Chapter-V
Chemistry of organic comounds
ALKANES
PREPARATION OF ALKANES WITH SPECIAL REFERENCE TO
METHANE AND ETHANE
Source of Methane: Methane is called Marsh gas because it is found in marshes, swamps,
muddy sediments wherever bacteria decompose organic matter in the absence of oxgyen. Methane
occurs in natural gas obtained from oil wells.
1.
PREPARATION OF METHANE:
(a)
Decarboxylation of Sodium acetate by sodalime:
When a mixture of sodium acetate(or sodium ethanoate) and sodalime is heated in a
tube, methane gas is produced. Sodalime is a mixture of sodium hydroxide(NaOH) and calcium
oxide(CaO).
CH3
O _
+
C O Na + NaOH
CaO
CH4 + Na2CO3
Here CaO oxide does not take part in the reaction. Look to the box shown in the equation. After
the removal of Na2CO3(contained inside the box), the CH3 group of methyl acetate joins with H
atom of NaOH to produce methane(CH4). Since the carboxylic acid group is lost from sodium
acetate in the form of sodium carbonate, the reaction is called decarboxylation of sodium acetate.
SAQ 1: What product will be obtained if a mixture of sodium propanoate and sodalime is heated?
2.
From Carbon monoxide:
When a mixture of CO and H2 gas is passed over finely divided nickel at 3000C, we get methane.
CO + 3 H2
Ni
3000C
CH4 + H2O
3.
Action of water on aluminium carbide.
Al4C3 + 12H2O ---------> 3CH4 + 4Al(OH)3
Methane can also be prepared by the action of water on aluminium carbide.
PREPARATION OF ETHANE:
Ethane is also found in natural gas alongwith methane. It can also be prepared by several
other ways.
1.
Kolbe's reaction: Electrolysis potassium acetate:
When patassium acetate solution is electrolysed, we get ethane at the anode and H2 gas
at cathode.
O
CH3
C
- +
OK
O
CH3
C
O
+
+ K
When dissolved in water potassium acetate dissociates to produce free acetate ion(anion) and
potassium ion(cation). When this solution is connected to a cathode and anode and electricity is
passed through the solution, ethane is evolved at the anode and hydrogen gas is produced at the
cathode.
CH3
At anode:
CH3
O
C
O
C
O
CH3 CH 3
-
+ 2 CO2 + 2 e
-
O
Two acetate ions move together to the anode and undergo oxidation at anode. 2 molecues of
CO2 molecules are evolved as shown in the rounded mark and a new bond is established between
the two CH3 groups(shown by broken line) to produce ethane. The two excess electrons present
in the two acetate ions are given to the anode. Note that anode is in the positive potential for
which it accepts the electron and brings about oxidation.
At cathode:
There are two competing ions for the cathode. One is K+ from
potassium acetate and the other H+ ion produced from H2O. Note that H2O produces H+ and
OH- due to dissociation. Out of the two cations, H+ and K+ ; H+ ion has a greater tendency to
undergo reduction i.e to gain electron. So it will be preferentially discharged at the cathode to
free hydrogen gas.
At cathode:
2 H+ + 2 e- -----------> H2
2.
Wurtz Reaction:
When methyl halide(chloride, bromide or iodide) react with metallic sodium in the presence
of ether solvent, we get ethane. This is called Wurtz reaction.
CH3
Cl + 2Na
+
Cl
CH3
ether
CH3
CH3 + 2 NaCl
In this case, two molecules of methyl chloride reacts with two atoms of sodium to produce
ethane and sodium chloride. In the above scheme, you find that after the removal of 2 NaCl
molecules(shown within the rectangula box), one CH3 group joins with the other CH3 group to
produce ethane. Note that ether is used only as a solvent in place of water.
In general, we can take any alkyl halide in place of methyl chloride and two such alkyl groups
will join to produce an alkane containing double the number of carbon atoms as compared to the
alkyl halide.
R Cl + 2Na
+
Cl
R
ether
R R
+
2 NaCl
R stands for any alkyl group like methyl, ethyl, propyl etc. The only demerit in Wurtz reaction is
that we always get alkane containing even number of carbon atoms(as the number of carbon
atoms is doubled). Answer the following SAQ.
SAQ 2: What alkane we shall get when ethyl bromide is treated with sodium in ether solvent.
3.
Hydrogenation of alkene and alkyne:
Ethane: Ethane can be prepared when a mixture of ethylene and H2 gases are passed
over heated nickel catalyst at 200-3000C. In this process the double bond gets saturated with
hydrogen atoms. This reduction is called Sabatier and Senderen's reduction.
H2C
CH2 + H2
Ni
200-3000C
CH3
CH3
Ethane can also be prepared from ethyne(acetylene) by the same method. In this case
2 moles of hydrogen are necessary. First, acetylene is reduced by one mole of hydrogen gas to
produce ethylene. Ethylene then reacts with the second mole of hydrogen to give ethane. The
triple bond of acetylene first changes to double bond in ethyelene and then the double bond
changes to single bond in ethane.
HC
CH +
H2
acetylene
H2C
CH2 + H2
Ni
200-3000C
Ni
200-300 0C
H2C
CH2
ethylene
CH3
CH3
SAQ 3: What product is formed when propene is treated with H2 gas in presence of nickel
catalyst at 3000C?
Methane from CO:
Methane is produced when a mixture of CO or CO2 and hydrogen is passed over finely
divided nickel at about 3000C. This is also a case of Sabatier and Senderens reduction.
CO + 3 H2
Ni
3000C
CH4 + H2O
PROPERTIES OF ALKANES:
Physical properties:
(i)
The first four alkanes namely methane, ethane, propane and butane are gases at room
temperature. The next 13 memebers(C5-C17) i.e from pentane onwards are liquids and higher
members(>C18) are wax like solids.
(ii)
They have lower boiling and melting points than any other organic compounds because
they are non-polar in nature and there is very weak Vander waal's forces existing between the
molecules. However, the boiling and melting point increases as the chain length of the alkane
increases. That means as we go from methane to ethane, then to propane and so on, the boiling
and metling points increases.
(iii)
They are not soluble in water because they are non-polar covalent compounds while
water is a polar solvent. You know the principle: LIKE DISSOLVES LIKE.
Chemical Properties:
Alkanes are inert. They do not react with acids, bases or common oxidising or reducing agents.
However few elements like halogens(Cl2, Br2 etc.) and oxygen gas react with the alkanes.
1.
Reaction with Halogens:
The reaction of flourine with alkane is explosive in nature and is not carried out. However
alkane reacts with Cl2, Br2 and I2 to produce haloalkanes. This is, in general, called halogenation
reaction. The chlorination of alkane is fastest among them and bromination is slower while
iodination is the slowest. Halogenation reactions are carried out in presence of sun-light and
heat. Chorination is so fast that it is carried out only in diffused sunlight(not direct sunlight) while
bromination is carried out in presence of both heat and light.
Chlorination of Methane:
Methane undergoes successive reactions with one, two, three and four moles of chlorine to give
mono-, di-, tri- and tetra chloro methane. These are commonly called substitution reactions. In
this type reaction, one H atom of alkane is substituted by one halogen atom(here Cl atom).
diff.
CH4 + Cl2 sunlight
CH3Cl
+
monochloromethane
or methyl chloride
HCl
diff.
CH2Cl2
+
CH3Cl + Cl2 sunlight
dichloromethane
HCl
diff.
CH2Cl2 + Cl2 sunlight
HCl
or methylene chloride
diff.
CHCl3 +Cl2 sunlight
CHCl3
trichloromethane
or chloroform
+
CCl4
+
tetrachloromethane
or carbon tetrachloride
HCl
SAQ 4: What main product will be formed when CH4 reacts with
(i)two moles of Cl2 gas (iii)four moles of Cl2 gas in presenc of diffused sun-light?
SAQ 5: What product is obtained when ethane reacts with one mole of Cl2 gas? How many
moles of chlorine will be required to remove all the hydrogen atoms of ethane?
SAQ 6: How many monochloro propanes will be produced when propane is treated with one
mole of chlorine gas in the presence of diffused sun-light.
Bromination: Bromination is slow and requires both direct sunlight and heat(1270C).
Due to larger size of Br atom, successive substitution to give dibromo, tribromo and tetrabromo
alkane is difficult and often not possible.
CH4 + Br2
sunlight
heat
CH3Br + HBr
We get monobromomethane or methylbromide.
In the same manner we can get methyl iodide but the reaction is very slow. We shall not discuss
the iodination reaction now.
2.
Reaction with Oxygen:
All alkanes burn in air or oxgyen to give usually carbon dioxde and water. This is called
combustion reaction. The gas fuel that we use in our kitchen(LPG gas) contains mostly butane
alongwith small quanities of propane and other gases(ethane and methane). They burn in air
giving heat, light, CO2 and H2O. These are highly exothermic reactions.
CH4 + 2O2 ----------> CO2 + 2H2O + heat
C2H6 + O2 ----------> 2CO2 + 3H2O + heat
Similar reactions can be written for propane, butane etc.
Incomplete combustion: When alkane burns in insufficient amount of oxygen then we
get other poisonous gase like CO and other gases like HCHO(formaldehyde), CH3COOH(acetic
acid) and also carbon particles which pollute our enviroment. The hundreds and thousands of
motor vehicles that ply in our roads produce these dangerous substances besides CO2 and has
been the major cause of air pollution.
