Download Organic Compounds containing Oxygen

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COO – 1
ORGANIC COMPOUND
CONTAINING OXYGEN
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COO – 2
C O N C E P T S (Monohydric Alcohols & Ether)
Alcohols :
C1A Physical Properties :
The lower alcohols are completely soluble in water but as the number of carbon atoms increases,
solubility decreases. This solubility in water is due to intermolecular H-bonding between water and
alcohols molecules due to their polar character.
Increase in C-chain increases organic part hence solubility in water decreases.
Isomeric 10, 20, 30 alcohols have solubility in order : 10 > 20 > 30.
C1B
Method of preparation of Alcohols :
Hydration of Alkenes :
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(i)
dil H SO
2
 
4 
Intermediate is carbonium ion that can change to more stable carbanion ion by hydride
shift, alkyl or phenyl shift.
(ii)
Hydrolysis of Alkyl Halides :
R – X + aq. NaOH (or H2O)  R – OH + NaX (or HX)
Reaction can be S N1 or S N 2 .
(iii)
Reduction of Carbonyl Compounds (including acid derivative) :
RCHO  4  
4  RCH 2 OH
10
(b)
 4  
4 
(c)
(iv)
LiAlH or NaBH
(a)
LiAlH or NaBH
Acid, acid halide, ester and anhydride are reduced to 10 alcohol.
Hydroboration and oxidation :
H O , OH 
BH in THF
2
CH 3  CH  CH 2 3 
(CH 3  CH 2  CH 2  ) 3 B 2 

 CH 3 CH 2 CH 2 OH  B(OH ) 3
The hydroboronation-oxidation (HBO) process gives product corresponding to
anti-Markownikoff addition of H2O to the carbon-carbon double bond.
(v)
Oxymercuration-Demercuration :
Hg ( OAc )
  
2 
H 2O
NaBH
 
4 
The alcohol obtained corresponds to Markownikov’s addition of water to an Alkene.
(vi)
(a)
Through Grignard Reagent :
Addition of Grignard Reagent on carbonyl compounds followed by hydrolysis
yields alcohols nature of which depends upon types of carbonyl compounds used.
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COO – 3
H O
3 
+ H2O
R1, R2 can be H, alkyl or aryl but R3 is not H.
Grignard reagent with epoxide and after hydrolysis gives 10 alcohol for e.g.,
(b)
H O
2



Practice Problems :
1.
Propene, CH3 – CH = CH2 can be converted into 1-propanol, Which of the reagent among the
following is ideal to affect the conversion :
alkaline KMnO4
(b)
B2H6 and alkaline H2O2
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(a)
(c)
O3/Zn
(d)
OsO4/CH2Cl2
(b)
2.
Glycol may be obtained :
(a)
by the oxidation of ethylene with cold, dilute, alkaline permanganate solution
(b)
by the hydrolysis of ethylene bromide under reflux with aqueous sodium carbonate
solution
(c)
by the hydrolysis of ethylene chlorohydrin on boiling with aqueous sodium bicarbonate
(d)
by any of the above methods
(d)
3.
Which of the following are the starting materials for the Grignard’s synthesis of tert. butyl alcohol
(a)
CH3MgI + CH3COCH3
(b)
CH3MgI + CH3CHOHCH3
(c)
CH3CH2MgBr + CH3COCH3
(d)
CH3CH2MgBr + CH3CHO
(a)
[Answers : (1) b (2) d (3) a]
C2
Chemical Properties :
(i)
Dehydration of Alcohol :
conc . H SO
2
  
4 
Alcohols leading to conjugated alkenes are more easily dehydrated then those of alcohols
leading to non-conjugated alkenes.
Dehydration of alcohol is in the order 30 > 20 > 10 as intermediate is carbocation.
(ii)
Reaction with Halogen Acids :
R – OH + HX  RX + H2O
Intermediate is carbonium ion. The order of reactivity of HX : HI > HBr > HCl for a given
alcohol.
(iii)
Reaction with Phosphorous Halides and Thionyl Chloride :
3R – OH + PX3  3RX + H3PO3
(X = Br, I)
R – OH + SOCl2  RCl + SO2 + HCl
(iv)
Acidic Character of Alcohol :
RO – H + Na  RO–Na+ + ½ H2
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COO – 4
alkoxide
The order of acidic character of alcohol with metal is :
CH3OH > 10 > 20 > 30.
The relative acidities is as follows :
RCOOH > C6H5OH > H2O > ROH > CH  CH > NH3 > RH
(v)
Ester Formation (alcohol can act as a nucleophile also).
Thus reactivity of alcohols for given acid is in order :
CH3OH > 10 > 20 > 30
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and those of acids for given alcohol is in order :
HCOOH > CH3COOH > RCH2COOH > R2CHCOOH > R3CCOOH
(vi)
Oxidation of Alcohol :
1.
Using Cu
Cu
(a)
CH 3 CH 2 OH 
CH 3CHO  H 2
0
300 C
0
1
Aldehyde
Cu

