carboxylic acid and its derivatives_chemistry
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BRINGiiT on – Study Pack By ASKIITIANS.COM – powered by IITians
SUBJECT – CHEMISTRY
TOPIC – CARBOXYLIC ACIDS & ITS DERIVATIVE
COURSE CODE – AISM-09/C/CAID
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Contents :- CARBOXYLIC ACIDS & ITS DERIVATIVE
Introduction……………………………………………………………………………...3
Nomenclature ………………………………….……………………………………...5
General methods of preparation……………………………………………...11
Physical properties…………………………………………………………………..24
Chemical reactions…………………………………………………………………..27
Reactions of RCOOH………………………………………………………………..28
Salt formation………………………………………………………………………….31
The mechanism of easterification reaction………………………………35
Carboxylic acid derivatives……………………………………………………….44
Chemical reactions of acid derivatives……………………………………..47
Base promoted hydrolysis of esters: Saponification………………….52
Claisen Condensation……………………………………………………………….55
Amides……………………………………………………………………………………..57
Answers to excersies…………………………………………………………………66
Miscellaneous…………………………………………………………………………..75
Solved examples……………………………………………………………………….83
IIT level questions………………………………………………………………….….87
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INTRODUCTION
Carboxylic acids are characterized by the presence of carboxyl group. The
–COOH group which itself is made up of a carbonyl group (C=O) and a
hydroxyl group ( OH) is called carboxyl group (carb from carbonyl and oxyl
from hydroxyl)
C
O
O H C O
O
H
Carbonyl Hydroxyl Carboxyl
Carboxylic acids may be aliphatic or aromatic
R C
O
O
H
Ar C
O
O
H
Aliphatic carboxylic acid
(Where R = H or any alkyl group)
Aromatic carboxylic acid
(Where Ar is any aryl group)
Comparison of resonating structures of carboxylic group and
carbonyl group
Carbonyl group has two resonance structures (I and II)
C O
(I)
C O
(II)
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However, for a carboxyl group, three resonance structures (A, B and C) can
be written.
C
O
O
H
(A)
C
O
O
H
(B)
C
O
O
H
(C)
In both structures (A) and (C), the C – atom and the two O – atoms have
eight electrons in their respective valence shells while in structure (B), C –
atom has only six electrons. Therefore, structure (B) is less stable than
structure (C), in other words the two important resonance structures of
carboxyl group are structures (A) and (C). In both these structures, carboxyl
carbon is electrically neutral. However in case of aldehydes and ketones,
only one structure i.e. I is electrically neutral. As a result, the carboxyl
carbon of the resonance hybrid is less positive and hence less electrophilic
than the carbonyl carbon of aldehydes and ketones. However, it may be
noted that like carbonyl group, carboxyl group is also polar due to resonance
structures (B) and (C).
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NOMENCLATURE
The aliphatic carboxylic acids are commonly known by their initial names,
which have been derived from the source of the particular acid.
Examples:
HCOOH Formic acid [Latin: Fermica = ant]
CH3COOH Acetic acid [Latin: acetum = Vinegar]
CH3–CH2–COOH Propionic acid [Greek: Proton = First; Pion = Fat]
CH3(CH2)2COOH Butyric acid [Latin: Butyrum = Butter]
CH3(CH2)3COOH Valeric acid
CH3(CH2)14COOH Palmitic acid
CH3(CH2)16COOH Stearic acid
Alternative system of nomenclature is naming the acids as the derivatives of
acetic acids. The only exception being formic acid.
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Example:
CH3 – CH2 – COOH Methyl acetic acid
(CH3)3C – COOH Trimethyl acetic acid
According to the IUPAC system of nomenclature, the suffix of the
monocarboxylic acid is ‘oic acid’, which is added to the name of the alkane
corresponding to the longest carbon chain containing the carboxyl group,
e.g.
HCOOH methanoic acid
CH3 – CH2 – CH2 – COOH butanoic acid
The positions of side-chains (or substituents) are indicated by numbers, the
numbering to be started from the side of the carboxyl group.
CH3 – CH – CHCH2 – COOH 3, 4-dimethylpentanoic acid | | CH3 CH3
1 1
3 2 4 5
NAMING OF ACYL GROUPS, ACID CHLORIDES AND ANHYDRIDES:
The group obtained from a carboxylic acid by the removal of the hydroxyl
portion is known as an acyl group. The name of an acyl group is created by
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changing the - ic acid at the end of the name of the carboxylic acid to –yl,
examples:
O||
C – OH
Benzoyl group
O||
H –C – O – H
O||
H –C –
Formic acid Formyl group
O||
C –
Benzoic acid
Acid chlorides are named systematically as acyl chlorides.
CH3
Cl
O
acetyl chloride
An acid anhydride is named by substituting anhydride for acid in the name of
the acid from which it is derived.
OH
OHO
O
succinic acid
O
O
O
dihydrofuran-2,5-dione
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COOH
COOH
phthalic acid
O
O
O
2-benzofuran-1,3-dione
Naming of Salts and Esters:
The name of the cation (in the case of a salt) or the name of the organic
group attached to the oxygen of the carboxyl group (in the case of an ester)
precedes the name of the acid.
O-Na
+
O
1-phenylethanone
O
O
CH3
CH3
ethyl 4-methylbenzoate
Sodium benzoate
Naming of Amides and Imides:
The names of amides are formed by replacing –oic acid (or –ic acid for
common names) by amide or –carboxylic acid by carboxamide.
CH3
NH2
O
acetamide
NH2
O
cyclohexanecarboxamide
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If the nitrogen atom of the amide has any alkyl groups as substitutents, the
name of the amide is prefixed by the capital letter N; to indicate substitution
on nitrogen, followed by the name(s) of alkyl group(s).
O2N
N
O
CH3
CH3
N-ethyl-N-methyl-4-nitrobenzamide
If the substituent on the nitrogen atom of an amide is a phenyl group, the
ending for the name of the carboxylic acid is changed to anilide
NH
O
CH3
N-phenylacetamide
O
NH
N-phenylbenzamide
Some dicarboxylic acids form cylic amides in which two acyl groups are
bonded to the nitrogen atom. The suffix imide is given to such compounds.
NH
O
O NH
O
O
pyrrolidine-2,5-dione 1N-isoindole-1,3(2H)-dione
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Illustration 1:
Give the IUPAC names and common names of the following
compounds
(i)
COOH
COOH
(ii) HOOC
COOH
(iii) CH3
COOH
OH
iv) HOOC
OH COOH
OH
Solution:
IUPAC Name General name
(i) Ethanedioic acid Oxalic acid
(ii) Butanedioic acid Succinic acid
(iii) 2-Hydroxy propanoic acid Lactic acid
(iv) 2, 3 Dihydroxybutanedioic acid Tartaric acid
Illustration 2:
Which is not a hydroxy acid?
(A) lactic acid (B) tartaric acid
(C) citric acid (D) succinic acid
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Solution:
(D)
GENERAL METHODS OF PREPARATION
1. Oxidation of alcohols, aldehydes and ketones
Refer AEP & AK
Illustration 3:
How will you prepare CH3CH2COOH from CH3 CH=CH2
Solution:
3 4
2 2
BH / THF MnO /H
3 2 3 2 2 3 2H O /OHCH CH CH CH CH CH OH CH CH COOH
Illustration 4:
Bring about the following transformations
CH2 OH
Ag NH OH3 2MnO2
HA B
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Solution:
CHOA : C OH
O
B :
2. Oxidation of alkyl benzenes
Although benzene and alkane are quite unreative towards the usual
oxidizing agents (KMnO4, K2Cr2O7 etc). The benzene ring renders an
aliphatic side chain quite susceptible to oxidation. The side chain is
oxidised down to the ring and only a carboxyl group ( COOH) remains
to indicate the position of the original side chain. Potassium
permanganate is generally used for this purpose, although potassium
dichromate or dilute nitric acid can also be used. (Oxidation of a side
chain is more difficult, however, than oxidation of an alkene and
requires prolonged treatment with hot KMnO4)
CH2CH2CH2CH3
hot KMnO 4
COOH
CO2
n - butyl benzene Benzoic acid
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This reaction is used for two purposes (a) synthesis of carboxylic acids
and (b) identification of alkyl benzenes.
3. Carbonation of Grignard reagents
The Grignard synthesis of a carboxylic acid is carried out by bubbling
gaseous CO2 into the ether solution of the Grignard reagent or by
pouring the Grignard reagent on crushed dry ice (solid CO2). In the
latter method dry ice serves not only as reagent but also as cooling
agent.
The Grignard reagent adds to the carbon – oxygen double bond of CO2
just as in the reaction with aldehydes and keotnes. The product is the
magnesium salt of the carboxylic acid, from which the free acid is
liberated by treatment with mineral acid.
R MgX C
O
O
R COO Mg XH
R COOH Mg2
X
Grignard's reagent
The Grignard’s reagent can be prepared from primary, secondary,
tertiary or aromatic halides. The method is limited only by the
presence of other reactive group in the molecule. The following
synthesis illustrate the application of this method.
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CH3 C
CH3
CH3
OHHCl
CH3 C
CH3
CH3
ClMg
CH3 C
CH3
CH3
MgClH
CH3 C
CH3
CH3
COOH
Trimethylacetic acid
CO2
Illustration 5:
(i)
CH3
CH3 CH3
CH3
COOH
CH3CH3
A BBr2 Mg CO2 H
EtherFe
Write A and B
(ii) Br
CH
C2H5
CH3
Mg CO 2 H
CH
C2H5
CH3
COOH
A B
p-bromo-sec-butyl benzene p-sec-butyl benzoic acid
Ether
Write A and B
Solution:
(i)
CH3
CH3CH3
Br2
CH3
Br
CH3CH3
Bromomesitylene
Mg
CH3
MgBr
CH3
CH3
CO 2 H
CH3 CH3
COOH
CH3Mesitylene
Mesitoic acid
( 2, 4, 6 - trimethyl - benzoic acid)
Ether
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(ii) Br
CH
C2H5
CH3
p-bromo-sec-butyl benzene
Mg
CH
C2H5
CH3
MgBr
CO2
CH
C2H5
CH3
COOMgBr
H
CH
C2H5
CH3
COOH
p-sec-butyl benzoic acid
Ether
Illustration 6:
Bring about the following transformations.
Br
C
O
OH
Solution:
Br Mg/Ether CO2 H O3 C
O
OHMgBr CO2MgBr
Bring about the following transformations
Exercise 1:
(i)
Br OH
O
(ii) CH CH
O
OH
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Exercise 2:
(i) Why we can not get anhydrous formic acid by fractional
distillation of the Fatty acids?
(ii) CH2=CH—CH=CH2
2 3
(i) HBr
(ii) Mg/ether+(iii) CO /H O
A. Identify A?
4. Hydrolysis of nitriles
Aliphatic nitriles are prepared by treatment of alkyl halides with
sodium cyanide in a solvent that will dissolve both reactants. In
dimethyl sulfoxide (DMSO), reaction occurs rapidly and exothermically
at room temperature. The resulting nitrile is then hydrolysed to the
acid by boiling with aqueous alkali or acid.
R C N H2O
H
OH
R COOH NH4
R COO NH3
DMSOR Cl NaCN
Illustration 7:
(i) n - C4H9Br NaCN
n - butyl bromide
NH4
A
BC
Write A, B and C.
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(ii) CH2Cl
Benzyl chloride
NaCN 70% H2SO4, reflux NH4A B
Write A and B.
Solution:
(i) n - C4H9Br NaCNn - C4H9CN
n - butyl bromide n - voleonitrile
(pentanenitrile)
n - C4H9COO NH3
n - C4H9COOH NH4
n - valeric acid
(pentanoic acid)
aq. alc. NaOH, reflux
H
(A)
(B)
(C)
(ii) CH2Cl
Benzyl chloride
NaCN
CH2CN
70% H2SO4, reflux
CH2COOH
NH4
Phenyl acetic acidPhenyl acetonitrile
Exercise 3:
Identify the missing reagents or products
+- i) BH /THFH O/H conc. H SOHCN/OH 32 2 4ii) CH COOH3
(A) (B) (C) (E)CH3
CH3
O
(D)
i) BH3/THF ii) H2O2/OH-
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Exercise 4:
Conc. H2SO4
ACl2/H2O
BCN
CH3O
D E2 - +
2 7Cr O /H
OH
Exercise 5:
A pleasant smelling optically active ester (A) has molecular weight
186. it does not react with Br2 in CCl4. Hydrolysis of (A) gives two
optically active compounds B and C. Compound (C) gives positive
iodoform test and on warming with conc. H2SO4 gives D (saytzeff
product) with no geometrical isomers. (C) on treatment with benzene
sulphonyl chloride gives (E) which on treatment with NaBr gives
optically active F. When Ag2+ salt of (B) is treated with Br2, racemic F
is formed. Give structures of A to F.