2CH4 + 3O2 --------->2 CO + 4H2O,
CH4 + O2 --------> C + 2H2O
CH4 + O2 --------> HCHO + H2O,
2CH4 + 3O2 ----->2 CH3COOH + 2H2O
SAQ 7:
Write the products of each reaction.
(i)Sodium acetate is subjected to electrolysis.
(ii)Sodium acetate is mixed with sodalime and the mixture heated.
(iii)2-chloropropane is treated with sodium metal in ether solvent.
(iv)One mole of methane is treated with three moles of chlorine gas in presence of
diffused
sunlight.
(v)Butane is treated with one mole of chlorine in presence of diffused sunlight.
(vi)How many monochloro compounds will be obtained when the following alkanes are
treated with one mole of chlorine in presence of diffused sunlight.
(a)2,2-dimethylpropane (b)2-methylbutane
SAQ 8: Solve the road map problem. Give complete equation in each step
Cl2
(i)
CH4
diff. sunlight
(ii)
CH3
CH3
A
Br2
sunlight / heat
Na
ether
A
B
Na
ether
B
Cl2
diff. sunlight
C + D
A and B are the organic products. The other inorganic product you shall have to write in complete
equations.
ALKENES
WITH A SPECIAL REFERECE TO ETHYLENE:
METHODS OF PREPARATION:
1.
Dehydrohalogenation of alkyl halide(Removal of HX):
H
Cl
CH2
CH2
alcoholic
H2C
KOH
CH2 + HCl
When ethyl chloride(chloroethane) is treated with alcoholic potassium hydroxide solution,
we get ethene(ethyelene) and HCl. KOH used as the reactant reacts with HCl formed to
convert it to KCl and H2O. This is called dehydrohalaogenation(removal of HX). In general, any
alkene can be prepare by this method.
H
Cl
CH
CH2
alcoholic
R
KOH
R
HC
CH2 + HCl
If we take 1-chloropropane(R=CH3), we get propene and if we take 1-chlorobutane, we get
but-1-ene and so on. This kind of reactions are called the elimination reactions. You have
already seen a sustitution reaction in the properties of alkane. In elimination reaction, two
groups leave from the adjacent carbon atoms in the form of a neutral molecule to result a double
bond. In this case, H from one carbon atom and halogen(X) from the adjacent carbon atom
leave as HX molecule resulting the formation of a double bond between those two carbon
atoms.
Saytzeff' Rule: If the halogen atom is located in an internal position and there is hydrogen atom
on either side of the halogen atom, then we get two alkenes. Out of the two, the one which is
more highly substituted is formed in greater quanity(major proudct).
H Cl
H2C CH CH2 CH3
Cl
alcoholic
KOH
H2C
CH CH2 CH3 + HCl
but-1-ene
CH3
CH CH2 CH3 + HCl
but-2-ene
H
H3C CH CH CH3
alcoholic
KOH
2-chlorobutane produces two alkenes i.e but-1-ene and but-2-ene. According to Saytzeff's rule
but-2-ene is formed in greater quantity than but-1-ene. This is because but-2-ene is more highly
substituted. But-2-ene has two branches(two CH3 groups) attached to the double bond while
but-1-ene has only one branch(ethyl). According to the Saytzeff's rule but-2-ene therefore is the
major product.The branches have been shown inside boxes for better understanding.
CH3
CH CH CH3
CH2
CH CH2 CH3
2.
Dedydration of Alcohol:
We get ethylene when a mixture of ethyl alcohol and conc. H2SO4 is heated to a
temperature of 1700C. Ethylene gas comes out from the reacting vessel and is collected in a gas
jar by the downward displacement of water. This is the laboratory method of preparation of
ethylene. Ethyl alcohol merely loses a molecule of water in this reaction. Note that in this reaction,
ethyl alcohol is not taken in large quantity. If the ethyl alcohol is taken in large quantity, we do not
get the alkene, rather we get different products, which we shall see in higher classes.
H
OH
CH2
CH2
conc. H2SO4
1700C
H2C
CH2 + H2O
This is also an elimination reaction in which H atom from one carbon and OH group from the
adjacent carbon are removed as H2O and double bond is formed in between the two carbon
atoms. Other alkene can also be prepared by this method. See the examples.
CH3
H
OH
CH
CH2
conc. H2SO4
1700C
CH3
HC
CH2 + H2O
(propene)
Actually dehydration of alcohol takes place in two steps. First when conc. H2SO4 is mixed with
alcohol, we get alkyl hydrogen sulphate. In the second step, this alkyl hydrogen sulphate is
heated to 1700C to give alkene and regenerate H2SO4.
Step1:
Step 2:
H 3C
CH2 O H + H O
H
OSO3H
CH2
CH2
H3C CH2 O SO 3H + H2O
ethyl hydrogen sulphate
SO 3H
1700C
H2C
CH2 + H2SO4
For simplicity, you better do not show these steps always while writing the reaction.
SAQ 9: Predict the products and write the equations.
(i)2-bromopropane + (alcoholic) KOH --------> ?
(ii) butan-2-ol + conc. H2SO4 -------heat-------> >
PROPERTIES OF ALKENES:
The first three members are gases(ethene, propene and but-1-ene/but-2-ene) are gases, the
next 13 members from C-5 to C-17 are liquids and higher members are solids at room temperature.
Like alkanes, alkenes too have lower boiling and melting points due to poor Van der Waals forces
between them. They are not also soluble in water like alkanes.
Chemical Reactions:
The most important reaction of alkenes is the addition reactions.
(A)
ADDITION REACTION:
R
CH
CH2 + A B
R
B
A
CH
CH2
When a molecule A-B is added to alkene, one atom(say A) adds to one carbon atom of the
double bond and the other atom(B) adds to the other carbon atom of the double bond and the
double bond is converted to a single bond. It is just the opposite of the elimination reaction by
which an alkene is prepared. In addition reaction, the alkene which is an unsaturated compound
becomes saturated. Let us see several cases.
(i)
Addition with H2:
We have already studied this in the preparation of alkanes(hydrogenation of alkenes).
When an alkene reacts with H2 at a higher temperature in presence of Ni catalyst, we get
alkane.
Ni
H2C
CH2 + H2
200-3000C
H3C
CH3
Similarly propene will add with H2 to give propane and butene gives butane and so on.
(ii)
Addition with halogens (Cl2, Br2 and I2 )
Addition of halogen with alkene is very fast. No sunlight or heat is necessary for this
reaction. The reaction can even take place even in dark. Alkene adds to halogen to give dihalo
alkane. See these examples.
Cl
H2C
CH2 + Cl2
H2C
Cl
CH2
1,2-dichloroethane
Br
CH3
HC
CH2 + Br2
CH3
HC
Br
CH2
(1,2-dibromopropane)
In the addition of bromine to alkenes, the red color of bromine is dicharged. This is used as test
for alkenes. The two halogen atoms lie on adjacent carbon atoms. Such dihalo alkanes are
generally called vicinal dihalides.
(iii)
Addition with HX(HCl, HBr, HI)
Ethylene adds with HX to give ethyl halide as shown below. H atom adds to one carbon
while X atom adds to the other carbon atom to to form the product.
H2C
CH2 + HCl
H3C CH2 Cl
(ethyl chloride)
Since ethylene is a symmetrical alkene, it forms one product as shown above, but unsymmetrical
alkenes like propene when adds to HX, we get two products, one of which is major and the other
minor. The major product in such case can be predicted from the Markonikoff's rule.
Markonikoff's rule:
The negative part of HX adds to that carbon atom of the double bond which bears less
number of H atoms with it. See this exmaple.
Cl
H3C
CH
+ -
CH2 + H Cl
H3C
CH
CH3
2-chloropropane
Propene is not a symmetrical alkene because, on one side of the double bond there is CH3 group
and on the other side, there is nothing. In such case, the negative part of HX will add to the
carbon which bears the less number of H atoms. You know that in HX, the negative part is X(as
it is more electronegative) and H is the positive part(less electronegative). According to
Markonikoff's rule, X will add to the carbon which is attached to less number of H atoms. In this
case X will add to the middle(C-2)carbon as it is attached to one H atom while H of HX will add
to the terminal carbon as it is attached to two H atoms. Thus we get 2-chloropropane(or
isopropyl chloride) as the major product. The other product, 1-chloropropane obtained by the
reverse addition of HCl to propene is formed to a very small extent and has therefore not been
written.
SAQ 10:
(i)What product is obtaiend when but-2-ene reacts with H2 gas in presence of
Ni
and at 3000C?
(ii)What product is obtained when but-1-ene reacts with Cl2?
(iii)What major product is obtaiend when but-1-ene reacts with HI?
(iv)Wht product is obtained when but-2-ene reacts with HCl? Is there any second
product in this reaction? Explain.
(iv)
Addition with HOX (hypohalous acid)
Alkene adds with HOX(HOCl, HOBr, HOI) to produce the addition product commonly
called halohydrin.
H2C
CH2 + (HO)Cl
OH
Cl
CH2
CH2
(ethylene chlorohydrin
or 2-chloroethanol)
Since ethylene is a symmetrical alkene, hypchlorous acid(HOCl) gives only one product as
shown above. If the the alkene is unsymmetrical, then the we get two products and the major
one is predicted from Markonikoff's rule.
