Problem 11.4 : The solubility of o-nitrophenol Ar - OH + NaOH(aq) Ar-O Na⊕(aq)
and p-nitrophenol is 0.2 g and 1.7 g/100 g of Ar - O Na⊕(aq) + HCl(aq) + H2O(l)
H2O respectively Explain the difference.
Ar - OH↓
Solution : + NaCl(aq)
δ- Do you know ?
O δ⊕
Sodium bicarbonate, sodium
H (intramolecular hydrogen hydroxide, sodium metal are
N⊕ O - bonding in o-nitrophenol) increasingly strong bases. Weak and strong
Oδ - acids can be distinguished from each other
qulitatively by testing their reactivity
δH⊕ -δH⊕ O⊕ O -δH⊕ δ- δH⊕ towards bases of different strengths. A
δ- N O weak acid does not react with a weak
O H base, it requires a stronger base instead.
O- Hence phenols react with NaOH but not
δ⊕ with NaHCO3. A strong acid shows high
reactivity towards weak as well as strong
(intermolecular hydrogen bonding in p-nitrophenol base. For example : HCl is a strong acid.
Its reacts with both NaHCO3 and NaOH as
and water) shown below:
p-Nitrophenol has strong intermolecular
hydrogen bonding with solvent water. On
the other hand, o-nitrophenol has strong
intramolecular hydrogen bonding and
therefore the intermolecular attraction
towards solvent water is weak. The stronger HCl(aq)+NaHCO3(aq)
the intermolecular attraction between solute H2O(l)+ NaCl(aq)+ CO2↑
and solvent higher is the solubility. Hence
p-nitrophenol has higher solubility in water HCl(aq)+NaOH(aq) NaCl(aq)+H2O(l)
than that of o-nitrophenol.
11.4.4 Chemical properties of Alcohols and • Alcohols show no acidic character in
Phenols aqueous solution, thus, alcohols do
a. Laboratory tests of alcohols and phenols : not react with either aqueous NaHCO3
or aqueous NaOH. Very weak acidic
i. Litmus test : Water soluble alcohols and character of alcohol is revealed in the
phenols can be tested with litmus paper. reaction with active metal. When alcohols
Aqueous solution of alcohols is neutral to are treated with very strong base like alkali
litmus (neither blue nor red litmus change metal Na or K they react to give sodium
colour). Aqueous solutions of phenols turn or potassium alkoxide with liberation of
blue litmus red. Thus, phenols have acidic hydrogen gas.
character.
2R - OH + 2Na 2R-O Na⊕ + H2↑
ii. Reaction with bases :
Liberation of H2 gas is used to detect the
• Acid strength of phenols being very low, presence of alcoholic -OH group in a molecule.
phenols cannot react with NaHCO3 but
react with NaOH. iii. Characteristic test for phenols : Phenols
reacts with neutral ferric chloride solution to
Ar - OH + NaHCO3(aq) No reaction give deep (purple/violet/green) colouration of
ferric phenoxide.
Phenols dissolve in aqueous NaOH by forming
water soluble sodium phenoxide and are 3Ar - OH + FeCl3 (Ar -O)3 Fe + 3HCl
reprecipitated/regenerated on acidification
with HCl. (neutral) (deep colour)
242
iv. Distinguishing test for alcohols (Lucas Electron donating inductive effect (+I effect) of
alkyl group destabilizes the alkoxide ion (the
test) : Primary, secondary and tertiary conjugate base of alcohol). As a result alcohol
does not ionize much in water, and behaves
alcohols can be distinguished from each other like neutral compound in aqueous medium.
• Ionization of phenol is represented by the
in the laboratory using Lucas reagent (conc.
equilibrium shown in Fig. 11.1.
HCl and ZnCl2). The reaction involved is :
R - OH HCl R - Cl
ZnCl2
Alcohols are soluble in Lucas reagent but
the product alkyl chloride is not. Hence, the O
+ H3O⊕
clear solution becomes turbid when product O-H
+ H2O (phenoxide)
starts forming. Tertiary alcohols reacts fast
Phenol OO
and the reagent turns turbid instantaneously. O
Secondary alcohols turn the reagent turbid
slowly. Primary alcohols turn the reagent
turbid only on heating.
b. Reactions due to breaking of O -H bond.
i. Acidic character of alcohols and phenols :
From the laboratory tests it is I II III
understood that in aqueous medium phenols
show weak acidic character while alcohols are OO
neutral. It is clear, therefore, that the reactivity
of alcohols and phenols towards ionization IV V
of O-H bond in them is different. The reason
behind this difference lies in the extent of Fig. 11.1 Ionization of phenol and resonance
stabilization of their respective conjugate stabilization phenoxide ion
bases by electronic effects as shown below.
Phenoxide ion, the conjugate base of
• Ionization of alcohols is represented by the phenol, is resonace stabilized by delocalization
following equilibrium of the negative charge.Therefore phenol ionizes
in aqueous medium to a moderate extent, and
R - OH + H2O R O + H3O⊕ thereby shows a weak acidic character.
(alcohol) (alkoxide)
Problem 11.5 : Arrange the following compounds in decreasing order of acid strength and
justify.
i. CH3-CH2-OH ii. (CH3)3C-OH
iii. C6H5-OH iv. p-NO2-C6H4-OH
Solution : Compounds (iii) and (iv) are phenols and therefore are more acidic than the alcohols
(i) and (ii). The acidic strenghts of compounds depend upon stabilization of the corresponding
conjugate bases. Hence let us compare electronic effects in the conjugate bases of these
compounds :
Alcohols : H3C
CH3 CH2 O (Conjugate base of (i)) and H3C C O (Conjugate base of (ii))
H3C
243
The conjugate base of the alcohol (i) is destabilized by +I effect of one alkyl group,
where as conjugate base of the alcohol (ii) is destabilized by +I effect of three alkyl groups.
Hence (ii) is weaker acid than (i)
O OO O O O
O N O O N O O N O O N O O N O O N O
⊕ ⊕ ⊕ ⊕ ⊕ ⊕
IV V VI
I II III
Phenols : The conjugate base of p-nitrophenol (iv) is better resonance stabilized due to six
resonance structures compared to the five resonance structures of conjugate base of phenol (iii)
(see Fig. 11.1). The resonance structure VI has -ve charge on only electronegative oxygens.
Hence the phenol (iv) is stronger acid than (iii). Thus the decreasing order of acid strength is
(iv) > (iii) > (i) > (ii)
Use your brain power O H⊕ O
R-OH + HO-C-R' R-O-C-R' + H2O
What are the electronic effects
(alcohol) (acid) (ester)
exerted by -OCH3 and -Cl ? predict
the acid-OstHrengrethlatoivfeH3tCo-Op-arent-OHpheannodl O H⊕ O
Cl - Ar-OH + HO-C-R' Ar-O-C-R' + H2O
-OH . (phenol) (acid) (ester)
ii. Esterification : Alcohols and phenols form Alcohols and phenols react with acid
esters by reaction with carboxylic acid, acid anhydrides in presence of acid catalyst to form
halides and acid anhydrides. The reaction ester.
between alcohol or phenol with a carboxylic
acid to form an ester is called esterification. OO H⊕ O
R-OH+R'-C-O-C-R' R-C-OR'+R'-COOH
Esterification of alcohol or phenol is carried
out in the presence of concentrated sulphuric (alcohol) (anhydride) (ester) (acid)
acid. The reaction is reversible and can be
shifted in the forward direction by removing OO H⊕ O
water as soon as it is formed. Ar-OH+R'-C-O-C-R' Ar-O-C-R'+R'-COOH
(phenol) (anhydride) (ester) (acid)
244
The reaction of alcohol and phenols with ii. Reaction with phosphorous halide :
acid chloride is carried out in the presence of Alcohols react with phosphorous pentahalide
pyridine (base), which neutralizes HCl. (PX5) and phosphorous trihalide (PX3) to form
alkyl halides. (refer to Chapter 10 section
OO 10.3.1).
R-OH + Cl-C-R' pyridine R'-C-OR + HCl
iii. Dehydration of alcohols to alkenes :
(alcohol) (acid chloride) (ester) Alcohol when treated with concerntrated
sulphuric acid or phosphoric acid or alumina
O O undergoes dehydration to form alkene and
Ar-OH+ Cl-C-R' pyridine R'-C-O-Ar + HCl water. (refer to Std. XI Chemistry Textbook
section 15.2) The reaction gives more
(phenol) (acid chloride) (ester) substituted alkene as the major product, in
accordance with Saytzeff rule.
Acetyl derivatives : The CH3-CO- group
is called acetyl group. The acetate esters of Problem 11.6 : Write the reaction showing
alcohols or phenols are also called ‘acetyl major and minor products formed on heating
derivatives’ of alcohols or phenols repectively. butan-2-ol with concentrated sulfuric acid.
The number of alcoholic or phenolic -OH
groups in the given compound can be deduced Solution : In the reaction described
from the number of acetyl groups introduced
in it as a result of acetylation. Aspirin, a butan-2-ol undergoes dehydration to give
well known generic medicine, is an acetyl
derivative of salicylic acid formed by its
acetylation using acetic anhydride.
COOH but-2-ene (major) and but-1-ene (minor) in
OH O O accordance with Saytzeff rule.
+ CH3-C-O-C-CH3
H⊕ OH CH3-CH=CH-CH3
But-2-ene (major)
(Salicylic acid) (Acetic anhydride) CH3-CH -CH2-CH3Conc H2SO4 CH2=CH-CH2-CH3
∆ But-1-ene (minor)
(Butan-2-ol)
COOH Do you know ?
O
According to the common accepted
-O-C-CH3 + CH3COOH mechanism dehydration involves
following three steps.
(Acetic acid)
1. Formation of protonated alcohols R-O⊕ H2
(Aspirin/
Acetyl salicyclic acid) 2. Its slow dissociation into carbocation
c. Reaction due to breaking of C-O bond in 3. Fast removal of hydrogen ion to form
alcohols : alkene.
i. Reaction with hydrogen halides : -C-C- H⊕ -HC-⊕CO-H2-Hsl2oOw -C-C- -H⊕ -C = C-
Alcohols reacts with hydrogen halides to form H OH fast H⊕ fast
alkylhalides (refer to Chapter 10 section 10.3.1)
In general, tertiary alcohols react rapidly with (Alcohol) (Protonated (Carbocation) (Alkene)
hydrogen halides; secondary alcohols react alcohol)
somewhat slower; and primary alcohols,
even more slowly. The order of reactivity of
hydrogen halides is
HI > HBr > HCl
HCl reacts only in the presence of anhydrous
ZnCl2. No catalyst is required in the case of
HBr and HI.
245
Problem 11.7 : Write and explain reactions R-CH2-OH KMnO4/K2Cr2O7/HNO3 [R-CHO]
to convert propan-1-ol into propan-2-ol ? (O)
(10 alcohol) (aldehyde)
Solution : The dehydration of propane-1-ol
to propene is the first step. Markownikoff O
hydration of propene is the second step to
get the product propan-2-ol. This is brought R-C-OH
about by reaction with concerntrated
H2SO4 followed by hydrolysis. (carboxylic acid)
CH3-CH2-CH2-OH Al2O3 CH3-CH=CH2 Tertiary alcohols are difficult to oxidise.
(Propan-1-ol) 623K (Propene) On oxidation with strong oxidising agents at
high temperature tertiary alcohol undergoes
breaking of C-C bonds and gives a mixture
of carboxylic acids containing less number of
carbon atoms than the starting 3° alcohol.
i. Con H2SO4 CH3-CH-CH3 Heating with Cu : When vapours of various
ii. H2O OH types of alcohols are passed over hot copper
the following reactions are observed.
(Propan-2-ol)
iv. Oxidation of alcohols : R-CH2-OH Cu/573K R-CHO
Oxidation
Can you recall ? (aldehyde)
What are the various definitions of (10 alcohol)
oxidation ?
R-CH-R' Cu/573K R-C-R'
OH Oxidation O
(20 alcohol) (ketone)
On reaction with oxidising agent H3C- CCH(O3H) -CH3 Cu/573K H3C- C = CH2
dehydration CH3
primary and secondary alcohols undergo
(alkene)
dehydrogenation to form carbonyl compounds, (30 alcohol)
namely aldehydes and ketones respectively
Seondary alcohol on oxidation with chromic
anhydride (CrO3) forms ketone. Problem 11.8 : An organic compound
gives hydrogen on reaction with sodium
R-CH-R' CrO3 R-C-R' metal. It forms an aldehyde having
(O) O molecular formula C2H4O on oxidation
OH with pyridinium chlorochromate Name the
(20alcohol) (ketone) compounds and give equations of these
reactions.
Primary alcohol on oxidation with CrO3
forms aldehyde. However, a better reagent to
bring about this oxidation is PCC (pyridinium
chlorochromate). Solution : The given molecular formula
C2H4O of aldehyde is written as
R-CH2-OH PCC R-CHO CH3 -CHO. Hence the formula
(O) of alcohol from which this is
(10 alcohol) (aldehyde) obtained by oxidation must be
CH3-CH2-OH. The two reactions can,
When common oxidizing agents like therefore, be represented as follows.
nitric acid, potassium permanganate or
potassium dichromate are used to oxidise 2CH3-CH2-OH 2Na 2CH3-CH2O Na⊕+ H2
primary alcohol, the oxidation does not stop (Ethyl alcohol) (Sodium ethoxide)
at aldehyde stage, but the aldehyde formed is
further oxidized to carboxylic acid containing CH3-CH2-OH [O] CH3-CHO + H2O
the same number of carbon atoms. PCC
(Ethyl alcohol) (Acetaldehyde)
246
d. Reactions of phenols : Phenol undergoes HO-SO3H OH
electrophilic substitution reactions more readily 293K SO3H
as compared to benzene. The -OH group in
phenol is ring activating and an ortho-/para- OH (o- Phenolsulfonic acid)
directing group. +
OH
i. Halogentaion of phenol: Phenol reacts
with aqueous solution of bromine to give (Phenol) HO-SO3H
2,4,6 -tribromophenol (chlorine reacts in the 373K
same way.)
SO3H
OH OH
+ 3Br2 H2O Br (p- Phenolsulfonic acid)
+ 3HBr
Br iv. Reimer-Tiemann Reaction : When phenol
is treated with chloroform in aqueous sodium
(Phenol) Br hydroxide solution followed by hydrolysis with
acid, salicylaldehyde is formed. This reaction
(2,4,6-tribromophenol) is known as Reimer-Tiemann reaction.
If the reaction is carried out in a solvent of OH ON⊕a NaOH
lower polarity than water, such as CHCl3, CHCl2
CCl4 or CS2, a mixture of ortho- and para- CHCl3
bromophenol is formed. aq. NaOH
OH OH OH (Phenol) Intermediate
+ Br2 Br ON⊕a
CS2 + + HBr CHO
Low OH
temperature H3O⊕ CHO
(Phenol) Br
(o- bromophenol)
(p- bromophenol)
Major product (Salicyaldehyde)
ii. Nitration of phenol : Phenol reacts with
dilute nitric acid at low temperature to give If carbon tetrachloride is used in place of
chloroform, salcylic acid is formed.
mixture of ortho- and para-nitrophenol.