5. Use of alkoxide
H
darkRO Na CO RCOONa RCOOH
3 3CH O Na CO CH COOH
0
8
210 C, pressure
3 3Co COCH OH CO CH COOH
6. Carbonylation of alkenes
3 40
H PO
2 2 2 3 2300 400 CCH CH CO H O CH CH COOH
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3 40
H PO
3 2 2 3300 400 C |CH CH CH CO H O CH CH COOH
CH3
7. Oxidative cleavage of alkenes, alkynes and cyclo alkenes
3 4
2
i O or KMnO
ii H ORC CR' RCOOH R'COOH
3
2
i O
3 3 3ii H OCH C CCH 2CH COOH
4i Alkaline KMnO
ii H /RCH CHR 2RCOOH
4KMnO /Δ/HCH2COOH
CH2COOH
Illustration 8:
Compound A, C5H8O3, when heated with soda lime gives B which
reacts with HCN to give C, C reacts with PCl5 to give D which reacts
with KCN to form E. E, on alkaline hydrolysis gives a salt which is
isolated and heated with soda lime to produce n-butane. A, on careful
oxidation with K2Cr2O7 gives acetic acid and malonic acid. Give
structural formulae of A to E.
Solution:
(C5H8O3) elimSoda
Compound B HCN
Compound (C)
Careful oxidation with K2Cr2O7
Compound E KCN Compound D,
Alk. hydrolysis
CH3COOH + CH2
COOH
COOH Salt
elimsodan-butane
PCl5
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On Oxidation only two –COOH groups can be introduced, i.e., one to
each carbon undergoing C–C fission, but in the resulting products we
have three –COOH groups. Hence, one –COOH group is already there
in compound A, the remaining portion C4H7O–, resembles with Keto
substituted alkyl group
C3H7 O
This indicates that the given organic compound A is Keto substituted
acid. To assign position to Keto group in carbon chain, we know that
keto acids on careful oxidation undergo C–C bond fission at a place
where -C -
O||
is situated, further Keto group is also converted into –COOH
and remains with acid having small number of carbon atoms. From the
above discussion it is clear that acetic acid is formed from:
3CH C arrangement
O||
C3H7CO–has 3 2 2CH C CH CH
O||
structure and compound A is
3 2 2CH C CH CH COOH
O||
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A, with soda-lime undergoes decarboxylation to give
CH3 – CO – CH2– CH3 (B)
B, being Ketone will give addition product with HCN to form
| CN
CH3 – C – CH2 – CH3
OH |
(C)
| CN
CH3 – C – CH2 – CH3
Cl |
(D)
| CN
CH3 – C – CH2 – CH3
CN |
| COONa
CH3 – C – CH2 – CH3 elimSoda CH3 – CH2 – CH2 – CH3
COONa |
(n-butane)
Alkaline hydrolysis
KCN
PCl5
Illustration 9:
How will you affect the following conversions?
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(i) COOH
Cl
CH3
(ii)
COOH
Solution:
(i) CH3
KMnO 4
OH
COOH
Fe/Cl 2
COOH
Cl
(ii)
CH3
OH2O/H
Hg2+
NaOH
I2COOH
Illustration 10:
An organic compound A (C4H8O3), acidic in nature, is oxidised to give B
which on gentle heating produces C, (C3H6O) and CO2. A on heating
yields D (C4H6O2) having an acid neutralization equivalent of 86. What
are A, B and C?
Solution:
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A on heating produces D (C4H6O2). It clearly tells us that it is
dehydration reaction and A must be a - hydroxyl acid.
So A is |
3 2
OH
CH HC CH COOH and D is CH3 CH = CH COOH
CH3C CH2
OH
COOH
H
Oxidation H3C C CH2
O
COOH
(B)
Δ decarboxylation
H3C C
O
CH3 CO2
(Because if a carbonyl group is present to carboxylic group, the
compound undergoes decarboxylation on heating)
Illustration 11:
Formic acid is obtained when
(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
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Solution:
(B)
Exercise 6:
A neutral organic compound A, C3H6O2 on hydrolysis gives a monobasic
acid B and a neutral compound C. The acid B reduces HgCl2 solution.
The compound C gives iodoform test. Write structures of A, B and C and
give equations to explain the reactions involved.
PHYSICAL PROPERTIES
Some important physical properties of carboxylic acids are given below,
1. Solubility
As the size of the alkyl group increases, the solubility of the acid
decreases and polarity is reduced.
2. Boiling points
Due to intramolecular hydrogen bonding dimerization of acid takes
place and boiling point of carboxylic acid is higher than expected.
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R C
O
O
H
RC
O
OH
3. Melting points
The melting points of aliphatic carboxylic acids do not show a regular
pattern. The first ten members show a alteration effect, i.e. the
melting point of an acid containing even number of carbon atoms is
higher than the next lower and next higher homologues containing odd
number of carbon atoms.
Illustration 12:
On the basis of H-bonding explain that the second ionization constant
K2 for fumaric acid is greater than for maleic acid.
Solution:
We know that H-bonding involving acidic H has an acid weakening
effect and H-bonding in conjugate base has an acid strengthening
effect.
Both dicarboxylic acids have two ionisable hydrogen atoms.
Considering second ionization step.
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Fumarate monoanion (no H-bonding)
C
C
C
H C
O
OH
O
O
C
C
H C
O
O
O
C
O
H
H
Maleate monoanion (H-bonding)
H
Since the second ionisable H of the Maleate ion participates in H-
bonding more energy is needed to remove this H because the H-bond
must be broken. The maleate mono anion is, therefore, the weaker
acid.
Exercise 7:
Explain the following
(i) Carbon-oxygen bond length in formic acid are 1.24 Å and 1.36
Å but in sodium formate both the carbon-oxygen bonds have
same value i.e.1.27 Å.
(ii) Acetic acid in the vapour state shows a relative molecular
weight of 120.
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CHEMICAL REACTIONS
The characteristic chemical behavior of carboxylic acids is, of course,
determined by their functional group, carboxyl, –COOH. This group is made
up of a carbonyl group (C = O) and a hydroxyl group (–OH). As we shall see,
it is the –OH that actually undergoes nearly every reaction. Loss of H+, or
replacement by another group but it does so in a way that is possible only
because of the effect of the C = O.
R C
H
H
C O H
O
(d)
(a)(b)
(c)
Carboxylic acids can also show various types of reactions.
(i) Removal of H+ (due to cleavage of O H bond) by reaction with a
base (at ‘a’ above)
(ii) C O bond breaks at b (by PCl5, PCl3, SOCl2, NH3/ )
(iii) Nucleophilic attack at point (c) in carboxyl carbon (Ester formation)
Reaction in which OH is replaced by NH2, Cl is SN type
R C OH
O
NH2 R C NH2
O
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(iv) Halogenation at –C by P/Br2 at point d (Hell – Volhard Zelinsky
reaction)
(v) Oxidation of - methylene group by SeO2
2
||SeO
2 2
O
RCH COOH RCCOOH H O Se
keto acid
Some reactions are summarized below:
LiAlH 4
R'OH/H
CH2N2
Na
S o d a lim e, Δ
NaOH
RCH 2OH
RCOOR'
RCOOCH 3
RCOONaH2
RH
RCOONa
ROCI
P2O5SOCl 2PCl 3PCl 5
(RCO) 2O
NaHCO 3NH3
3NH , Δ
RCOONH 4 RCONH 2 RCOONa CO 2
R C
O
OH
X
Y
Δ
3N H
2 2 2N CO R NH
Reactions of RCOOH
1. Acidity and salt formation
Acidity of Carboxylic Acids
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The acidity of a carboxylic acid is due to the resonance stabilization of its
anion.
R
O
O+
H
R
O
O H
R
O
O
R
O
O
H
H
(I) (II) (III) (IV)
Because of the resonance, both the carbon oxygen bond in the carboxylate
anion have identical bond length. In the carboxylic acid, these bond lengths
are no longer identical.
The acidity of carboxylic acid depends very much on the substituent
attached to – COOH group. Since acidity is due to the resonance stabilization
of anion, substituent causing stabilization of anion increases acidity whereas
substituent causing destabilization of anion decrease acidity. For example,
electron withdrawing group disperses the negative charge of the anion and
hence makes it more stable causing increase in the acidity of the
corresponding acid, on the other hand, electron-releasing group increases
the negative charge on the anion and hence makes it less stable causing the
decrease in the acidity. In the light of this, the following are the orders of a
few substituted carboxylic acids.
(a) Increase in the number of Halogen atoms on -position increases the
acidity, eg.
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Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH
(b) Increase in the distance of Halogen from COOH decreases the acidity
e.g.
CH3
COOH
Cl
CH3
COOH
Cl
COOH
Cl
This is due to the fact that inductive effect decreases with increasing
distance.
(c) Increase in the electronegativity of halogen increases the acidity.
FCH2COOH > BrCH2COOH > ICH2COOH
Illustration 13:
Which one of the following would be expected to be most highly ionised
in water?
(A) CH2Cl–CH2–CH2COOH (B) CH3CHCl –CH2–COOH
(C) CH3–CH2–CHCl–COOH (D) CH3CH2–CCl2 –COOH
Solution:
(D)
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SALT FORMATION
Carboxylic acids are weak acids and their carboxylate anions are strong
conjugate bases and are slightly alkaline due to the hydrolysis of carboxylate
anion compared to other species, the order of acidity and basicity of
corresponding conjugate bases are as follows:
Acidity RCOOH > HOH > ROH > HC CH > NH3 > RH
Basicity RCOO– < HO– < RO– < HC C– < 2NH < R–
1. The carboxylic acids react with metals to liberate hydrogen and are
soluble in both NaOH and NaHCO3 solutions. For example.
2CH3COOH + 2Na 2CH3COO–Na+ + H2
CH3COOH + NaOH CH3COO–Na+ + H2O
CH3COOH + NaHCO3 CH3COO–Na+ + H2O + CO2
Examples:
3 3 222CH COOH Zn CH COO Zn H
Aceticacid Zinc acetate
3 2 3 2 210 10CH CH COOH NaOH CH CH COO Na H O
lauricacid Sodiumlaurate
COOH
NaHCO 3
Benzoic acid
COO Na
CO2 H2O
Sodium benzonate
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Illustration 14:
Arrange the following illustration in order of increasing acidity
(i) HCOOH, ClCH2COOH, CH3COOH
(ii) CH3COOH, (CH3)2CHCOOH, (CH3)3CCOOH
(iii) ClCH2COOH, Cl2CHCOOH, Cl3CCOOH
(iv) ClCH2COOH, CH3CH2COOH, ClCH2CH2COOH, (CH3)2CHCOOH,
CH3COOH
(v) CH3COOH, Cl2CHCOOH, CH3CH2COOH, Cl3CCOOH, ClCH2COOH
Solution:
(i) CH3COOH HCOOH ClCH2COOH
(ii) (CH3)3CCOOH (CH3)2CHCOOH CH3COOH
(iii) ClCH2COOH < Cl2CHCOOH < Cl3CCOOH
(iv) (CH3)2CHCOOH CH3CH2COOH CH3COOH ClCH2CH2COOH
ClCH2COOH
(v) CH3CH2COOH CH3COOH ClCH2COOH Cl2CHCOOH Cl3CCOOH
Illustration 15:
Acetic acid reacts with chlorine in the presence of catalyst, anhydrous
FeCl3 to give
(A) acetyl chloride (B) methyl chloride
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(C) tri-chloro acetic acid (D) chloral hydrate
Solution:
(C)
2. Conversion into functional derivatives
R C
OH
O
R C
Z
O
(Z = -Cl, -OR', -NH 2)
(a) Conversion into acid chlorides
R C
OH
O
R C
Cl
OSOCl 2
PCl 3
PCl 5
Acid chloride
Examples:
COOH PCl 5
Benzoic acid
COCl
01 0 0 CPOCl 3 HCl
Benzoyl chloride
reflux
17 35 2 17 35 2Thionyl chlorideStearic acid Stearoyl chloride
n C H COOH n SOCl n C H COCl SO HCl
050 C
3 3 3 3 3Acetic acid Acetyl chloride
CH COOH PCl 3CH COCl H PO
Illustration 16:
Benzoyl chloride is prepared from benzoic acid by
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(A) Cl2,h (B) SO2Cl2
(C) SOCl2 (D) Cl2, H2O
Solution:
(C)
(b) Conversion into esters
Conversion into Esters (Esterification):
Carboxylic acid on reacting with alcohols in presence of
dehydrating agent (H2SO4 or dry HCl gas) gives esters. The
reaction is known as esterification.
