Me2CH–OH, Me2CH–OH
(iii)
MeCH2–OH, Me–CH2OH
(iv)
No reaction , Me–CH2–OH + EtOH
(v)
No reduction, MeCH2–OH
(vi)
No reaction, MeCH2–OH + MeOH
(vii)
No reduction, MeCH2NH2
(viii)
No reaction, Me–CH2–NH–Me
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Page # 24
OXIDATION & REDUCTION OF ORGANIC COMPOUND
Me
Q.5
(ix)
No reduction, MeCH2–N
(xi)
No reaction, No reaction
(i)
OH ,
Me–CH2–NH2 , Me–CH2–NH2
(ii)
NH
Me
OH
(iii)
No reaction,
(v)
No reaction,
(vii)
No reaction, Me–CH2–N
(ix)
Ph–CH=CH–CH2–OH, Ph–CH2–CH2–CH2–OH
(x)
Q.6
(x)
CH2OH
OH
,
HO
(a)
(iv)
NH
OH , NH
NH
,
OH
NH
CH2OH
OH
(vi)
No reaction,
(viii)
CH2=CH–CH2–OH, CH2=CH–CH2–OH
HO
(b)
(c)
CH3
Q.7
(i)
OH CH OH
2
(ii)
N
(iii) CH 3– CH – CH 3 , CH3–CH2– CH 2OH
OH
HO
OH
CH2OH
(iv)
+ MeOH (B) + Et–OH (C)
(v)
CH 2OH
(A)
CHO
CH2OH
Q.8
(i)
OH
CH
NaBH4
COOEt
O
CH2OH
OH
COOEt
CH3
CH2OH + EtOH
CH
CH3
OD
H
(ii)
CH2OH + EtOHNaOI
 +Iodoform test
OH
(iii)
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Page # 25
OXIDATION & REDUCTION OF ORGANIC COMPOUND
Q.9
O
C—CH3
NBS


Zn
( Hg

) 
(a)
HCl
(b)
(c)
O
H
CH2—CH3
CH=CH2
Br—C—CH3
HCl


alc
.KOH


Ph–CH2–CH3
(d)
Cl—CH–CH3
R 2O 2
(e)
O
(f)
,
Q.10
(A) Ni will reduce alkene, aldehyde and all it is not specific ;(B) NH2 – NH2 / H2O2 ;(C) LiAlH4
EXERCISE - 3
OBJECTIVE
Q.1
D
Q.2
Q.8
C
Q.9
C
C
Q.3
Q.10
C
D
Q.4
Q.11
C = CH – CH3
Br2
A
A
Q.5
Q.12
B
C
Q.6
Q.13
B
C
Q.7
Q.14
A
CD
SUBJECTIVE
CH3
CH3
Q.1
CH3
CH3
CH3
(i)
(ii O3
)H
2O
cold &
/Z
n
dil. KMnO4
CH3
C = CH – CH3
CH3
CH3
OH OH
C – CH – CH3
Br Br
C = O + OHC – CH3
Propanone
Ethanol
CH3 – CH – CH – CH2 – CH3
CH3 Cl
(A)
Alc. KOH
Q.2
CH3 – C = CH – CH2 – CH3
(B)
CH
3
O3
CH3
CH3
C = O + OHC – C2H5
+ CH3 – CH – CH = CH – CH3
(C)
CH
3
O3
CH3 – CH – CHO + OHC – CH3
CH3
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OXIDATION & REDUCTION OF ORGANIC COMPOUND
CH2
Q.3
Q.4
cold dil. KMnO4/H
CH2 – OH
+
CH2
CH2 – OH
O
3


Gas ( A )
KOH
KO3(B) + O2(C)
Pottasium ozonide (Deep Red Coloured)
O3 ( A )
 
CH3 – CH = CH – CH3 
( ii ) Zn / H 2 O 2CH3 – CHO
(A)  O3 ; (B)  KO3 ; (C)  O3
CH2
Q.5
O
CH3 – C – CHO
CH3 – C – CH = CH2
(E)
H2
+
2-Keto propanol
CH2O
Formaldehyde
CH3
CH3 – CH – CH2 – CH3
Q.6
CH3 – CH2 – CH = CH – CH2– CH2
(A)
CH3 – CH2 – CH2 – CH2 – CH2– CH2
KMnO4 / 
2CH3 – CH2 – COOH
CH3
Q.7
(a)
O
CH2 Ozonolysis
CH2
H
CH3
(b)
+
H
C
CH2
H
O
CH3
H
CH2
CH2
C
CH3
O
CH3
H
C
C
CH2
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Page # 27
OXIDATION & REDUCTION OF ORGANIC COMPOUND
NOTES
BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
Page # 28
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