OH OH OH v. Kolbe reaction : The treatment of sodium
+ HO-NO2 NO2 phenoxide with carbondioxide at 398 K under
+ pressure of 6 atm followed by acid- hydrolysis,
salicylic acid (o-hydroxybenzoic acid) is
(Phenol) (o- nitrophenol) NO2 formed. This reaction is known as Kolbe’s
reaction
(p- nitrophenol)
Phenol reacts with concerntrated nitric acid to
form 2, 4, 6-trinitrophenol (picric acid) ON⊕a OH
OH OH COONa
+ 3HO-NO2 ConcO. 2N NO2 + CO2 398K H3O⊕
+ 3H2O 6 atm
H2SO4
(Sodium (Sodium OH
(Phenol) NO2 Phenoxide) salicylate) COOH
(picric acid)
iii. Sulfonation of phenol : Phenol reacts (Salicylic acid)
with concerntrated sulfuric acid at room
temperature to give o-phenolsulfonic acid and
at 373K, p-phenol sulfonic acid
247
Do you know ? C2H5OH H2SO4/443K CH2= CH2
-H2O
Sodium phenoxide is more reactive (ethanol) (ethene)
than phenol towards electrophilic
substitution. Hence it is able to react Symmetrical ethers can be obtained from
with a weak electrophile like CO2 at high primary alcohols by this method. Use of higher
temperature and pressure in Kolbe reaction. temperature or 2°/3° alcohols gives alkene as
the major product.
vi. Oxidation of phenol : Phenol on oxidation Do you know ?
with chromic anhydride or sodium dichromate
in presence of H2SO4 gives p-benzoquinone. Dehydration of alcohols to form
ether is a SN2 reaction. Protonated
OH CrO3 OO alcohol species undergoes a backside
attack by second molecule of alcohol in a
or Na2Cr2O7/H2SO4 (p-Benzoquinone) slow step. Subsequent fast deprotonation
results in formation of ether.
(Phenol) i. Protonation : C2H5-O-H H⊕ C2H5-O⊕ -H
Phenol oxidizes slowly giving a dark coloured H
mixture in presence of air.
vii. Catalytic hydrogenation of phenol: ii. SN2: C2H5- O H + CH2-O⊕-H -H2O
Phenol on catalytic hydrogenation gives H3C H ⊕
cyclohexanol. In this reaction a mixture of
vapours of phenol and hydrogen is passed over C2H5-O-CH2-CH3
nickel catalyst at 433 K. H
OH OH iii. Deprotonation : ⊕
+ 3H2 Ni C2H5-HO-C2HC5 2-HH⊕5-O-C2H5
433 K
b. Williamson Synthesis : Simple as well as
(Phenol) (Cyclohexanol) mixed ethers can be prepared in laboratory
by Williamson Synthesis. In this method
viii. Reduction of phenol : Phenol is reduced alkyl halide is treated with sodium alkoxide
to benezene on heating with zinc dust. or sodium phenoxide to give dialkyl ethers or
alkyl aryl ethers.
OH
+ Zn + ZnO R-X + Na⊕ O -R R-O-R + NaX
(Phenol) (Benzene) R-X + Na⊕ O -Ar R-O-Ar + NaX
11.5 Ethers This reaction follows SN2 mechanism. Ether
is formed as a result of backside attack by
11.5.1 Preaparation of ethers alkoxide/ phenoxide ion (a nucleophile) on
alkyl halide. The alkyl halide used in this
a. Dehydration of alcohols : When alcohol is reaction must be primary. For example : t-butyl
heated withdehydratingagentlikeconcentrated methyl ether can be synthesised by reaction of
H2SO4 or H3PO4 two products, either an ether methyl bromide with sodium t-butoxide.
or an alkene, can form depending upon the
temperature. For example : dehydration of (CH3)3C-O Na⊕ + CH3-Br
ethanol by H2SO4 gives ethoxyethane at 413
K, while ethene is formed at 443 K. (sodium t-butoxide) (methyl bromide)
2C2H5OH H2SO4/413K C2H5-O-C2H5 (CH3)3C-O-CH3 + NaBr
-H2O
(ethanol) (Ethoxyethane) (t-butyl methyl ether)
248
If secondary or tertiary alkyl halides are used, C2H5-Cl + Na⊕O -CH-CH3 C2H5-O-CH-CH3
ch(Elotrhi ydle) (S CoHdiu3 m (ethyl iCsoHpr2opyl
the reaction leads mainly to alkene formation
isopropoxide) ether)
(elimination reaction). For example :
CH3-CCH=3CH2
CH-CCH- 3Cl + Na⊕O -C2H5 + NaCl
CH3
( (sodium ethoxide) (isobutene) 11.5.2 Physical properties :
t-butyl chloride)
+ C2H5OH + NaCl
(ethanol) a. Physical states and boiling points
Aryl halides do not give Williamson’s i. Dimethyl ether and ethyl methyl ether are
gases. Other ethers are colourless liquids with
synthesis. pleasant odour.
Can you think ? ii. Lower ethers are highly volatile and highly
Williamson synthesis is effectively inflammable substances.
a method of preparation of ethers
from two hydroxy compounds. The iii. Boiling points of ethers show gradual
two substrates of Williamson synthesis, increase with the increase in molecular mass.
namely the nucleophile and alkyl hadlides
are obtained from hydroxy compounds as Ether B.P. / K
shown below.
CH3-O-CH3 248
OH ONa C2H5-O-CH3 284
C2H5-O-C2H5 308
+ NaOH
Phenol (Phenoxide) b. Polarity and solubility : Since -C-O-C-
bond angle is 110° and not 180°, the bond
C2H5-OH HX C2H5X (alkyl halide) dipole moments of the two C-O bonds donot
Na C2H5ONa (alkoxide) cancel each other; therefore ethers posses a
small net dipole moment (For example, dipole
Problem 11.9 : Ethyl isopropyl ether does moment of diethyl ether is 1.18 D)
not form on reaction of sodium ethoxide
and isopropyl chloride.
C2H5-ONa + Cl-CCHH-3CH×3 C2H5-O-CH-CH3 R O net dipole
moment
CH3 110°
R
i. What would be the main product of this
reaction ? Weak polarity of ethers does not affect their
boiling points, which are about the same as
ii. Write another reaction suitable for the those of alkanes having comparable molecular
preparation of ethyl isopropyl ether. mass. (see table 11.6).
Solution : i. Isopropyl chloride is a Table 11.6 Comparative boiling points of
secondary chloride. On treating with sodium alkane, ether and alcohol
ethoxide it gives elimination reaction to
form propene as the main product . n-Hep- Methyl n-Hexyl
tane n-pentyl alcohol
C2(HSo5d- OiumNa + C(ils-oCC pHHro-3pCylH3 CH3-CH=CH2 Name
ether
etho xide) chloride) (Propene)
Molecular 100 102 102
+ C2H5OH + NaCl mass
(Ethyl alcohol)
ii. Ethyl isopropyl ether can be prepared as
Boiling
follows using ethyl chloride (10chloride) as point / K 371 373 430
as substrate.
249
The intermolecular hydrogen bonding that R-O-R + H-O-H H3O⊕ 2R-OH
holds alcohol molecules together strongly, is ∆
not present in ethers and alkanes. However,
solubility/miscibility of ethers in water is R-O-R' + H-O-H H⊕ R-OH + R'-OH
similar to that of alcohols of comparable ∆
molecular mass. This is because ethers can
form hydrogen bonds with water through the Ar-O-R + H-O-H H⊕ Ar-OH + R-OH
ethereal oxygen. ∆
R-δO δH⊕-O ii. Reaction with PCl5 : Ethers react with PCl5
RH to give alkyl chlorides
R-O-R' + PCl5 D R-Cl + R'-Cl + POCl3
iii. Reaction with hot concentrated acid :
Alkyl ethers react with hot and concentrated
For example diethyl ether and n-butyl alcohol HI and HBr to give an alcohol and an alkyl
have respective miscibilities of 7.5 and 9g per
100 g of water. halide.
R-O-R + HX R-X + R-OH HX R-X
11.5.3 Chemical properties of ethers : R-OH HX R-X + H2O
The order of reactivity of HX is HI>HBr>HCl
a. Laboratory test for ethers : Ethers are
neutral compounds in aqueous medium.
Ethers do not react with bases, cold dilute Do you know ?
acids, reducing agents, oxidizing agents and Mechanism of first stage : Reaction
of ether with hot concentrated
active metals. However, ethers dissolve in HI involves formation of oxonium ion by
protonation in the first step and subsequent
cold concerntrated H2SO4 due to formation of nucleophilic substitution reaction brought
oxonium salts. about by the powerful nucleophile I . The
least substituted carbon in oxoinium ion is
R-O-R' + H2SO4 H HSO4 attacked by I following SN2 mechanism.
R-O-R'
⊕
This property distinguishes ethers from
hydrocarbons.
b. Reaction involving alkyl group of ether : CH3-O-CH2-CH3 + H-I ∆ ⊕
CH3-O-CH2-CH3
H + I
i. Formation of peroxide : Ethers combine
with atmospheric oxygen to form peroxide. 1 O21 ⊕-CH2-CH3
O-OH I CH3-O⊕-CH2-CH3 I2 CH3
H
CH3-CH2-O-C2H5 + O2 Long CH3-CH-O-C2H5 H
contact
with air (peroxide of
(diethyl ether) CH3- I + CH3-CH2-OH
(oxygen) diethyl ether)
For example :
All ethers which have been exposed to • Use of excess HI converts the alcohol into
the atmosphere contain peroxide. This is very
undesirable reaction. Peroxides are hazardous alkyl iodide.
because they decompose violently at high
temperature. • In case of ether having one tertiary alkyl
c. Reaction involving C-O bond group the reaction with hot HI follows SN1
i. Reaction with hot dilute sulphuric acid mechanism, and tertiary iodide is formed
(Hydrolysis) : Ethers when heated with
dilute sulfuric acid undergo hydrolysis to give rather than tertiary alcohol. (CH3)3C⊕ +
alcohols/phenols. Step 1 : (CH3)3C-O⊕-CH3 slow CH3OH
H
Step 2 : (CH3)3C⊕ + I fast (CH3)3C-I
250
Aryl alkyl ethers have stronger and shorter OCH3 OCH3 OCH3
bond between oxygen and the aromatic ring. + CH3
Hence an aryl alkyl ether undergoes cleavage + CH3Cl Anhydrous
of oxygen - alkyl bond and yields a phenol AlCl3
and an alkyl halide on reaction with HI.
(Anisole) (4-MCethHo3xy (2-Methoxy
O CH3 OH toluene) toluene)
(major) (minor)
+ HI 398K + CH3I OCH3 OCH3
(Anisole) (Phenol) (Methyl iodide) + CH3COCl Anhydrous +
AlCl3
(4-CMOetChoHxy3
d. Electrophilic substitution in aromatic (Anisole)
ethers : The alkoxy group in aromatic acetophenone)
ether is a ring activating and ortho-, para- (major)
directing group toward electrophilic aromatic
substitution. This is evident from the resonance OCH3
structures: COCH3
OR ⊕ R O⊕ R O⊕ R OR (2-Methoxy
acetophenone)
O
(minor)
(I) (II) (III) (IV) (V)
iii. Nitration :Anisole reacts with concentrated
+R Effect of -OR group results in increased nitric acid in presence of concentrated sulfuric
electron density at the para- and two ortho- acid (Nitrating mixture) to give a mixture of
positions (see resonance structures II, III and o-nitro anisole and p-nitro anisole.
IV).
OCH3 OCH3 OCH3
conc HNO3 + NO2
conc H2SO4
i. Halogenation : Anisole undergoes
bromination with bromine in acetic acid even (Anisole) NO2
in the absence of ferric bromide catalyst. It (4-Nitroanisole) (2-Nitroanisole)
is due to activation of benzene ring by the
methoxy group. (major) (minor)
11.6 Uses of alcohols, phenols and ethers
OCH3 OCH3 OCH3 Alcohols :
+ Br
Br2 1. Methyl alcohol is used as a solvent for
paints and varnishes.
CH3COOH
(Anisole) Br 2. Ethyl alcohol is used as antifreeze agent
in automobile radiators. It is also used as
90% 10% solvent.
(p-bromo anisole)(o-bromo anisole)
ii. Friedel Crafts reaction : Anisole reacts Ethers :
with alkyl halide and acyl chloride in presence
of anhydrous AlCl3 (Lewis acid) as catalyst. 1. Earlier diethyl ether was used as a general
anaesthetic in surgical operations.
2. Diethyl ether is used as a solvent for
Grignard reagents, fats, waxes, oil, etc.
251
Phenols :
1. Phenol is used in preparation of phenol formaldehyde resin For example : bakelite.
2. Phenols are used as antiseptic in common products like air freshners, deodarants, mouthwash,
calamine lotions, floor cleaners, etc.
Exercises
1. Choose the correct option.
i. Which of the following represents the v. Resorcinol on distillation with zinc dust
increasing order of boiling points of (1), gives
(2) and (3)?
A. Cyclohexane B. Benzene
(1) CH3-CH2-CH2-CH2-OH C. Toluene D. Benzene-1, 3-diol
(2) (CH3)2CHOCH3 vi. Anisole on heating with concerntrated HI
gives
(3) (CH3)3COH
A. (1) < (2) < (3) B. (2) < (1) < (3) A. Iodobenzene
C. (3) < (2) < (1) D. (2) < (3) < (1) B. Phenol + Methanol
ii. Which is the best reagent for carrying out C. Phenol + Iodomethane
following conversion ? D. Iodobenzene + methanol
CH3 CH3 vii. Which of the following is the least acidic
OH compound ?
OH
A. OH
A. LiAlH4 B. Conc. H2SO4, H2O
B.
C. H2/Pd D. B2H6, H2O2-NaOH CH3
OH
iii. Which of the following substrate will give OH
C.
ionic organic product on reaction ?
A. CH3-CH2-OH + Na D.
B. CH3-CH2-OH + SOCl2 NO2 NO2
C. CH3-CH2-OH + PCl5 viii. The compound incapable of hydrogen
bonding with water is ......
D. CH3-CH2-OH + H2SO4
A. CH3-CH2-O-CH3
iv. Which is the most resistant alcohol
towards oxidation reaction among the B. CH3-CH2-CH2-CH3
follwoing ?
A. CH3-CH2-OH B. (CH3)2CH-OH OH
C.
C. (CH3)3C-OH D. C2H5 CH -OH
CH3 D. CH3-CH2-CH2-OH
252
ix. Ethers are kept in air tight brown bottles 4. An ether (A), C5H12O, when heated
because with excess of hot HI produce two
alkyl halides which on hydrolysis form
A. Ethers absorb moisture compound (B)and (C), oxidation of (B)
gave and acid (D), whereas oxidation
B. Ethers evaporate readily of (C) gave a ketone (E). Deduce the
structural formula of (A), (B), (C), (D)
C. Ethers oxidise to explosive peroxide and (E).
D. Ethers are inert 5. Write structural formulae for
x. Ethers reacts with cold and concentrated a. 3-Methoxyhexane
H2SO4 to form b. Methyl vinyl ether
A. oxonium salt B. alkene c. 1-Ethylcyclohexanol
C. alkoxides D. alcohols d. Pentane-1,4-diol
2. Answer in one sentence/ word. e. Cyclohex-2-en-1-ol
i. Hydroboration-oxidation of propene 6. Write IUPAC names of the following
gives.....
ii. Write the IUPAC name of alcohol having i. HO CH3
molecular formula C4H10O which is
resistant towards oxidation.
iii. Write structure of optically active alcohol ii. CH3-CH -CH -CH2-OH
having molecular formula C4H10O OH CH3
iv. Write name of the electrophile used in
Kolbe’s Reaction.
3. Answer in brief. OH iv. O-CH3
iii. NO2
i. Explain why phenol is more acidic than
ethyl alcohol.
ii. Explain why p-nitrophenol is a stronger
acid than phenol.
iii. Write two points of difference between Activity :
properties of phenol and ethyl alcohol.