R – C
O
OH
R OH R –C
O
OR
+ H2O
Acid
H+
+
This reaction is reversible and the same catalyst, hydrogen ion,
that catalyzes the forward reaction, esterification, necessarily
catalyzes the reverse reaction hydrolysis.
The equilibrium is particularly unfavourable when phenols
(ArOH) are used instead of alcohol; yet if water is removed
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during the reaction, phenolic esters [RCOOAr] are obtained in
high yield.
The presence of bulky group near the site of reaction, whether in
alcohol or in the acid, slows down esterification (as well as its
reverse, hydrolysis).
Reactivity CH3OH > 1° > 2° > 3°
In esterification HCOOH > CH3COOH > RCH2COOH > R2CHCOOH
> R3CCOOH
The Mechanism of the Esterification Reaction:
The step in the mechanism for the formation of an ester from an acid and an
alcohol are the reverse of the steps for the acid-catalyzed hydrolysis of an
ester, the reaction can go in either direction depending on the conditions
used. A carboxylic acid does not react with an alcohol unless a strong acid is
used as a catalyst, protonation makes the carbonyl group more electrophilic
and enables it to react with the alcohol, which is a weak nucleophile.
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O || C
CH3 O – H
+ H+
Protonation of the carbonyl group
O – H || C
CH3 O – H
+
+
O – H | CH3 – C – O – H
B
O
CH3 H
O – H | CH3 – C – O – H
O
CH3
O – H || C
CH3 O – H
O
CH3 H
+
H+
-BH
Nucleophilic attack at the carbonyl group
Deprotonation of the intermediate
Tetrahedral intermediate being protonated at another site
O || C
CH3 O – CH3
Deprotonation of the carbonyl group
O – H || C
CH3 O – CH3
+
O – H | CH3 – C – O – H
O
CH3
H
+
O
H H
B H – B+
+
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Examples:
COOH CH3OH
Benzoic acid
COOCH 3 H2O
Methyl benzoate
Methanol
H
CH2OH
Benzyl alcohol
HCH3COOH
Acetic acid
CH3 C
O
O CH2
2 52 C H OHSOCl
3 3 3 2 53 3 3Trimethyl acetic acid Ethyl trimethyl acetate
CH CCOOH CH CCOCl CH CCOOC H
Illustration 17:
Assign a structure to each compound indicated by a letter in the
following equations.
OH C
O
(CH2)9 C
O
OH 3
2 4
CH OH (excess)
H SO , ΔA
Solution:
OCH 3(CH2)9O
CH3
O O
A =
Exercise 8:
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Assign a structure to each compound indicated by a letter in the
following equations.
OH
OH
O
3 2 2CH CH -OH SOCl
HCl(g), Δ ΔA B
(c) Conversion into amides
R C
OH
O
R C
Cl
O
SOCl 2
An acid chloride
NH3 R C
NH2
O
An amide
Example:
32 NHSOCl
6 5 2 6 5 2 6 5 2 2Phenyl acetic acid Phenyl acetyl chloride Phenyl acetatamide
C H CH COOH C H CH COCl C H CH CONH
Illustration 18:
Discuss the reason for that a characteristic reaction of aldehydes and
Ketones is one of nucleophilic addition while Acyl compounds yield
Nucleophilic substitution product.
Solution:
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R
R
C = O
Nu–
Nu
R
C – O
R
Nu
R
C – O – H
R
H – Nu
Aldehyde or Ketone
Nucleophilic addition
An acyl compound
R
L
C = O
Nu – H
Nu+
R
C – O
L
H
Nu
R
C – O
HL+
Nu
R
C = O + HL
Another acyl compound
Nucleophilic addition
(L = –OH, –Cl, –NH2, OR etc.) Elimination
Nucleophilic substitution
The initial step in both reactions involves nucleophilic addition at the
carbonyl carbon atom. It is after the initial nucleophilic attack has
taken place that the two reactions differ. The tetrahedral intermediate
formed from an aldehyde or ketone usually accepts a proton to form a
stable addition product. By contrast, the intermediate formed from an
acyl compound usually eliminates a leaving group: this elimination
leads to regeneration of the carbon oxygen double bond and to a
substitution product. The overall process in the case of acyl
substitution occurs, therefore, by a nucleophilic addition–elimination
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mechanism. Acyl compounds react as they do so because they all have
good leaving groups attached to the carbonyl carbon atom.
3. Reduction
4
0
LiAlH
21 Alcohol
R COOH R CH OH
Examples:
4(CH3)3CCOOH 3LiAlH 4Ether [(CH3)3CCH 2O]4AlLi 2LiAlO 2 4H2
(CH3)3CCH 2OH
Neopentyl alcohol
( 2, 2 - dimethyl - 1 - propanol)
COOH
CH3
LiAlH 4
CH2OH
CH3
m - toluic acid m - methyl benzyl alcohol
4. Substitution in alkyl or aryl group
(a) Halogenation of Aliphatic Acids (Hell-Volhard-Zelinsky
Reaction)
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In the presence of phosphorus, aliphatic carboxylic acids react
smoothly with chlorine or bromine to yield a compound in which -
hydrogen has been replaced by halogen.
2 2 2Cl /,P Cl ,/ P Cl ,/ P
3 2 2 3CH COOH Cl CH COOH Cl CH COOH Cl CCOOH
The function of the phosphorus is ultimately to convert a little of the acid
into acid halide so that it is the acid halide, not the acid itself, that
undergoes this reaction.
P + X2 PX3
R – CH2 – COOH + PX3 RCH2 – COX
R COX2X R
X
COX
2H OHX R
X
CO2H
The halogen of these halogenated acid undergoes nucleophilic displacement
and elimination same as it does in the simple alkyl halides. Halogenation is
therefore the first step in the conversion of a carboxylic acid into many
important substituted carboxylic acid.
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R
Br
COOH
3NH R
NH2
COOH
An halogenated acid
An hydroxy acid
R
Cl
COOH
HNaOH R
OH
COOH
An amino acid
Examples:
CH3COOH
Acetic acid
Cl 2/PClCH 2COOH
Chloroacetic acid
Cl 2/PCl 2CHCOOH
Dichloroacetic acid
Cl 2/P
Cl 3CCOOHTrichloroacetic acid
CH3CHCH 2COOH
CH3
Isovaleric acid
CH3CH
CH3
CHCOOH
Br
α Bromoisovaleric acid
Br2/P
(b) Ring substitution in aromatic acids:
–COOH deactivates and directs incoming electrophilic to meta
position.
Example:
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COOH
HNO 3/H2SO4/Heat
NO2
COOH
benzoic acid m - nitrobenzoic acid
Illustration 19:
Hydrolysis of a compound A (C7H3Cl5) gives an acid B of the formula
C7H4Cl2O2. Decarboxylation of acid yields a neutral substance (C), the
nitration of which forms only one mono derivative (D). Identify A, B, C
and D.
Solution:
A =
Cl
Cl
CCl 3
B =
Cl
Cl
COOH
C =
Cl
Cl
D =
Cl
Cl
NO2
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Illustration 20.
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
Solution:
(A)
CARBOXYLIC ACID DERIVATIVES
There are four carboxylic acid derivatives. These are generally represented
as||O
R C Z , where Z is halogen (usually Cl), OCOR , OR or NH2 (or NHR or
NR2 ).
(a) When Z is halogen (usually Cl), the derivatives are called as acid
chlorides.
R C Cl
O
(b) When Z is OR , the derivatives are called as esters.
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(c) When Z is||O
O C R' , the derivatives are called carboxylic anhydrides.
(d) Where Z is NH2, the derivatives are called amides. When Z is NHR
or NR2 they are called N – substituted amides.
Synthesis of acid derivatives
Carboxylic acid derivatives are exclusively prepared from carboxylic acids.
The preparation methods of carboxylic acid derivatives are already discussed
under the chemical reactions of carboxylic acids.
Illustration 21:
(a) 21. Excess CO
2. HPhLi A
(b) 32 eq. CH Li
3 3 HCH COOH B 2
2
Ag O
3 Br ,3CH COOH C
Solution:
(a)
Ph O C O Ph C O
O
Ph C OH
O
H
(A)
(b) ||
3 3 33 3
O
CH CCOOH CH C C OCH B
2
2
Ag O
3 3 2Br ,3 3CH CCOOH CH CBr CO
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Illustration 22:
Suggest a likely mechanism for each of the following reactions.
(a) OH R C
O
CX3
H2ORCOO CHX 3
(b)
C
O
F
NH2
NH3
F
C
O
NH2
Solution:
(a)
OH R C CX3
O
R C OH
O
CX3
(b) C
O
F
NH2
NH3
C NH2
O
C
O
FNH2
F
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Exercise 9:
(i)
O
O
O
AlCl3A(C10H10O3)
Zn(Hg)
HClB(C10H12O2)
SOCl2
CAlCl3D(C10H10O)
LiAlH4E(C10H12O)
H2SO4F(C10H10)
Galc. KOH
H(C10H8)
NBS
(ii) When heated with acid (e.g. conc. H2SO4), O – benzoyl benzoic
acid yields a product of the formula C14H8O2. What is the
structure of this product?
CHEMICAL REACTIONS OF ACID DERIVATIVES
(i) Acyl chloride
We have already seen that acyl chlorides are the most reactive of all
acid derivatives. As a result, acyl chlorides are often selected as the
starting material for the preparation any other acid derivative. Let us
see how this is done.
R C Cl
O
R C O Na
O
R C O
O
C
O
R'
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R C Cl
O
R OH R C O
O
R'
R C Cl
O
R C NH2
O
NH3
R C Cl
O
R C NHR'
O
R'NH 2
R C Cl
O
R C NR'R"
O
R'R"NH
CH3 C Cl
O
H2NC6H5 C6H5NHC
O
CH3HCl
Acetanilide
OH
COOH
CH3COCl
O
COOH
C CH3
O
Acetyl salicylic acid (aspirin)
Reaction of acetyl chloride with olefins
Acetyl chlorides add on to the double bond of an olefin in the presence of a
catalyst (AlCl3 or ZnCl2) to form a chloro ketone which on heating, eliminates
a molecule of hydrogen chloride to form an unsaturated ketone.
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CH3 CH CH2 CH3COClZnCl 2 CH3 CH CH2
Cl
C
O
CH3
Δ
CH3 CH CH C
O
CH3HCl
Conversion into acids: Hydrolysis.
R C
O
Cl H2O R C
O
OH HCl
An acid
Illustration 23:
Acetyl chloride reacts with water more readily than methyl chloride.
Explain.
Solution:
Alkyl halides are much less reactive than acyl halides in nucleophilic
substitution because nucleophilic attack on the tetrahedral carbon of
RX involves a hindered transition state. Also, to permit the attachment
of the nucleophile a bond must be partly broken. In CH3COCl, the
nucleophile, attack on > C = O involves a relatively unhindered
transition of acyl halides occurs in two steps. The first step is similar to
addition to carbonyl compound and the second involves the loss of
chlorine in this case.
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+H2O
O
C
CH3 Cl
O-
C
CH3 Cl
+OH2
O–
C
CH3 Cl
+OH2
Transition state Intermediate –
O
C
CH3 OH2
Cl
O
C
CH3 OH
–H+
+Cl–
+
Transition state Product
–
Illustration 24:
Hydrogenation of benzoyl chloride in the presence of Pd and BaSO4
gives
(A) benzyl alcohol (B) benzaldehyde
(C) benzoic acid (D) phenol
Solution:
(B)
(ii) Carboxylic acid anhydrides
Carboxylic acid anhydrides can be used to prepare esters and amides.
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R C
O
O C
O
RR'OH
R C
O
OR' R C
O
OH
NH3 excess
R C
O
NH2 R C
O
O NH 4
Anhydrides can be hydrolysed to get back acids.