0
(b)
300 C
Cu

0
(c)
300 C
2.
Oxidation using KMnO4/K2Cr2O7
(a)
( i ) KMnO
RCH 2OH   
4  RCOOH

( ii ) H
(b)
KMnO , CrO in glacial acetic , acid or K Cr O
(c)
3.
C3
4
  
3       2 2 7 
Other reagents used for oxidation :
(a)
PCC (pyridinium chlorochromate) to oxidise 10 alcohols to aldehydes.
(b)
MnO2 selectively oxidises the OH group of allylic and benylic 10 and 20
alcohols to give aldehydes and ketones respectively.
Test to distinguish 10, 20 and 30 Acohols :
(a)
Lucas Reagent (anhydrous ZnCl2/conc. HCl)
(b)
Oxidation
(c)
Victor Meyer Test
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COO – 5
Practice Problems :
1.
Identify (Z) in the following reactions series :
PBr
Alc . KOH
H SO , room temp .
Ethanol  3  ( X )    ( Y ) 
 2 4    ( Z )
H 2 O , heat
(a)
2.
(b)
CH3CH2OH
(c)
CH3CH2OSO3H (d)
C2H5OC2H5
An organic compound gives hydrogen on reacting with sodium metal. It also gives iodoform test and
forms an aldehyde of molecular formula C2H4O on oxidation with acidified dichromate. The
compound is :
(a)
CH3OH
(b)
CH3COOH
(c)
CH3CHO
(d)
C2H5OH
The order of reactivity of the following alcohols,
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3.
CH2 = CH2
towards conc. HCl is
(a)
I > II > III > IV
(b)
I > III > II > IV
(c)
IV > III > II > I
(d)
IV > III > I > II
[Answers : (1) b (2) d (3) c]
C4
Ethers
Nomenclature of Ethers :
Cyclic ethers can be named in severay ways :
In another system, a cyclic three membered ether is named as oxirane and a four membered ether is
called oxetane.
C5
Methods of Preparation of Ether :
(a)
Williamson Synthesis :
Example :
(i)
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COO – 6
(ii)
Discussion : Phenol in Williomson’s synthesis :
Because phenols are stronger acids than alcohols they can react with NaOH to change into sodium
phenoxide ion. But alcohols can be converted into sodium alkoxide ion by reaction with sodium
metal only.
[Alkyl halide used here can not be aryl halide as it does not undergo nucleophilic substitution easily]
R – X  can be X = –Cl, –Br, I,
(b)
– OSO3CH3 etc.
Ethers by Intermolecular Dehydration of Alcohols :
Alcohols can dehydrate to form alkenes.
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10 Alcohols can also dehydrate to form ethers.
The formation of ether takes places by S N 2 mechanism mainly, with one molecule acting as the
nucleophile and with another protonated molecules of the alcohol acting as the substrate.
Finally this method is not useful for the preparation of ether with 30 alcohol because they form
alkene too easily. This method is not useful for the praparation of unsymmetrical ethers from
primary alcohols because the reaction leads to a mixture of products.
(c)
Ethers may be prepared by the addition of alcohols to alkenes in the presence of acid e.g.
Practice Problems :
1.
Which route provides a better synthesis of ether :
(a)
I
(b)
II
(c)
equal
(d)
none
[Answers : (1) b]
C6
Reactions of Ethers :
(a)
Ethers are comparitavely unreactive compounds. The ether linkage are quite stable
towards bases, oxidizing agent, reducing agents. Ether can undergo just one kind of
reaction, cleavage by acids :
HX
R  O  R   HX 
 RX  R OH 
 RX  R X
Reactivity of HX : HI > HBr > HCl
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COO – 7
Cleavage takes place only under quite vigourous conditions : concentrated acids (usually HI or HBr)
and high temperature.
Oxygen of the ether is basic, like the oxygen of alcohol. The initial reaction between an ether and an
acid is undoubtedly formation of the protonated ether.
Cleavage then involves the nucleophilic attack by halide ion on this protonated ether, with
displacement of the weakly basic alcohol molecule.
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As we might expect primary alkyl group tend to undergo S N 2 and 30 tend to undergo S N1 .
(b)
Acid Hydrolysis : C 2 H 5  O  C 2 H 5  H 3O   2C 2 H 5OH .
(c)
Acid Hydrolysis of Epoxide :
(d)
Formation of Halohydrin :
H O
3 
HX