SAQ 11: What major product is obtained when propene is treated with hypobromous acid(HOBr)?
(v)
Addion with Ozone(O3): Ozonolysis
This is a very important reaction of alkene and is called ozonolysis. Alkene reacts with ozone to
form an ozonide first. In alkene ozonide, the three O atoms of O3 are linked at the double bond
site, two O atoms on one side and one on the other side of the double bond. The bond between
C-C vanishes to account for the tetravalency of carbon.
H2C
CH2 + O3
O
O
CH2
CH2
(ethylene ozonide)
O
The alkene ozonide is unstable and reacts with water to produce aldehydes or ketones or both.
This reaction is catalysed by Zn. See the case of ethylene ozonide.
O
H
O
H
C
H
O
H C H + H C H + H2O2
(methanal)
O
C
O
+
H2O
Zn
H
The ozonide bridge breaks down in this process and each carbon atom of the formerly double
bond takes up one O atom and the remaining O atom reacts with H2O to form H2O2. Thus in this
case two methanal(formaldehyde) molecules are formed. Here two same aldehdydes are formed.
In other cases, we may get two different aldehydeds or two same or different ketones or a
mixture of aldhehyde and ketone depending on the structure of alkenes. The formation of ozonide
and its hydrolysis to give aldehydes/ketones together is called ozonolysis.
SAQ 12:
What products are are obtained by the ozonolysis of propene.
SAQ 13:
What products ar obtained by the ozonolyis of
(i)but-2-ene and (ii) 2-methylbut-2-ene
(iv)
Addition with H2O:
Alkenes react with H2O to give alcohols. This is the reverse reaction of dehydration of
alcohols to prepare alkene. However, alkene does not add with H2O as such. First alkene is
treated with conc. H2SO4 to form the alkyl hydrogen sulphate. On diluting this with water,
hydrolysis of alkyl hydrogen sulphate takes place to form alcohol.
OH
H2 C
CH 2 + H
H3 C
OSO 3H
CH2
H
OSO3H
(ethyl hydrogen sulphate)
+ H2O
CH3
CH2
OH + H2SO4
(ethyl alcohol)
Ethylene forms ethyl alcohol. For unsymmetrical alkenes(like propene), Markonikoff's
rule is applicable. In H2SO4, H is the positive part and OSO3H is the negative part. Accordingly
addition of H2SO4 to alkene will take place.
(B)
OXIDATION REACTIONS:
(i)
With Dilute Alkaline KMnO4 solution:(Baeyer's Reagent):
When alkene is passed through dilute alkaline KMnO4 solution called the Baeyer's reagent, the
pink colour fo KMnO4 is discharged. We get a diol by the addition of two OH groups.
H2C
CH2 +
H2O + [O]
alk.
KMnO4
OH
OH
CH2
CH2
(ethane-1,2-diol
or ethylene glycol)
KMnO4 supplies the nascent O atom which reacts with H2O to form H2O2 and this
provides two OH groups which are added to the two carbon atoms of the double bond to form a
diol. Note that all alkenes react with Baeyer's reagent in the similar manner as ethylene .
(ii)
Combustion in air:
Alkene burns in air or O2 to give CO2 and H2O like alkane.
SAQ 14: What products are obtained when propene reacts with conc. H2SO4 and then diluted
with water.
SAQ 15: What products are obtained when propene reacts with Baeyer's reagent.
ALKYNES
(WITH SPECIAL RERFERENCE TO ACETYLENE)
Do you know that the gas is used for artificial ripening of fruits like banana, mango etc. is
acetylene ? Also acetylene gas is used for making oxyacetylene flame to provide very high
temperature needed for welding of metals.
PREPARATION:
1.
Dehydrohalogenation of vicinal dihaloalkanes:
Two Halogen atoms attached to adjacent carbon atoms are called vicinal dihaloalkanes.
When such compounds are treated with alcoholic KOH, we get alkyne.
H
Cl
CH
CH
Cl
H
alcoholic
KOH
CH
CH + 2 HCl
(acetylene)
When 1,2-dichloroethane reacts with alcoholic KOH, elimination of two HCl molecules takes
place in the manner shown above, and two new bonds are produced between the two carbon
atoms and we get a triple bond. We get acetylene. This is similar to dehydrohalogenation of alkyl
halides(haloalkanes) to produce alkenes. In haloalkanes, one HX molecule is eliminated to form
alkene; while in dihaloalkanes, two HX molecules are eliminated to form alkyne. The halogen
and H atoms are shown face to face so as to show the two elimnation by means of two boxes.
We can get different alkynes depending on the structure of the dihaloalkanes. See the SAQ
below.
SAQ 16: What product we get when 1,2-dichloropropane is treated with alcoholic KOH.
2.
Hydrolysis of calcium carbide:
CaC2 + H2O ---------> Ca(OH)2 + C2H2
H
-
OH
Ca
++
C
+ 2H2O
CH
CH + Ca(OH)2
C
OH
H
-
CaC2 is an ionic compound containing Ca2+ and C22-. In the carbide(C22-) there is a triple bond
between the two carbond atoms. When water reacts with calcium carbide, hydrolysis occurs at
two carbon atoms, and we get acetyelene and Ca(OH)2. This is the simplest method of preparing
acetylene in the laboratory.
PROPERTIES:
Acetylene is a colourless gas and very less soluble in water.
A.
ADDITION REACTIONS:
(i)
With H2: When alkyne reacts with H2 in presence of Ni catalyst at high
temperature(200-3000C), first alkene is fomred with one mole of H2 and then alkane is formed
with the second mole of H2. This has been already discussed in the topic alkane(hydrogenation
of alkenes and alkynes).
(ii)
With X 2(halogen): Alkyne reacts with two moles of halogen to form
tetrahaloalkane.
CH
CH + 2 Cl2
Cl
Cl
CH
CH
Cl
Cl
(1,2,3,4-tetrachloroethane)
When two molecules are added, the triple bond is converted to a single bond.
(iii)
With HOX (hypohalous acid):
Cl
CH
CH + 2 (HO) Cl
OH
CH
CH
Cl
OH
- H2 O
Cl
O
CH
C
Cl
(2,2-dichoroethanal)
H
Two OH and two Cl groups are added repeated to the same carbon atoms and form an unstable
comound shown within the bracket. Two OH groups cannot be attached to one carbon atom.
A molecule of H2O is eliminated from the unstable compound to form a carbonyl group. Thus we
get dichloro acetaldehyde when acetylene reacts with two moles of hypocholorous acid.
(iv)
With Water:
Alkynes react with H2O in presence of catalyst HgSO4 and
H2SO4 at 60-800C to form aldehyde or ketone. Acetylene forms aldehyde(acetaldehyde) and all
other alkynes form ketones.
CH
CH + 2 H2O
HgSO 4
H2SO4
H
OH
CH
CH
H
OH
- H 2O
O
CH3 C H
(acetaldehyde)
Two H2O molecules are added successively to form an unstable compound shown within the
bracket. One H2O molecule is eliminated from this unstable compound to form acetaldeyde.
(v)
With Ozone: Alkynes also react with ozone to form first the ozonide like alkene.
Subsequently the ozonide is hydrolysed to form two molecules of carboxylic acid. Acetylene
forms two molecules of methanoic acid(formic acid).
(vi)
With Alkaline KMnO4: Acetylene and all other alkynes also decolorises the
pink colour of alkaline KMnO4(Bayer's reagent) like alkenes. Acetylene forms oxalic acid(HOOCCOOH) when reacts with Baeyer's reagent.
SAQ 17: Write the products of the following reactions.
(i)
Propyne + 2 Br2
?
(ii) Propyne + H2O
HgSO 4
H2SO4 heat
?
(iii)
?
Propyne + 2 HOBr
SAQ 18: Solve the road map problem.
CH4
E
(B)
Cl2
diff. sunlight
alcoholic
F
KOH
A
Na
Ether
H2O
HgSO 4/ H2 SO4
B
Br2/light
heat
C
alcoholic
KOH
D
Cl2
G
ACIDIC PROPERTIES
(i) With metallic sodium:
H atom attached to carbon atom bonded with a triple bond is an acidic hydrogen atom.
When sodium reacts with acetylene, hydrogen gas is evolved and the salt of acetylene called
acetylide is formed. First monosodium acetylide is formed with one mole of sodium and then
disodium acetylide is formed with the second mole of Na.
CH
CH + Na
CH
C Na + 1/2 H2
monosodium acetylide
CH
C Na
+
Na
Na C
C Na + 1/2 H2
disodium acetylide
This test is used to distinguish between ethylene and acetylene. With sodium, ethylene does not
produce H2 gas while acetylene produces H2 gas.
HALOALKANES(ALKYL HALIDES)
PREPARATION:
1.
From alkanes:
R-H + X2 ---------R-X + HX
(Where R=alkyl group and X=Cl,Br,I)
We know that when an alkane reacts with a halogen(Cl2,Br2,I2) at the appropriate
conditions, produces haloalkane. One H atom of alkane is substituted by a halogen atom.
2.
From Alcohols:
We can convert an alcohol(R-OH) to the corresponding alkyl halide(R-X) by treating
alcohol with selective reagents. For chlorination we use PCl3, PCl5, SOCl2(thionyl chloride) and
for bromination we use PBr3 and iodiation PI3.