R C O C R
OO
H2O2R C OH
O
Illustration 25:
MeO CHO 3
3
CH COONa
H O(X)
COOH
The compound (X) is
(A) CH3COOH (B) BrCH2 – COOH
(C) (CH3CO)2O (D) CHO – COOH
Solution:
(C)
Exercise 10:
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O
O
O
3 2CH -CH -OH
TsOH ΔA
(iii) Esters
Ester hydrolysis
Acid catalysed esterification is an essentially reversible reaction. If you
follow the backward course of reactions of esterification it gives you
the mechanism for ester hydrolysis.
R C OR'
O
H2OR C OH
O
HR'OH
Base promoted hydrolysis of esters: Saponification
Esters undergo base promoted hydrolysis also. This reaction is known as
saponification, because it is the way most of the soaps are manufactured.
Refluxing an ester with aqueous NaOH produces an alcohol and the sodium
salt of the acid.
R C OR'
O
H2ONaOH R C O Na
O
R'OH
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This reaction is essentially irreversible because carboxylate ion is inert
towards nucleophilic substitution.
Mechanism
R C OR'
O
OH
R C OR'
O
OH
R C OH
O
R'O
R C O
O
ROH
If an ester is hydrolysed in a known amount of base (taken in excess), the
amount of base used up can be measured and used to calculate the
saponification equivalent; the equivalent weight of the ester, which is
similar to the neutralization equivalent of an acid.
Transesterification
Esters can also be prepared by transesterification (an alcohol displacing
another from an ester)
C
OR'R
O
R" OHHA
C
OR"R
O
heatR' OH
The mechanism of this reactions is quite similar to esterification.
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H2C CHCOCH 3
O
CH3CH2CH2CH2OH H2C CHCOCH 2CH2CH2CH3
O
Transesterification is an equilibrium reaction. To shift the equilibrium to
right, it is necessary to use a large excess of the alcohol whose ester we
wish to make or else to remove one of the products from the reaction
mixture.
Reduction of esters
(i) Catalytic hydrogenation
20
||H , CuO, Hg/H
3 2 2 3 3 2 2 2 3150 C high pressure
O
CH CH CH C O CH CH CH CH CH OH CH OH
(ii) Chemical reduction
4
2 5
||LiAlH /H
3 2 2 3 3 2 2 2 3or Na / C H OH
O
CH CH CH C O CH CH CH CH CH OH CH OH
Reaction of esters with Grignard reagents
The reaction of carboxylic esters with Grignard’s reagent is a good method
for the preparation of 3 alcohols.
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R C
O
OR' R"MgXR C
O
R" R"MgXR C
OMgX
R"
R"
R C
OH
R"
R"
H
Initially ketones are formed. However, as we know, ketones themselves
readily react with Grignard reagent to yield teriary alcohols.
Claisen Condensation
When ethyl acetate reacts with sodium ethoxide, it undergoes a
condensation reaction. After acidification, the product is a - keto ester,
ethyl aceto acetate (commonly known as – aceto acetic ester)
2CH3C
O
OC2H5
NaOC 2H5 CH3 C CH
O
C
O
OC2H5
Na
C2H5OHHCl
CH3C
O
CH2C
O
OC2H5
Condensation of this type is known as Claisen Condensation. For esters, it is
the exact counterpart of the Aldol Condensation. Like the Aldol
Condensation, the Claisen Condensation involves nucleophilic attack by a
carbanion on an electron – deficient carbonyl compound. In the Aldol
Condensation, nuclephilic attack leads to addition (the typical reaction of
aldehydes and ketones). In the Claisen Condensation, nucleophilic attack
leads to substitution (the typical reaction of acyl compounds)
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Mechanism
Step-1 R CH
H
C
O
OC2H5 OC2H5 C
O
OC2H5RCH C
O
OC2H5RCH
CH2C
OC2H5
O
R HC C
R
O
OC2H5 RCH 2 C
O
OC2H5
CH
R
C
O
OC2H5
RCH 2 C
O
CH C
O
OC2H5
R
C2H5O
Step- 2
Step-3 RCH 2C C
H
COC 2H5
R
OO
C2H5O C CRCH 2C
O
R
O
OC2H5 C2H5OH
This step is highly favourable and draws the overall equilibrium toward the
product formation
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Step-4 C CRCH 2C
O
R
O
OC2H53H O
rapidRCH 2C
O
CH
R
C
O
OC2H5RCH 2C C
OH
R
COC 2H5
O
Keto form Enol form
When planning a claisen condensation with an ester it is important to use
alkoxide ion that has the same alkyl group as the alkoxyl group of the ester.
This is to avoid the possibility of transesterification.
An intramolecular claisen condensation is called Dieckmann condensation.
In general, the Dieckmann condensation is useful only for the preparation of
five and six membered rings.
Amides
Preparation:
In the laboratory amides are prepared by the reaction of ammonia with acid
chlorides or, acid anhydrides.
Basic Character of Amides:
Amides are very feebly basic and form unstable salts with strong inorganic
acids. e.g. RCONH2HCl. The structure of these salts may be I or II
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R – C
NH2
O – H
Cl–
+
R – C
NH3
O
Cl–
+
or
I II
Acidic Character of Amides:
Amides are also feebly acidic e.g. they dissolve mercuric oxide to form
covalent mercury compound in which the mercury is probably linked to the
nitrogen.
2RCONH2 + HgO (RCONH)2Hg + H2O
Exercise 11:
PhHC CH CH CH3
O
C O
Optically active
PhHC CH CH CH3
OH
(Optically active)
OH
Exercise 12:
An ester A (C4H8O2) on treatment with excess methyl magnesium
chloride followed by acidification gives an alcohol B as the only
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organic product. Alcohol B on oxidation with NaOCl followed by
acidification, gives acetic acid. Deduce the structures of A & B. Show
the reactions involved.
(a) Hydrolysis of amides
Amides undergo hydrolysis when they are heated with aqueous acid or
aqueous base.
Acidic hydrolysis
R C
O
NH22
Δ
H OH3O R C
O
OH NH4
Basic hydrolysis
R C
O
NH22
Δ
H OOH R C
O
O NH3
(b) Reduction of amides
Amides are reduced by Na/C2H5OH or by LiAlH4 to a primary amine.
R C
O
NH22 5Na / C H OH
4H R CH2 NH2 H2O
(c) Reaction with P2O5
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When heated with P2O5 amides are dehydrated to cyanides.
R C
O
NH22 5
2
P O
H OR C N
Amides may also be converted to cyanides by PCl5.
R C
O
NH25PCl
R C NRCCl 2NH22HCl
(d) Hofmann rearrangement
Amides with no substitutents on the nitrogen react with solutions of
Br2 or Cl2 in NaOH to yield amines through a reaction known as
Hofmann rearrangement.
R C
O
NH2Br2 4NaOH 2NaBr Na2CO3RNH 2
2H2O
Mechanism
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R C
O
N
H
H
OHR C
O
N
H
H
R C N H
O
Br Br
R C N Br
O H
Br
OH
R C N Br
O
-BrR N C O
OH
R N C O
OH
R N C
OH
O
R N C
O
O
H
Carbamate ion
H OH
2 2
3
RNH CO OH
HCO
Illustration 26:
Acetamide reacts with NaOBr in alkaline medium to form
(A) NH3 (B) CH3NH2
(C) CH3CN (D) C2H5NH2
Solution:
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(B)
Illustration 27:
A(C8H8O)NH2OH/HCl
B and C
H H
D E
F G
Alc. KOHCH3 C Cl
O
alc. KOH
(C7H6O2) (C6H7N)
Keeping in mind the fact that B, C, D and E are all isomers of
molecular formula (C8H9NO), identify A to G.
Solution:
(A)
C
CH3
O
(B) C N
Ph
CH3
OH
(C) C N
Ph
CH3 OH (D) Ph C
O
NH CH3
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(E) CH3 C
O
NH Ph (F) Ph COOH
(G) Ph NH2
Illustration 28:
Formic acid and acetic acid may be distinguished by reaction with
(A) sodium
(B) dilute acidic permanganate
(C) 2, 4 dinitrophenylhydrazine
(D) sodium ethoxide
Solution:
(B)
Exercise 13:
(a) CH2 CH CH CH2
CHCOOH
CHCOOH
A
(b) 3NH
2 Δ2CH COOH B
Exercise 14:
Identify A, B, C and D?
(i) R—COOH 3ND /Δ A 2
KOH
Br /ΔB
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(ii) R—COOH 3NH /Δ C 2
KOD
Br /Δ D
Exercise 15:
(i)
Ph C OH
OSOCl2
CH2N2
A
B
H2/Pd(BaSO4)
H
1 eq. DIBALH
C
Find out A, B and C
(ii)
Ph C
O
OHConvert to PhNH2
(iii) (a)
CH3C OH
O
3
+
i 2 eq. CH Li
ii H A
(b)
C
O
C
O
O
1 eq. CH3CH2OH
1. LiAlH4
2. H
B
C
What are A, B and C?
Exercise 16:
An organic compound (A), C8H14O forms an oxime and gives positive
haloform reaction. On ozonolysis it gives acetone and a compound
(B), C5H8O2 (B) forms a dioxime and on subjecting to haloform
reaction gives an acid (C), C4H6O4. On treatment with excess of
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ammonia and strong heating (C) gives a neutral compound (D),
C4H5O2N (D) on distillation with Zn dust forms pyrrole. Suggest
structures for (A), (B), (C) and (D). Give the IUPAC names of
compounds (A) to (D).
Exercise 17:
An acidic compound (A), C4H8O3 loses its optical activity on strong
heating yielding (B), C4H6O2 which reacts readily with KMnO4. (B)
forms a compound C with SOCl2, which on reaction with (CH3)2NH
gives (D). The compound (A) on oxidation with dilute chromic acid
gives a compound (E) which on gentle heating readily gives (F),
C3H6O. Give structures of (A) to (F) with proper reasoning.
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ANSWERS TO EXERCISES
Exercise 1:
(i) Br MgBr
H O3
O
OMgBrMg/Ether
OHKMnO /H4
OH
O
(ii) CH CH
O
OHi COHBr Mg/Ether 2
2 2ii H O3
CH CHBr CH CH MgBr
Exercise 2:
(i) Boiling points of formic acid (100.3°C) and water (100°C) are
almost equal.
(ii) CH2
CH2
HBrCH3 Br
Mg/ ether
CH3Mg
Br2
3
CO
H O CH3COOH
Exercise 3:
A = CH3
CH3
OH
CN
B = CH3
CH3
OH
OH
O
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C = CH3
CH2
OH
O
D =
CH3
OH
O
OH
E =
CH3
CH3 OH
O
Exercise 4:
A = B =
OH
Cl
C =
OH
CN
D =
OH
COOH
E =
O
COOH
Exercise 5:
(A)
(H3C)2HC CH
CH3
C
O
O CH
CH3
CH(CH 3)2
(B) CH3 CH
CH3
CH COOH
CH3
(C) CH3 CH
OH
CH CH3
CH3
(D) H3CHC C(CH3)2
(E) CH3 CH
OSO 2C6H5
CH(CH 3)2
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(F) CH3 CH
Br
CH CH3
CH3
Exercise 6:
Since neutral compound A on hydrolysis gives an acid B and neutral
compound C, it must be an ester.
H
A B CEster Acid Alcohol
Further since the acid B reduces mercuric chloride, it must be formic
acid (HCOOH) which is the only reducing carboxylic acid. So A must be
HCOOC2H5.
H
2 5 2 5BA C
HCOOC H HCOOH C H OH
2 2 2 2HCOOH 2HgCl Hg Cl 2HCl CO
2
NaOH
2 5 3IC
C H OH HCOOH CHI
Exercise 7:
(i) In formate ion resonance gives rise to identical bond lengths
H—C—O– H—C = O
O O–
H—C O
O
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Whereas no such resonance is noticed in formic acid (H — COOH) and
thus C—O bonds are different in HCOOH
(ii) Acetic acid undergoes intermolecular H–bonding to form dimeric
state (CH3COOH)2 and thus results in double molecular weight.