Practice Problems :
1.
Ethoxy ethane does not react with
(a)
2.
HI
(b)
conc. H2SO4
(c)
PCl5
(d)
Na
An ether, (A) having molecular formula, C6H14O, when treated with excess of HI produced two alkyl
iodides which on hydrolysis yield compounds (B) and (C). Oxidation of (B) gives an acid (D), whereas
oxidation of (C) results in the formation of a mixed ketone, (E). Thus structures of (A) is
(a)
(b)
(c)
(d)
CH3CH2CH2CH2OCH2CH3
[Answers : (1) d (2) c]
Aldehydes & Ketones
C7A 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.
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COO – 8
C7B
Method of Preparation :
1.
Oxidation :
(a)
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(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)
(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
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H O, H 
CH 3 C  CH 2  CH 3 COCH 3
HgSO 4
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COO – 9
7.
Hydroboration Oxidation :
Hydroboration of a non-terminal alkyne followed by oxidation of the intermediate yields a ketone
but terminal alkyne yield aldehyde.
8.
Use of Grignard Reagent :
(b)
9.
With HCN aldehyde is formed.
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(a)
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
12.
Reduction of acid chloride with organocopper compounds :
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COO – 10
13.
Friedel-Crafts Acylation :
(a)
(b)
14.
From Cadmium and Lithium Salts :
1.
2RMgX + CdCl2 
 R – Cd – R + MgCl2 + MgX2
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 RCOR’ + RCdCl
R – Cd – R + R’COCl 
R 
 should be 10 alkyl or aryl (–C6H5)
15.
 Aldehyde/Ketone
Ozonylysis of Alkene (Zn/H+) 
C8
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.
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COO – 11
Addition of cyanide :
(b)
Addition derivatives of ammonia :
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(a)
(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)
(ii)
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COO – 12
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.
Other Reactions of Aldehyde and Ketones :
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C9
(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]
(c)
Haloform Test :
(i)
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COO – 13
(ii)
Hypohalite NaOX (NaOH + X2) cannot only halogenate can also oxidise alcohols
(d)
Reduction :
Reduction to alcohol
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(a)
(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)
(h)
Crossed Cannizzaro reaction :
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COO – 14
(i)
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
Test
RCHO
RCOR
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S.No.
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 :
1.
HBr
(a)
2.
3.
4.
C2H 5I
(b)
C2H5OH
(c)
CHI3
2
(d)
(a)
aldehyde
(b)
secondard alcohol
(c)
alkene
(d)
tert. alcohol
CH3CHO
Which statement is incorrect in the case of acetaldehyde and acetone
(a)
both react with hydroxylamine
(b)
both react with NaHSO3
(c)
both react with hydrazine
(d)
both reduce ammonical silver nitrate
Which of the following undergoes Cannizzaro’s reaction
CH3CHO
(b)
CH3CH2CHO
(c)
(CH3)2CHCHO (d)
HCHO
(c)
acetone
phenol
Urotropine is formed by the action of ammonia on
(a)
6.
NaOH
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)
5.
Hydrolysis
 ( X )    
 ( Y )) I( excess
 ) ( Z )
Identify (Z) in the reaction series, CH 2  CH 2  
acetaldehyde
(b)
formaldehyde
(d)
Hydrocarbons are formed when aldehydes and ketones are reacted with amalgamated zinc and
conc. HCl. The reaction is called
(a)
Cannizzaro’s reaction
(b)
Clemmensen’s reduction
(c)
Rosenmund’s reaction
(d)
Tischenko reaction
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COO – 15
7.
8.
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
(a)
9.
(b)
acetaldehyde
benzaldehyde
Which of the following reagent reacts differently with HCHO, CH3CHO and CH3COCH3
(a)
10.
formaldehyde
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
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
/ 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.
17.
methane
(b)
methanol
(a)
CH3CHO > CH3COCH3 > CH3COC2H5
(b)
C2H5COCH3 > CH3COCH3 > CH3CHO
(c)
CH3COCH3 > CH3CHO > C2H5COCH3
(d)
CH3COCH3 > C2H5COCH3 > CH3CHO
(c)
(c)
To distinguish between 2-pentanone and 3-pentanone which reagent can be used
(c)
19.
C2H5OH
The correct order of reactivity in nucleophilic addition reaction CH3CHO, CH3COC2H5 and
CH3COCH3 is
(a)
18.
(b)
If formaldehyde and KOH are treated together, we get
(a)
16.
C2H 5I
NaOH/I2
K2Cr2O7/H
+
(b)
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
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COO – 16
20.
(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)
CH3CH2CH2CHO
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
21.
The reagent which can be used to distinguish acetophenone from benzophenone is :
22.
X and Y can be distinguished by :
(a)
23.
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]
Carboxylic Acids and Derivatives
C10
Common name of carboxylic acids :
1.
CH3CH2CH2CH2COOH
valeric acid
3.
Malonic acid;
5.
maliec acid
6.
Einstein Classes,
2.
oxalic acid
4.
;
Succinic acid
Fumaric acid
adipic acid
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COO – 17
7.
or
8.
9.
12.
14.
16.
18.
19.
20.
C11
O-Toluic acid
11.
Phathalic acid
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
10.
CH2 = CH – COOH (acrylic acid)
Isophthalic acid
Salicylic acid
13.
Terphthalic acid
15.
Anthranilic acid
17.
CH3 – CH = CH – COOH (crotonic acid)
(Pyruvic acid)
Methods of preparation of carboxylic acid :
(a)
By acid hydrolysis of cyanidies
(b)
By the use of Grigrand Reagent
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COO – 18
(c)
By oxidation method
(i)
Oxidation of alcohols (10)
(ii)
Oxidation of aldehyde
(iii)
Oxidation of ketones
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
Iodoform Test :
(d)
i ) Alkaline KMnO 4
Oxidation of alkenes and alkynes R  CH  CH  R (
   2RCOOH
( ii )  / H
(i )O
R  CH  CH  R 3  2RCOOH
( ii ) H 2 O
Practice Problems :
1.
Cyanohydrin of which of the following gives lactic acid on hydrolysis
(a)
acetone
(b)
acetaldehyde
(c)
propanal
(d)
HCHO
[Answers : (1) b]
C12
Chemical Properties of Carboxylic Acids :
(a)
Carboxylic acid undergo nucleophilic substitution reaction
(i)
Formation of amide from carboxylic acid
(ii)
Formation of Acyl chloride
(iii)
Formation of esters
(b)
Formation of anhydride :
(c)
Schmidt reaction :
(d)
HVZ reaction :
(e)
Reduction of caboxylic acid : CH 3 COOH LiAlH

4  CH 3 CH 2 OH
(f)
 CaO
Formatin of alkane : CH 3 COOH NaOH
   CH 4
(g)
( OH )
Formatin of Salts : CH 3 COOH Ca
 2 (CH 3 COO ) 2 Ca

effect of heating on salts : (CH 3 COO ) 2 Ca 
CH 3 COCH 3
(h)
i ) KOH
Kolbe electrolysis : CH 3 COOH (

 CH 3 CH 3
( ii ) Electrolysis
(i)
Acidic character : CH 3 COOH NaHCO
 3  CH 3 COONa  CO 2  H 2 O
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COO – 19
Practice Problems :
1.
2.
Which of the following will not undergo HVZ reaction
(a)
2, 2-dimethyl propanoic acid
(b)
propanoic acid
(c)
acetic acid
(d)
2-methyl propanoic acid
Amongst the acids (a) CH  CCOOH (b) CH2 = CHCOOH and (c) CH3CH2COOH, the acid strength
follows the sequence
(a)
3.
(a) > (b) > (c)
(b)
(a) < (b) < (c)
(c)
(a) = (b) = (c)
(d)
(a) = (b) < (c)
Cl 2
CH 3CH 2 COOH 
(A) Alc
.KOH