• Collect information about
iv. Give the reagents and conditions necessary production of ethanol as byproduct
to prepare phenol from in sugar industry and its
importance in fuel economy.
a. Chlorobenzene
• Collect information about phenols
b. Benzene sulfonic acid. used as antiseptics and polyphenols
having antioxidant activity.
v. Give the equations of the reactions for
the preparation of phenol from isopropyl
benezene.
vi. Give a simple chemcial test to distinguish
between ethanol and ethyl bromide.
253
12. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS
Can you recall ? 12.2.1 Classification of aldehydes :
(Aldehydes are classified as aliphatic and
• Draw the structures of the aromatic aldehydes)
following compounds and
classify them on the basis of a. Aliphatic aldehydes : The compounds
C-O single bond and C = O double in which the –CHO group (formyl group) is
bond present in them, Ethyl alcohol, attached directly to sp3 hybridized carbon
acetaldehyde, o-nitrophenol, Diethyl atom that is saturated carbon atom are
ether, isopropyl alcohol, acetone. called aliphatic aldehydes. (Exception :
Formaldehyde, H-CHO is also classified as
• What are carbonyl compounds? aliphatic aldehyde though –CHO group is not
attached to any carbon ). For example :
12.1 Introduction : In the previous chapter,
you learnt about the organic compounds CH3 - CHO CH3 - CH2 - CHO
which contain carbon –oxygen single bond. In
this chapter, we are going to study the organic (Acetaldehyde) ( Propionaldehyde)
compounds containing carbon - oxygen double
bond (>C=O ) called carbonyl group, which O
is one of the most important functional group
in organic chemistry. C
RH
O carbonyl oxygen
General formula
C carbonyl carbon
R R' (R= H or alkyl group)
Both aldehydes and ketones contain b. Aromatic aldehydes : The compounds in
a carbon –oxygen double bond (-C-) as their which –CHO group is attached directly to an
functional group. Therefore theyO are called aromatic ring are called aromatic aldehydes.
carbonyl compounds. In aldehydes , carbonyl For example :
carbon is bonded to at least one hydrogen apart
from an alkyl or aryl group. The functional HO HO HO
group of aldehydes, therefore, is - CHO which
is called formyl group or aldehydic carbonyl OH
group. On the other hand in ketones, carbonyl
carbon is bonded to two alkyl or aryl groups NO2
either identical (R = R) or different (R ≠ R').
It is called ketonic carbonyl group. The (Benzaldehyde) (Salicylaldehyde)(p-Nitrobenzaldehyde)
functional group of carboxylic acids is -COOH
called carboxyl group. Due to the -OH group Use your brain power
bonded to (>C=O ) group, carboxylic acids are
distinct from aldehydes and ketones. Classify the following as aliphatic
and aromatic aldehydes.
12.2 Classification of aldehydes, ketones
and carboxylic acids : Aldehydes, ketones O O
and carboxylic acids are classified as per the H
nature of carbon skeleton bonded to (>C=O ). H
CHO CHO
CH3
254
12.2.2 Classification of ketones : Ketones are Use your brain power as
classified as aliphatic and aromatic ketones: butanone,
• Classify the followings
a. Aliphatic ketones : The compounds in simple and mixed ketones.
which >C=O group is attached to two alkyl
groups are called aliphatic ketones. • Benzophenone, acetone,
acetophenone.
O Oxo or ketonic carbonyl group Do you know ?
C • Aldehydes and ketones are
R R' responsible for many flavours and
odours that you will readily recognize :
General formula
CHO
(Where R , R’ = alkyl group, identical or different)
On the basis of types of alkyl groups bonded to CHO
carbonyl carbon, aliphatic ketones are further
classified as simple and mixed ketones. OH OCH3
i. Simple or symmetrical ketones : The Vanillin
(Vanilla flavour)
ketones in which both the alkyl groups bonded Benzaldehyde
(Bitter almond flavour)
to carbonyl carbon are identical, are called
CH = CH - CHO
simple ketones or symmetrical ketones. For
example :
OO
H3C - C - CH3 H5C2 - C - C2H5 O
(Dimethyl ketone) ( Acetone) (Diethyl ketone) Cinnamaldehyde Camphor
(Cinnamon flavour) (Camphor fragance)
ii. Mixed or unsymmetrical ketones : The • Structures of many important biological
compounds contain carbonyl moiety. For
ketones in which two alkyl groups bonded
to carbonyl carbon are different, are called example progesterone and testosterone,
mixed ketones or unsymmetrical ketones. For the female and male sex hormones
example : O respectively.
O
H5C2 - C - CH3 H5C2 - C - CH2 - CH2 - CH3 Do you know ?
(Ethyl methyl ketone) (Ethyl n-propyl ketone) • Butyraldehyde is used in
margarine and food for its
b. Aromatic ketones : The compounds in buttery odour.
which a >C=O group is attached to either two
aryl groups or one aryl and one alkyl group are • Acetophenone has smell of pistachio
called aromatic ketones. For example : and is used in ice-cream. Muscone has
musky aroma and is used in perfumes.
OO Popcorn has butter flavour which
CC contains butane-2,3-dione.
CH3 12.2.3 Classification of carboxylic acids :
Carboxylic acids are classified as aliphatic
Benzophenone Acetophenone and aromatic carboxylic acids :
(Diphenyl ketone) (Methyl phenyl ketone)
255
a. Aliphatic carboxylic acids : The organic Remember...
compounds in which carboxyl (-COOH)
group is bonded to an alkyl group are called The aromatic compounds in which
aliphatic carboxylic acids or fatty acids. the –COOH group is not attached
(Exception : Formic acid, H-COOH is also directly to the ring are called side-chain
classified as aliphatic carboxylic acid though aromatic acids. For example :
–COOH group is not attached to any carbon).
For example : CH2 - COOH
H3C - COOH H3C - CH2 - COOH (Phenyl acetic acid)
(Acetic acid) (Propionic acid ) Carboxylic acids are widely distributed
in nature; they are found in both the plants and
O Carboxyl animals. L-lactic acid is present in curd , citric
functional group acid is found in citrus fruit (Lemons). Acetic
RC acid is the key ingredient of vinegar.
OH
12.3 Nomenclature of aldehydes, ketones
General formula and carboxylic acids :
(R= H or alkyl group)
12.3.1 Nomenclature of aldehydes and
Depending on the number of –COOH groups carboxylic acid : The names of aldehydes
and carboxylic acids are related to each other.
present carboxylic acids are classified as There are two systems of naming aldehydes
and carboxylic acids : trivial and IUPAC.
mono, di, tri carboxylic acids and so on. For
a. Trivial names of aldehydes and carboxylic
example : COOH acids : Trivial names of aliphatic aldehydes
are derived from the corresponding trivial
H3C - CH2 - COOH COOH names of carboxylic acids. Here the ending
‘ic acid’ of carboxylic acid is replaced by
Propionic acid Oxalic acid the ending ‘aldehyde’. In case of substituted
( a monocarboxylic acid) (a dicarboxylic aliphatic carboxylic acids and aldehydes the
position of substituent is indicated by labeling
acid) the carbon serially as α, β, γ and so on. The
carbon atom adjacent to carbonyl carbon is
CH2 - COOH labeled as α and next one is β and so on. (See
HO - C - COOH Table 12.1).
CH2 - COOH
Citric acid
(a tricarboxylic acid)
b. Aromatic carboxylic acids : These are the
compounds in which one or more carboxyl
groups (-COOH ) are attached directly to the
aromatic ring. For example :
COOH COOH
(Benzoic acid) COOH
(Phthalic acid)
OCOCH3
COOH
(Acetyl salicylic acid) ( Aspirin)
256
Do you know ? When two –CHO groups are present at
the two ends of the chain the ending ‘e’of
A series of straight chain alkane is retained and the suffix –‘dial’ is
added to the name of parent aldehyde. In case
dicarboxylic acids are of dicarboxylic acids, ‘dioic acid’ is added
to the name of the parent alkane. In IUPAC
commercially known by the following nomenclature an alicyclic compound in
which –CHO group is attached directly to the
common names: COOH COOH ring is named as a carbaldehyde. The suffix
COOH COOH COOH ‘carbaldehyde’ is added after the full name
(CH2)2 (CH2)3 of parent cycloalkane structure. Similarly an
COOH CH2 COOH COOH (CH2)4 alicyclic compound having a carboxyl group
(oxalic COOH COOH directly attached to alicyclic ring is named as
(glutaric (adipic cycloalkane carboxylic acid.
acid) (malonic (succinic acid) acid)
acid) Substituted aromatic aldehydes and
acid) carboxylic acids : When two or more different
functional groups are attached to a ring ,
Aseries of lower fatty acids are commercially the higher priority group (std. XI Chemistry
Textbook, Chapter 14, sec.14.4.7) is given
known by the following common names. lower number. When –CHO group, appears
as substituent prefix ‘formyl’ is used in the
CH3 CH3 CH3 CH3 CH3 IUPAC name.
COOH CH2 (CH2)2 (CH2)3 (CH2)4
(Acetic COOH COOH COOH COOH CHO
acid) (valeric (caproic
(propionic (butyric acid) acid) OH CH3
acid) acid) (4-Hydroxy-3-methylbenzaldehyde)
b. IUPAC names of aldehydes and carboxylic COOH
acids :
OH
Can you recall ? CH3
Which suffix do appear in the (3-Hydroxy-4-methylbenzoic acid)
IUPAC names of aldehydes and
carboxylic acids ? CHO
According to IUPAC system, the name of an COOH
aliphatic aldehyde is derived from the name of
the corresponding alkane by replacing ending (2-Formylbenzoic acid)
‘e’ of alkane with ‘ al ’. Aldehyde is named
as alkanal (Table 12.1). The IUPAC name of Trivial and IUPAC names of some aldehydes
aliphatic carboxylic acid is derived from the and carboxylic acids are given in Table 12.1.
name of the corresponding alkane by replacing
ending ‘ e’ of alkane with ‘ oic acid ’. (Refer to
Std. XI Chemistry Textbook sec. 14.4.7).
Alkane Alkanal
Alkane Alkanoic acid
The longest chain including –CHO or –
COOH group is identified as the parent chain.
Numbering of the chain is done by giving
number 1 to the –CHO or –COOH carbon. The
name of substituent is included along with its
locant.
4CH3-3CH2-2CH2-1CHO ; 4CH3-3CH2-2CH2-1COOH
Aldehyde (-CHO) group and carboxyl
(-COOH) group are always present at the end
of the parent straight chain.
257
Table 12.1 Trivial and IUPAC names of carboxylic acids and aldehydes
Structure Carboxylic acids IUPAC name Structure Aldehydes IUPAC name
H - COOH Trivial name Methanoic acid H-CHO Trivial name Methanal
CH3 - COOH Ethanoic acid CH3-CHO Formaldehyde Ethanal
CH3 - CH2 - COOH Formic acid Propanoic acid CH3-CH2-CHO Acetaldehyde Propanal
CH3 - CH2 - CH2 - COOH Acetic acid Butanoic acid CH3-CH2-CH2-CHO Butanal
(CH3)2CH - COOH 2-Methylpropanoic (CH3)2CH-CHO Propionaldehyde 2 - Methylpropanal
Propionic acid acid
CH2 = CH - COOH CH2=CH-CHO Butyraldehyde Prop-2-enal
HOOC - COOH n-Butyric acid Prop-2-enoic acid CHO-CHO Ethanedial
Ethanedioic acid Isobutyrlaldehyde
Isobutyric acid Benzoic acid (a -
( a - methylpropionic methylpropionaldehyde)
acid) 2-Methylbenzoic acid Acrolein
Acrylic acid
Oxaldehyde (Glyoxal)
Oxalic acid
258 COOH Benzoic acid CHO Benzaldehyde Benzaldehyde
COOH (Benzenecarboxylic (Benzenecarbaldehyde)
acid)
o-Toluic acid CHO o-Tolualdehyde 2-Methylbenzaldehyde
CH3 Salicyclic acid 2-Hydroxybenzoic acid CH3 Salicylaldehyde 2-Hydroxybenzaldehyde
COOH CHO
OH Phthalic acid Benzene-1,2- OH Phthalaldehyde Benzene-1,2-
COOH dicarboxylic acid CHO dicarboaldehyde
COOH Cyclohexylcarboxylic Cyclohexanecarboxylic CHO Cyclohexane aldehyde Cyclohexanecarbaldehyde
COOH acid acid CHO
Do you know ? In case of substituted aliphatic ketones the
The trivial names of carboxylic position of substituent is indicated by labelling
the carbon serially as α, β, γ and so on. The
acids are often derived from Latin carbon atom adjacent to carbonyl carbon
names of their original natural source. is labelled as α and next one is β and so on.
Names of aromatic ketones are based on a
For example, Formic acid is obtained phenone. (see Table 12.2)
from red ants (Formica means ant), acetic
acid is obtained from acetum (acetum b. The IUPAC names of aliphatic ketones are
means vinegar), propionic acid is from basic derived from the name of the corresponding
fat (propion means first fat), butyric acid is alkanes by replacing ending ‘e’ of alkane
from butter (butyrum means butter). with ‘one’. They are named as alkanone. The
longest chain of carbon atoms containing the
12.3.2 Trivial and IUPAC names of ketones: ketonic carbonyl group is numbered from the
end closer to the carbonyl carbon.
a. The trivial names of aliphatic ketones are
based on the names of alkyl groups or aryl Alkane Alkanone
groups attached to carbonyl carbon .Names of
alkyl or aryl groups are written in alphabetical When two >C=O groups are present,
order followed by the word ketone. then ending ‘e’of alkane is retained and
the suffix –‘dione’ is added to the name of
parent ketone indicating the locants of ketonic
Table 12.2 Trivial and IUPAC names of some ketones
Sr.No. Compound Trivial name IUPAC name
1 CH3-CO-CH3 Dimethyl ketone (Acetone) Propanone
2 CH3-CO-CH2-CH3 Butanone
3 CH3-CO-CH2-CH2-CH3 Ethyl methyl ketone Pentan-2-one
Methyl n-propyl ketone Pentan-3-one
4 CH3-CH2-CO-CH2-CH3
Diethyl ketone 2-Bromohexan-3-one
5 CH3-CHBr-CO-CH2-CH2-CH3
α-Bromoethyln-propyl 4-Methylpent-3-en-2-
6 (CH3)2C=CH-CO-CH3 ketone one
Mesityl oxide Hexane-2,4-dione
3-Oxobutanal
7 CH3-CO-CH2-CO-CH2-CH3 --------------------- Benzophenone
8 CH3-CO-CH2-CHO ---------------------
9O Diphenyl ketone
C (Benzophenone)
10 O Methyl phenyl ketone Acetophenone
CH3 (Acetophenone)
11 O ----------------------- 2-Chloro-4-
Cl methylcyclohexanone
259
carbonyl groups. In case of polyfunctional ii. By dehydrogenation of alcohols : This
ketones, higher priority group is given lower method has industrial application. Aldehydes
number. When ketonic carbonyl is a lower and ketones are prepared by passing the
priority group it is named as 'oxo', preceded vapours of primary and secondary alcohols
by the locant. In alicyclic ketones, carbonyl respectively over hot copper powder. (See
carbon is numbered as 1. (Refer Table 12.2). Chapter 11)
Use your brain power b. From hydrocarbons :
Write IUPAC names for the
following compounds. Can you recall ?
i. H • What is ozonolysis ?