Exercise 8:
OH
O
O
CH3(b) A =
Cl
O
O
CH3B =
Exercise 9:
(i)
A =
C CH2 CH2 COH
OO
B =
CH2 CH2 CH2 COH
O
C =
CH2 CH2 CH2 C
O
Cl
D =
O
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E =
OH
F =
G =
Br
H =
(ii)
C
O
C O
OH
C
O
C
O
(Friedel - crafts alkylation)
Exercise 10:
O
O
CH3
O
OH(c) A =
Exercise 11:
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CH CH CH CH3
O
CPh O
Ph
OH
CH CH CH CH3
O
CPh O
Ph
O
C OHPh
O
CH CHPh CH
O
CH3C OPh
O
CH CHPh CH
OH
CH3
Exercise 12:
A = H C
O
O HC
CH3
CH3
B = CH3 CH
OH
CH3
Exercise 13:
(a) COOH
COOH(A)
(Diel's - Alder Reaction)
(b)
H2C
H2C
C
C
NH
O
O
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Exercise 14:
(i) R—COOH 3ND ,2RCOND
(A) 2
KOH
Br2RND
(B)
(ii) R—COOH 3NH ,2RCONH
(C)
2
KOD
Br2R NH
(D)
Exercise 15:
(i)
Ph C
O
ClA = Ph C
O
OCH 3B =
Ph C
O
HC =
(ii)
Ph C
O
OHSOCl 2
Ph C
O
ClNH3
ExcessPh C
O
NH2
OH Br2
Ph NH2
(Hoffmann rearrangement)
(iii) (a)
CH3C OH
O
CH3LiH3C C
O
OLi CH4
CH3LiH3C C
OLi
OLi
CH3
H
H3C C
OH
OH
CH3
H3C C
O
CH3
(A)
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(b)
C
C
O
O
O CH2 CH3
OH
CH2OH
CH2OH
(B) (C)
Exercise 16:
A is (CH3)2 C = CH–CH2–CH2–COCH3
6–methyl–5–hepten–2–one
B is CH3–CO–CH2–CH2–CHO
4–oxopentanal
C is HO2 CCH2 CH2 CO2 H
1, 4–butandioic acid
D is succinimide
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CH2 – CH2
O = C C = O
N | H
Exercise 17:
|
3 2 3
|| ||
3 3 3 2
|| || ||
3 2 3 3
OH
A. CH C H CH COOH B. CH CH CH COOH
O O
C. CH CH CH C Cl D. CH CH CH C N CH
O O O
E. CH C CH C OH F. CH C CH
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MISCELLANEOUS EXERCISES
Exercise 1:
Suggest a suitable oxidising agent for the conversion.
3 3 3 22 2CH C = CHCOCH CH C = CHCO H
Exercise 2:
Arrange the following compounds in order of increasing acid strength:
3 2 3 2 3 2 3 2 22
i CH CH CH Br COOH,CH CH Br CH COOH, CH CHCO H, CH CH CH COOH
(ii) Benzoic acid, 4-nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4-
methoxybenzoic acid.
Exercise 3:
Why does benzoic acid not undergo Friedel-Crafts reaction?
Exercise 4:
Why carboxylic acids do not form oximes?
Exercise 5:
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Arrange the following in decreasing order of the boiling points:
CH3CH2CH2CH2OH, CH3CH2OCH2CH3.CH3CH2CH2COOH.
Exercise 6:
Why is an amide more acidic than amines?
Exercise 7:
Identify (A) and (B) in the following sequence of reactions.
(i) 3
4
CH COOH excess distil
3HgSOCH CH A B CH CHO
(ii) 34 CH COOHAIPO
3 1075KCH COOH A B
Exercise 8:
Why excess amine is necessary in the reaction of an acyl chloride with
an amine?
Exercise 9:
Write the structure of the products
(a) acetic acid + benzyl alcohol
(b) propionic acid + isobutanol
Exercise 10:
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Complete the following:
H5C6
O
OH
heat
5PCl
CH3
O
OK
Electrolysis
2H O
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ANSWERS TO MISCELLANEOUS EXERCISES
Exercise 1:
Alkaline KMnO4, acidified K2Cr2O7 or HNO3 can not be used since all of
these will cleave the molecule at the site of double bond giving a
mixture of ketone/acids. The most suitable reagent for this oxidation is
NaOI (I2/NaOH) since methyl ketones on treatment with NaOH
undergo Iodoform reaction to give Iodoform alongwith the Na salt of a
carboxylic acid having one carbon atom less than the starting methyl
ketone.
CH3 C
CH3
CH C
O
CH3 3NaOI CH3 C
CH3
CHCOONa 3CHI 2NaOH
2H /H O
CH3 C
CH3
CHCOOH
4 Methylpent 3 en 2 one
(Mesityl oxide)
3 Methylbut 2 en 1 oic acid
( , Dimethylacrylic acid)
Exercise 2:
(i) We know that +I, effect decreases while I effect increases the acid
strength of carboxylic acids. Since +I effect of isopropyl group is more
than that of propyl group, therefore, (CH3)2CHCOOH is a weaker acid
than CH3CH2CH2COOH.
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Further since I effect decreases with distance, therefore,
CH3CH2CH(Br)COOH is a stronger acid than CH3CH(Br)CH2COOH. Thus,
the overall acid strength increases in the order:
CH3 CH
CH3
COOH < CH2 COOHH2C
< CH3 CH CH2
Br
COOH < CH3 CH2 CH COOH
Br
CH3
(ii) Since electron donating groups decrease the acid strength,
therefore, 4–methoxybenzoic acid is a weaker acid than benzoic
acid.
Further since electron–withdrawing groups increase the acid
strength, therefore, both 4–nitrobenzoic acid and 3,
4–dinitrobenzoic acids are stronger acids than benzoic acid.
Further due to presence of an addition NO2 at m – position w.r.t
COOH group, 3, 4–dinitrobenzoic acid is a little stronger acid
than 4–nitrobenzoic acid. Thus the overall acid strength
increases in the order:
4 methoxybenzoic acid benzoic acid 4 nitrobenzoic acid
3,4 dinitrobenzoic acid
Exercise 3:
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Due to deactivation of the benzene ring by electrons withdrawing
effect of the COOH group.
Exercise 4:
Due to resonance between lone pairs of electrons on the O-atom of the
OH group and C = O, the carboxyl carbon is less electrophilic than
carbonyl carbon in aldehyde and ketones. Therefore, nucleophilic
addition of NH2OH to the C = O group of carboxylic acid does not occur
and hence carboxylic acids do not form oximes.
Exercise 5:
CH3CH2CH2COOH > CH3CH2CH2CH2OH > CH3CH2OCH2CH3
Hint: Due to hydrogen bond
Exercise 6:
R NH2
O
R NH2
O
Firstly due to delocalization of lone pair of electrons of N atom over the
C = O group, amino group gets a positive charge which makes N–H
bond weak. Secondly the anion left after removal of a proton is
stabilized by resonance.
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R NH
O
R NH
O
In amines no such stabilization is possible.
Exercise 7:
(i) 3
4
CH COOH excess distil
3 3 3 3HgSO 2 2Acetylene AcetaldehydeEthylidene diacetate Acetic anhydride
CH CH CH CH OOCCH CH CO O CH CHO
(ii) 34
2
CH COOHAIPO , 1075K
3 2 3H O 2KetoneAcetic acid Acetic anhydride
CH COOH CH C O CH CO O
Exercise 8:
R Cl
O
2R NH
R NHR'
O
HCl
The products of this reaction are an amide and HCl. HCl formed in this
reaction protonates unreacted amine and since protonated amines are
not nucleophiles, the reaction with acyl chloride stops.
2 3
No nucleophilic site
R NH HCl RNH Cl
R Cl
O
3RNH No reaction
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Hence, reaction must be carried out with twice as much as amine as
acyl chloride.
Exercise 9:
(a) CH3—COOCH2—Ph
(b) CH3—CH2COOCH2—CH(CH3)2
Hint: Esterification
Exercise 10:
(a) COOH
5PCl
COCl
3HCl POCl
(b) electrolysis
3 2 3CH COOK H O CH COO K
At anode:
CH3
CH3
22CO 2e32CH COO
At cathode:
2 22e 2H O 2OH H
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SOLVED PROBLEMS
Subjective:
Board Type Questions
Problem 1:
Write chemical reactions to effect the following transformation.
(a) Butan-1-ol to butanoic acid
(b) Benzyl alcohol to phenylacetic acid
(c) Bromobenzene to benzoic acid
(d) p-methylacetophenone to benzene-1, 4-dicarboxylic acid
Solution:
(a) 4 2 4Aq. KMnO Dil. H SO
3 2 2 2 3 2 2 3 2 2CH CH CH CH OH CH CH CH COOK CH CH CH COOH
Butan-1-ol Pot. Butanoate Butanoic acid
(b) 3
2
PBr KCN Hydrolysis
6 5 2 6 5 2 6 5 2 6 5 2or HBr H /H O
Benzyl alcohol Benzyl bromide Phenylacetic acid
C H CH OH C H CH Br C H CH CN C H CH COOH
(c)
2
3H OMg Dry ice
6 5 6 5 6 5 6 5dry ether
Bromo benzene Phenyl mag. bromide Benzoic acid
i.e. CO (s)
C H Br C H MgBr C H COOMgBr C H COOH
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(d) CH3 COCH3 4KMnO /KOH, KOOC COOK 2 4Dil. H SO HOOC COOH
P-methylacetophenone Dipotassium benzene- 1, 4-
dicarboxylate
Benzene-1, 4-dicarboxylic acid
Problem 2:
Arrange the following compounds in increasing order of their boiling
points:
acetic acid, methyl formate, acetamide, propan-1-ol
Solution:
(i) Out of all these compounds methyl formate does not undergo H-
bonding therefore, its boiling point is the lowest.
(ii) Amongst the remaining three compounds intramolecular H-
bonding is most extensive in acetamide, followed by acetic acid
and least in propan-1-ol. Therefore, their boiling points decrease
in the same order i.e. acetamide>acetic acid>propan-1-ol. Thus,
the overall increasing order of their boiling points is methyl
formate<propan-1-ol<acetic acid<acetamide.
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Problem 3:
CH3 C
O
OHSOCl2 A
H2/Pd
BaSO 4B
HCNC
H3OD E
Δ
Solution:
A = CH3 C
O
Cl B = CH3 C
O
H C = CH3 C
OH
CN
H
D = CH3 C
OH
COOH
H
E =
O
CH
C
C
CH
O
O
CH3
O
CH3
Problem 4:
A dicarboxylic acid (A), C4H6O4, gave a compound (B), C6H10O4 upon
treatment with excess of methanol and a trace of H2SO4. Subsequent
treatment of (B) with LiAlH4 followed by usual work up gave C,
C4H10O2. Heating of A yielded D, C4H4O3. Assign structures to A, B, C
and D.
Solution:
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A =H2C
H2C
COOH
COOH
B =H2C
H2C
C
C
OCH 3
OCH 3
O
O
C =CH2CH2OH
CH2CH2OH
D =H2C
H2C
C
C
O
O
O
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IIT Level Questions
Problem 5:
Two organic compounds A and B have vapour densities 15 and 30
respectively. (A) reduces Fehling solution, but does not react with
Na2CO3. (B) does not reduce Fehling solution but gives effervescences
with Na2CO3. B when treated with a concentrated solution of base
followed by prolonged heating gives a compound C(C3H6O). Identify A,
B and C.
Solution:
(A) H C
O
H
(B) CH3 C
O
OH
(C) CH3 C CH3
O
Problem 6:
An organic acid A, C3H4O3 is catalytically reduced in presence of
ammonia to give B, C3H7NO2. B reacts with acetyl chloride,
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hydrochloric acid and alcohols. It can also react with nitrous acid to
give another compound C, C3H6O3, along with the evolution of
nitrogen. What are A, B and C. Give reasons?
Solution:
Compound A is acid having one –COOH group only, the remaining part
C2H3O can be
O || CH3 – C – only. Hence, structural formula of A is
CH3 COOH
O
on catalytic reduction keto group is converted into secondary alcohol
which with ammonia will give amino acid, i.e.,
O || CH3 – C – COOH reduction CH3 – CH – COOH 2HNH
CH3 – CH – COOH –H2O
OH |
NH2
|
with nitrous acid, B, react to give
NH2
| CH3 – CH – COOH + HNO2 CH3 – CH – COOH + N2 + H2O
OH |
(B) (C)
Problem 7:
Compound A (C6H12O2) on reduction yields two compounds B and C.
The compound B on oxidation gave D, which on treatment with
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aqueous alkali and subsequent heating furnished E. The latter on
catalytic hydrogenation gave C. The compound D was oxidised further
to give F which was found to be a monobasic acid (mw = 60). Deduce
the structures of A to E.
Solution:
ReductionA B C
OHD E
H2/Pt[O]
O
monobasic acidmol. wt. 60
F
From the molecular weight data F comes out to be CH3COOH. If we
work back B is CH3CH2OH and D is CH3CHO.