3 R Cl + H3PO3
R Cl + POCl3 + HCl
R Cl + SO2 + HCl
3 R OH + PCl3
R OH + PCl5
R OH + SOCl2
3 R OH + PBr3
3 R OH + PI3
Br2/P4
I 2/P4
3 R Br + H3PO3
3 R I + H3PO3
In each case the OH group is substituted by a halogen atom.
SAQ 19:
Write the products of the following reaction.
(i)
Ethyl alcohol is treated with phosphorous pentachloride.
(ii)
Propan-2-ol(isopropyl alcohol) is treated with PBr3.
(iii)
Methyl alcohol is treated with I2 in presence of phosphorous.
PROPERTIES:
1.
With aqueous KOH (Preparation of Alcohol)
R
X +
K OH(aq.)
R
OH + KX
When alkyl halide reacts with aqueous KOH, we get the corresponding alcohol. In this
reaction, the halogen atom(X) is substituted by OH group.
2.
With alcoholic KOH(Preparation of Alkene)
We have discussed this in the chapter, Alkenes that when alkyl halide reacts with alcoholic
KOH, dehydrohalogenation(-HX) takes place and we get an alkene. You noticed that mere
change of the solvent from water(aqueous) to alcohol(alcoholic), the KOH performs different
function. Aqueous KOH brings about substitution reaction to give an alcohol while alcoholic
KOH brings about elimination reaction to give an alkene.
3.
With Potassium cyanide(KCN):
R
X
+
K CN
R
CN + KX
When alkyl halide reacts with potassium cyanide(deadly poison) gives alkyl cyanide(alkane nitrile).
Here also the halogen atom is substituted by cyanide(CN) group.
4.
with Ammonia(NH3):
R
X
+
H NH2
R
NH2 + HX
Alkyl halide reacts with ammonia under high pressure to give amine(primary amine). Here also
the halogen atom is substituted by amino(NH2) group.
SAQ 20:
Write the products with equations for the following reactions.
(i)Methyl choride reacts with aqueos KOH.
(ii)2-bromopropane reacts with alcoholic KOH
(iii)Ethyl iodide reacts with potssium cyanide
(iv)n-propyl choride reacts with ammonia under high pressure.
ALCOHOLS
(WITH SPECIAL REFERENCE TO METHANOL AND ETHANOL)
METHODS OF PREPARATION
Types of Alcohols:
There are three types of alcohols namely;
(i)Primary(10) (ii)Secondary(20)
and
(iii)Tertiary(30):
0
(i)
Primary alcohol(or 1 alcohol) is the alcohol in which the OH group is attached to
that carbon which bears 2 or 3 H atoms.
(ii)
Secondary alcohol(20 alcohol) is the alcohol in which the OH group is attached to that
carbon which bears 1 H atoms.
(iii) Tertirary alcohol (30 alcohol) is the alcohol in which the the OH group is attached
to that carbon which bears no hydrogen atoms.
R
R
R CH2 OH , CH3OH
R''
CH OH
R'
primary(10)
C
OH
R''
secondary(20)
Tertiary(30)
1.
From Alkyl halides:
When alkyl halides(haloalkanes) reacts with aqueous alkali(KOH), we get an alcohol. This is a
substitution reaction in which the halogen is substituted by OH group.We have already studied
this in the chapter, alkyl halides.
R X
+
K OH(aq)
R
OH
+
KX
Where R is any alkyl group. Note that if you use alcholic KOH instead of aqueous KOH, you
shall get a different product, alkene due to elimination of HX. We have done this in alkene
chapter.
Examples:
2.
CH3
Cl +
CH3
CH2
K O H(aq)
Br
+
K OH(aq)
CH3 OH + KCl
(methyl alcohol)
CH3
CH2
OH
+
KBr
(ethyl alchohol)
From Alkenes:
We already know from alkene chapter that alcohol is produced when alkene is added
with water in presence of sulphuric acid. We can prepare all alcohols excepting methyl alcohol
by this method. (Refer chapter, alkenes for details).
3.
From Aldehydes and Ketones:
When aldehydes and ketones are reduced by reducing agents like LiAlH4(lithium
aluminium hydride) we get alcohols.
Aldehdye gives a primary alcohol while a ketone gives a secondary alcohol.
O
R
C
H + 2 [H]
LiAlH4
C
CH2
OH
prim. alcohol
O
R
R
R'
R' + 2 [H]
LiAlH4
R
CH OH
(sec. alcohol)
If we take acetaldehyde(R=CH3-), we get ethyl alcohol and if we take formaldehyde(R=H) we
get methyl alcohol. If we take propanone(R=CH3- and R'=CH3-), then we shall get isopropyl
alcohol(20 alcohol).
SAQ 21:
(i)
Indicate which type of alcohols are the following(prim. sec. or tert.):
(a)methyl alcohol
(b)ethanol
(c)butan-1-ol
(d)isopropyl alcohol(propan-2-ol)(e)tert.butyl alcohol(2-methylpropan-2-ol).
(ii)
What happens when:
(a)2-iodopropane is treated with aqueous KOH
(b)Acetaldehyde is treated with lithium aluminium hydride
MANUFACTURE OF METHYL ALCOHOL(CH3OH)
Long back, methyl alcohol was produced from wood by distillation called destructive distillation
of wood. Therefore it was called wood alcohol. The word methyl originates from the Greek
word methy: wine and yle: wood i.e the wine produced from wood. When wood is distilled in
the absence of air(O2), it produces a distillate called, pyroligneous acid which contains about
3% methyl alcohol, about 10% acetic acid, 0.5% acetone and rest water. Methyl alcohol was
obtained by separating it from the other two constituents of pyroligneous acid. Note that acetic
acid(vinegar) was also manufactured from pyroligneous acid by this method. Methyl alcohol is
manufactured nowadays from CO and H2(water gas) as follows.
CO + 2 H 2
ZnO/ Cr 2O3
4000C, 150 atm.
CH3OH
When a mixture of CO and H2 in the ratio 1:2 is passed over heated catalyst(mixture of zinc
oxide and chromic oxide) at 4000C and a high pressure of 150 atm., we get methanol.
USES: Methanol is primarily used to prepare formaldehyde. It is also used as a solvent. It is
surprising to know that recently methanol is used to prepare(synthesise) biologically important
proteins. Do you know the laboratory where this synthesis is carried out? It is inside a living
being, a single celled species such as bacteria and years. When these microorganisms are fed
with methanol and other aqueous nuitrient salt solutions containing P, S and N, the bacteria/yeast
synthesise very useful proteins from these. This is indeed marvellous!!!
MANUFACTURE OF ETHYL ALCOHOL(C2H5OH)
This is the narcotic alcohol which some people people drink in the name liqour, brandy, beer,
wine, whisky etc.It is prepared by the fermentation of carbohydrates such as molasses,fruit
juices, rice, potatato, barely etc.
(a)Fermentation of molasses and fruit juice: (C12H22O11)
Molasses or cane sugar(obtained from sugar cane) and different fruit juices contain the
carbohydrate called disaccharides(mainly sucrose, maltose etc.). The chemical formula of all
disaccharides is C12H22O11. When such substances are kept in the absence of air(O2) for a long
period, microorganisms such as yeast develop in it and bring about chemical transformation of
the sugar(disaccharides). This chemical reaction catalysed by microorganisms is called
fermentation. The catalysts produced by microorganisms are called enzymes. The enzyme
produced by yeast in this case is invertase which converts disaccharides to simple
sugars(monosaccarides).
C 12H 22O 11
sucrose
invertase
C 6H12O6 + C 6H 12O 6
glucose
fructose
Glucose and fructose have the same formula and are monosaccharides(simple sugars).
In the next step, the simple sugar is converted to ethyl alochol and carobn dioxide by another
enzyme called zymase.
zymase
C 6H12O 6
monosaccharides
2 C 2H5 OH
+ 2 CO 2
(b)Fermentaion of starch: (rice, potatato, barley etc.)
Starch is a complex sugar available in rice, wheat, potatao, barley etc. This has a chemical
formula (C6H10O5)n. It is called a polysaccharide. It is a long chain molecule(polymer). First
starch is converted to disachharides(maltose) by one enzyme called diastase. Then another
enzyme called maltase converts maltose to simple sugar(glucose). Then the third enzyme, zymase
converts simple sugar into ethyl alcohol and carbon dioxide.
(C 6H10O 5) n + H2O
(starch)
C12 H22O1 1 + H2O
(maltose)
diastase
maltase
C12H22O 11
(maltose)
2 C 6H12 O6
(glucose)
zymase
C 6H12O 6
monosaccharides
2 C 2H5 OH
+ 2 CO 2
zymase
C 6H12O 6
(glucose)
2 C 2H5 OH
+ 2 CO 2
USES: Commercial alcohol is a constant boiling mixture(azeotrope) containing 95% ethyl alcohol
and 5% water and it cannot be further purified by distillation. To remove the remaining alocohol
to obtain 100% pure alcohol(absolute alcohol), the commercial alcohol is treated with quick
lime(CaO) which removes water.** Ethyl alcohol is mostly used as a solvent and as an antiseptic.
Many other chemicals are prepared from it. Unfortunately, ethyl alcohol is consumed in large
scale as liqour by human beings. Is it its use or misuse???