(B) The compound (B) is
P
(a)
CH3CH2OH
(b)
CH3CH2CN
(c)
CH2 = CHCOOH
(d)
CH3CHClCOOH
Formic acid
(a)
is immiscible with water
(b)
reduces ammonical silver nitrate
(c)
is a weak acid nearly three and a half times weaker than acetic acid
(d)
is prepared by heating potassium formate
5.
Strongest acid out of CH3CO2H, CH2 = CHCO2H,
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
4.
(a)
CH3CO2H
(c)
—CO2H
—CO2H
(b)
CH2 = CHCO2H
(d)
equal Ka
[Answers : (1) a (2) a (3) c (4) b (5) c]
CARBOXYLIC ACID DERIVATIVES
C13A Structures : Closely related to the cabroxylic acids and to each other are a number of chemical
families known as functional derivative of carboxylic acids : Acid chlorides, anhydrides, amides and
esters. Acid chlorides, anhydrides, amides and esters. These derivatives are compounds in which the
–OH of a carboxylic group has been replaced by –Cl, –OOCR, –NH2 or  OR 
They all contain acyl group
C13B Physical properties : The presence of
group makes the acid derivatives polar compounds.
Acid chlorides and anhydrides and esters have boiling point is same as of aldehydes and ketones of
corresponding molcular weight.
Amides have quite high boiling points because they are capable of strong intermolecular hydrogen
bonding.
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COO – 20
The borderline for solubility in water ranges from three to five carbons for the esters to five to six
carbons for the amides. The acid derivatives are soluble in usual organic solvents.
Volatile esters have pleasant, rather chracteristic odours; they are often used in the praparation of
perfumes and artificial flavoring.
Acid chlorides have sharp, irritating odours, at least partly due to their ready hydrolysis to HCl and
carboxylic acids
C14A Nucleophilic acyl substitution :
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
Each derivative is nearly always prepared - directly or indirectly from carboxylic acids, and can be
readily converted into the carboxylic acid by simple hydrolysis.
The derivatives of carboxylic acids, like the acids themselves, contain the carbonyl group,
.
Here, too, as in aldehydes and ketones, the carbonyl group performs two functions :
(a)
It provides a site for nucleophilic attack
(b)
It increases the acidity of hydrogents attached to the alpha carbon.
Acyl compounds i.e. Carboxylic acids and their derivatives, typically undergo nucleophilic
substitution in which – OH, – Cl, – OOCR, or – OR is replaced by some other basic group.
Substitution takes place much more readily than at a saturated carbon atom; many of these
substitution do not take at all in the absence of carbonyl group e.g. replacement of – NH2 by – OH.
– W = – OH, – Cl, – OOCR, – NH2, – OR 
Let us discuss now properties of acyl compounds.
Both electronic and steric factors make the carbonyl carbon suceptible to nucleophilic attack.
(a)
The tendency of oxygen to acquire the negative charge.
(b)
The relatively unhindered transition state leading from the trigonal reactant to the
tetrahedral intermediate. These factors make acyl compounds, too suceptible to
nucleophilic attack.
It is the second step of reaction that acyl compounds differ from aldehydes and ketones.
The tetrahedral intermediate from an aldehyde/ketone compound gains a proton and the result is
addition.
The tetrahedral intermediate from an acyl compound ejects the :W group, returning to a trigonal
compound and thus result in substitution.
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COO – 21
C15A Acid Chlorides : Acid chlorides are prepared from the corresponding acids by reaction with SOCl2,
PCl5 or PCl3 as already discussed.
C15B Reactions : Acid chlorides are the most reactive of the derivatives of carboxylic acids.
1.
Like other derivatives it also undergo nucleophilic substitution reaction. Chlorine is expelled as
chloride and its place is taken by some other basic group.
Conversion into acids, hydrolysis :
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
(a)
(b)
Conversion into amides :
(c)
Conversion into Esters :
Acid chloride is added in portions to a mixture of the hydroxy compound and a base usually aq.
NaOH or pyridine. Base serves not only to neutrilize the HCl evolved but also to catalyze the
reaction.
2.
Formation of Ketones. Friedel - Craft’s acylation :
3.
Formation of ketones, Reaction with organo copper compounds :
4.
Formation of aldehydes by reduction : RCOCl or ArCOCl    3  RCHO or ArCHO
1.
Sodium acetate + acetyl chloride gives
LiAlH ( OBU  t )
(a)
CH3COOH
(b)
sodium formate
(c)
acetic anhydride
(d)
acetone
(c)
C16A Acid Anhydrides
Preparation :
(i)
Acetic anhydride is prepared by the reaction of acetic acid with ketene CH2 = C = O, which
itself is prepared by high teperature dehydration of acetic acid.
(ii)
In contrast to monocarboxylic acids, certain dicarboxylic acids yield anhydrides on simple
heating. In those cases where five or six membered ring is produced.
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COO – 22
(a)
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
(b)
Ring size is crucial : with adipic acid, e.g. anhydride formation would produce seven-membered ring
and does not take place. Instead carbon dioxide is lost and cyclopentanone is produced (a five
membered ring is formed) :
C16B Reactions of Acid Anhydrides :
(i)
Conversion into acids. Hydrolysis :
Example :
(ii)
Conversion into amides. Ammonolysis :
Example :
(iii)
Conversion into esters :
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COO – 23
(iv)
Formation of Ketone : Friedel craft acylation
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
Amides
C17A Preparation : In the laboratory method of preparation of amide includes its preparation from acid
chloride with ammonia, acid anhydride with ammonia.
In industry they are prepared by heating ammonium salts of carboxylic acids.
C17B Reaction of Amides :
1.
Hydrolysis :
2.
Conversion into nitrile :
3.
Conversion into imides :
Example :
4.
Hoffmann degradation of amides :
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COO – 24
Practice Problems :
1.
The end product in the followin series of reaction is
2O5
CH 3COOH NH