2H /PtOH
3 3 3 2 2 2D E C
CH CHO CH HC CH CHO CH CH CH CH OH
Aldol condensation
From here we can deduce that reduction of A with LiAlH4 gives two
alcohols. So A must be an ester. Hence (A) comes out to be
CH3CH2CH2COOC2H5.
Problem 8:
An organic compound (A) on treatment with acetic acid in the
presence of sulfuric acid produces an ester (B). (A) on mild oxidation
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gives (C). (C) with concentrated KOH followed by acidification with dil.
HCl regenerates (A) and produces (D). (D) with phosphorous
pentachloride followed by reaction with ammonia gives (E). (E) on
dehydration produces hydrocyanic acid. Identify compounds A, B, C, D
and E.
Solution:
E-H2O
HCN So (E) must be H C
O
NH2
H C
O
OH
(D)
PCl 5H C
O
ClNH3
H C
O
NH2
(E)
(A) reacting with acetic acid is giving an ester (B). So (A) must be an
alcohol. (A) alcohol on mild oxidation gives (C). Hence (C) is an
aldehyde. (C) with conc. KOH followed by acidification is yielding (A)
alcohol and | |O
H C OH (D). So this reaction is cannizzaro reaction. So A
must be CH3OH.
||
3 3 3 2
O
CH OH CH C O CH HCHO HCOOH HCONH
A (B) C D E
Problem 9:
An aromatic compound A on treatment with CHCl3 and KOH gives B &
C, both of which, in turn give the same compound D when distilled
with Zn dust.Oxidation of D yields E of formula C7H6O2. The sodium
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salt of E on heating with soda lime gives F which can also be obtained
by distilling A with Zn dust. Identify A, B, C, D, E and F.
Solution:
Molecular formula of (E) is C7H6O2 and reaction of its sodium salt with
soda lime (decarboxylation) to form (F) indicates that (E) and (F)
should be C6H5COOH and C6H6 respectively. Since (F) is also obtained
from (A) by reaction with Zn dust, it indicates that (A) should be
phenol. Nature of (A) as phenol is confirmed by the fact that it
explains all the given reactions.
OH
(A)
CHCl 3
OH
CHO
OH
CHOB & C
Zn dust
CHO
(D)
[O]
COOH
(E)
NaOH
Soda lime
(F)
KOH
Zin
c dust
dis
tilla
tion
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Problem 10:
Five isomeric para – disubstituted aromatic compounds A to E with
molecular formula C8H8O2 were given for identification. Based on the
following observations, give structures of the compounds?
(i) both A and B form a silver mirror with Tollen’s reagents; B gives
a positive test with FeCl3 solution also.
(ii) C gives positive iodoform test.
(iii) D is readily extracted in NaHCO3 solution.
(iv) E on acid hydrolysis gives 1, 4 dihydroxy benzene.
Solution:
(A) and (B) gives tollen’s test. Hence must contain aldehyde group.
But (B) shows phenolic test also. Hence (A) & (B) are
OCH 3
CHO
CH2OH
CHO
or
(A)
OH
CH2CHO
(B)
(C) shows Iodoform test show it must contain ||
3
O
C CH group.
Therefore (C) is
COCH 3
OH
(C)
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(D) shows presence of carboxylic group. Hence (D) and (E) are
COOH
CH3
OH
O CH CH2
(D) (E)
Problem 11:
An organic compound A, C8H4O3 in dry benzene in the presence of
anhydrous AlCl3 gives compound B. The compound B on treatment
with PCl5, followed by reaction with H2/Pd (BaSO4) gives compound C,
which on reaction with hydrazine gives cyclised compound D
(C14H10N2). Identify A, B, C and D.
Solution:
Nature of reagents in the conversion of A to B indicates that the
reaction must be Friedal – crafts reaction.
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C
C
O
O
O
C
C C6H5
OH
O
O(A)
C6H6
AlCl 3
PCl 5
C
C C6H5
Cl
O
O
H2/Pd BaSO 4
CHO
C C6H5
O
(C)
N
N
C6H5
(D)
(B)
NH2NH2
Problem 12:
An organic compound A, C6H10O on reaction with CH3MgBr followed by
acid treatment gives compound B. The compound B on ozonolysis
gives compound C, which in presence of a base gives 1- acetyl
cyclopentene. The compound B on reaction with HBr gives compound
D. Write the structures of A, B, C and D.
Solution:
As B undergoes ozonolysis to form C, if must have a double bond. C in
the presence of base gives , unsaturated ketone, so it is aldol
condensation.
O
(A)
(i) CH 3MgBr
(ii) H
CH3
(B)
O3CH3
H
O
O
(C)
OH
COCH 3
OH
-H2O
COCH 3
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HBr
CH3 CH3 Br
(B) (D)
Problem 13:
A phenolic compound (A), C7H6O2 on mild oxidation gives a highly
volatile oil (B). A forms (C) on reaction with dimethyl sulphate in
alkali. Oxidation of (C) with hot KMnO4 gives (D), the silver salt of
which reacts with bromine water followed by heating gives (E)
containing about 72% bromine. Give structures A to E.
Sol.
A can be
OH
CHO (ortho, meta or para)
O H
C H
O
But it must be ortho because on oxidation, it gives a highly volatile oil
indicating that there is intramolecular hydrogen bonding.
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OH
CHO
(A)
OH
CHO
(A)
OH (CH3)2SO4
OCH 3
CHO
(C)
Oxidation
OH
COOH
Oxidation
OCH 3
COOH
(D)
Ag2O
Br2/H2O
OCH 3
BrBr
Br
(E)
(B)
240 100% of bromine in E 69.56%
345
Problem 14:
An aromatic compound (A) gave a mixture of two isomeric products B
and C on reaction with NH2OH. C rearranged to D (C8H9NO) on heating
with H2SO4. D on hydrolysis produced E and F. A was oxidised with per
benzoic acid to G. Hydrolysis of G gave H and E. An anhydride of E and
its sodium salt on condensation with PHCHO produced cinnamic acid. H
on reaction with phthalic anhydride in H2SO4 gave phenolphthalein.
Suggest structures for A to H.
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Sol.
C
CH3H5C6
O
NH2 OH C N
H5C6
CH3
OH
(A)(B)
per benzoic acid
C
OH3C
O
C6H5
H
CH3 C
O
OH
(G)
(E)
C6H5OH
(H)
C N
H5C6
CH3 OH
(C)
2 4H SO , Δ
CH3CONHC 6H5
(D)
H2O
C6H5CO 3H
CH3CONHC 6H5
OH
H2OCH3COOH C6H5NH2
(D) (E) (F)
(CH3CO) 2O C6H5CHO
CH3COONa H5C6
COOH
Anhydride of (E) (Perkin Reaction)
Formation of cinnamic acid proves that the anhydride is acetic
anhydride. Hence E is acetic acid.
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Objective:
Problem 1:
End products of the following sequence of reactions is
O
C CH3
O
2+
1. I +NaOH
2. H , Δ
(A) CHI3 ,
O
COOH
(B) CHI3 ,
O
CHO
(C) CHI3 ,
O
(D) CHI3 ,
COOH
COOH
Solution:
Intermediate is
O
COOH which loses CO2 on heating.
(C)
Problem 2:
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End product of the reaction || ||
3 2 2 2
O O
H C C CH CH CH COH 4
2
1. NaBH
2. H O, H is
(A)
O
CH3
O
(B
)
O
O
CH3
(C) H3CHC
OH
CH2CH2CH2CH2OH
(D
) CH3C
O
CH2CH2CH2CH2OH
Solution:
NaBH4 reduces ketonic group but not acidic group
CH3 CH
OH
CH2 CH2 CH2 COOH
is the intermediate which forms cyclic ester.
O
CH3
O
(A)
Problem 3:
R CH2 CH2OH can be converted into R CH2CH2COOH. The
correct sequence of reagents is
(A) 3 2PBr , KCN, H /H O (B) 3 2PBr , KCN, H
(C) KCN, H (D) HCN, H
Solution:
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3
2
PBr KCN H
2 2 2 2 2 2 2 2H OR CH CH OH R CH CH Br R CH CH CN R CH CH COOH
(A)
Problem 4:
The ease of alkaline hydrolysis is more for
(A) COOCH 3
NO2
(B) COOCH 3
Cl
(C) COOCH 3
(D) COOCH 3
OCH 3
Solution:
There is more electron deficiency on carbonyl carbon in
COOCH 3
NO2
(A)
Problem 5:
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Which of the following undergoes HVZ reaction?
(A) HCOOH (B) CCl3COOH
(C) C6H5COOH (D) none
Solution:
None of these contains - hydrogen.
(D)
Problem 6:
The product C in the reaction 5 2 4PCl H /Pd, BaSO dil. NaOH
3CH COOH A B C
(A) CH3CH = CHCHO (B) CH3CH = CHCOOH
(C) CH3CH2CHO (D) CH3CH2CH2COOH
Solution:
5 2 4
||PCl H /Pd, BaSO OH
3 3 3 3
O
CH COOH CH COCl CH C H CH CH CHCHO
Problem 7:
The end product of the reaction is
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Br2A
H2, Pt KCN
H3OCB
C is
(A) propanoic acid (B) adipic acid
(C) malonic acid (D) succinic acid
Solution:
Br
Br
Br
Br
COOH
COOH
Adipic acid(A) (B) (C)
Problem 8:
Find out D in the following sequence
OH
conc.
H2SO4
A NBS B KCN CH3O D
(A)
CH3
COOH
(B)
COOH
CH3
(C)
COOH
COOH
(D)
CH3
COOH
Solution:
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OH
conc.
H2SO4
NBS
Br
KCN
CNCOOH
Problem 9:
The order of the acidic strength of following acids is
(i) CH3CH2COOH (ii) CH2 = CHCOOH (iii) CH CCOOH
(A) i > ii > iii (B) iii > ii > i
(C) ii > i > iii (D) iii > i > ii
Solution:
sp hybridized carbon is most electronegative and it stabilizes the
carboxylate ion. to the maximum extent.
Problem 10:
CH3
COOH
OH
CrO3 A Δ
B
A and B are
(A) CH3COCH2COOH, CH3COCH3
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(B) CH3CH(OH)CH3, CH3COCH3
(C) CH3COCH3, CH3COOH
(D) CH3CH = CHCOOH, CH3CH = CH2
Solution:
COOH
OH
CrO3 COOH
O
Δ
O
(A) (B)
(A)
Problem 11:
*
3
3
i Me MgXNaHCO
3 2 ii H OCH CH COOH A a gas B
B is *
14C is C
(A) *
3CH COOH (B) *
3CH COOH
(C) *
3 2 3CH CH COCH (D) *
3 3CH COCH
Solution:
*
3
*NaHCO
3 2 3 2 2 2CH CH COOH CH CH COONa H O CO
3
* *H O
2 3CO MeMgX CH COOH
(A)
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Problem 12:
OC2H5
CH2CO2C2H5
O
C2H5OA
A is
(A) O
OC2H5
(B) O
CO2C2H5
(C)
O
O
(D) O
O
Solution:
OC2H5
CH2CO2C2H5
O
C2H5OOC2H5
CH
O
C O
O
C2H5
O
C
O
O C2H5
Problem 13:
The carboxylic acid which has maximum solubility in water is
(A) phthalic acid (B) succinic acid
(C) malonic acid (D) salicylic acid
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Solution:
Salicylic acid has intramolecular hydrogen bonding. So it is least
soluble. Out of the first three acids malonic acid has smallest non –
polar alkyl part. So it is most soluble.
(C)
Problem 14:
Br
CH3
H
O
O
2
3
i Mg/ether
ii CO
iii H O
The final product is
(A) COOH
CH3
H
O
O
(B) MgBr
CH3
CHO
(C) COOH
CH3
CHO
(D) None of these
Solution:
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Br
CH3
H
O
O
Mg/ ether
MgBr
CH3
H
O
O
2CO
COO
CH3
H
O
O
H3O
COOH
CH3
CHO
Problem 15:
In the reaction 3 2 5NH P O
3 2CH CH COOH A B C,
Identify C.
(A) CH3CH2CN (B) CH3CH2CH2NH2
(C) CH3CH2CH2OH (D) (CH3CH2CO)2O
Solution:
CH3CH2COOHNH3 H2CH3C C
O
ONH 4Δ CH2 C
O
NH2CH3
P2O5
CH2 CNCH3
Fill in the blanks
Problem 16:
Benzoic acid has a………….pKa value than phenol.