• What is the role of zinc dust in
Br Br O
ozonolysis process?
ii. O i. By ozonolysis : Alkene reacts with ozone to
give ozonide which on decomposition with zinc
O dust and water gives aldehyde and/or ketones.
iii. H (See Std. XI Chemistry Textbook, Chapter 15)
O ii. By hydration of alkynes : Alkynes react
with water in presence of 40% sulfuric acid
iv. CHO v. O and 1% mercuric sulfate to give aldehydes
or ketones. (See Std. XI Chemistry Textbook,
Try this... Chapter 15)
Draw structures for the following
a. 2-Methylpentanal 12.4.2 Other methods of preparation of
b. Hexan-2-one aldehydes and ketones : Some methods of
preparation of aldehydes and ketones involve
12.4 Preparation of aldehydes and ketones : common starting functional groups but
12.4.1 General methods of preparation of different types.
aldehydes and ketones :
a. From acyl chlorides (Acid chlorides) :
a. By oxidation of alcohols : i. Aldehydes Aldehydes and ketones both can be obtained
and ketones are prepared by the oxidation of from acyl chloride, but the reactions involved
primary and secondary alcohols respectively. are different.
(See Chapter 11 )
• Preparation of aldehyde from acyl
Can you tell ? chloride
What is the reagent which
oxidizes primary alcohols to Acyl chloride is reduced to corresponding
only aldehydes and does not oxidize aldehyde by hydrogen using a palladium
aldehydes further into carboxylic acid ? catalyst poisoned with barium sulfate. This
reaction is known as Rosenmund reduction.
O H2 O
Pd-BaSO4 R - C - H + HCl
R - C - Cl
(Aldehyde)
(Acyl chloride)
260
Use your brain power hydrochloric acid which on acid hydrolysis
Write the structure of the give corresponding aldehydes. This reaction
is called Stephen reaction.
product formed on Rosenmund
reduction of ethanoyl chloride and benzoyl R - C ≡ N + 2[H] SnCl2, HCl R - HC = NH.HCl
chloride.
(Alkane nitrile) (imine hydrochloride)
Can you think ?
H3O⊕ R - CHO + NH4Cl
What is the alcohol formed when
benzoyl chloride is reduced with (Aldehyde)
pure palladium as the catalyst ?
For example,
H3C - C ≡ N + 2[H] SnCl2, HCl
(reduction)
(Ethanenitrile)
b. Preparation of ketone (aliphatic and C(etHha3n-imHiCne = NH.HCl H3O⊕ C(EHth3-aCnaHl)O+NH4Cl
aromatic) from acyl chloride :
hydrochloride)
i. Preparation of aliphatic ketones from
C6H5 - C ≡ N + 2[H] SnCl2, HCl
acyl chloride: ketones are obtained from acyl (Benzonitrile)
chloride by reaction with dialkyl cadmium
which is prepared by the treatment of cadmium C6H5 - HC = NH.HCl H3O⊕ C6H5-CHO+NH4Cl
chloride with Grignard reagent.
(Benzanimine hydrochloride) (Benzaldehyde)
2R - MgX + CdCl2 R2Cd + 2Mg(X)Cl
Alternatively, nitriles are also reduced by
diisobutylaluminium hydride (DIBAl-H)
2R' - cChOlorCidle) + R2Cd 2 R' - CO - R or AlH (i-Bu)2 to imines followed by acid
hydrolysis to aldehydes. An advantage of this
(Acyl (Ketone) method is that double or triple bond present in
+ CdCl2
For example, the same molecule is not reduced..
2CH3 - COCl + (CH3)2Cd H3C - CH = CH - CH2 - C ≡ N AIH (i-Bu)2
(Ethanoyl chloride) (Dimethyl cadmium) (Pent-3-enenitrile) H3O⊕
2CH3-CO-CH3 + CdCl2 H3C - CH = CH - CH2 - CHO
Propanone (Acetone) (Pent-3-enal)
2C6H5 - COCl + ( CH3)2Cd • Preparation of ketones from nitriles :
(Benzoyl chloride) (Dimethyl cadmium) Ketones are prepared by reacting nitriles
C6H5 - CO - CH3 + CdCl2 with Grignard reagent in dry ether as solvent
followed by acid hydrolysis.
(Acetophenone)
ii. Preparation of aromatic ketones from H3C - C ≡ N + H3CMgCl dry ether
acyl chloride : Aromatic ketones are prepared
by Friedel Craft's acylation reaction (Std. XI (Ethanenitrile)
Chemistry Textbook, Chapter 15, sec. 15.4.6)
CH3 - C = NMgCl H3O⊕ CH3 - CO - CH3
b. From nitriles : Aldehydes and ketones both H3C
can be obtained from nitriles but by different (Acetone)
reaction.
+ NH3 + Mg(Cl)OH
• Preparation of aldehydes from nitriles :
C6H5 - C ≡ N + C6H5 - MgBr dry ether
Nitriles are reduced to imine hydrochloride
by stannous chloride in presence of (Benzonitrile)
C6H5 - C = NMgBr H3O⊕ C6H5 - CO - C6H5
C6H5
(Benzophenone)
+ NH3 +Mg(Br)OH
261
c. From aromatic hydrocarbons : Aromatic iii. By side chain chlorination of toluene:
aldehydes and ketones are both prepared Side chain chlorination of toluene gives
from aromatic hydrocarbons but by different benzal chloride which on acid hydrolysis at
methods. 373K gives benzaldehyde. Benzaldehyde, is
manufactured commercially by this method.
• Preparation of aromatic aldehydes from
hydrocarbon CH3 Cl - Cl CHCl2 H3O⊕
∆ 373K
Strong oxidizing agents transform -CH3 group hv
bonded to aromatic ring into carboxyl group
(-COOH). For obtaining aromatic aldehyde (Toluene) (Benzal chloride)
from methyl arene the following special
methods are used. CHO
i. Etard reaction : Methyl group in methyl (Benzaldehyde)
benzene (or methyl arene) is oxidized by
oxidizing agent chromyl chloride in carbon iv. Gatterman –Koch formylation of arene:
disulfide as solvent, to form a chromium Benzene or substituted benzene is treated
complex, from which the corresponding under high pressure with carbon monoxide and
benzaldehyde is obtained on acid hydrolysis. hydrogen chloride in presence of anhydrous
This reaction is known as Etard reaction. aluminium chloride or cuprous chloride to give
benzaldehyde or substituted benzaldehyde.
H CO,HCl CHO
H
CS2 Anhyd.AlCl3
CH high pressure
+ CrO2Cl2 (Benzene)
(Benzaldehyde)
(Toluene) (Chromyl chloride) • Preparation of Aromatic ketones from
hydrocarbons :
CHO
H OCr(OH)Cl2 H3O⊕ By Friedel-crafts acylation of arene : (Refer
OCr(OH)Cl2 to sec. 12.4.2 (a) ii and Std. XI Chemistry
Textbook, Chapter 15, sec. 15.4.6).
(Chromium complex)
(Benzaldehyde)
ii. By oxidation of methyl arene using CrO3 Use your brain power
: Methylarene is converted into a benzyllidene
diacetate on treatment with chromium Name the compounds which
oxide in acetic anhydride at 273-278 K. The are used for the preparation
diacetate derivative on acid hydrolysis gives of benzophenone by Friedel-Crafts
corresponding aldehyde. acylation reaction. Draw their structures.
H 12.4.3 Preparation of aldehydes only from
H esters :Aliphatic or aromatic esters are reduced
to aldehydes by using diisobutylaluminium
C H + CrO3 + (CH3CO)2O 273 - 283 K hydride DIBAl-H or AlH (i-Bu)2. The
reaction is usually carried out at 195 K to
(Methylbenzene) (Acetic anhydride) prevent further reduction of the aldehyde
produced.
CHO
H OCOCH3 ∆ O AIH (i-Bu)2 O
OCOCH3 H3O⊕ R - C - O - R' H3O⊕ R-C-H
(Benzaldehyde) (Ester) (Aldehyde)
262
Try this... b. Anhydrides on hydrolysis with water give
Draw the structure of the product carboxylic acids.
formed by the combination of
carbon monoxide and HCl. OO O
2R - C - O - H
Use your brain power R - C - O - C - R + H2O
(Carboxylic acid)
Identify the reagents necessary (Anhydride)
to achieve each of the following
transformation 12.5.3 From esters : Carboxylic acids can be
obtained from esters either by acid hydrolysis
OO or alkaline hydrolysis.
Cl H
a. Acid hydrolysis of ester : Esters on
O hydrolysis with dilute mineral acid like dilute
HCl or dilute H2SO4 give the corresponding
carboxylic acid .
O dil.H2SO4 O
∆ R-C-O-H
R-C-O- R' + H2O
(Carboxylic acid)
(Ester)
OO + R' - OH
O2N OCH3 H (Alcohol)
O2N b. Alkaline hydrolysis of ester using dilute
alkali like dilute NaOH or dilute KOH form
12.5 Preparation of carboxylic acids : solution of water soluble sodium or potassium
salt of the acid (carboxylate). On acidification
12.5.1 From nitriles and amides : Alkyl with concentrated HCl, free acid is formed.
nitriles or aryl nitriles on acid hydrolysis give O
amides . Amides on further acid hydrolysis give H5C2 - C - O - CH3 + dil.NaOH ∆
corresponding carboxylic acids. Hydrolysis is
carried out by using dilute mineral acids like (Methyl propanoate)
dilute sulfuric acid or dilute hydrochloric acid.
O
OH H5C2 - C - O - N⊕ a + CH3 - OH
R - C ≡ N + H2O [R - C = NH] (Sodium propanoate)
(Nitrile) O N⊕a H⊕
Conc. HCl
O H5C2 - C - O - + H2O
R - C - NH2 H2O R - COOH + NH3 (Sodium propanoate) O
dil.HCl
(Amide) (carboxylic acid) H5C2 - C - O - H + NaOH
R - C ≡ N + 2H2O + dil.HCl ∆ R - COOH (Propanoic acid)
+ NH4Cl
The sodium or potassium salts of higher fatty
12.5.2 From acyl chloride and anhydrides : acids are known as soaps. Hence alkaline
hydrolysis of esters is called saponification
a. Acyl chlorides on hydrolysis with water give (Std XI Chemistry Textbook, Chapter 16).
carboxylic acids. This method is useful for
preparation of aliphatic as well as aromatic 12.5.4 From alkyl benzene : Aromatic
acid. carboxylic acids can be prepared by oxidation
of alkyl benzene with dilute HNO3 or alkaline
R - COCl + H2O R - COOH + H - Cl /acidic KMnO4 or chromic acid. The entire
(Acyl chloride) (Carboxylic acid)
263
alkyl chain , regardless of its length, is oxidized Internet my friend
to a carboxyl group. (Tertiary alkyl substituent
on benzene, however, is not oxidized). Collect information of dry ice and
ice from internet. Draw chemical
For example, structures of dry ice and regular ice.
Prepare a chart of uses of dry ice.
H
H KMnO4.KOH COOK⊕
∆
CH
(Toluene) (Potassium benzoate)
COOH
H3O⊕
(Benzoic acid)
CH3 Use your brain power
CH
H3C COOK⊕ Predict the products (name and
structure) in the following reactions.
KMnO4.KOH
∆ CH3CH2CN ∆ ?
dil. HCl ?
(Cumene) (Potassium benzoate)
CH3-CONH2 ∆
COOH dil. HCl
H3O⊕ C6H5-CH2-CH3 alk.KMnO4 ?
∆ ?
(Benzoic acid) C6H5-COO-C2H5 ∆
dil. H2SO4
12.5.5 From alkenes : Carboxylic acids can CH3MgBr (i) dry ice/dry ether ?
also be prepared by the oxidation of alkenes (ii) dil. HCl
by KMnO4 in dilute H2SO4.
12.6 Physical properties :
H5C6 - CH = CH KMnO4.dil.H2SO4 H5C6 - COOH 12.6.1 Nature of intermolecular forces :
2∆
(Phenyl ethene) (Benzoic acid) The carbonyl bond ( C=O) in aldehydes and
COOH ketones is a polar covalent bond. As a result,
COOH
KMnO4.dil.H2SO4 these compounds contain dipole-dipole forces
∆
of attraction. (Fig. 12.1) The molecules orient
(Cyclohexene) (Adipic acid) in such a way as to have oppositely polarized
12.5.6 From Grignard reagent : Grignard atoms facing each other.
reagent in dry ether solvent is added to solid
carbon dioxide (dry ice) to give a complex δ δC⊕
which on acid hydrolysis gives corresponding O
carboxylic acid.
CO
R - Mg - X + O = C = O dry ether R - COOMgX δ⊕ δ
(Alkyl (dry ice) (complex) Fig. 12.1 Dipole-dipole attraction in carbonyl
magnesium halide) compounds
HOH R - COOH + Mg(X)OH Carboxyl group of carboxylic acid contains
dil.HCl O-H bond which is responsible for formation
of hydrogen bonding. Thus, carboxylic acids
(carboxylic acid) have the strongest intermolecular forces of
attraction. (Fig.12.3 in section 12.6.4).
264
Table 12.4 Boiling points of aldehydes and ketones
Number of carbon atoms Aldehyde Boiling point Ketone Boiling point
1 --------
2 Methanal 252 K ----- ---------
3 329 K
4 Ethanal 294 K ------- 353 K
5 375 K
6 Propanal 319 K Propanone 400 K
Butanal 348 K Butan -2-one
Pentanal 376 K Pentan-2-one
Hexanal 392 K Hexan-2-one
12.6.2 Physical state and boiling points of 12.6.4 Physical state, boiling points and
aldehydes and ketones : solubilities of carboxylic acids : Lower
aliphatic carboxylic acids upto nine carbon
Formaldehyde is a gas at room temperature atoms are colourless liquids with irritating
and has irritating odour. Acetaldehdye is odours. The higher homologues are colourless,
extremely volatile, colourless liquid. Higher odourless wax like solids, have low volatility.
aldehydes have pleasant odour. Acetone is a Boiling points of lower carboxylic acids are
liquid at room temperature and has pleasant listed in Table 12.5.
odour but most of the higher ketones have
bland odours. Carboxylic acids have higher boiling
points than those of alkanes, ethers, alcohols
Increasing boiling points in the homologous aldehydes and ketones of comparable mass
series of aldehydes and ketones are listed in (Table 12.6). The reason is that , in liquid
Table 12.4. phase, carboxylic acids form dimer in which
two molecules are held by two hydrogen
12.6.3 Solubility of aldehydes and ketones : bonds. Acidic hydrogen of one molecule form
hydrogen bond with carbonyl oxygen of the
The oxygen atom of ( C=O) can involve other molecule (Fig.12.3). This doubles the
in hydrogen bonding with water molecule size of the molecule resulting in increase in
(Fig 12.2). As a result of this, the lower intermolecular van der Waals forces, which in
aldehydes and ketones are water soluble (For turn results in high boiling point. In the case of
example : acetaldehyde, acetone). As the acetic acid dimers exist even in the gas phase
molecular mass increases, the proportion of (Fig.12.3).
hydrocarbon part of the molecule increases
which cannot form hydrogen bond; and the δ O δH⊕ O
water solubility decreases.
δ δ⊕ δ H3 C - C C - CH3
δ⊕ C = O H - O Hδ⊕
OHO
Fig. 12.2 : Hydrogen bonding in carbonyl δ⊕ δ
Fig. 12.3 : Dimer of acetic acid (Two molecules
compound and water
held by two hydrogen bonds)
Table 12.5 Increasing boiling points of carboxylic acids
Name Formula Boiling point in K
Formic acid HCOOH 373 K
Acetic acid CH3COOH 391 K
Propionic acid CH3CH2COOH 414 K
Butyric acid 437 K
CH3CH2CH2COOH 460 K
Valeric acid CH3CH2CH2CH2COOH
265
Table 12.6 : Variation of boiling point with functional group
Compound Family Molecular mass Boiling point Strength of intermolecular
forces
CH3-CH2-CH2-CH3 Alkane 58 272 K increases
CH3-O-CH2-CH3 Ether 60 281 K
CH3-CH2-CHO Aldehyde 58 322 K
CH3-CO-CH3 Ketone 58 329 K
CH3-CH2-CH2-OH Alcohol 60 370 K
CH3-COOH Carboxylic acid 60 391 K
Remember... solubility of carboxylic acids in water decreases
with increase in molecular mass. Higher
Relative strength of carboxylic acids are practically insoluble
in water due to the increased hydrophobic
intermolecular force : H-Bond (water hating) interaction of hydrocarbon part
with water. Aromatic acids like benzoic acid
> dipole-dipole attraction > van der are also practically insoluble in water at room
temperature. Water insoluble carboxylic acids
Waals force. Hence, Boiling points : are soluble in less polar organic solvents like
ether, alcohol, benzene, and so on.