3 ( A ) heat
( B) P
(C)
2.
(a)
CH4
(b)
CH3OH
(c)
acetonitrile
(d)
ammonium acetate
NH
PO
2 5
In a reaction, ( A ) 
3 (B) heat
(C) 
 C 2 H 5 CN (A), (B) and (C) are :
(a)
CH3COOH, CH3COONH4 and CH3CONH2
(b)
CH3COCl, CH3CONH2 and CH3COONH4
(c)
C2H5COOH, C2H5COONH4 and C2H5CONH2
(d)
C2H5COONH4, C2H5CONH2 and C2H5COOH
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
[Answers : (1) c (2) c]
Esters :
C18A Preparation of Esters :
1.
From acids :
Reactivity of R OH is 10 > 20 > 30
similarly
> 10 > 20 > 30
Example
2.
From Acid Chloride or anhydride :
3.
Formation of cyclic esters (Lactone)
A hydroxy acid is both alcohol and acid. In those cases where five or six membered ring can form
intermolecular esterification occurs. Thus  or  hydroxy acid looses water spontaneously by yield of
cyclic ester known as lactone. Treatment with base (OH–) rapidly opens the lactone ring.
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COO – 25
C18B Reactions of Esters :
Conversion into acids and acid derivatives
2.
Conversion into amides
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
1
CH3COOC2H5 + NH3  CH3CONH2 + C2H5OH
Treatment of an ester with ammonia, generally in ethylalcohol solution yield the amides. This
reaction involved the nucleophilic attack by a base, ammonia, on the electron-deficient carbon; the
alkoxy group,  OR' is replaced by – NH2.
3.
Conversion into esters : Transesterfication :
Acid is  H2SO4 or dry HCl
Base  alkoxide ion generally used.
4.
Reactions with Grignard Reagent :
5.
Reduction to alcohols :
(a)
Catalytic hydrogenation : Hydrogenolysis
CuO, CuCr O
2 4
RCOOR   2H 2   

 RCH 2OH  R' OH
0
250 C
(b)
6.
10 alcohol
. ether
Chemical Reduction : R  COOR   LiAlH 4 Anhyd
   RCH 2 OH  R OH
Reaction with carbanions : Clasien Condensation :
R
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COO – 26
Example :
7.
Crossed Clasien Condensation :
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
Like a crossed aldol condensation, a crossed clasien condensation is also possible only when one of
the reactants has no -hydrogens and thus is incapable of undergoing self condensation.
PHENOLS
C19A Structure and Nomenclature of Phenols :
Compounds that have a hydroxyl group directly attached to benzene ring are called phenols. Thus
phenol is specific name of hydroxy benzene
Compounds that have a hydroxyl group attached to a polycyclic benenoid ring are chemically
similar to phenols, but they are called napthols and phenanthrols, e.g.
C19B Physical Properties of Phenols : The presence of hydroxy groups in the molecules of phenols means
that phenols are like alcohols in being able to form strong intermolecular hydrogen bonds.
This hydrogen bonding causes phenols to be associated and therefore to have higher boiling points
than hydrocarbons of the same molecular weight.
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COO – 27
C20A Synthesis of Phenols :
Laboratory Synthesis : The most important laboratory synthesis of phenols is by hydrolysis of
arenediazonium salts.
This method is highly versatile and the conditions required for the diazotisation step and the
hydrolysis step are mild.
1.
General Reaction :
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
Specific Example :
Industrial Synthesis :
2.
Hydrolysis of Cholorobenzene (Dow Process) :
(The mechanism for the reaction probably involves benzyne intermediate)
3.
Alkali Fusion of Sodium benzene sulfonate : Sodium benzene sulfonate is melted (fused) with sodium hydroxide at 3500C to produce sodium phenoxide acidification then yields phenol.
4.
From Cumene Hydroperoxide :
I
II
This cumene is oxidized to cumene hydrolperoxide
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COO – 28
Finally, when treated with 10% sulfuric acid, cumene hydroperoxide undergoes a hydrolytic
rearrangement that yields phenol and acetone.
C20B Chemical Properties of Phenol :
1.
Reaction of Phenol as Acid : Strength of phenols as acids :
Although phenols are structurally similar to alcohols, they are much stronger acids. The pKa values
of most alcohols are of the order of 18. However the pKa values of phenols are smaller than 11.
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
Let us compare two superficially similar compounds cyclohexanol ane phenol.
Although phenol is weak acid when compared with carboxylic acid such as acetic acid (pKa = 4.75)
phenol is much stronger than cyclohexanol by a factor of 8.
Phenols are more acidic than cyclohexanol because of following reasons :
Phenoxide ion is more resonance stabilized than phenol. Resonance structures of phenoxid ion do
not involve charge separation. No resonance structure can be written for cyclohexanol and its anion
of course. The benzene ring of phenol acts as if it were as electron withdrawing group when we
compare it with cyclohexanol. That causes –OH oxygen to be more positive.
Reason : Carbon atom that bears hydroxyl group in phenol is sp2 hybridized, whereas in cyclohexanol
sp3 bybridised. Greater the S– characted more electronegative the carbon. Thus carbon of benzene
is more electronegative than cyclohexanol.
Because phenols are more acidic than water, the following reaction goes almost completion.
The reaction between cyclohexanol and NaOH does not occur to significant extent as H2O is stronger
acid than 1-wexanol.
Acidity order of Phenols :
(a)
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COO – 29
(c)
(d)
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
(e)
Distinguishing and separating phenols from alcohols and carboxylic acids :
1.
Phenols dissolve in aqueous sodium hyroxide whereas most alcohols with six carbon atoms or more
do not. Thus we can distinguish them.
2.
Alcohols with five carbon atoms are more or less soluble in NaOH but do not form appreciable
amount of sodium alkoxide.
3.
Most phenols are not solible in aqueous sodium bicarbonate, but carboxylic acids are soluble.
C21
Other reactions of the O – H Group of Phenols :
1.
Phenols react with carboxylic acid anhydrides and acid chlorides to form esters.
Thest reactions are similar to alcoholic as we have already discussed in alcohols.
2.
Phenols in the Williamson Synthesis :
Because phenols are more acidic than alcohols they can be converted to sodium phenoxide through
the use of sodium hydroxide (rather than use of sodium metal, the reagent that convert alcohols to
alkoxide ion).
(a)