Solution:
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Lower
Problem 17:
The acid derivatives are hydrolysed…………..readily in either alkaline or
acidic medium than in neutral medium.
Solution:
More
Problem 18:
O-nitrobenzoic acid is…………acid than p-nitrobenzoic acid.
Solution:
Stronger
Problem 19:
Malonic acid on heating with phosphorous pentoxide gives…………………..
Solution:
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Carbon suboxide (C3O2)
Problem 20:
Cinnamic acid can be prepared from benzaldehyde
using…………….reaction and shows…………..isomerism.
Solution:
Perkin, geometrical
True and False
Problem 21:
Methanoic acid like ethanoic acid can be halogenated in the presence
of red phosphorous and Br2.
Solution:
False:
Due to the absence of - hydrogen, methanoic acid can not be
halogenated by using Br2/red P.
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Problem 22:
As the number and size of the substituents around the COOH group
increases the rate of esterification slows down.
Solution:
True
Problem 23:
The boiling point of propionic acid is less than that of n-butanol, an
alcohol of comparable molar mass.
Solution:
False:
Due to extensive intermolecular hydrogen bonding, and the cyclic
dimer formation, propanoic acid has higher b.p. than n-butanol.
Problem 24:
The conjugate base of chloro acetic acid is more resonance stabilized
than the conjugate base of acetic acid.
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Solution:
True:
Due to the presence of chlorine atom.
Problem 25:
The correct order of acid strength of the given acid is
2 4 3 2 2 5H SO > CH COOH > H O > HC ≡ CH > C H OH
Solution:
False:
The correct order is2 4 3 2 2 2 5H SO CH CO H H O C H OH HC CH .
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ASSIGNMENT PROBLEMS
Subjective:
Level-O
1. Describe the following:
(a) Saponification (b) Decarboxylation
2. How is acetic acid prepared from?
(i) Acetylene (ii) Ethyl alcohol
3. How is Grignard reagent employed to prepare a carboxylic acid?
4. Explain the following about acetic acid:
(i) Its boiling point is higher than that of n – propanol.
(ii) It is weaker acid than formic acid.
(iii) It is stronger acid than phenol.
5. Acid anhydrides have higher boiling points than the corresponding
carboxylic acids.
6. Give IUAPC name for each of the following compounds.
(a) H3CHC CHCOOH (b) COOH
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(c) CHCOOH
CH3
(d) Cl2CHCOOH
(e)
HOC
O
(CH2)3COOH
7. HOCH2CH2CH2CH2COOH can be prepared from HOCH2CH2CH2CH2Br in
two ways. One is through Grignard synthesis and the other through
nitrile synthesis. Which method would you prefer?
8. Why the bond length of C = O in carboxylic acids is bit longer than in
aldehydes?
9. Acetic acid can be halogenated in presence of phosphorus and chlorine
but formic acid cannot be halogenated in the same way.
10. During the preparation of esters from a carboxylic acid and an alcohol
in the presence of an acid catalyst, the water or the ester should be
removed as fast as it is formed. Explain.
11. Identify the products in the following reactions
(a) 3 2
3
i) CH CH MgBr /ether
3 ii) H OHCOOCH (b) dil. NaOH
6 5 3 2C H CHO CH CH CHO
12. Explain why benzamide is less easily hydrolysed than methyl benzoate.
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13. There are two –NH2 groups in semicarbazide. However only one is
involved in the formation of semicarbazones. Explain.
14. Carbon-oxygen bond length in formic acid are 1.24 Å and 1.36 Å but in
sodium formate both the carbon-oxygen bonds have same value
i.e.1.27 Å.
15. Acid and acid derivatives although contain >C=O group, do not
undergo the usual properties of carbonyl group explain.
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Level – I
1. Convert the following:
(a) Acetaldehyde Butane 1, 3 diol
(b) Acetaldehyde But 2 enoic acid
2. Discuss the action of heat on - hydroxy carboxylic acids.
3. Compare the Ka1 and Ka2 values of maleic and fumaric acids.
4. R NH2 are soluble in aqueous acids like HCl, H2SO4 etc but ||
2
O
R C NH
are almost insoluble in those acids. Explain.
5. What happens when:
(i) Dry chlorine is passed through acetic acid in presence of
sunlight.
(ii) Formic acid is reacted with ammonical silver nitrate solution.
6. Complete the following equations by writing the missing A, B, C.
3 4H O KMnO
3 2 3 2 2 2CH CH MgBr A CH CH CH CH OMgBr B C
7. What is A and B?
3CrO warm
4 8 3C H O A BCH3 CH3
O
2CO
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8. (a) ||
3 2
O
CH CH CH CH C OH A
(b) 2 3 2|| ||HO C CH C CH C OH B
O O
9. Cite 2 reasons why malonic acid does not form a cyclic anhydride.
10. 6 62 4 2
4 3
O C HH SO SOCl
HgSO anhy AlClCH CH A B C D
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Level – II
1. A hydrocarbon A, (C8H10), on ozonlysis gives compound B (C4H6O2)
only. The compound B can also be obtained from the alkyl bromide, C
(C3H5Br) upon treatment with magnesium in dry ether, followed by
CO2 and acidification. Identify A, B and C and also gives equations for
the reactions.
2. Predict the product of the reaction given below, given that it is formed
under high – dilute conditions. EtO cat
2 2 24EtO C CH CO Et
3. Compound (A), C5H8O2, liberates CO2 on reaction with sodium
bicarbonate. (A) exists in two forms neither of which is optically active.
It yields compound (B), C5H10O2 on hydrogenation. Compound (B) can
be separated into two enantiomorphs. Write the structural formulae of
A and B.
4. An organic compound (A) reacts with PCl5 to form a compound (B). A
also reacts with NaOH to form a compound (C). Both B and C react
together to form a compound D (C6H10O3). What are A, B, C and D?
5.
a Δ
CH3
CH3
O
CH3
H
`
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b 2 6B H
HCCH3
O
CH2OH
c 2
2
1 Ag O
2. BrO2N Br
6. The active ingredient of an insect repellent is N, N–diethyl–m–
toluamide. Outline a synthesis of this compound starting with benzene.
7. A carboxylic acid (A) of unknown structure was found to contain only
C, H and O. 154.5 mg of this acid required 11.9 mL of 0.22 N NaOH to
reach equivalence point.
Gentle heating of A led to evolution of CO2 and formation of new
carboxylic acid (B) with nutralisation equivalent of 74. Suggest
structures for A & B.
8. Compound (A), C5H11NO is not soluble in cold dil. alkaline or acidic
solutions. When (A) is refluxed with NaOH solution, a gas (B) is
evolved and a salt (C) is formed. Acetyl chloride reacts with (B) to give
(D), C4H9NO. (B) reacts with HNO2/HCl to give a yellow oil (E). Give
structures of A to E.
9. Compound (A) (C6H12O2) on reduction with LiAlH4 yielded two
compounds B and C. The compound B on oxidation gave (D) which on
treatment with aqueous alkali and subsequent heating furnished (E).
The latter on catalytic hydrogenation gave C. The compound D was
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oxidised further to give F which was found to be monobasic acid with
molecular weight 60. Deduce the structures of A, B, C, D, E and F.
10. Suggest a mechanism for each of the following tranforamations.
O
CO 2Et
CO 2Et t - BuO
CO 2Et
O
CO 2Et
t - BuOH
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Objective:
Level – I
1. Which of the following will not give cyclic anhydride on heating?
(A)
H2C
COOH
COOH
(B) H2C
H2C
COOH
COOH
(C)
H2C
CH2COOH
CH2COOH
(D) COOH
COOH
2. In the reaction
2Cl / red P alc. KOH
3 2CH CH COOH A B
the compound B is
(A) CH3CH2OH (B) CH3CH2COCl
(C) CH2 = CH COOH (D) CH3CHClCOOH
3. Give the structure of the product of the following reaction
O
CH3
O
CH3NH2
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(A)
O
CH3
N CH3
(B)
O
CH3
NHCH 3OH
(C)
CH3 NH C
O
CH2 CH2 CH
OH
CH3
(D) None
4. Malonic and succinic acids are distinguished by
(A) heating (B) NaHCO3
(C) both A and B (D) none
5. Consider the following acids
1. MeCH2COOH
2. Me2CHCOOH
3. Me3CCOOH
4. Et3CCOOH
Correct order of rate of esterification.
(A) 1 > 3 > 2 > 4 (B) 1 > 2 > 3 > 4
(C) 4 > 3 > 2 > 1 (D) 3 > 2 > 1 > 4
6. Which of the following compound on heating with caustic potash gives
ammonia?
(A) CH3CONH2 (B) CH3CH2NH2
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(C) (CH3)2C = NOH (D) ||
2 3 2
O
CH C N CH
7. Identify the final product in the following sequence of reactions
CH3CH2MgBrH2C CH2
O
XH
YKMnO 4
Z
(A) CH3CH2CH2COOH (B) CH3CH(CH3)COOH
(C) CH3CH2COOH (D) CH3CH2CH2CH2OH
8. The typical reaction of carbonyl group is suppressed the most in
(A) CH3 CHO (B) CH3CONH2
(C) CH3COCH3 (D) CH3COOCH3
9. Which of the following carboxylic acids undergoes decarboxylation
easily?
(A) C6H5COCH2COOH (B) C6H5COCOOH
(C) C6H5 HC
OH
COOH
(D) CH
NH2
COOHH5C6
10. Benzoic acid gives benzene on being heated with X and phenol gives
benzene on being heated with Y. Therefore X and Y respectively are
(A) sodalime and copper (B) Zn dust and NaOH
(C) Zn dust sodalime (D) sodalime and Zn dust
11. 2 5 2
||P O Br
2 OH
O
X R C NH Y
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X and Y respectively are
(A) R CN, RCH2NH2 (B) RNH2, RCN
(C) RCN, RNH2 (D) RCN, RCH2NH2
12. 2 5 3Al OC H
32CH CHO
Product and the name of the reaction respectively are
(A) CH3COOCH2CH3, Tisckenko
(B) CH3COOH, CH3CH2OH, cannizzaro
(C) 2CH3CH2OH, perkin
(D) none
13. Which of the following reactions sounds correct?
1.
COOH
COOH Br2
BrHOOC
Br COOH
2.
COOH
COOH alk.KMnO 4
COOHOH
OH COOH
Cold
(meso)
3. Br2
COOHBr
Br COOH
HOOC
COOH
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4. alk.KMnO 4
COOHOH
OH COOH
HOOC
COOH
(meso)
(A) 1 and 2 (B) 2 & 3
(C) 3 & 4 (D) 1 & 4
14.
CH3ONa
C
O
OH
(A)
C O CH3
O
(B)
C O Na
O
CH3OH
(C)
C
O
CH3
(D)
C
O
OH
15. The reaction of ||
2
O
R C NH with a mixture of Br2 and KOH gives R NH2
as a product. The intermediate involved in this reaction is
(A) ||O
R C NHBr (B) R N C O
(C) R NHBr (D) ||
2
O
R C NBr
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16. Among the carboxylic acids:
CH3
O O
OH
OHOO
H
OH
O
and
COOH
NO2
(a) (b) (c) (d)
the, one or more than one undergoing decarboxylation readily is (are)
(A) Only a (B) Both a and b
(C) a, b and d (D) a, b and c
17. In the following reaction,
+
i) DIBALH
ii) Hhydroxy carboxylic acid
A is
(A) lactone (B) Lactide
(C) a hemiacetal (D) , - unsaturated acid
18. Which of the following can form an anhydride on heating?
(A) H
OH
O
(B)
O
OH O
OH
(C)
O
OH OH
O
(D)
OH
O
O
OH
19. i) ND , 3
ii) KOH, Br , H O2 2Product is
OH
O
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(A)
NH2
(B)
ND 2
(C)
NH2O
(D)
ND2O
20. Acetamide is treated separately with the following reagents. Which one
of these would give methylamine?
(A) PCl5 (B) Sodalime
(C) NaOH + Br2 (D) None of these
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Level-II
1. ||
2 2
O
Ph C OH CH N Product
(A) PhCOOCH2N2 (B) PhCOOCH3
(C) PhCH3 (D) none of these
2.