PROPERTIES:
Alcohols are highly soluble in water due to formation of hydrogen hydrogen bonding(Refer the
chapter chemical bond). They have higher boiling and melting points than alkanes and alkene
because of the same reason.
1.
Oxidation of Alcohols:
(a) Primary alcohol on oxidation gives first an aldehyde which on further oxidation
forms a carboxylic acid.
Look to the structure below. The H atom attached to O and another H atom attached to the
adjacent carbon are removed with the help of a nascent oxgyen as H2O. Thus C=O is formed.
The most common oxidising agent used for the purpose is chromic anhydride(CrO3) dissolved in
sulphuric acid and acetone(called the Jone's reagent). Other oxidising agents can also be used.
OH
[O]
CH 3
CH H
+ [O ]
CH3
- H 2O
O
O
[O]
CH3
C
H
CH3
C
OH
Since ethyl alcohol is a primary alcohol, we first get acetaldeyde(aldehyde) and then on further
oxidation get acetic acid(carboxylic acid). In the first step H2O is formed which has been shown
below the arrow mark(-H2O).
(b) Secondary alcohol on oxidation gives a ketone.
O
OH
CH 3
[O]
C H + [O]
CH 3
C
CH3 + H2O
CH3
Thus acetone(a ketone) is produced by the oxidation of secondary alcohol(isopropyl alcohol).
Note that the oxidation of alcohols is just the reverse process of reduction of aldehydes and
ketones that we studied in the preparation of alcohols.
(c) Tertiary alcohols are very much resistant to usual oxidation.
2.
Catalytic dehydrogenation:
This is analogous to oxidation of alcohols. Here we pass the alcohol vapours over
heated copper at 3000C. A primary alcohol gives an aldehyde and a secondary alcohol gives a
ketone. Here H2 is formed in stead of water, formed in oxidation reaction. Note that in this case,
the aldehyde formed from primary alcohol does not further oxidise to carboxylic acid.
O
OH
CH3
CH H
(ethyl alcohol)
Cu
3000C
CH3
C
O
OH
CH3
C H
CH3
(isopropyl alcohol)
2.
Formation of Alkyl halides:
H
(acetaldehyde)
Cu
3000C
CH3
C
CH3
(acetone)
In the chapter Alkyl halides, we have studied how alocohol reacts with various reagents
to produce haloalkanes(alkyl halides). Revise it.
3.
Reaction with Sodium/Potassium:(Test for alcohols):
When a piece of sodium is dropped into an alcohol, vigorous effervescence takes place with the
evolution of H2 gas. This is used as a test to distinguish alcohols from others. Although carboxylic
acids also give this test, acohol is detected from its typical alcohol smell(have you come near an
alcoholic person or a drunkard and experienced the smell of alcohol?). Carboxylic acids give a
different smell(pungent).
CH3-OH + Na ----------> CH3-O- Na+ + H2
The other product is sodium methoxide(in general metal alkoxide). Other alcohols react in the
similar manner. Remember that H atom attached to oxygen atom is acidic in nature and is easily
displaced by active metals like Na, K etc.
ALDEHYDES AND KETONE
Aldehydyes and ketones are commonly called carbonyl comounds as each of them contains a
carbonyl group(C=O) in it. That is why they have many similarities in their methods of preparation
and properties. We shall therefore study them together.
ALDEHYDES:
(with a special reference to formaldehyde and acetaldehyde)
Methods of Preparation:
1.
Oxidation of Alcohols:
10 alcohol -----------[O]---------> Aldehyde
We know from the chapter alcohols that primary alcohol on oxidation produces first aldehyde
and subsequently carboxylic acid. To get aldehyde from primary alcohol, we shall carry out mild
oxidation and arrest the reaction at the aldehyde stage.
Formaldehyde (Methanal):
methyl alcohol on mild oxidation gives formaldehyde.
O
CH 3 OH
methyl alcohol
Ag
3000C
H
C
H +
formaldehyde
H2
Heated siliver(Ag) at 3000C is used as the the catalyst to carry out the oxidation(dehydrogenation).
If the reaction is carried in presence of air(O2), we get water instead of H2 gas. Other oxidising
agent such as MnO2 in acetone can also be used for the purpose.
Formaldehyde is a gas having boiling point -210C. But it cannot be stored in a free state because
it changes to a plastic mass. Therefore it is preserved and marketted as a 37% aqueous solution
called formalin. In this form it is also used as a disinfectant and preservative.
Acetaldehyde(Ethanal):
Ethyl alcohol on mild oxidation gives acetaldehyde.
O
Ag
CH2 OH
CH3
3000C
ethyl alcohol
CH3
C
H +
acetaldehyde
H2
Heated silver(Ag) at 300 0 C is used here also as the catalyis to bring about this
oxidation(dehydrogenation). If the reaction is carried out in presence of air(O2), H2O is produced
in stead of H2. MnO2 in acetone can also be used as oxidising agent for the purpose. Acetaldehyde
is also a gas at room temperature but liquefies at 200C(b.p).
2.
Dry distillation of a mixture of cacium salt of carboxylic acid with calcium formate:
When a mixture of calcium salt of carboxylic acid and calcium salt of formic acid(calcium
formate) is strongly heated(distilled), aldehyde is distilled out leaving behind CaCO3 as residue.
See the following scheme. Acetaldehyde and other higher aldhedydes are parepared by this
method. Formaldehyde is not prepared by this method.
O
CH3
O
C
CH3
O
C O
O
O
Ca
calcium acetate
+
Ca
O
C H
O
O
2 CH3
C H + 2 CaCO3
acetaldehyde
C H
calcium formate
Two molecules of CaCO3 are removed as shown in the above scheme.
SAQ 22:
(i)What happens when propan-1-ol is treated with MnO2 in acetone.
(ii)What happens when a mixture of cacium propanoate and calcium formate is heated.
PREPARTION OF KETONES
[With a special reference to ACETONE(PROPANONE)]
1.
Oxidation of Secondary(20) alcohol:
A secondary alcohol on oxidation gives a ketone. We have studied this already in the
chapter, alcohols. Acetone(propanone), the first memeber of ketone family and is prepared by
the oxidation of isopropyl alcohol(propan-2-ol). Refer the chapter, alcohols.
2.
Dry distallation of salt of carboxylic acid:
When calcium salt of carboxylic acid is distilled, we get a ketone.
O
CH3
C
CH3
O
C O
O
O
Ca
CH3
C
acetone
CH3
+
CaCO3
calcium acetate
cacium acetate, on dry distillation produces acetone. A molecule of CaCO3(as rounded off) is
also formed.
COMMON PROPERTIES OF ALDEHYDES AND KETONES:
(A)
ADDITION REACTION:
An aldehyde or ketone undergoes addtion reaction at the carbonyl group as per the following
scheme.
H
R
H
C
O
+
-
+
-
+
X
R
Y
C
Y
O ( If H=R', it becomes a ketone)
X
XY is the compound which adds on the aldehyde or ketone at the C=O double bond. XY is
called the addendum. The positve part of the addendum(X) adds to the negative part of aldehyde/
ketone i.e the oxygen(O) atom while the negative part of addendum(Y) adds to the positve part
of aldehyde/ketone i.e carbon(C) atom. Thus double bond vanishes and we get the product as
shown above.
(a)
Addition with NH3:
H
CH3
C
+
O
-
+
+
H
-
NH2
H
CH3 C OH
(acetaldehyde ammonia)
1-aminoethanol
NH2
In NH3, H is the positve part and NH2 is the negative part. A ketone adds with NH3 in the same
manner as aldehdye.
(b)
With NaHSO3(Sodium bisulphite):
-
OH
O
CH3
C
+
CH3 +
+
-
H
SO 3Na
CH 3 C CH 3
(acetone bisulphite)
SO 3Na
In sodium bisulphite, H is the positve part and SO3Na is the negative part. So H adds to O atom
and SO3Na adds to the cabon atom of the carbonyl group to give the product which is named as
bisulphite of the parent aldehyde or ketone. Aldehdye reacts with sodium bisulphite in the
same manner as ketones.
(c)
With HCN:
-
OH
O
+
CH3
C
+
H +
H
-
CN
CH3 C H (acetaldehyde cyanohydrin)
CN
In HCN, H is the positve part and CN is the negative part. Thus an aldehyde or ketone adds to
HCN to give a compound called the cyanohydrin of the parent compound.
(B)
CONDENSATION REACTIONS:
Adehydes and ketones react with reagents like hydroxyl amine(NH2OH), hydrazine(NH2-NH2)
or phenyl hydrazine(NH2NHC6H5) to give oxime, hydrazone and phenyl hydrazone of the parent
compound respectively. In these reactions, the O of the carbonyl group of aldehyde or ketone
condenses with two H atoms of the reagent to eliminate a molecule of water. Such reactions in
which water molecule is removed by the reaction between two substances are called
condensation reactions. C atom of aldehyde and ketone is linked with the N atom of the
reagent by a double bond.