Carboxylic acids > Alcohols > Ketones >
12.7 Polarity of carbonyl group : The
Aldehydes > ether > Alkanes polarity of a carbonyl group originates from
higher electronegativity of oxygen relative to
Do you know ? carbon as well as resonance effects as shown
Commercially available forms of in Fig. 12.4.
formaldehyde and acetaldehyde:
O Oδ
i. Formaldehyde is available commercially
as solid polymer called paraformaldehyde C C δ⊕
HO [CH2 - O]n H and trioxane (CH2O)3
(Trioxane has cyclic structure). These are (A)
convenient for use in chemical reactions as
source of formaldehyde. O O
ii. Aqueous solution of formaldehyde C C
gas is called formalin, which is used for ⊕
preservation of biological and anatomical major
specimens. (B) minor
(C)
iii. When dry formaldehyde is required, it
is obtained by heating paraformaldehyde Oδ
or trioxane.
iv. Acetaldehyde is also conveniently used
as solid trimer (paraldehyde) and tetramer
(metaldehyde).
Lower aliphatic carboxylic acids C δ⊕
containing upto four carbons are miscible
with water due to formation of intermolecular resonance
hydrogen bonds between carboxylic acid hybrid
molecules and solvent water molecules. The (D)
Fig. 12.4 : Polarity of carbonyl group
266
The carbonyl carbon has positive polarity Try this...
(see structures (A) and (D)). Therefore, it is
electron deficient. As a result, this carbon Draw structure of propanone and
atom is electrophilic (electron loving) and is indicate its polarity.
susceptible to attack by a nucleophile (Nu: ).
12.8 Chemical properties of aldehydes and
12.7.1 Reactivity of aldehydes and ketones ketones :
: Reactivity of aldehydes and ketones is
due to the polarity of carbonyl group which 12.8.1 Laboratory tests for aldehydes and
results in electrophilicity of carbon. In general, ketones : Aldehydes are easily oxidized to
aldehydes are more reactive than ketones carboxylic acids and therefore, act as reducing
toward nucleophilic attack. This can be well agents toward mild oxidizing agents. Ketones,
explained in terms of both the electronic effects do not have hydrogen atom directly attached to
and steric effect. carbonyl carbon. Hence, they are not oxidized
by mild oxidizing agents. On the basis of this
1. Influence of electronic effects : Alkyl difference in the reactivity, aldehydes and
groups have electron donating inductive effect ketones are distinguished by the following
(+I). A ketone has two electron donating alkyl tests:
groups bonded to carbonyl carbon which are
responsible for decreasing its positive polarity a. Tests given by only aldehydes :
and electrophilicity. In contrast, aldehydes
have only one electron donating group bonded 1. Schiff test : When alcoholic solution of
to carbonyl carbon. This makes aldehydes aldehyde is treated with few drops of Schiff 's
more electrophilic than ketones. reagent, pink or red or magenta colour appears.
This confirms the presence of aldehydic
2. Steric effects : Two bulky alkyl groups (-CHO) group.
in ketone come in the way of incoming
nucleophile. This is called steric hindrance to 2. Tollens' test or silver mirror test : When
nucleophilic attack. an aldehyde is boiled with Tollens' reagent
(ammonical silver nitrate), silver mirror
On the other hand, nucleophile can easily is formed. The aldehyde is oxidized to
attack the carbonyl carbon in aldehyde because carboxylate ion by Tollens' reagent and Ag⊕
it has one alkyl group and is less crowded or ion is reduced to Ag.
sterically less hindered . Hence aldehyde are
more easily attacked by nucleophiles. R - CHO + 2 Ag (NH3)2⊕ + 3OH ∆
(aldehyde) Tollens reagent
Oδ Oδ R - COO + 2 Ag↓ + 4NH3↑ + 2H2O
C C (carboxylate) (Silver mirror)
R δ⊕ R R δ⊕ H
3. Fehling test : When a mixture of an aldehyde
(Ketone) (Aldehyde) and Fehling solution is boiled in hot water, a
red precipitate of cuprous oxide is formed.
Remember...
An aldehyde is oxidized to carboxylate ion
Aromatic aldehydes are less by Fehling solution and Cu2⊕ ion is reduced
reactive than aliphatic aldehydes to Cu⊕ ion. It may be noted that α-hydroxy
in nucleophilic addition reactions. This is ketone also gives this test positive.
due to electron-donating resonance effect
of aromatic ring which makes carbonyl
carbon less electrophilic.
267
R - CHO + 2Cu2⊕ + 5OH boil a. Addition of hydrogen cyanide (H-CN) :
Hydrogen cyanide (weak acid) adds across
(Aldehyde) (Fehling solution) the carbon-oxygen double bond in aldehydes
and ketones to produce compounds called
R - COO + Cu2O↓ + 3H2O
Do you know ?
(carboxylate ion) red ppt
1. Schiff 's reagent is a colourless
Can you tell ? solution obtained by passing sulfur
dioxide gas (oxidant) through magenta
Simple hydrocarbons, ethers, coloured solution of p-rosaniline
ketones and alcohols do not hydrochloride .
get oxidized by Tollens' reagent.
Explain, Why ? NH2⊕Cl
Use your brain power
Why is benzaldehyde NOT oxidized
by Fehling solution ?
b. Laboratory test for ketonic group : H2N NH2
Sodium nitroprusside test :
(p-rosaniline hydrochloride)
When a freshly prepared sodium nitroprusside
solution is added to a ketone, mixture is shaken aldehyde SO2/H2O
well and basified by adding sodium hydroxide
solution drop by drop, red colour appears in NH2
the solution, which indicates the presence of
ketonic ( >C=O) group.
CH3 - CO - CH3+ OH CH3 - CO - CH2 SO3H
(Acetone)
[Fe( CN)5NO]2 + CH3- CO- CH2 + H2O H2N NH2
(Nitroprusside ion) (Schiff 's reagent)
[Fe(CN)5 NO (CH3- CO- CH2)]3 2. Tollens' reagent is prepared by mixing
Red colouration a few drops of AgNO3 solution and a few
mL of dilute sodium hydroxide solution.
The anion of ketone formed by alkali reacts
with nitroprusside ion to form a red coloured A brown precipitate is formed which is
complex which indicates the presence of
ketonic group. then dissolved by adding dilute ammonium
hydroxide.
12.8.2 Chemical reactions of aldehydes 3. Fehling solution is a mixture of two
and ketones with nucleophile : In all these solutions Fehling A and Fehling B. Fehling
reactions the nucleophilic reagent brings about A is prepared by dissolving crystals of
reactions by attacking on positively polarized copper sulfate in concentrated sulfuric
electrophilic carbonyl carbon in aldehydes acid. Fehling B is prepared by dissolving
and ketones. sodium potassium tartarate in sodium
hydroxide solution.
268
cyanohydrins. The negative part of the reagent H3C CH3
( CN) attacks the electrophilic carbon of C = O + NaHSO3 H3C − C − OH
carbonyl group. The reaction requires either
acid or base as catalyst. H3C SO3Na
(Acetone) (Acetonesodium
R R bisulfite adduct)
C = O + HCN R' − C − OH
Use your brain power
R' CN
Sodium bisulfite is sodium salt
(Aldehyde when R' = H (cyanohydrin) of sulfurous acid, write down its
Ketone : R' = alkyl/aryl group) detailed bond structure .
For example ,
H3C R Do you know ?
C = O + H - CN
H3C − C − OH Sodium bisulfite addition product
H CN so formed can be split easily to
regenerate aldehydes and ketones on
(Acetaldehyde) (acetaldehyde cyanohydrin) treatment with dilute acid or base. Thus,
this reaction is used to separate and purify
H3C CH3 the aldehydes and ketones from other
C = O + H - CN H3C − C − OH organic compounds.
H3C CN c. Addition of alcohols : Aldehyde reacts
with one molecule of anhydrous monohydric
(Acetone) (Acetone cynohydrin) alcohol in presence of dry hydrogen chloride
to give alkoxyalcohol known as hemiacetal,
Remember... which further reacts with one more molecule
of anhydrous monohydric alcohol to give a
i. Cyanohydrin formation is a geminaldialkoxy compound known as acetal
'step-up' reaction as a new carbon as shown in the reaction.
- carbon single bond is formed.
Step 1 :
ii. The - C ≡ N group can be converted to
–COOH, - CH2 - NH2 and so on. R H
C = O +R' - OH dry HCl R − C − OR'
iii. Therefore, cyanohydrins are used as
intermediate in step up synthesis. H dil. HCl OH
b. Addition of NaHSO3 (Sodium bisulphite) (Aldehyde) (Hemiacetal) unstable
: Aldehydes and ketones react with saturated
aqueous solution of sodium bisulfite to give Step 2 :
crystalline precipitate of sodium bisulfite
adduct (addition compound). For example ,
H3C H H H
C = O + NaHSO3 R − C − OR' +R' - OH dry HCl R − C − OR'
H3C − C − OH
H SO3Na OH dil. HCl OR'
(Acetaldehyde) (Acetaldehyde (Hemiacetal) (Acetal) stable
sodium bisulfite
adduct) + H2O
269
For example , Do you know ?
Step 1 : + H5C2 - OH Cyclic ketal is used as protecting
group for ketone in reactions of
H3C multifunctional compound to which
C=O ketone is sensitive.
H Use your brain power
Predict the product of the
(Acetaldehyde) H following reaction :
dry HCl H3C − C − OC2H5 Br O HO-CH2-CH2-OH
dil. HCl OH
Mg in dry ether
(Hemiacetal) unstable
Acetals and ketals are hydrolysed with
Step 2 : aqueous mineral acids to give corresponding
aldehydes and ketones respectively.
H H5C2 - OH dry HCl
dil. HCl Remember...
H3C − C − OC2H5 + Organic molecule containing
OH
an alcohol and carbonyl group can
(Hemiacetal) (unstable) undergo intramolecular reaction with dry
HCl to form cyclic hemiacetals/hemiketals.
H
d. Addition of Grignard reagent : Aldehydes
H3C − C − OC2H5 + H2O and ketones on reaction with alkyl magnesium
OC2H5 halide followed by acid hydrolysis give
alcohols.(Refer to Chapter 11, sec. 11.4.1 d.)
(1,1-Diethyoxyethane)
(stable) e. Nucleophilic addition –elimination of
aldehydes and ketones with ammonia
Similarly, Ketones react with alcohol in derivatives : Aldehydes and ketones undergo
presence of acid catalyst to form hemiketal addition elimination with some ammonia
and ketal. derivatives (NH2-Z ) to give product
containing C = N bonds (imines). The reaction
Ketones react with 1,2- or 1,3- diols in is reversible and takes place in weakly acidic
presence of dry hydrogen chloride to give five- medium. The substituted imine is called a
or six -membered cyclic ketals . Schiff 's base.
R + HO -CH2
HO -CH2
C=O
R (ethane-1,2-diol)
(ketone)
dry HCl RO CH2
dil. HCl C CH2
RO
(cyclic ketal)
The reaction can be reversed by treating the C = O + NH2 Z C N Z
cyclic ketal with aqueous HCl to regenerate
the ketone. (Aldehyde OH H -H2O
or Ketone)
C N Z + H2O
(imine)
270
Table 12.7 Nucleophilic addition – elimination reactions of aldehydes and ketones with
ammonia derivatives
Sr. Aldehyde(R'=H)/ + NH2 - Z -H2O imine (a crystalline
No. Ketone(R'≠H) -H2O derivative)
-H2O
R NH2 OH -H2O R
1. C = O + Hydroxyl amine -H2O
R' C N OH
R' -H2O
oxime
R NH2 NH2
2. C = O + Hydrazine R
R' R' h Cyd ra Nzo n eNH2
R
R NH2 NH C6H5
3. C = O + Phenyl hydrazine R' p Che ny Nlh yd NraHzo n Ce 6H5
R
R'
R NH2 NH CONH2 R' Cs em Ni ca NrbHaz o CneONH2
4. C = O + Semicarbazide
R'
R H NO2 R H NO2
C=O
5. +H2N N NO2 C = N N NO2
R
R' 2, 4 - Dinitrophenyl
2, 4 -
hydrazine Dinitrophenylhydrazone
Where Z = -R, -Ar, -NH2, -NHC6H5, Remember...
-NHCONH2 , -NHC6H3(NO2)2 In strong acidic medium,
For example,
nitrogen atom of ammonia
H3C H3C C N OH derivative H2N-Z is protonated to form
HO H (H3N+ – Z) ion which is no longer a
C = O + NH2 OH nucleophile.
H H
(Hydroxyl Use your brain power
(Acetaldehyde) amine)
Draw the structures of
-H2O H3C C N OH i. The semicarbazone of cyclohexanone
ii. The imine formed in the reaction
(Acetaldoxime)
between 2-methylhexanal and ethyl
All aldehydes and ketones give similar amine
reactions .Some important reactions are iii. 2, 4 - dinitrophenylhydrazone of
listed in Table 12.7. The resulting products acetaldehyde.
have high molecular mass and are crystalline
solids. These reactions are, therefore, useful
for characterization of the original aldehydes
and ketones.
271
f. Haloform reaction : This reaction is g. Aldol condensation :
given by acetaldehyde, all methyl ketones
(CH3-CO-R) and all alcohols containing CH3- Try this...
(CHOH)- group. When an alcohol or methyl When acetaldehyde is treated
ketone is warmed with sodium hydroxide and with dilute NaOH, the following
iodine, a yellow precipitate is formed. Here reaction is observed.
the reagent sodium hypoiodite is produced
in situ. During the reaction, sodium salt of 2CH3-CHO dil. NaOH CH3 -CH-CH2-CHO
carboxylic acid is formed which contains OH
one carbon atom less than the substrate. The
methyl group is converted in to haloform. • What are the functional groups in the
For example : Acetone is oxidized by sodium product ?
hypoiodite to give sodium salt of acetic acid
and yellow precipitate of iodoform. • Will there be another product formed
during the same reaction ? (Deduce
O NaOH, I2 the answer by doing atomic audit of
∆ reactant and product)
H3C C CH3 + 3 Na OI
• Is this an addition reaction or
condensation reaction ?
(Acetone) (Sodium
hypoiodite)
O⊕ Aldehydes containing at least one
CHI3 ↓ + H3C C ONa + 2 NaOH α –hydrogen atom undergo a reaction in
presence of dilute alkali ( dilute NaOH, KOH
(Iodoform) (Sodium acetate) or Na2CO3) as catalyst to form β-hydroxy
aldehydes (aldol). This reaction is known
Remember... as aldol reaction. Formation of aldol is an
addition reaction. Aldol formed from aldehyde
1. If C=C bond is present in a given having α-hydrogens undergoes subsequent
aldehyde or ketone or methyl elimination of water molecule on warming,
ketone, it is not attacked by hypohalite. giving rise to a, b - unsaturated aldehyde.