X
General Reaction : ArOH NaOH
 ArO  N a R
 ArOR  NaX
[ X  Cl , Br , I ]
(b)
Cleavage of Alkyl-Aryl Ether :
conc . HX
HX
C6 H 5  O  R   C6 H 5  OH 
 no rxn.  RX
heat
3.
Reactions of the Benzene ring of Phenol :
(a)
Bromination : The hydroxyl group is a powerful activating group and an ortho-para
director in electrophilic substitution. Phenol itself react with Br2 in aqueous solution to
yield 2, 4, 6-tribromophenol. Note that a Lewis acid is not required for the bromination of
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COO – 30
this highly activated ring.
Monobromination of phenol can be achieved by carrying out the reaction in carbon
disulfide (CS2) at low temperature. Conditions that reduce the electrophilic activity of
bromine. The major product is the para isomer.
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
(b)
4.
(a)
Nitration : Phenol react with dilute nitric acid to yield a mixture o- and p-nitrophenol
Although the yield is relatively low (because of oxidation of ring). The ortho and para isomers can be
separated by steam distillation. o-Nitrophenol is more volatile isomer because its intramolecular
hydrogen bonding. p-Nitrophenol is less volatile because intermolecular H-bonding causing
association among molecules. Thus o-nitrophenol passes over steam and p-Nitrophenol remain in
the distillation flask.
(b)
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COO – 31
Sulfonation : Phenol reacts with concentrated sulfuric acid to yield mainly o-sulphonated product if
the reaction is carried out at 250C and mainly the para substituted product if the reaction is at 1000C.
6.
Kolbe’s Reaction : The phenoxide ion is even more suceptible to electrophilic aromatic substitution,
then phenol itself.
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
5.
High reactivity of phenoxide ion is used in a reaction called as kolbe reaction. In kolbe reaction
carbon dioxide act as the electrophile.
Reaction of salicylic acid with acetic anyhydride yields widely used pain reliver aspirin.
7.
The Clasien rearrangement :
Clasien rearrangement can also take place when allyl vinyl ether are present
8.
Diel Alder reaction is also pericyclic reaction :
9.
Quinones : Oxidation of hydroquinone gives quinone
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COO – 32
Reimer-Tiemann Reaction :
11.
Fries rearrangement :
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
10.
1.
2.
3.
4.
Rearrangement involves RCO+, which then attacks the ring.
Practice Problems :
When phenol is distilled with zinc dust, the main product is
(a)
biphenyl
(b)
benzene
(c)
benzaldehyde
(d)
phenolphtalein
When sodium benzene sulphonate is fused with sodium hydroxide (solid), the product formed is
(a)
benzene
(b)
phenol
(c)
benzene triphenol
(d)
none of these
Which of the following acids is strongest
(a)
C6H5SO3H
(b)
CH3COOH
(c)
C6H5COOH
(d)
(COOH)2
Phenol is less acidic then
(a)
p-nitrophenol
(b)
ethanol
(c)
cresol
(d)
benzyl alcohol
(i ) O 2
(
 (X) and (Y)
ii ) H O , H 
5.
Cumene
6.
(X) and (Y) respectively are
(a)
toluene, propene
(c)
phenol, acetone
In the following compounds
2
7.
toluene, propylchloride
phenol, acetaldehyde
the order of acidity is
(a)
III > IV > I > II
(b)
I > IV > III > II
(c)
II > I > III > IV
(d)
IV > III > I > II
When phenol reacts with benzene diazonium chloride, the product obtained as
(a)
phenyl hydrazine
(b)
p-amino azobenzene
(c)
phenol hydroxylamine
(d)
p-hydroxy azobenzene
Zinc
8.
(b)
(d)
Conc . HNO 3
Zn
Phenol distillati


 ( A ) Conc

 ( B )) NaOH
(C) . In the above reaction, compounds (A),
on
. H SO
at 60 0
2
4
(B) and (C) are
(a)
benzene, nitrobenzene and aniline
(b)
benzene, dinitrobenzene and m-nitroaniline
(c)
toluene, nitrobenzene and m-toluidine
(d)
benzene, nitrobenzene and hydrazobenzene
[Answers : (1) b (2) b (3) a (4) a (5) c (6) d (7) d (8) d]
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COO – 33
INITIAL STEP EXERCISE
1.
Identify (Z) in the series,
6.
. H 2SO 4
Br2
C3H 7 OH Conc


(X) 
(Y )
0
In the following sequence of reactions, the end
product is
on
g agent
CH 3CHO Condensati


(A ) Dehydratin


(B)
170 C
mild alkali
of
Excess


( Z)
Alc. KOH
(a)
7.
(b)
(d)
CH3 – C  CH
8.
2.
When ethyl benzoate is hydrolysed with aqueous
alkali, the products present in the medium are
(a)
(b)
(c)
(d)
3.
(c)
4.
C6H5COOH and C2H5O—
C6H5COO— and C2H5OH
C6H5COO— and C2H5O—
(b)
(c)
(d)
crotonaldehyde
(c)
paraldehyde
(d)
metaldehyde
Which of the following reaction is a condensation
reaction
(a)
HCHO  Paraformaldehyde
(b)
CH3CHO  Paraldehyde
(c)
CH3COCH3  Mesityl oxide
(d)
CH2 = CH2  Polyethylene
9.
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
(a)
chloroform
(b)
chloral
(c)
trichloro ethanol
(d)
trichloro acetic acid
Salicylic acid on heating with soda lime forms
(b)
4
(a)
benzene
(b)
phenol
3
(d)
2
(c)
benzyl alcohol
(d)
benzoic acid
10.
Out of butane, butanol-1, butanal and butanone,
the decreasing order of their boiling point is
CH3CH2CH2OH
(a)
butane > butanol > butanal > butanone
CH3CH2CH(OH)CH3
(b)
butanol > butane > butanal > butanone
CH3CH(OH)CH2CH2CH3
(c)
butanone > butanal > butanol > butane
(CH3)2C(OH)CH2CH3
(d)
butanol > butanal > butanone > butane
Which aromatic acid among the following is weaker
than simple benzoic acid
(a)
(b)
5
An alcohol on oxidation is found to give CH3COOH
and CH3CH2COOH. The structure of the alcohol is
(a)
5.
C6H5COOH and C2H5OH
How many isomers of C5H11OH will be primary
alcohols :
(a)
aldol
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
(c)
(a)
heat
11.
(b)
12.
(c)
Einstein Classes,
(d)
The alkene which on ozonolysis yields only acetone
is
(a)
CH2 = CH2
(b)
CH3 – CH = CH2
(c)
(CH3)2C = C(CH3)2
(d)
CH3CH = CHCH3
In the reaction CH3CHO + HCN  CH3CHOHCN,
a chiral centre is produced. The product is
(a)
meso compound
(b)
laevorotatory
(c)
dextrorotatory
(d)
racemic mixture
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COO – 34
13.
Leaving tendency of the following in increasing
order :
19.
Which of the following reactions is expected to
readily give a hydrocarbon product in good yield
(a)
is
RCOOK Electrolys
 