CH 3
CH 3
C O
+
C N OH + H2O
CH3 (acetone oxime)
H 2 N OH
CH3
Here acetone reacted with NH2OH(hydroxyl amine) to form the acetone oxime. Aldehyde
reacts in the same manner as ketones. Hydrazine(NH2NH2) and phenyl hydrazine(NH2NHC6H5)
react in the same manner as hydroxyl amine(NH2OH). In each case two H atoms of the reagent
reacts with O atom of carbonyl group to form the condensation products. Phenyl hydrazine
gives a yellow precipitate when reacts with aldehydes and ketones. This reaction is
commonly used as a test for aldehydes and ketones. More about condensation reaction shall
be taken in higher classes.
(C)
REDUCTION OF ALDEHYDES AND KETONES:
You know that on reduction with reducing agents like LiAlH4(lithium aluminium hydride), or
H2/Ni, an aldehydes gives a primary alcohol and a ketone gives a secondary alcohol. This
we have already discussed in the chapter, alcohols.
REACTIONS ONLY FOR ALDEHYDES:
Although aldehydes and ketones show many common properties as explained before, aldehydes
show some different reactions which are not shown by ketones. These reactios are used to
distinguish between an aldehyde from a ketone.
(i)
With Tollen's reagent: Tollen's reagent is ammoniacal AgNO3. When aldehydes
react with Tollen's reagent, a silver coating in the form of mirror is produced in the glass tube.
This is called silver mirror. Formation of silver mirror is a sure test for all aldehydes.
(ii)
With Fehling solution: Fehling solution is a mixture of CuSO4 solution and
alkaline solution of sodium potassium tartarate. When aldehdyes react with Fehling solution a
red precipitate of Cu2O is formed. This is another test for all aliphatic aldehdyes. The details of
these reactions will not be given now.
REACTIONS ONLY FOR FORMALDEHYDE:
Formaldehyde being the first member of the aldehyde family show a few unique properties
which are not shown by others. We shall know only one of them now.
(i)
Formaldehyde reacts with ammonia in a different manner(not in a way other aldehydes
and ketones react) to form hexamethylene tetramine[(CH2)6N4].
6HCHO + 4NH3 ----------> (CH2)6N4 + 6H2O
SAQ 23:
Solve the road map problem.
(i)
2-chloropropane
aq. KOH
A
CrO3
B
H2NNHC6H5
C
MnO2
(ii)
propan-1-ol
(iii)
calcium propanoate
NaHSO3
A
heat
B
LiAlH4
A
B
conc. H2S O4
1700C
C
(iv)
acetone
LiAlH4
PCl5
A
B
alcoholic
KOH
O3
C
H2O/Zn
D + E
SAQ 24: You are given acetone and ethyl alcohol both of which are colourless liquids. Without
taste or smell, how can you know by any chemical test which is which?
SAQ 25: How can you distinguish between acetaldehyde and acetone by a chemical test.
CARBOXYLIC ACIDS
(With a special reference to formic and acetic acid)
1.
Oxidation of Primary alcohols and aldehydes:
We already know from the chapter alcohols that primary alcohol on oxidation first
gives an aldehyde which on further oxidation gives carboxylic acid. In the chapter aldehydes, we
have also studied that aldhehyde on oxidation gives carboxylic acid. Methyl alcohol gives
formaldehyde first and then formic acid(methanoic acid). If you start from formaldehyde, we
also get formic acid.
H
H
O
O
CH2 O + [O]
(methyl alcohol)
CrO3
H C H
(formaldehyde)
[O]
H C OH
(formic acid)
Ethyl alcohol on oxidation first gives acetaldehyde and then acetic acid. If we start from
acetaldehyde, we shall also get acetic acid(ethanoic acid).
H
H
O
O
CH3 CH O
ethyl alcohol
+
[O]
CrO3
CH 3 C H
acetaldehyde
[O]
CH3 C OH
acetic acid
2.
Hydrolysis of alkyl cyanides:
When an alkyl cyanide(alkane nitrile) is treated with dilute mineral acids and boiled,
hydrolysis occurs to form carboxylic acid and ammonia.
OH
CH3 C
OH
OH
H
N + 2 H 2O
H
H
O
H+
CH 3
C
OH
+
NH3
In this reaction, methyl cyanide(ethanenitrile) reacts with water in presence of acid(dil. HCl, dil.
H2SO4 etc.) to form acetic acid. To understand how the reaction occurs, let us consider that
three molecules of H2O react at first and then one H2O molecule is removed as shown in the
structure. So virtually two H2O molecules react with the alkyl cyanide to to form carboxylic
acid and ammonia gas. Acid(H+) acts as a catalyst in this reaction. Formic acid is not prepared
by this method.
3.
From salt of Carboxylic acid:
Salt of carboxylic acids such as sodium acetate, potassium formate etc. react with dilute
mineral acids like H2SO4, HCl to form carboxylic acid.
O
CH3 C O Na +
sodium acetate
O
CH 3 C OH + Na2SO4
acetic acid
H2SO 4
Sodium acetate(sodium ethanonate) forms acetic acid(ethanoic acid).
3.
Hydrolysis of Ester:
(a)Acidic hydrolysis: When ester reacts with water in presence dilute mineral
acids like HCl, H2SO4 etc., hydrolysis occurs to form carboxylic acid and alcohol.
O
CH3
O
+
H
C
O CH 2 CH 3 +
ethyl acetate
OH
H 2O
CH 3 C OH
acetic acid
H
+ CH3 CH2 OH
ethyl alcohol
Here C-O single bond breaks to form a mixture of carboxylic acid and alcohol. Ethyl acetate(ethyl
ethanonate) hydrolyses to form acetic acid(ethanoic acid) and ethyl alcohol(ethanol). Here
acid(shortly written as H+) acts as catalyst.
(b)alkaline hydrolysis: (saponification of ester)
When ester is boiled with NaOH or KOH solution, hydrolysis occurs. We get salt of carboxylic
acid and alcohol. The salt of carboxylic acid is then treated with mineral acid like H2SO4 to get
free carboxylic acid.
O
O
H
C
ONa
H
O
C
O CH2 CH3 +
ethyl formate
H C ONa + CH3 CH 2 OH
ethyl alcohol
sodium formate
N aO H
H
ON a +
sodium formate
H2 SO 4
H
O
C
OH + Na2 SO 4
formic acid
In the above example, ethyl formate(ethylmethanoate) is boiled with alkali(NaOH) to form sodium
formate(sodium methanoate) and ethyl alcohol first. Sodium formate is then treated with dilute
H2SO4 to give formic acid. Hydrolysis of ester by an alkali is called saponification of ester.
The term saponification has come from the word soap. Soap is prepared by the alkaline hydrolysis
of natural esters(fats and oils). Soaps are sodium or potssium salts of higher carboxylic acids(fatty
acids). The details of formation of soap will not be discussed here.
Manufacture of Formic acid:
Formic acid is prepared industrially by the reaction of CO with NaOH at pressure of 6-10
atmosphere and 2100C. Sodium formate is obtained first. On treatement with dilute H2SO4,
sodium formate forms formic acid.
6-10 atm
2100C
CO + NaOH
H
O
C ONa
sodium formate
O
H
C
O
ONa
+
H2 SO4
C OH + Na2SO4
formic acid
H
Manufacture of acetic acid:
6-10% aqueous solution of acetic acid is called vinegar. You know that vinegar is used in food
material for enhancing taste. This is industrially prepared by quick vinegar process. Ethyl
alcohol is oxidised by air in presence of bacteria, mycoderma aceti grown naturally on wood
shavings to form acetic acid(vinegar).
CH3
CH2
mycoderma
O 2 aceti
OH +
CH 3
O
C
OH + H 2O
PROPERTIES:
Carboxylic acids are corrosive and are weak acids. The R-COOH hydrogen is acidic and
undergoes weak ionisation to form H+ and RCOO- ions.
1.
With active metal and metal hydroxide:
Active metals like Na, K and their hydroxides(NaOH and KOH) react with carboxylic
acids to form salt. Metals form H2 gas while metal hydroxides form water besides salt of carboxylic
acid.
O
CH3
O
C OH + N a
CH 3
O
CH3
C ONa + H2
O
C OH + KOH
CH3
C OK
+ H2O
Acetic acid reacts with sodium to give sodium acetate(sodium ethanoate) and H2 gas. While any
alkali(KOH) reacts with acid to form the salt(potassium acetate) and H2O.
2.
Formation of Ester (Esterification):
A caroxylic acid reacts with an alcohol in presence of conc. H2SO4 to form ester and
water. The OH of carboxylic acid reacts with H of alcohol to form H2O. Thus an ester is
formed. Acetic acid reacts with methyl alcohol to form methyl acetate and water. Conc. H2SO4
acts as a dehydrating agent.
O
CH3
C
acetic acid
3.
O H
+
HO
CH3
methyl alcohol
conc. H2SO4
heat
CH3
O
C O CH3 + H2O
methyl acetate
Formation of acid chloride:
Carboxylic acids react with PCl5 or PCl3 or SOCl2(thionyl chloride) to form the
corresponding acid chloride in the same way as alcohols give alkyl chloride.
O
O
CH 3
CH3 C Cl + POCl3 + HCl
acetyl chloride
C OH + PCl5
If we use PCl3 in place of PCl5 , we shall get H3PO3 in stead of POCl3 and HCl. If we use
SOCl2, we shall get SO2 and HCl gas alongwith the acid chloride.
4.
Formation of acid amides:
When NH3 gas reacts with carboxylic acid, first the ammonium salt of carboxylic acid is
formed which on heating forms acid amide and water.