2. Non methyl ketones do not give a 2R-CH2-CHO aq. NaOH R-CH2-CH-CHR-CHO
positive iodoform test. OH
(aldehyde)
3. Secondary alcohols having CH3- (aldol)
CHOH- group give positive iodoform
test because the reagent first oxidizes it R-CH2-CH-CHR-CHO -H2O R-CH=CR CHO
to a CH3-CO- group which subsequently warm
forms iodoform.
OH
(aldol) (a, b - unsaturated aldehyde)
For example :
Try this... (i)
H
Write chemical reactions taking H C C H + H CαH2 C H
place when propan-2-ol is treated dil. NaOH
with iodine and sodium hydroxide.
HO O
(Ethanal) (Ethanal)
272
H3C b CH aCH CHO-H∆2O H3C b CH a CH CHO Use your brain power
• Observe the following
OH H (But-2-enal)
reaction.
(3-Hydroxybutanal)
(an Aldol) OO
CC
The overall reaction is called aldol
condensation. It is a nuclephilic addition- H + H3C
elimination reaction.
Ketones containing at least two a- hydrogen (Benzaldehyde) (1-Phenylethanone)
also undergo aldol condensation reaction (acetophenone)
and give an a, b - unsaturated ketone. For dil.base
example: ∆ O
(ii) CH3 H3C β CCH 3 Ca H O CH C
2 H3C C C CH
Ba(OH)2 CH3
O OH H (1,3-Diphenylprop-2-en-1-one)
(Benzal acetophenone)
(Propanone) (4-Hydroxy-4-methylpentan-2-one)
(a Ketol) • Will this reaction give a mixture of
products like a cross aldol reaction ?
∆ H3C β CCH 3 Ca H O α-hydrogen atom. Aldehydes undergo self
-H2O C CH3 -oxidation and reduction reaction on heating
with concentrated alkali. This is an example of
(4-methylpent-3-en-2-one) disproportionation reaction. In cannizzaro
reaction one molecule of an aldehyde is
Cross aldol condensation : Cross aldol reduced to alcohol and at the same time second
condensation refers to the aldol condensation molecule is oxidized to carboxylic acid salt.
that takes place in between two different For example, Formaldehyde and benzaldehyde
aldehydes or ketones . If both aldehydes or
ketones contain two α-hydrogen atoms each, (i) O
then a mixture of four products, is formed.
For example, a mixture of ethanal and propanal 2H C H + NaOH ∆
on reaction with dilute alkali followed by
heating gives a mixture of four products (Formaldehyde) (sodium hydroxide)
(Fig.12.5).
(50%)
O H
C OH
Ketones can also be used as one of the H C O Na⊕ + H
components in cross aldol condensation .
(Sodium formate) H
h. Cannizzaro reaction : This reaction
is given only by aldehydes having no (Methanol)
O Self H HO H3C O
H3C- C-H condensation
H3C - C = C - C - H + H3C - CH2 - CH = C - C - H
(Ethanal) i. dil.NaOH (Crotonaldehyde / But-2-enal) (2-Methylpent-2-enal)
+ O ii. heat H3C O H HO
H3C-CH2-C-H Cross H3C - CH = C - C- H + H3C - CH2 - C = C - C - H
condensation
(Propanal) (2-Methylbut-2-enal) (Pent-2-enal)
Fig. 12.5 : Cross aldol condensation
273
(ii) H R CHO K2Cr2O7 R COOH
C dil. H2SO4
(Aldehyde) (Carboxylic acid)
2 O KOH in MeOH
CH2 OH Ketones resist oxidation due to strong
∆ CO-C bond ,but they are oxidized by strong
oxidizing agents such as CrO3, alkaline
(Benzaldehyde) KMnO4 or hot concentrated HNO3 to a mixture
of carboxylic acids having less number of
O carbon atoms than the starting ketone. Thus,
C Oxidation of ketones is accompanied by
breaking C - C bond.
O K⊕ +
O
(Potassium benzoate) (Phenylmethanol /
benzyl alcohol)
Use your brain power R C R' CrO3 R COOH + R' COOH
Can isobutyraldehyde undergo (Ketone) (Carboxylic acids)
Cannizzaro reaction ? Explain.
For example,
Cross Cannizzaro reaction : When a mixture O
of formaldehyde and non-enolisable aldehyde (i) H3C C CH3 CrO3 H3C COOH
(aldehyde with no α-hydrogen) is treated with (Acetone) (Acetic Acid)
a strong base, formaldehyde is oxidized to
formic acid while the other non-enolisable is O
reduced to alcohol. Formic acid forms sodium
formate with NaOH. On acidification sodium (ii) H3C C C2H5 CrO3 H3C COOH
formate is converted into formic acid. For (Butan-2-one) (Ethanonic acid)
example : + H5C2 COOH
H (Propanoic acid)
O C b. Clemmensen and Wolf-Kishner
H C H + O i. conc. NaOH reduction: The carbonyl group of aldehydes
and ketones is reduced to methylene group
ii. H3O⊕ (-CH2- ) on treatment with zinc –amalgam and
concentrated hydrochloric acid (Clemmensen
(Formaldehyde) (Benzaldehyde) reduction) or hydrazine followed by heating
with sodium or potassium hydroxide in high
CH2 OH O boiling solvent like ethylene glycol (Wolf-
Kishner reduction).
+ H C OH
(Phenylmethanol) (formic acid)
12.8.3 Oxidation and reduction reactions of In both the reactions, oxygen is replaced by
aldehydes and ketones ; two hydrogen atoms.
a. Oxidation of aldehydes and ketones Clemmensen reduction :
by dilute HNO3, KMnO4 and K2Cr2O7 :
Aldehydes are oxidized to the corresponding C O + 4[H] Zn-Hg, conc. HCl C H2 + H2O
carboxylic acids by oxidant such as dilute nitric
acid, potassium permanganate and sodium or ∆
potassium dichromate in acidic medium.
(Carbonyl group (Methylene group)
in aldehydes and
ketones)
274
For example : Wolf-Kishner reduction is used to
CH3 Zn-Hg, conc. HCl synthesize straight chain alkyl substituted
(i) H3C C O + 4[H] ∆ benzenes which is not possible by Friedel-
Crafts alkylation reaction.
(Acetone) CH3
12.8.4 Electrophilic substitution reactions:
H3C C H2 + H2O Aromatic aldehydes and ketones undergo
electrophilic substitution reactions such as
(propane)
(ii) CH3 CH2 CHO + 4[H] Zn-Hg, conc. HCl nitration ,sulfonation and halogenation. The
∆ aldehydic ( -CHO) and ketonic (>C=O)
(Propanal) groups are electron-withdrawing by inductive
CH3 CH2 CH3 + H2O as well as resonance effects. They deactivate
(Propane) the benzene ring at ortho- and para- positions.
Wolf-Kishner reduction : This results in the formation of meta-product.
For example ,
C O H2N-NH2 C N NH2 CHO conc. H2SO4
-H2O
(Hydrazone) + HO NO2 CHO
(Carbonyl group + H2O
(conc. HNO3)
in aldehydes and (Benzaldehyde)
ketones)
KOH, HO-CH2-CH2-OH CH2 + N2
∆
(Methylene group) NO2
For example : H2N-NH2 C2H5 CH N NH2 (m-Nitrobenzaldehyde)
-H2O
(i) C2H5 CHO (Hydrazone) 12.9 Chemical properties of carboxylic
acids :
(Propanal)
12.9.1 Acidic character of carboxylic acids:
KOH, HO-CH2-CH2-OH CH3 CH2 CH3 + N2 The carboxyl group (- COOH) imparts acidic
character to carboxylic acids. A carboxyl
∆ (Propane) group is made of -OH group bonded to a
carbonyl group. In aqueous solution the H
C2H5 atom in OH of carboxyl group dissociates as
C O proton and carboxylate ion is formed as the
conjugate base,
(ii) H2N-NH2 R - COOH + H2O R(ca-rCboOxyOlate+ioHn)3O⊕
-H2O
(Ethyl phenyl ketone) KOH, HO-CH2-CH2-OH Carboxylate ion is resonance stabilized by
∆ two equivalent resonance structures as shown
C2H5 below.
C N-NH2
(Hydrazone) C2H5 O O O
CH2 + N2 RC RC RC
(n-Propyl benzene) O O O
(i) (ii) resonance
hybrid
275
Carboxylate ion has two resonance structures Table 12.8 : pKa values of haloacetic acids
(i) and (ii) and both of them are equivalent to
each other (Refer to Std. XI Chemistry Textbook Acid pKa Acid
Chapter 14). This gives good resonance strength
stabilization to carboxylate ion, which in turn
gives acidic character to carboxylic acids. F-CH2-COOH 2.56 decreases
Cl-CH2-COOH 2.86
Can you recall ? Br-CH2-COOH 2.90
3.18
What is the numerical parameter I-CH2-COOH 4.76
to express acid strength?
CH3-COOH
Remember... Halogens are electronegative atoms and exert
Lower Ka value, higher pKa: electron withdrawing inductive effect (-I
Weaker acid. effect). The negatively charged carboxylate ion
in the conjugate base of haloacetic acid gets
Higher Ka value, lower pKa : stronger acid. stabilized by the -I effect of halogen. Which is
responsible to diffuse the native charge.
Influence of electronic effects on acidity
of carboxylic acids : All the carboxylic O
acids do not have the same pKa value. The
structure of 'R' in R-COOH has influence on X CH2 C
the acid strength of carboxylic acids. Various O
haloacetic acids illustrate this point very well
(Tables 12.8 and 12.9). Higher the electronegativity of halogen greater
is the stabilization of the conjugate base,
stronger is the acid and smaller is the pKa
value.
Problem 12.1
Alcohols (R-OH), phenols (Ar-OH) and carboxylic acids (R-COOH) can undergo ionization
of O-H bond to give away proton H⊕; yet they have different pKa values, which are 16, 10 and
4.5 respectively. Explain
Solution : pKa value is indicative of acid strength. Lower the pKa value stronger the acid.
Alcohols, phenols and carboxylic acids, all involve ionization of an O-H bond. But their
different pKa values indicate that their acid strength are different. This is because the resulting
conjugate bases are stabilized to different extent.
Acid(HA) Conjugate base(A ) Electronic effect Stabilization/destabilization
R-O-H RO +I effect of R destabilization of conjugate base
group
Ar-O-H Ar-O -R effect or Ar stabilization of conjugate base is
group moderate because all the resonance
structures are not equivalent to each
other
O O -R effect of C = O stabilization is good because all the
R - C - O-H R-C-O group resonance structures are equivalent
to each other
As the conjugate base of carboxylic acid is best stabilized, among the three, carboxylic acids
are strongest and have the lowest pKa value. As conjugate base of alcohols is destabilized,
alcohols are weakest acids and have highest pKa value. As conjugate base of phenols is
moderately stabilized phenols are moderately acidic and have intermediate pKa value.
276
Try this... two ring exerts electron withdrawing inductive
effect (-I effect) which stabilizes the conjugate
Compare the following base and increases the acid strength of
conjugate bases and answer. aromatic acids.
O O Table 12.9 illustrates that more the number of
C electron withdrawing substituents higher is the
CH3 C Cl - CH2 acid strength.
O O
(b) Try this...
(a)
Arrange the following acids in
• Indicate the inuctive effects of CH3 - order of their decreasing acidity.
group in (a) and Cl - group in (b) by CH3-CH-CH2-COOH, CCl3-CH2-COOH,
putting arrowheads in the middle of Cl
appropriate covalent bonds.
CH3COOH
• Which species is stabilized by
inductive effect, (a) or (b) ?
• Which species is destabilized by Electron–withdrawing groups like -Cl, -CN,
and -NO2 increase the acidity of substituted
inductive effect, (a) or (b) ? benzoic acids while electron –donating group
like –CH3, - OH , - OCH3 and -NH2 decrease
Use your brain power the acidity of substituted benzoic acids .
• Compare the pKa values and COOH COOH COOH
arrange the following in an
increasing order of acid strength.
Cl3CCOOH, ClCH2COOH, CH3COOH, NO2 (Benzoic acid) CH3
Cl2CHCOOH (pKa = 4.2)
(4-Nitrobenzoic (4-Methylbenzoic
• Draw structures of conjugate bases acid) acid)
of monochloroacetic acid and (pKa = 3.41) (pKa = 4.4)
dichloroacetic acid. Which one is more
stabilized by -I effect ? Try this...
Acidity of aromatic carboxylic acids : Arrange the following carboxylic
Benzoic acid is the simplest aromatic acid. acids in order of increasing acidity.
From the pKa value of benzoic acid (4.2) we
understand that it is stronger than acetic acid m-Nitrobenzoic acid, Trichloroacetic acid,
(pKa 4.76). The sp2 hybrid carbon of aromatic benzoic acid, α-Chlorobutyric acid.
Table 12.9 pKa values of chloroacetic acids
Name Structure pKa Acid strengthincreases
Monochloroacetic acid 2.86
Cl - CH2 - COOH
Dichloroacetic acid Cl - CH - COOH 1.26
Cl
Trichloroacetic acid Cl 0.6
Cl C - COOH
Cl
277
12.9.2 Laboratory tests for carboxyl 12.9.3 Formation of acyl chloride
(-COOH) group : The presence of -COOH
group in carboxylic acids is identified by the Reaction with PCl3, PCl5, SOCl2 : Carboxylic
following tests: acids on heating with PCl3, PCl5, SOCl2 give
the corresponding acyl chlorides. Thionyl
a. Litmus test : (valid for water soluble chloride (SOCl2) is preferred because the
substances) byproducts formed are in gaseous state so they
can easily escape from the reaction mixture.
Aqueous solution of Organic compound In this reaction –OH group of –COOH is
containing –COOH group turns blue litmus replaced by –Cl .
red which indicates the presence of acidic
functional group. (It may be noted that R COOH + SOCl2 ∆ R - COCl + SO2↑
aqueous solutions of water soluble phenols (Carboxylic acid) (acyl chloride)
also turn blue litmus red.)
+HCl ↑
b. Sodium bicarbonate test : 3 R COOH + PCl3 ∆ 3 R - COCl + P(OH)3
R COOH + PCl5 ∆ R - COCl + POCl3
When sodium bicarbonate is added to an + HCl
organic compound containing –COOH
group, a brisk effervescence of carbon Use your brain power
dioxide gas is evolved. Water insoluble acid
goes in solution and gives precipitate on Fill in the blanks and rewrite the
acidification with conc.HCl. This indicates balanced equations.
the presence of –COOH group.
R - COOH + NaHCO3(aq) -CO2 • CH3COOH + thionyl chloride ∆
-H2O
(water insoluble) ++
R - COONa(aq) HCl R - COOH↓ • CH3-CH2-COOH + ∆
+ NaCl(aq) + + H3PO3
(Phenol does not evolve CO2 gas with • C6H5-COOH + ∆ +
sodium bicarbonate. Hence, carboxylic
acid and phenol are distinguished by this phosphorous oxylchloride + HCl
test.)
• CH3-COOH + phosphorous trichloride
∆+
c. Ester test : One drop of concentrated • CH3-COOH NH3
sulfuric acid is added to a mixture of given ∆
organic compound containing –COOH
group and one mL of ethanol, the reaction • NH3 ∆ C6H5-CONH2
mixture is heated for 5 minutes in hot water
bath. After this , hot solution is poured in 12.9.4 Reaction with ammonia : Formation
a beaker containing water, fruity smell of of amide : Carboxylic acids react with
ester confirms the presence of carboxylic ammonia to from ammonium carboxylate
acid. salt which on further strong heating at high
temperature decomposes to give acid amide.