H C
O
O
H
A
B
H C
O
OC
D
(C O) bond lengths designated by A, B, C, D are in order
(A) A = C < B = D (B) A < B < C = D
(C) A < C = D < B (D) all are equal
3. A (monobasic acid) 3NH /Δ
BKBrO
ΔC 2HNO
(H3C)3C OH
or
(H3C)3C Br
or
H3C C CH3
CH2
A is
(A) 3 3CH C COOH (B)
CH3 CH CH2 COOH
CH3
(C) CH3 CH2 CH2 COOH (D)
CH3 C CH COOH
CH3
CH3
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4. O
O
O
O
on saponification forms
(A) OH
OH
O
H
H
O
(B) 2CH3OH, 2HCOOH
(C) OH
OH
O
OH
OH
O
(D) OH
OH
H C
O
OH
5.
4 10P OA
3
MeMgBr
H OBC NH2
O
2
2 3
i Ca OH
ii I , H O C
(A)
C CH3
O
(B) COOH
(C)
C
O
(D)
6. 2 5 2 5 3C H O / C H OH i) H O
ii)(A) (B)
O
COOC 2H5
COOC 2H5
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The compound B is
(A)
O
COOH
(B) O
COOH
COOH
(C) O
COOEt
COOEt
(D) O
COOEt
7. Which of the following reaction/s will yield carboxylic acid?
1. RMgX + CO2
2. PhOH + CHCl3 + NaOH
3. PhCN + H3O+
4. PhOH + NaOH + CO2 H
(A) 3 and 4 (B) 2 and 4
(C) 2 only (D) 4 only
8. An ester (A) with molecular formula C9H10O2 was treated with excess
of CH3MgBr and the complex so formed was treated with H2SO4 to
give an olefin (B). Ozonolysis of (B) gave a ketone with molecular
formula C4H8O which shows positive iodoform test. The structure of
(A) is
(A) 6 5 2 5C H COOC H (B)
||
2 5 6 5
O
C H C O C H
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(C) C4H9COOC4H9 (D) ||
5 11 3 7
O
C H C O C H
9. An ester (A) with molecular formula C11H14O2 was treated with LiAlH4
to give two compounds C9H12O (B) and C2H6O (C). B on heating with
an acid forms C9H10(D). Compound (D) on oxidation with KMnO4 forms
phthalic acid. Compound A is
(A) CH2COOC 3H7
(B)
H2C
CH3
C
O
O C2H5
(C)
H2C C
O
O C2H5
CH3
(D)
H2C C
O
O C2H5
CH3
10. O
CH3CO2OHP. P is
(A)
O
O
(B)
O
O
O
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(C)
O
O
(D) O
OHOH
11. Identify the product ‘A’ in the following reaction.
H2C
COOH
COOH
ΔCH3COOH A
(A) CO2 (B) CH3CHO
(C) CH3OH (D) none of theses
12. Which carboxylic acid of equivalent mass 52 g losses CO2 when heated
to give an acid of equivalent mass of 60 g?
(A) CH3COOH
(B) CH2(COOH)2
(C) CH2(COOH) CH2(COOH)
(D) HOOC(CH2)3COOH
13. CH3CH2COOHNaN3, conc. H2SO4
ΔX by reaction R 1
Y by reaction R 2
Br2/P
Choose the correct alternate
X Y R1 R2
(A) CH3CH2NH2 3|
CH C HCOOH
Br
Curtius HVZ
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(B) CH3CH2CONH2 CH3CH2OBr HVZ Hoffmann
(C) CH3CH2NH2 CH3CH2OBr Curtius HVZ
(D) none of these
14.
O
O
O
NaBH 4X LiAlH4Y X & Y respectively are
(A)
O
OH
OH
and
(B)
O
O
OH
andOH
OH
(C) O
O
both are
(D) OH
OHboth are
15.
Prolonged
O
COOH
COOH
COOH
Heating
(A) O
COOH
COOH
(B) COOH
COOH
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(C)
C
C
O
O
O
(D)
C
C
O
O
O
O
16. When acetic acid reacts with ketene, product formed
(A) ethyl acetate (B) aceto - acetic ester
(C) acetic anhydride (D) no reaction
17. R—CH2—CH2OH can be converted in R—CH2CH2COOH. The correct
sequence of reagents is
(A) PBr3, KCN, H+ (B) PBr3, KCN, H2
(C) KCN, H+ (D) HCN, PBr3, H+
18. The pKa of acetylsalicylic acid (aspirin) is 3.5. The pH of gastric juice in
human stomach is about 2-3 and pH in the small intestine is about 8.
Aspirin will be.
(A) Unionized in the small intestine and in the stomach
(B) Completely ionized in the stomach and almost unionized in the
small intestine.
(C) Ionized in the stomach and almost unionized in the small
intestine
(D) Ionised in the small intestine and almost unionised in the
stomach
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19. On subjecting mesityl oxide to the iodoform reaction, one of the
products is the sodium salt of an organic acid. Which acid is obtained?
(A) (CH3)2C=CH—CH2COOH (B) (CH3)2CH—COOH
(C) (CH3)2C=CH—COOH (D) (CH3)2C=CH—CO—COOH
20. The ease of alkaline hydrolysis is more for
(A)
COOCH 3
NO 2
(B) COOCH 3
Cl
(C) COOCH 3
(D) COOCH 3
OCH 3
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ANSWERS TO ASSIGNMENT PROBLEMS
Subjective:
Level – O
1. See the text
2. (i) 4
o2 4
1% HgSO
dil. H SO , 80 CCH CH
CH3 H
O
4
[O]
KMnO
CH3 OH
O
(ii)
2 2 7 2 4
[O]
3 2 K Cr O /H SO ,CH CH OH
CH3 H
O
4
[O]
KMnO
CH3 OH
O
3.
2
(i) Dry ether
3 (ii) H /H OCH MgBr O C O
CH3 OH
O
4. (i) In acetic acid O–H bond is more polar so H-bonding will be
stronger and boiling point will be high.
(ii) Due to the presence of +I effect of –CH3 group acetic acid is
weaker than formic acid.
(iii) It is stronger acid than phenol due to the presence of
symmetrical resonating structure whereas in case of phenoxide
ion negative charge comes on less electronegative carbon atom
and contributions of resonating structures are less.
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5. Due to the high molecular mass of anhydride van der Waals’ force
increases hence boiling point increases.
6. (a) But-2-enoic acid
(b) 2-Cyclohexenecarboxylic acid
(c) 2-Cyclopropylpropanoic acid
(d) 2, 2 – Dichloroethanoic acid
(e) Pentane-1, 5-dioic acid
7. A nitrile synthesis. Preparation of Grignard reagent from
HOCH2CH2CH2CH2Br is not possible because of the presence of the
acidic hydroxyl group.
8. Due to resonance.
9. This is HVZ reaction. It occurs only in those carboxylic acids which
have -hydrogen atoms. Acetic acid posseses three -hydrogen atoms
but formic acid does not have even a single -hydrogen atom. Thus,
formic acid does not undergo this reaction.
10. Because water form can again hydrolyse the ester into acid and
alcohol.
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11. (a) H
O
OCH3
3 2
3
(i) CH CH MgBr / ether
(ii) H O
H5C2 C2H5
OH
(b) H5C6 H
O
CH3
H
O
dil. NaOH
6 5 2C H CH CH CH CHO
12. In Benzamide carbonyl group is in resonance with lone pair of N so
positive charge decreases therefore nucleophilic attack inhibit.
13. Lone pair on nitrogen atom adjacent to the carbonyl group will
participate in the resonance with the carbonyl group so this lone pair is
not available for the attack at carbonyl group.
14. In formate ion resonance gives rise to identical bond lengths
H—C—O– H—C = O
O O–
H—C O
O
Whereas no such resonance is noticed in formic acid (H — COOH) and
thus C—O bonds are different in HCOOH.
15. It is because of the possibility of resonance which compensates the
electron deficiency of carbonyl carbon to some extent for example.
R C
O
NH2 R C
O
NH2
or
R C
O
OH R C
O
OH
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Level – I
1. (a) CH3
O
H2dil. NaOH
(aldol) CH3
OH
O
H
4LiAlH
CH3 OH
OH
(b) CH3
O
H2dil. NaOH
(aldol) CH3
OH
O
H CH3 H
O
Tollen's reagent
CH3 OH
O
2. - Hydroxy acids are not capable of forming lactones because of
presence of strain in three – membered ring lactone. But they undergo
intramolecular self-esterification to yield a cyclic product containing
two carbonyl groups and is called a lactide.
CHOH
COOH
CH3 COOH
CHCH 3HO22H O
O
O
CH3
O
O
CH3
3. Ka1 of maleic acid is more than that of fumaric acid. This is explained
by the fact that conjugate base of maleic acid is stabilized by hydrogen
bonding
C C
CC
OO
O OH
H H
But Ka2 of maleic acid is lesser than that of fumaric acid because it is
difficult to remove hydrogen bonded H in the conjugate base of maleic
acid.
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4. This can be explained by the fact that amines are more basic than
amides. The less basicity of amides are due to cross conjugation.
5. (i) Cl
3 2sunlightCH COOH CH ClCOOH
(ii) 3 2 3 4 3 2HCOOH Ag(NH ) NO Ag 2NH NO CO
6.
3 2CH CH MgBr
O
H2C CH23H O
3 2 2 2 3 2 2 2CH CH CH CH OMgBr CH CH CH CH OH(A) (B)
4KMnO
3 2 2CH CH CH COOH(C)
7.
4 8 3(C H O )
CH3
OH
O
OH
(A)
CH3
O O
OH
(B)
warm
CH3
O CH3
2CO3CrO
8. (a) CH3 CH = CH CH3 (A) (decarboxylation)
(b)
O
CH3
CH3
O
O
(B)
(cyclic anhydride formation)
9. 1. The cyclic anhydride of malonic acid would have a strained four –
membered ring.
2. The heating causes decarboxylation instead.
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HOC
O
CH2C
O
OHΔ
HO CH3C
O
CO 2
10.
A = CH3 C
O
H B = CH3 C
O
OH C = CH3 C
O
Cl
C
O
CH3
D =
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Level – II
1. A = COOHB =
BrC =
2. High dilute conditions favour intramolecular condensation over inter –
molecular process.
EtO
2 2 2 2 2 2 2 24EtO C CH CO Et EtO CCHCH CH CH CO Et
CO2Et
OEt
O
CO2Et
O
3.
A = C C
CH3
H COOH
CH3
(or) C C
CH3
H CH3
COOH
B = (or)H5C2 C COOH
CH3
H
HOOC C C2H5
CH3
H
4. ||
3 2
O
A. CH CH COH
||
3 2
O
B. CH CH C Cl
||
3 2
O
C. CH CH C O Na
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3 2 2 3|| ||
D. CH CH C O C CH CH
O O
5. (a) CH3
CH3
O
CH3
COOH
(b)
CCH3
O
COOH
(c) O2N COOH
6.
CH3Cl
AlCl 3 anhy
CH3
KMnO 4
OH, heat
COOH
CH3Cl
Anhy AlCl 3
COOH
CH3
SOCl 2
C
CH3
O
Cl
(C2H5)2NH
CH3
C N(C2H5)2
O
7. The neutralization equivalent of A can be determined as follows
= 154.5
59 gm11.9 0.22
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The fact that A on gentle heating is giving another carboxylic acid
suggests that A is a dicarboxylic acid and that too it must be malonic
acid or its derivative.
OH C
O
C C
R'
R O
OH Δ H C C
R'
R O
OH CO2
Acid (B) has equivalent weight 74. So its molecular weight is 74. It can
be deduced that it must be propanoic acid. So
OH C
O
CH C
CH3 O
OH Δ CH3 CH2 COOH CO2
(A) (B)
8. ||
2 5 3 2
O
A. C H CN CH 3 2B. CH NH
2 5C. C H COONa 3 32
D. CH NCOCH
3 2E. CH NNO
9. ||
3 2 2 2 3
O
A. CH C O CH CH CH CH 3 2B. CH CH OH
3 2 2 2C. CH CH CH CH OH
3D. CH CHO
||
3
O
E. CH CH CH C H 3F. CH COOH
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10.
O
CO 2Et
O2C 1. t - BuO
O
CO2Et
O2C
CO 2Et
O2C O
CO 2Et
O
O2C
CO 2Et
O
O2C
Et Et
EtEt Et
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Objective:
LEVEL - I
1. A 2. C 3. C
4. C 5. B 6. A
7. A 8. B 9. A
10. D 11. D 12. A
13. C 14. A 15. B
16. C 17. C 18. D
19. D 20. C
LEVEL – II
1. B 2. B 3. A
4. C 5. B 6. A
7. A 8. B 9. D
10. B 11. B 12. B
13. A 14. B 15. D
16. C 17. A 18. D
19. C 20. A