CH 3
O
O
O
C OH
CH 3
NH3
+
C
O(NH4)
heat
CH3 C NH2
+
H2O
acetamide
ammonium acetate
Acetic acid first forms ammonium acetate which on heating forms acetamide(ethanamide).
5.
Reduction of Carboxylic acids (formation of alcohols):
When carboxylic acid is reduced by LiAlH4, we get a primary alcohol.
O
CH3
C
LiAlH4
OH + 6 [H]
CH 3 CH2 OH +
2 H 2O
Acetic acid is reduced to ethyl alcohol by the help of nascent hydrogen produced by LiAlH4.
SAQ 26: Solve the following road map problems.
CrO3
1.
CH3 OH
2.
CH 3 CH 2 CH 2 Cl
n-propyl alcohol
A
H2SO4
KCN
A
H2O/H+
B
PCl3
C
O
3.
CH 3 CH 2 C
O CH2 CH3
H2O/H+
A + B
SAQ 27: What happens when:
1.
Formic acid is treated with ammonia and the product formed is heated.
2.
Propanal is treated with an oxidising agent like KMnO4.
3.
Methyl propionate is heated in water in presence of dil NaOH.
PRACTICE QUESTIONS
1.
Solve the following road map problems.
Br2
(a)
CH4
(b)
CH3CH2CH2OH
A
he at/light
Na/ether
SOCl2
conc. H2SO 4
A
B
alcoholic
KOH
O3
(c)
propan-2-ol
2.
Make the following conversions.
(i)Methyl alcohol to Ethyl alcohol
heat
A
Cl2/light
aq. KOH
C
HCl
B
B
+
D
MnO2
E
C
C
H2O /Zn
(ii) Ethyl alcohol to methyl
alcohol
(iii)Methane to Ethane
(v)Formaldehyde to acetaldehdye
formaldehyde
(vii)ethyl alcohol to acetic acid
alcohol
3.
(iv)Ethane to methane
(vi)Acetaldehyde
to
(viii)Acetic acid to methyl
What happens when. Write the equations.
(i)Calcium acetate is dry distilled and the product is treated with LiAlH4.
(ii)Propene is treated with HI.
(iii)Ethyl alcohol is treated with conc. H2SO4 at a temperature of 1700C.
(iv)Acetone reacts with hydroxyl amine.
(v)Acetaldehyde reacts with Tollen's reagent
(vi)1,2-dichloroethane is treated with alcoholic KOH.
(vii)But-1-ene is subjected to ozonolysis.
(viii)Ethyl formate is treated with dilute H2SO4 and heated.
(ix)Propyne is treated with sodium metal
(x)Propionic acid reacts with ammonia and the product formed is heated.
RESPONSE TO SAQs
SAQ 1:
CH 3
CH2
O _
+
C O Na
+
NaOH
CaO
CH 3
CH 3 + Na2 CO 3
When Na2CO3 is removed(shown in the box), CH3-CH2-(ethyl) group joins with H to produce
ethane(CH3-CH3 or C2H6). Thus decarboxylation of sodium propanoate gives ethane. Note
that this method is not a good practical method to prepare ethane as we get other side products
such as methane, hydrogen, ethylene besides ethane. So this method is best suitable to prepare
pure methane gas.
SAQ 2:
H3 C CH2 Cl + 2 Na
H2C
CH
Cl
CH3 +
CH2 CH3
ether
H3 C CH2 CH2 CH3 + 2 NaCl
butane
The joining of one ethyl group with another ethyl group gives butane(or n-butane) in this case.
SAQ 3: We get propane by the reduction of propene.
+
H2
Ni
H3C
3000C
CH2 CH3
SAQ 4: . (i)CH4 + 2 Cl2 ---------> CH2Cl2 + 2HCl
When we take two moles of chlorine, two H atoms are replaced by two Cl atoms as shown by
the first two steps in the text and we get dichloromethane as the main product.
(ii) CH4 + 4Cl2 ----------> CCl4 + 4HCl
In this case we get tetrachloromethane as the main product by the substitution of four H atoms
by 4 chlorine atoms. All the four steps as shown in the test will be taking place.
SAQ 5:
CH3-CH3 + Cl2 ---------> CH3-CH2-Cl + HCl
We get monochloroethane or ethyl chloride as the product. For substituting all the six hydrogen
atoms of ethane, we have to take 6 moles of Cl2 so that we get hexachloroethane(C2Cl6) and six
molecules of HCl.
SAQ 6: CH3-CH2-CH3 has two types of carbon atoms. The first and last methyl carbon atoms
are of identical type and the middle CH2 carbon atom is of different type. In CH3-CH3 however
all the two carbon atoms are of identical type. Therefore we get only one monochloroethane
(SAQ 5). In propane however we shall get two monochloropropanes which are the isomers of
each other. See this.
H3C
CH2 CH2 Cl
1-chloropropane
or n-propyl chloride
H3C
CH2
CH3
+
Cl2
+
HCl
Cl
H3C
CH
CH3
2-chloropropane
isopropyl chloride
+ HCl
When one H atom is replaced from CH3 group we get 1-chloropropane(or n-propyl chloride)
and when one H atom of the middle CH2 group is replaced, we get 2-chloropropane(or isopropyl
chloride). In fact we get a mixture these two monochloropropanes.
SAQ 7:
(i)Ethane (refer text: Kolbe's reaction)
(ii) Methane(refer text: decarboxylation of methyl acetate)
CH3
H3C
CH
(iii)
Cl + 2 Na +
Cl
CH
ether
CH CH
CH3
H3C
2,3-dimethylbutane
CH3
H3C
CH3
H3C
The joining of two isopropyl groups in the Wurtz reaction gives 2,3-dimethylbutane.
(Trichloromethane or chloroform will be formed)
(iv) CH4 + 3Cl2 -------> CHCl3 + 3HCl
H3C CH2 CH2 CH 2 Cl + HCl
1-chlorobutane
(v)
Cl
H3C CH 2 CH 2 CH 3 + Cl2
H3C CH 2 CH CH 3 + HCl
2-chlorobutane
Butane has two types of carbon atoms, namely CH3- and -CH2- and will give two
monochloro butanes(1-chlorobutane and 2-chlorobutane).
CH3
H 3C
(vi)
(a)
C
CH3
H 3C
CH3 + Cl2
C
CH2 Cl +
HCl
CH3
CH3
1-chloro-2,2-dimethylpropane
or neopentyl chloride
2,2-dimethylpropane has one type of hydrogen atoms. All the four CH3 groups are equivalent.
So Cl substitutes one H atom from any one CH3 group to give one product i.e 1-chloro-2,2dimethylpropane or neopentyl chloride.
CH3
Cl
CH3
H3 C
(b)
CH CH2 CH3 +
Cl2
H2C CH CH2 CH3 + HCl
CH3
H3C C CH2 CH3 + HCl
Cl
CH3 Cl
H3C CH CH
CH3
+ HCl
CH3
H3C CH CH2 CH2
Cl + HCl
In 2-methylbutane there are four different types of carbon atoms hence we get four different
monochloro compounds. When a H atoms from the left terminal CH3 group is replaced we get
1-chloro-2-methylbutane(shown in the top). When a H is replaced from the second carbon
from left(CH), we get 2-chloro-2-methylbutane(second from top), and when one H is replaced
from the 3rd carbon atom(CH2) we get 3-chloro-2-methylbutane(second from bottom) and
when H is replaced from the fourth carbon(right terminal CH 3), we get 1-chloro-3methylbutane(shown at the bottom). Thus all the four products are obtained in this reaction.
SAQ 8: (i) CH4 +
CH3
Cl2
diff. sunlight
CH3 Cl + HCl
(A)
Cl + 2Na + Cl
CH3
ether
(Wurtz reaction)
CH3 + 2NaCl
CH3
(B)
In the first step, monochlorination occurs to give methyl choride(A) which reacts with Na
in presence of ether(Wurtz reaction) to produce ethane(B).
(ii) H3 C
CH3 + Br2
Br 2
H3 C
light/heat
H3C CH2 Br +2Na + Br CH2 CH3
CH2 Br + HBr
(A)
ether
H3C CH2 CH2 CH3 + HBr
(B)
(Wurtz reaction)
H3C
CH2 CH2 CH2 Cl +
HCl
(1-chlorobutane)
(C)
Cl
H3C CH2 CH2 CH3 + Cl2
H3C
CH2 CH CH3 +
(2-chlorobutane)
HCl
(D)
In the first step ethane reacts with Br2 to give ethyl bromide(A). In the second step, ethyl
bromide undergoes Wurtz reaction to give butane(B). Butane, in the third step, undergoes
monochlorination(substitution reaction) to give two monochloroproducts, 1-chlorobutane and
2-chlorobutane(C and D).
H
alcoholic
SAQ 9: (i)
(ii)
H2C
H
OH
H2 C
CH
CH2
OH
H
CH
CH
H3C
Br
CH
CH3
CH3
CH3
H2C
KOH
conc. H2SO4
heat
conc. H2SO4
heat
CH
(propene)
H2 C
CH3
CH3 + HBr
CH CH2 CH3 + H2 O
but-1-ene
CH
CH CH3 +H2 O
but-2-ene
In this case we get two alkenes: but-1-ene and but-2-ene but Saytzeff's rule holds good
here also. But-2-ene, therefore,is the major product.