R - COOH + C2H5OH H⊕
warm
R - COO - C2H5 + H2O R COOH + NH3 ⊕
( Carboxylic acid)
(ester) R COONH4
(ammonium carboxylate)
∆ R CONH2
-H2O
(Acid amide)
278
Do you know ? Acyl chloride and sodium salt of acid
are prepared by reacting carboxylic acid
COOH COONH4 separately with thionyl chloride and sodium
hydroxide respectively.
+ 2NH3 COONH4
O R COOH SOCl2 R COCl ∆
COOH C R COOH NaOH R COON⊕a
Phthalic acid NH
C
∆ CONH2 O
-2H2O -NH3
Phthalimide
CONH2 OO
R C O C R + NaCl
b. Acid amides can also be prepared by
reacting acid chloride with ammonia. (Anhydride)
R COCl + NH3 R CONH2 + HCl Can you recall ?
Which molecule is eliminated in a
(Acyl chloride) (Acid amide) decarboxylation ?
12.9.5 Formation of acid anhydride :
Can you tell ? 12.9.6 Decarboxylation of carboxylic acids :
What is the term used for Sodium salts of carboxylic acids on heating
elimination of water molecule ? with soda lime give hydrocarbons which
contain one carbon atom less than the
Mono carboxylic acids on heating with carboxylic acid. For example,
strong dehydrating agent like P2O5concentrated
H2SO4 give acid anhydrides. The reaction is ⊕ CaO R H + Na2CO3
reversible. Anhydrides are readily hydrolyzed ∆
back to acids on reaction with water. R COONa + NaOH
O O ⊕ CaO CH4 + Na2CO3
R C ∆
CH3 COONa + NaOH
O
R C R C (Sodium acetate) (Methane)
+ O H ∆ P2O5 O 12.9.7 Reduction of carboxylic acids :
R C +H2O
O H (Acid anhydride) Carboxylic acids are reduced to primary
alcohols by powerful reducing agent like
O lithium aluminium hydride. Carboxylic acid
can also be reduced by diborane (diborane
(Carboxylic acid) does not reduce –COOR , -NO2 , -X).
Better yield of acid anhydride is obtained (Note : Sodium borohydride (NaBH4 )does
not reduce-COOH group).
by heating sodium carboxylate with acyl
chloride.
⊕ + R COCl
R COONa
(sodium carboxylate) (acyl chloride) R COOH + LiAlH4 dry R CH2OH
ether
OO
∆ R C O C R + NaCl
(acid anhydride)
279
Exercises
1. Choose the most correct option. d. COOH
i. In the following resonating structures OCH3
A and B, the number of unshared v. Diborane reduces
electrons in valence shell present on
oxygen respectively are
OO
C C⊕ a. ester group b. nitro group
(A) (B) c. halo group d. acid group
a. 2, 4 b. 2, 6
vi. Benzaldehyde does NOT show positive
c. 4, 6 d. 6, 4 test with
ii. In the Wolf -Kishner reduction, alkyl a. Schiff reagent
aryl ketones are reduced to alkyl
benzenes. During this change, ketones b. Tollens' ragent
are first converted into
c. Sodium bisulphite solution
a. acids b. alcohols d. Fehling solution
c. hydrazones d. alkenes 2. Answer the following in one sentence
iii. Aldol condensation is i. What are aromatic ketones?
a. electrophilic substitution reaction ii. Is phenyl acetic acid an aromatic
carboxylic acid ?
b. nucleophilic substitution reaction iii. Write reaction showing conversion of
ethanenitrile into ethanol.
c. elimination reaction
d. addition - elimination reaction iv. Predict the product of the following
reaction:
iv. Which one of the following has lowest
acidity ? i. AlH (i-Bu)2
ii. H3O⊕
a. COOH CH3 CH2 COOCH3 ?
b. NO2 v. Name the product obtained by reacting
COOH toluene with carbon monoxide and
hydrogen chloride in presence of
anhydrous aluminium chloride.
vi. Write reaction showing conversion of
Benzonitrile into benzoic acid.
Cl vii. Name the product obtained
c. COOH by the oxidation of
1,2,3,4-tetrahydronaphthalene with
acidified potassium permanganate .
viii.What is formalin ?
280
ix. Arrange the following compounds in the 4. Answer the following
increasing order of their boiling points :
i. Write a note on –
Formaldehyde, ethane, methyl alcohol.
a. Cannizaro reaction
x. Acetic acid is prepared from methyl
magnesium bromide and dry ice b. Stephen reaction.
in presence of dry ether. Name the
compound which serves not only reagent ii. What is the action of the following
but also as cooling agent in the reaction. reagents on toluene ?
3. Answer in brief. a. Alkaline KMnO4 , dil. HCl and heat
i. Observe the following equation of b. CrO2Cl2 in CS2
reaction of Tollens' reagent with
aldehyde. How do we know that a redox c. Acetyl chloride in presence of
reaction has taken place. Explain.
anhydrous AlCl3.
iii. Write the IUPAC names of the following
structures :
R CHO + 2 Ag(NH3)2+ + OH- ∆ a. O b. COOH
R COO- + 2 Ag + 4 NH3 + 2 H2O COOH
ii. Formic acid is stronger than acetic acid. iv. Write reaction showing conversion
Explain. of p- bromoisopropyl benzene into
p-Isopropyl benzoic acid ( 3 steps).
iii. What is the action of hydrazine on
cyclopentanone in presence of ---. v. Write reaction showing aldol
condensation of cyclohexanone.
KOH in ethylene glycol ?
iv. Write reaction showing conversion Activity :
of Acetaldehyde into acetaldehyde
dimethyl acetal. Draw and complete the
following reaction scheme which
v. Aldehydes are more reactive toward starts with acetaldehyde. In each empty
nucleophilic addition reactions than box, write the structural formula of
ketones. Explain. the organic compound that would be
formed.
vi. Write reaction showing the action of the
following reagent on propanenitrile – HCN CH3CHO reduction
a. Dilute NaOH dilute H2SO4, heat Tollens' reagents
b. Dilute HCl ?
vi. Arrange the following carboxylic acids
with increasing order of their acidic
strength and justify your answer.
COOH COOH Conc. H2SO4 heat
O CH2 = CHCO2H Cold dilute KMnO4/H+
COOH
O
281
13 AMINES
Can you recall ? Use your brain power
• Write some examples of nitrogen Classify the following amines
containing organic compounds. as simple/mixed; 1°, 2°, 3° and
aliphatic or aromatic.
• What are the types of amines?
(C2H5)2NH, (CH3)3N, C2H5-NH-CH3,
Amines are nitrogen containing organic
compounds having basic character. Amines C6H5-NH2, CH3-CCHH-3NH2, NH-C6H5
are present in structure of many natural ,
compounds like proteins, vitamins, hormones
and many plant products like nicotine. CH3-CC-HN3H2, N(CH3)2 N
CH3 ,
13.1 Classification of Amines : Amines are
classified as primary (1°), secondary (2°) Remember...
and tertiary (3°) amines. Their structures are
obtained in simple way by replacing one, two Other organic compopunds
or three hydrogen atoms of NH3 molecule by like alkyl halides or alcohols are
alkyl/aryl groups (see Table 13.1). classified as 1°, 2°, 3° depending upon
the nature of the carbon atom to which
Secondary and tertiary amines are further functional group is attached where as
classified as simple / symmetrical amines and amines are classified depending upon the
mixed / unsymmetrical amines. When all the number of alkyl or aryl groups directly
alkyl or aryl groups on nitrogen are same, it is attached to the nitrogen atom. Thus,
a simple amine. If these groups are different, isopropyl amine is 10 amine, but isopropyl
then the amine is a mixed amine. alcohol is 20 alcohol.
Amines are also divided into two major
classes, namely, aliphatic and aromatic amines
on the basis of nature of the groups attached to
the nitrogen atom.
Table 13.1 Types of amines
Type Functional group Examples
Formula Common Name
Name Formula
Primary amine, (10) Amino -NH2 C2H5-NH2 Ethylamine
Secondary amine, (20) Imino NH Dimethylamine
CH3 NH
N CH3 Trimethylamine
Tertiary amine, (30) Tertiary CH3
nitrogen CH3 N
CH3
282
13.2 Nomenclature of Amines : 13.3 Preparation of Amines :
13.2.1 Common names : Common names 13.3.1 : By ammonolysis of alkyl halides :
of aliphatic amines are given by writing the
name of alkyl group followed by suffix-amine, When alkyl halide is heated with alcoholic
that is, ‘alkyl amine’. In the case of mixed solution of excess ammonia it undergoes
amines, the names of alkyl groups are written nucleophilic substitution reaction in which the
in alphabetical order. If two or three identical halogen atom is replaced by an amino (-NH2)
alkyl groups are attached to nitrogen atom, the group to form primary amine. This process of
prefix ‘di-’ or ‘tri-’ is added before the name of breaking of C-X bond by ammonia is known
alkyl group. The parent arylamine, C6H5-NH2, as ammonolysis. The reaction is also known
is named as aniline. Other aromatic amines as alkylation of ammonia. The reaction is
are named as derivatives of aniline (see Table carried out in a sealed tube at 373 K. It may be
13.2). noted that the primary amine obtained in the
1st step is stronger nucleophile than ammonia.
13.2.2 IUPAC names : In IUPAC system, Hence, it further reacts with alkyl halide to
primary amines are named by replacing form secondary and tertiary amines and finally
the ending ‘e’ of the parent alkane by suffix quaternary ammonium salt if NH3 is not used
-amine (alkanamine). A locant indicating the in large excess.
position of amino group is added before the
suffix amine. When two or more amino groups R-X +NH3(alc). ∆ R-NH2
are present, the prefix ‘di-’, ‘tri-’ etc. are used
with proper locant. In this case the ending ‘e’ (excess) 10 amine
of parent alkane is retained.
The order of reactivity of alkyl halides
Secondary or tertiary amines are named with ammonia is R-I > R-Br > R-Cl.
as N-substituted derivatives of primary
amines. The largest alkyl group attached to Use your brain power
nitrogen is taken as the parent alkane and other
alkyl groups as N-substituents. While naming • Write chemical equations for
arylamines ending ‘e’ of arene is replaced by
‘amine’. The common name of aniline is also 1. reaction of alc. NH3 with C2H5I.
accepted by IUPAC (see Table 13.2).
2. Ammonolysis of benzyl chloride
Remember... followed by the reaction with two
The name of amine moles of CH3-I.
(common or IUPAC) is always • Why is ammonolysis of alkyl halide not
written as one word. For example : a suitable method for the preparation of
C2H5-NH2 Ethylamine (Ethanamine) primary amine ?
Try this...
Draw possible structures of all
the isomers of C4H11N. Write their
common as well as IUPAC names.
283
Table 13.2 : Common and IUPAC names of some alkyl and arylamines
Amines Common names IUPAC names
a. Primary amines :
CH3-NH2 Methylamine Methanamine
n-Propylamine Propan-1-amine
CH3-CH2-CH2-NH2
Isopropylamine Propan-2-amine
CH3-CH-CH3 Ethylenediamine Ethane-1, 2-diamine
NH2 Prop-2-en-1-amine
Allylamine
H2N-CH2-CH2-NH2 Aniline or Benzenamine
CH2=CH-CH2-NH2 Aniline/ Phenylamine
NH2
H2N-(CH2)6-NH2 Hexamethylenediamine Hexane-1, 6-diamine
CH3 p-Toluidine 4-Methylaniline
H2N Cyclohexylamine Cyclohexanamine
NH2
CH2-NH2 Benzylamine Phenylmethanamine
Br p-Bromoaniline 4-Bromoaniline or
4-Bromobenzenamine
H2N Dimethylamine
b. Secondary Amines Ethylmethylamine N-methylmethanamine
Methylphenylamine N-methylethanamine
CH3-NH-CH3 N-methylaniline or
Diphenylamine N-methylbenzenamine
CH3-CH2-NH-CH3 N-Phenylbenzenamine
NH-CH3
Trimethylamine N, N-Dimethylmethanamine
C6H5-NH-C6H5
c. Tertiary Amines Ethyldimethylamine N, N-Dimethylethanamine
CH3-N-CH3 Ethylmethyln-propylamine N-Ethyl-N-methyl propan-1-
CH3 Ethylmethylisopropylamine amine
C2H5-N-CH3 N-Ethyl-N-methyl
CH3 propan-2-amine
CH3-N-C3H7 N, N-Dimethylaniline N, N-Dimethylbenzenamine
C2H5
CH3-CH-CH3
C2H5-N-CH3
N-CH3
CH3
284
Do you know ? Solution : Methyl bromide can be converted
into ethyl amine in two stage reaction
When tert-butyl bromide sequence as shown below.
is treated with alcoholic NH3,
isobutylene is formed. This is the result CH3-Br + KCN CH3-CN + KBr
of elimination reaction preferred over
nucleophilic substitution through the stable CH3-CN Na/C2H5OH CH3-CH2-NH2
tertiary butyl carbocation intermediate. (reduction)
The starting compound methyl bromide
H3C-CC-HC3H3 H3C-CC⊕-HC3H3 contains one carbon atom while the product
Br
-Br -H⊕ ethylamine contains two carbon atoms.
NH3(alc)
A reaction in which number of carbons
(tert-Butyl bromide) increases involves a step up reaction. The
overall conversion of methyl bromide into
CH2 ethyl amine is a step up conversion.
H3C-C-CH3
(isobutylene)
13.3.2 Reduction of nitrocompounds : Use your brain power
Aliphatic and aromatic nitrocompounds Use your brain power:
can be reduced to primary amines by using
metal-acid mixture (Sn/HCl or Fe/HCl or Zn/ Identify ‘A’ and ‘B’ in the following
HCl) or catalytic hydrogenation (H2/Ni or Pt conversions.
or Pd) or LiAlH4 in ether. (i) CH3-I KCN A Na/C2H5OH B
R-NO2 + 6[H] Sn/HCl R-NH2 + 2H2O (ii) CH3-Br AgNO2 A Sn/HCl B
(iii) C2H5-I AgCN A Na/C2H5OH B
13.3.3 Reduction of alkyl cyanide
(alkanenitriles) : 13.3.4 By reduction of amides :
Can you recall ? Primary amines having same number of
carbon atoms can be obtained by the reduction
• How is alkyl halide converted of amides by LiAlH4 in ether or by Na/C2H5OH.
into alkyl cyanide ?
O
Primary amines can be obtained by the 4[H] LiAlH4 / ether CH3-CH2-NH2
reduction of alkyl cyanide with sodium and CH3-C-NH2 + or Na/C2H5OH (Ethylamine)
ethanol. This is known as Mendius reduction. (Acetamide)
The reaction can also be brought about by
lithium aluminium hydride. 13.3.5 Gabriel phthalimide synthesis : This
method is used for the synthesis of primary
amine. It involves the following three stages.
R-C N + 4[H] Na/C2H5OH R-CH2-NH2 i. Formation of potassium salt of phthalimide
or LiAlH4 1° amine from phthalimide on reaction with alcoholic
potassium hydroxide.
Problem 13.1 : Write reaction to convert ii. Formation of N-alkyl phthalimide from
methyl bromide into ethyl amine ? Also, the potassium salt by reaction with alkyl
comment on the number of carbon atoms halide.
in the starting compound and the product.
iii. Alkaline hydrolysis of N-alkyl phthalimide
to form the corresponding primary amine.
285
OO O
The overall result is removal of the -C-
i. C N-H alc.KOH C NK⊕ group from the amide. As the product contains
-H2O C one carbon atom less than the original amide.
C
OO of It is a step down reaction.
(Potassium salt
(Phthalimide) phthalimide) Use your brain power
O
O Write the chemical equations for
the following conversions :
ii. C NK⊕ R-X C
C -KX N-R i. Methyl chloride to ethylamine.
C ii. Benzamide to aniline.