(b)
Br2
RCOOAg 
(c)
Cl 2
CH 3CH 3 
hv
2 H 5 OH
(CH 3 ) 3 C  Cl C


(d)
15.
16.
17.
VI < V < III < II < IV < I
20.
(b)
I < II < III < VI < V < IV
When propionic acid is treated with aqueous
NaHCO3, CO2 is liberated. The ‘C’ of CO2 comes
from
(c)
I < III < V < II < IV < VI
(a)
methyl group
(d)
can’t be decided
(b)
carboxylic acid group
(c)
methylene group
(d)
bicarbonate
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
14.
(a)
Formylchloride has not been prepared so far. Which
can function as formylchloride in formylation
(a)
HCHO + HCl
(b)
HCOOCH3 + HCl
Which will liberate ammonia when boiled with
NaOH solution
(c)
CO + HCl
(a)
ethyl amine
(b)
acetamide
(d)
HCONH2 + HCl
(c)
aniline
(d)
all
21.
Formic acid is obtained when
22.
(a)
calcium acetate is heated with conc.
H2SO4
(b)
glycerol is heated with oxalic acid
(c)
acetaldehyde is oxidised with K2Cr2O7
and H2SO4
(d)
calcium formate is heated with calcium
acetate
Formic acid and formaldehyde can be distinguished
by treating with
(a)
Benedict’s solution
(b)
Tollen’s reagent
(c)
Fehling’s solution
(d)
NaHCO3
23.
24.
25.
In the reaction
PI 3
C 2 H 5 OH 
(A ) KCN

(B) Hydrolysis
 (C)
The product (C) is
18.
(a)
acetic acid
(b)
formic acid
(c)
oxalic acid
(d)
propionic acid
26.
Acetyl chloride is not obtained from acetic acid by
the action of
(a)
CHCl3
(b)
SOCl2
(c)
PCl3
(d)
PCl5
Main product of the reaction CH 3 CONH 2 +
HNO2  is
(a)
CH3NH2
(b)
CH3CH2NH2
(c)
CH3COOH
(d)
CH3OH
Urea is a
(a)
monobasic acid (b)
dibasic acid
(c)
monoacid base (d)
diacid base
Al ( OC H )
5 3

 CH3COOCH2CH3. This
2CH3CHO  2 
reaction is called
(a)
Cannizzaro’s reaction
(b)
Aldol condensation
(c)
Claisen’s reaction
(d)
Tischenko reaction
Explain the reactivity of acyl compounds in the
order
(a)
acid chloride > amide > anhydride >
ester
The above reaction is called
(b)
acid chloride > anhydride > ester > amide
(a)
HVZ reaction
(c)
ester > acid chloride > anhydride > amide
(b)
Hunsdiecker reaction
(d)
ester > anhydride > acid chloride > amide
(c)
Schmidt reaction
(d)
Decarboxylation reaction
H SO ( conc .)
2
4


 RNH2 + CO2 + N2
RCOOH + N3H 
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COO – 35
28.
29.
30.
Artificial oil of bitter almonds or oil of Mirabane is
the name given to
(a)
chlorobenzene
(b)
benzaldehyde
(c)
aniline
(d)
nitrobenzene
35.
 OAc,  OMe,  OSO2 Me,  SO 2 CF3
( I)
Which does not have a carboxyl group
(a)
picric acid
(b)
ethanoic acid
(c)
aspirin
(d)
benzoic acid
Phenols do not react with
(a)
sodium bicarbonate
(b)
sodium hydroxide
(c)
potassium hydroxide
(d)
ferric chloride
36.
In the reaction,
37.
C 6 H 5CH 2 Cl Aq
. NaOH

(A) Oxidation

(B) ,
Cu ( NO 3 ) 2
(B) is
(a)
(c)
31.
(b)
(c)
(d)
32.
(b)
(c)
(d)
34.
(b)
phenol
toluene
(d)
benzaldehyde
aldol condensation
38.
39.
benzoin condensation
( II )
( III )
( IV )
the order of leaving group ability is
(a)
I > II > III > IV
(b)
IV > III > I > II
(c)
III > II > I > IV
(d)
II > III > IV > I
Benzyl alcohol is obtained from benzaldehyde by
(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
(a)
C6H5CHO
(b)
C6H5COOH
(c)
C6H5CH2Cl
(d)
C6H5CH3
Which of the following is the weakest acid
(a)
benzene sulphonic acid
(b)
benzoic acid
(c)
benzyl alcohol
(d)
phenol
In the Cannizzaro’s reaction given below

2Ph  CHO OH

 Ph  CH 2OH  PhCO 2
Perkin reaction
the slowest step is
(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
Wurtz reaction
Benzaldehyde when refluxed with aqueous alcoholic
KCN forms
(a)
33.
benzoic acid
The reaction involving condensation of acetic anhydride with an aromatic aldehyde by a carboxylate ion is an example of
(a)
In the following groups,
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
27.
benzoin
benzene
phenyl cyanide
phenyl isocyanide
The reaction,
C6H5CHO + CH3CHO  C6H5CH = CHCHO
is called :
(a)
benzoin condensation
(b)
Claisen condensation
(c)
Perkin reaction
(d)
Cannizzaro’s reaction
Cannizzaro’s reaction is not given by
(a)
trimethyl acetaldehyde
(b)
acetaldehyde
(c)
benzaldehyde
(d)
formaldehyde
Einstein Classes,
40.
C6 H 5OH ClOCCH