OO iii. 1, 4 - Dichlorobutane to hexane - 1, 6 -
diamine.
(N-Alkylphthalimide)
iv. Benzamide to benzylamine.
OO
iii. C 13.4 Physical properties of Amines :
N-R NaOH (aq) C-ONa 13.4.1 Intermolecular forces, boiling points
C and solubility : The N-H bond in amines
C-ON⊕a is polar because the electronegativities of
Nitrogen (3.0) and Hydrogen (2.1) are different.
OO Due to the polar nature of N-H bond primary
and secondary amines have intermolecular
(sodium phthalate) hydrogen bonding. The intermolecular
+ R-NH2
(1° amine)
Aromatic amines cannot be prepared
by this method because aryl halides do not
undergo nucleophilic substitution with the
anion formed by phthalimide.
13.3.6 By Hofmann degradation (Hofmann hydrogen bonding is to greater extent in
rearrangement / Hofmann bromamide primary amine than in secondary amines,
degradation / Hofmann hypobromite because primary amines have two hydrogen
degradation ) : atoms bonded to nitrogen for hydrogen bond
This is a good laboratory method for the formation (see Fig 13.1).
conversion of an amide into primary amine R-HNδ⊕-Hδ⊕ δ NR-Hδ⊕ δ Hδ⊕
containing one carbon less. The reaction is N-R
brought about by warming the amide with
bromine and concentrated aqueous KOH Hδ⊕ H
solution. H-Nδ-R
Hydrogen bond
O H
R-C-NH2 + Br2 + 4KOH(aq) ∆
(Amide) Fig. 13.1 : Intermolecular hydrogen bonding in
R-NH2 + 2KBr+ K2CO3 + 2H2O primary amines
(1° amine) Tertiary amines do not have intermolecular
hydrogen bonding as there is no hydrogen
For example : atom on nitrogen of tertiary amine. But due
to polar N-C bonds, tertiary amines are polar
O (aq)(m∆ethCylHa3m-NinHe)2 molecules, and have intermolecular dipole-
dipole attractive forces. Thus intermolecular
CH3-C-NH2 + Br2 + 4KOH forces of attraction are strongest in primary
(Acetamide) amines and weakest in tertiary amines.
+2KBr + K2CO3 + 2H2O
286
The observed order of boiling points of Table 13.3 Boiling points of alkane, alcohol
isomeric amines is : primary amine > secondary and amines of similar molar masses
amine > tertiary amine (see Table 13.3 serial
numbers 1, 2, 3). It can be explained on the Sr. Compound Molar B.P. (K)
basis of the intermolecular forces in them. No mass
The lower aliphatic amines are gases 1 n-C4H9NH2 73 350.8
with fishy odour, middle members are liquids
and higher members are solids under ordinary 2 (C2H5)2NH 73 329.3
temperature and pressure.
3 C2H5N(CH3)2 73 310.5
Aniline and other arylamines are usually
colourless liquids but get coloured as they are 4 C2H5COOH 74 414.4
easily oxidised by air.
5 n-C4H9OH 74 390.3
Due to their ability to form hydrogen
bond with water molecule, lower aliphatic 6 (CH3)3C-NH2 73 318.15
amines are soluble in water (see Fig. 13.2).
Solubility of amines decreases with increase 7 C2H5CH(CH3)2 72 300.8
in molar mass of amines due to increase in size
of hydrophobic alkyl group. Aromatic amines 13.5 Basicity of Amines
and higher aliphatic amines are insoluble in
water. The basic nature of amines is due to
presence of a lone pair of electrons on the
Hydrogen bond nitrogen atom. In terms of Lewis theory, amines
are bases because they can share a lone pair of
δ δ⊕Rδ δ⊕ δ electrons on ‘N’atom with an electron deficient
H-O H-N H-O species. For example : Trimethylamine shares
its lone pair of electrons with the electron
HHH deficient boron trifluoride.
Fig. 13.2 : Hydrogen bonding between Me3N + BF3 Me3N⊕-BF3
amine and water molecule
Basic strength of amines is expressed
Since N-H bonds in amines are less polar quantitatively as Kb or pKb value. In terms of
than O-H bond in alcohol, water solubilities of Lowry-Bronsted theory, the basic nature of
alcohols, amines and alkanes of comparable amines is explained by writing the following
molar mass in water are in the decreasing equilibrium.
order: alcohols > amines > alkanes.
N + H2O N⊕-H + OH ........... (13.1)
The order of boiling points of alkanes,
amines, alcohols and carboxylic acid of (amine) (conjugate acid)
comparable molar mass is as follows :
In this equilibrium amine accepts H⊕,
Alkanes < Amines < Alcohols < Carboxylic hence an amine is a Lowry-Bronsted base.
acid. (Table 13.3, serial number 4, 5, 6, 7) For stronger base, this equilibrium shifts
towards right, thereby the Kb value is larger
Use your brain power and pKb value is smaller and vice versa (refer
Arrange the following : to Chapter 3). Table 13.4 gives pKb values of
a. In decreasing order of the b.p. some amines.
C2H5-OH, C2H5-NH2, (CH3)2NH
b. In increasing order of solubility in water:
C2H5-NH2, C3H7-NH2, C6H5-NH2
287
Table 13.4 : pKb Values of some amines in aqueous medium
Amine Structural Formula pKb value
Primary alkanamines : CH3-NH2 3.38
Methanamine CH3-CH2-NH2 3.29
Ethanamine (CH3)2-CH-NH2 3.40
Propan-2-amine 4.70
Phenylmethanamine CH2-NH2
3.27
Secondary alkanamines : (CH3)2NH 3.00
N-Methylmethanamine (CH3CH2)2NH
N-Ethylethanamine 4.22
(CH3)3N 3.25
Tertiary alkanamines : (CH3CH2)3N 4.75
N, N-Dimethylmethanamine NH3
N, N-Diethylethanamine
Ammonia
Arylamines :
Benzenamine (aniline) NH2 9.38
N-Methylaniline NHCH3 9.30
N, N-Dimethylaniline N(CH3)2 8.92
13.5.1 : Basic strength of aliphatic amines : Can you recall ?
• What is meant by +I effect ?
The trend in the observed pKb values (see
table 13.4) and basic strength of 1°, 2°, 3° • Which of the following species is
amines and NH3 can be represented as shown better stabilized and by which effect ?
below :
⊕⊕
Order of pKb values:
NH3 > R-NH2 > R2NH < R3N CH3-CH2 and CH3-CH-CH3
Order of basic strength :
Basicity of amines is related to the structural
NH3 < R-NH2 < R2NH > R3N (13.2) effects which influence stabilization of various
species. Greater is the stabilization of the
Thus as per the observed pKb values of the protonated amine, that is, the conjugate acid,
aliphatic amines, secondary amines are the greater is the basicity of the amine.
strongest bases. Basic strength increases as Use your brain power
Refer to pKb values from Table
we move from NH3 to R-NH2 and from R-NH2 13.4 and answer which compound
to R2NH, but basic strength decreases as we from the following pairs is stronger base ?
move from R2NH to R3N (Table 13.4).
i. CH3-NH2 and (CH3)2NH
The basic strength and the corresponding ii. (C2H5)2NH and (C2H5)3N
iii. NH3 and (CH3)2CH-NH2
pKb value depends upon the position of the
equilibrium shown in Eq. (13.1). Greater the
stabilization of the conjugate acid more on
right side the equilibrium will lie and stronger
will be the base and smaller will be its pKb
value.
288
a. Influence of +I effect on stabilization of HOH , R
conjugate acids of aliphatic amines and NH3 H R-N⊕-R
can be represented as shown below :
R-N⊕-R H
H H H H H
H N⊕H R N⊕H R N⊕ R HOH
⊕ HOH
RNR
H HHR
An alkyl group exerts electron releasing The solvent water stabilizes the conjugate
acid by hydrogen bonding through the ‘H’
inductive effect (+I) which stabilizes positive bonded to the ‘N⊕’. The number of ‘H’ atoms
bonded to the ‘N⊕’ decreaes from 4 in NH4⊕
charge on atom bonded to it. As we move to 1 in R3NH⊕. As a result NH4⊕ is best
stabilized by solvation while the stabilization
from conjugate acid of ammonia (NH4⊕) to by solvation is very poor in R3NH⊕.
that of tertiary amine (R3NH⊕), the number of
alkyl groups (R) bonded to Nitrogen goes on c. Combined influence of +I effect and
solvation on stabilization if conjugate acids
increasing steadily. This results in increasing of aliphatic amines decides the observed basic
strength and pKb value. These two influencing
stabilization of the conjugate acids and factors operate in opposite directions.
thereby an increasing order of basic strength
is expected.
Order of stabilization :
N⊕ H4 < R-N⊕ H3 < R2N⊕ H2 < R3N⊕ -H
Expected order of basic strength : Solvation increases
+I effect
NH3 < R-NH2 < R2NH < R3N N⊕ H4 R-N⊕ H3 R2N⊕ H R3N⊕ H
The expected order of basic strength on the increases
basis of +I effect differs from the observed order
(Eq.13.2). It is seen that the observed increasing The net results is that as we move from
basic strength from ammonia to amine and NH3 to RNH2 to R2NH, the basic strength
from 1° amine to 2° amine is explained on the increases due to better stabilization of the
basis of increased stabilization of conjugate corresponding conjugate acids. But 3° amine
acids by +I effect of increased number of is weaker base than 2° amine because the
alkyl (R) groups. However, decreased basic stabilization of conjugate acid of 3° amine by
strength of 3° amine implies that the conjugate solvation is very poor.
acid of 3° amine is less stabilized even
though the +I effect of three alkyl groups in 13.5.2 Basicity of arylamines :
R3N⊕ H is expected to be large. This is suggestive
of existance of another influencing factor in Can you recall ?
stabilization of conjugate acids of amines. Refer to Table 13.4 and answer :
b. Influence of solvation by water on Are the pKb values of aniline,
stabilization of conjugate acids of aliphatic N-methylaniline and N, N-dimethylaniline
amines and ammonia can be represented as larger or smaller than those of NH3 and
shown below : CH3NH2 ?
Which one of the two, aniline or CH3NH2, is
HOH HOH stronger base ?
H O H O H , H O H ,
H H-N-H H R-N⊕-H H
H⊕ H
HOH HOH
289
From the pKb values we understand that As a result the equilibrium (13.3) is
arylamines in general are weaker bases than shifted towards left side. This makes aniline
ammonia and aliphatic amines. Strength (and also other arylamines) weaker bases than
of arylamines is explained in accordance aliphatic amines and ammonia.
with Lowery Bronsted theory by writing the
following equilibrium (Eq. 13.3) For aniline Use your brain power
(similar to eq. 13.1). Arrange the following amines
NH2 ⊕ NH3 in decreasing order of their basic
strength -
+ H2O + OH ....... (13.3)
NH3, CH3-NH2, (CH3)2NH, C6H5NH2.
(Base) (Conjugate acid) 13.6 Chemical properties of amines
Here, both the species base and conjugate 13.6.1 Laboratory test for amines :
acid, are resonance stabilized but to different
extent. a. Test for amines as the ‘base’ : All amines
1°, 2° and 3° are basic compounds. Aqueous
In arylamines, the -NH2 group is attached solution of water soluble amines turns red
directly to an aromatic ring. The lone pair litmus blue.
of electrons on nitrogen is conjugated to
the aromatic ring and is less available for The ‘basic’ nature of amines is detected in
protonation. Aniline is resonance stabilized by laboratory by reaction with aqueous solution
the following five resonance structures. of strong mineral acid HCl.
⊕ ⊕ N + HCl(aq) N⊕ -H(aq) + Cl (aq)
NH2
(amine) (a substituted
ammonium chloride)
NH2 NH2
N⊕ -H(aq)+ Cl (aq) + NaOH(aq) N +
(excess)
NaCl(aq) + H2O
(I) (II) (III)
⊕NH2 NH2 Water insoluble amine dissolves in
aqueous HCl due to formation of water
(IV) (V) soluble substituted ammonium chloride,
which on reaction with excess aqueous NaOH
On the other hand anilinium ion obtained regenerates the original insoluble amine.
by accepting a proton does not have lone
pair of electrons on nitrogen. Hence it can be b. Diazotization reaction/ Orange dye test:
stabilized by only two resonance structures In a sample of aromatic primary amine,
and therefore less stabilized than aniline. 1-2 mL of conc. HCl is added. The aqueous
solution of NaNO2 is added with cooling. This
⊕ NH3 ⊕ NH3 solution is transfered to a test tube containing
solution of b naphthol in NaOH. Formation
(I) (II) of orange dye indicates presence of aromatic
primary amino group. (It may be noted that
temperature of all the solutions and reaction
mixtures is maintined near 0°C throughout the
reaction).
290
The reaction involved in this test will be 13.6.3 Hofmann Elimination :
discussed in section 13.6.5. When tetraalkylammonium halide is
13.6.2: Alkylation of amines : Hofmann’s hetated with moist silver oxide, it gives
exhaustive alkylation : When a primary amine quanternary ammonium hydroxide which is
is heated with excess of primary alkyl halide it a deliquescent crystalline solid, and strongly
gives a mixture of secondary amine, tertiary basic like NaOH or KOH. Quaternary
amine along with tetraalkylammonium halide ammonium hydroxides on strong heating
(also refer to sec. 13.3.1). undergo b-elimination to give an alkene, the
reaction is called Hofmann elimination. The
R-NH2 R-X R2NH R-X R3N R-X R4N⊕ X least substituted alkene is obtained as major
-HX -HX -HX product (in contrast to Saytzeff elimination)
(Tetraalkyl
(1° Amine) (2° Amine) (3° Amine) For example :
ammonium
halide) CH3CH2CH2-N⊕ (CH2CH3)3 I + AgOH
If excess of alkyl halide is used tetraalkyl (N,N,N-triethylpropylammonium iodide)
ammonium halide is obtained as major
product. The reaction is known as exhaustive ∆ moist
alkylation of amines. Ag2O
The tetraalkylammonium halides are ab
called quaternary ammonium salts which are
crystalline solids. They are the derivatives CH3-CbH' 2-CaH' 2⊕-CNH-C2-HC2H-C3 H3 OH
of ammoium salts in which all the four CH2-CH3
(N,N,N-triethylpropanammonium hydroxide)
hydrogen atoms attached to nitrogen in ∆ -H2O
N⊕ H4 are replaced by four alkyl groups (same
or different). Primary, secondary and tertiary CH2= CH2 + CH3CH2CH2-N-CH2CH3
amines consume three, two and one moles of (ethene)
alkyl halide respectively to get converted into CH2CH3
(N,N-diethylpropylamine)
quaternary ammonium salt. The reaction is
carried out in presence of mild base NaHCO3, Use your brain power
to neutralize the large quantity of HX formed.
If the alkyl halide is methyl iodide, the reaction Complete the following reaction:
is called exhaustive methylation of amines.
CH3-CH2-N⊕ (CH3)3I Moist ?∆
For example : When methylamine is heated Ag2O
with excess methyl iodide, it gives tetramethyl Do you know ?
ammonium iodide.
CH3-NH2 + CH3-I ∆ (CH3)2NH + HI • Acetylcholine is a quaternary
(CH3)2NH + CH3-I ∆ (CH3)3N + HI ammonium salt which occurs
(CH3)3N + CH3-I ∆ (CH3)4N⊕ I in nervous system and functions as
neurotransmitter
Use your brain power CH3-CO-O-CH2-CH2-N⊕ (CH3)3
C2H5-NH2 + C2H5-I ∆ ?
excess • Quaternary ammonium salts are also
(C2H5)2NH + CH3-I ∆ ? present in cationic detergents.
excess
C6H5-NH2 + CH3-I ∆ ?
excess
291
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