3  C 6 H 5OOCCH 3
NaOH ( aq .)
41.
42.
The above reaction is an example of
(a)
acetylation
(b)
benzoylation
(c)
Schotten-Baumann reaction
(d)
Reimer-Tiemann reaction
Phenol on standing in air develops a red colour due
to formation of
(a)
cyclohexane
(b)
resorcinol
(c)
phenoquinone (d)
quinol
Benzoic acid gives benzene on being heated with X
and phenol gives benzene on being heated with Y.
Therefore X and Y are respectively
(a)
soda lime and copper
(b)
zinc dust and sodium hydroxide
(c)
zinc dust and soda lime
(d)
soda lime and zinc dust
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COO – 36
43.
( i ) CHCl / NaOH

 Salicyladehyde
Phenol  3

44.
( ii ) H
This reaction is known as
(a)
Gattermann aldehyde synthesis
(b)
Duff reaction
(c)
Perkin reaction
(d)
Reimer-Tiemann reaction
Phenol on treatment with dil HNO3 gives
(a)
picric acid
(b)
o- and p-nitro phenols
(c)
o- and m-nitro phenols
(d)
p- and m- nitro phenols
FINAL STEP EXERCISE
2.
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
(b)
(a)
(c)
CH3CHOHCH2CHO
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
1.
(b)
HOCH2CH2CHO
(c)
CH3COCH2CHO
(d)
CH3COCH2CH2OH
(d)
none is correct
In a cannizzaro’s reaction, the intermediate that will
be the best hydride donar is
4.
cannizzair
 o  X and Y
(Y is alcohol, D is deuterium)
X and Y will have structure :
(a)
(a)
(b)
(b)
(c)
(c)
(d)
5.
(d)
none is correct
Consider reduction of 2-butanone
NaBD 4
B 
 2  butanone NaBD
4  A
D 2O
H 2O
NaBD4
3.
 product X
CH3CH = CH2 HO
/ OH 
2
2
X is :
(a)
A, B and C are :
(a)
Einstein Classes,
in all cases
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COO – 37
(b)
,
8.
,
(c)
propane
(d)
ethyl methyl ether
The major product obtained when 3-phenyl-1,
2-propane-diol is heated with H2SO4 is
(a)
C6H5 – CH2 – CO – CH3
(b)
C6H5 – CH2 – CH2 – CHO
(c)
C6H5 – CH2 – CH = CH2
(d)
9.
,
(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
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
(c)
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
,
10.
In the Cannizzaro’s reaction given below,

(d)
2Ph  CHO OH

 PhCH 2 OH  PhCOO
in all cases
the slowest step is
6.
The reaction of CH3CH = CH —
(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
—OH
with HBr gives :
(a)
CH3CHBrCH2 —
—OH
11.
(b)
(c)
(d)
7.
CH3CH2CHBr—
—OH
CH3CHBrCH2—
—Br
CH3CH2CHBr—
The reaction of
with a mixture of
Br2 and KOH gives R — NH2 as a product. The
intermediates involved in this reaction are
(a)
(b)
R—N=C=O
(c)
R — NHBr
—Br
The product (D) in the following sequence of reaction is :
(d)
( aq )
Na
C 2 H 4 HBr

 (A) NaOH

(B) 
(
3I
(C) CH


(D)
(a)
butane
(b)
ethane
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COO – 38
Which of the following carboxylic acid undergo decarboxylation easily
(a)
C6H5COCH2COOH
(b)
C6H5COCOOH
(c)
(d)
ANSWERS (INITIAL STEP
EXERCISE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
c
c
b
c
c
b
c
b
b
d
c
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
d
a
c
b
d
d
c
a
d
b
a
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
c
c
d
b
d
a
a
d
c
a
b
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
12.
b
b
b
a
c
b
a
c
d
d
b
ANSWERS (FINAL STEP EXERCISE)
1.
2.
3.
4.
5.
6.
Einstein Classes,
a
c
b
a
b
b
7.
8.
9.
10.
11.
12.
d
d
b
b
b
a
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COO – 39
TEST YOURSELF
2.
3.
4.
In alcohol 10, 20 and 30 cann’t be distinguish by
CH CHO
hydrolysis
R  Mg  X  3( A )   (B) .
(a)
Catalyst dehydrogenration
Compound B in above sequence of reaction is
(b)
Victor meyer’s test
(a)
Carboxylic acid
(c)
Lucas reagent test
(b)
Primary alcohol
(d)
can be distinguish by all
(c)
Secondary alcohol
(d)
Tertiary alcohol
The number of isomers represented by molecular formula C4H10O is
(a)
3
(b)
4
(c)
7
(d)
10
10.
The reaction,
H SO
RCOOH  N 3 H 2 
4  RNH 2  CO 2  N 2
Conc.
LiAlH4 converts acetic acid into
is called
(a)
acetaldehyde
(b)
ethane
(a)
(c)
ethyl alcohol
(d)
methyl alcohol
(b)
Hoffmann’s reaction
RONa + XR  
 ROR   NaX is
(c)
Schmidt reaction
(a)
(d)
Decarboxylation reaction
(b)
(c)
(d)
5.
9.
HVZ reaction
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
1.
Wurtz reaction
Williamson synthesis
Reimer-tiemer reaction
Tischenko reaction
The ether
. When
treated with cold HI produces
(a)
(b)
(c)
(d)
6.
2C6H5OH
C6H5I and C6H5CH2OH
C6H5CH2I and C6H5OH
General formula of both aldehydes and ketones is
(a)
(c)
7.
2C6H5CH2I
CnH2n + 2O
(b)
CnH2nO
CnH2n – 2O
(d)
CnH2n + 4O
The appropriate reagent for the following
transformation is :


(a)
(c)
8.
Zn(Hg)/HCl
(b)
NH2NH2/OH–
H2/Ni–
(d)
NaBH4
The Cannizzaro’s reaction is not given by
(a)
trimethyl acetaldehyde
(b)
acetaldehyde
(c)
benzaldehyde
1.
d
6.
b
(d)
formaldehyde
2.
c
7.
b
3.
c
8.
b
4.
b
9.
c
5.
d
10.
c
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ANSWERS
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