SAQ 10:
(i)
H3 C
CH
CH
CH3 + H2
Ni
H3 C CH2 CH2
200-3000C
(butane)
CH3
Cl Cl
(ii)
H2C
CH CH2
CH3 +
H2C CH CH2 CH3
(1,2-dichlorobutane)
Cl2
I
(iii)
+ -
H 3 C CH CH2 CH3
(2-iodobutane)
This is an unsymmetrical alkene and the negative part of HI i.e I adds on the C-2 which
bears one H atom according to Markonikoff's rule. So the mojor product is 2-iodobutane.
H2 C
CH CH2
CH3 + H I
(Markonikoff' rule)
Cl
(iv)
H 3C
CH
CH
CH3 + HCl
H3C CH CH2 CH3
(2-chlorobutane)
But-2-ene is a symmetrical alkene. The additon of Cl can take place on any carbon atom of the
double bond and we get the same compound 2-chlorobutane. Note that there is nothing called
3-chlorobutane. This is a wrong name. If you number the carbon chain in the reverse way, it is
actually 2-chlorobutane.
SAQ 11:
-
H3 C
CH
OH
Br
CH
CH2 or 1-bromopropan-2-ol)
+
CH2 + (HO) Br
H3 C
(propene bromhydrin
Note that the positve part in HOX is X and negatvie part OH although X is written to the right of
HO. The negative part, OH adds on to the middle carbon which bears less number of H atoms
to give 1-bromopropan-2-ol or called propene bromohydrin.
SAQ 12:
O
O
CH
CH 2
H2O/Zn
CH 3
CH
CH2
+
H3C
O3
O
O
CH 3
C
O
H
(ethanal)
+H C H
(methanal)
+ H2O 2
First propene ozonide will be formed which will by hydrolysed to form
ethanal(acetaldehyde), methanal(formaldehyde) and H2O2. Thus two different aldehydes are
formed in this case.
SAQ 13: (i)
CH3
CH
CH
CH 3
O3
+
H3C
O
O
CH
CH
CH3
O
O
CH3
O
C
H + CH3
H + H2O 2
C
(ethanal)
(ethanal)
In this case we get two same aldehyde(ethanal) molecules.
(ii)
CH3 C
O
H3C O
CH3
CH
CH3
O3
+
C
O
CH3
CH
H3C
O
C
CH3 + CH3
(propanone)
CH3
O
C H + H2O2
(ethanal)
In this case we get one ketone(propanone or acetone) and one aldehyde(ethanal).
SAQ 14: The OH being negative part adds to the carbon bearing one H atom according to
Marknikoff's rule. We get propan-2-ol as the main product.
OH
H OH
+
H3C CH
CH2
+
H 2SO 4 /H 2O
CH3 CH CH3
propan-2-ol
SAQ 15: We get propane-1,2-diol by the addition of one OH group on either side of the double
bond.
SAQ 16: We get propyne by the removal of 2HCl molecules from 1,2-dichloropropane.
CH3
H
Cl
C
CH
Cl
H
alcoholic
KOH
CH3
C
CH + 2 HCl
(propyne)
SAQ 17:
(i)
CH 3
C
CH3
CH + 2 Br2
Br
Br
C
CH (1,1,2,2-tetrabromoproane)
Br
Br
O
OH H
(ii)
CH3
C
CH
+
2 H2O
+
-
H
OH
HgSO 4
H 2 SO 4
-
CH3
C
CH
OH H
H2 O
CH3 C CH3
Propyne is an unsymmetrical alkyne. Markonikoff's rule will be applied here. The negative
part of H2O i.e OH group will add to the middle carbon which bears no H atom. Since two OH
groups cannot attach to the same carbon atom, a molecule of H2O is eliminated to form a
ketone(acetone or propanone). Excepting acetylene, any other alkyne will form a ketone when
reacts with water in presence of warm HgSO4/H2SO4.
OH Br
-
(iii)
CH3
C
CH
+
CH3
2 (HO ) Br
C
O Br
-H O
+
2
CH
CH3 C CH
Br
(1,1-dibromopropanone)
OH Br
Markonikoff's rule is applied here. The negative part of HOBr i.e OH group repeatedly
adds to the middle carbon which does not bear any H atom. After the elimination of H2O, we get
1,1-dibromopropanone.
CH3 Cl + HCl
(i) CH4 + Cl2
SAQ 18:
(ii) CH3 Cl
(iii) CH3
H
+
CH3
(A)
Na + Cl
+
CH3
CH3
CH3 CH2 Br
Br 2
(B)
CH3 + NaCl
HBr
+
(C)
Br
alc. KOH
(iv) CH2 CH2
CH2
CH2
+
HBr
(D)
Cl
(v) CH2
CH2
+
Cl2
Cl
CH2 CH2
(E)
(vi)
H
Cl
CH
CH
Alc. KO H
CH
H
(vii)
CH
+
2 HCl
(F)
H
Cl
CH
CH + 2 H2O
OH
CH
CH
H
OH
O
- H2O
CH3
(G)
C
H
In the first step, CH4 undergoes monochlorination to give chloromethane or methyl chloride(A)
which reacts with Na/ether(Wurtz reaction) to give ethane(B). Ethane undergoes
monobromination to give monobromoethane or ethyl bromide(C) which undergoes
dehydrobrimination(-HBr) by alcoholic KOH to give ethene or ethylene(D). Ethylene undergoes
addition reaction with Cl2 to form 1,2-dichloroethane(E) which undergoes dehydrochorination
(-2HCl) to form ethyne or acetylene(F). Acetylene reacts with water in presence of HgSO4/
H2SO4 to give ethanal or acetaldehyde(G).
SAQ 19:
(i)
CH3 CH2 OH + PCl5
CH3
CH2
OH
(ii)
3 H 3C
(iii)
Cl + POCl3 + HCl
(ethyl chloride)
Br
CH
CH3 + PBr 3
3 CH 3
OH + PI 3
3 H 3C CH CH3 + H 3PO 3
(2-bromopropane)
I2/P4
3 CH3
I + H 3PO 3
SAQ 20:
(i)
(ii)
CH3
CH 3
Cl +
Br
H
CH
CH 2
OH + KCl (We get methyl alcohol)
CH3
K OH
alc. KOH
CH3 CH CH 2 + HBr
(propene)
Here a molecule of HBr is eliminated and we get propene.
(iii)
CH3
CH2
I +
K CN
CH3 CH2
CN + KI
(ethyl cyanide)
We get ethyl cyanide(propanenitrile) in this reaction.
(iv)
H3 C
CH 2 CH 2 Cl +
H NH 2
H3 C CH 2 CH2 NH2 + HCl
(n-propyl amine)
We get n-propyl amine(1-chloropropane) in this reaction.
SAQ 21:
(i)
Write the structures of the alochols and count the number of H atoms attached to the
carbon atom to which OH group is attached. If it is 2 or 3 it is a primary alcohol, if it is 1, a
secondary alcohol and if none it is tertiary.
(a)prim.
(b)prim.
(c)prim.
(d)sec.
(e)tert.
I
(ii)
(a)
CH 3
OH
CH
CH3
+ KOH(aq)
CH 3 CH CH3
(propan-2-ol)
O
(b)
CH3
C
H + 2 [H]
O
CH3
CH2 OH
(ethyl alcohol)
O
H
SAQ 22: (i) CH 3 CH2 CH H
propan-1-ol
LiAlH4
[O]
MnO 2
CH 3
CH 2
C H
propanal
+ H2 O
Propanal-1-ol is a primary alcohol and it is oxidised by MnO2 in acetone to form propanal and
water.
O
O
CH3
CH 2
C
Ca + Ca
O
(ii)
CH 3 CH 2 C
calcium propanoate
C H
heat
O
O
O
O
O
C H
O
2 CH 3 CH 2 C H + 2 CaCO3
propanal
calcium formate
When calcium propanoate and calcium formate mixture is distilled, we get propanal.
SAQ 23:
CH3
(i) CH 3
CH
CH3
Cl +
CH3
CH3
C
H
CH 3 CH OH + KCl
2-propanol(A)
K OH(aq.)
O
OH
+
CrO3
[O]
CH3 C CH 3 + H2O
acetone(B)
N NHC6 H5
O
CH 3
C
CH3
+ H2 NNHC 6H5
CH3
C
CH3
+
H2O
acetone pheny hydrazone(C)
In the first step, an alkyl halide reacts with aq. KOH to give an alocohl. Since this is a secondary
alcohol, on oxidation with chormic oxide, in the second step, we get a ketone(B). Ketone reacts
with phenyl hydrazine in the third step to give a yellow precipiate(phenyl hydrazone). This is a
condensation reaction.
H
H
(ii)
CH 3
CH2
CH O H +
[O]
CH3
CH2 C
O +
H2O
propanal(A)
H
+ -
CH3 CH 2 C O
+
MnO 2
-
+
H SO3Na
H
CH3 CH2 C OH propanal bisulphite(B)
SO3Na
In the first step, propan-1-ol, which is a primary alcohol, on mild oxidation gives an
aldehyde(propanal,B). Propanal reacts with sodium bisulphite to give a white precipitate of
propanal bisulphite. This is an addition reaction of aldehyde.
O
(iii)
CH3 CH2 C O
O
O
heat
Ca
H3C
CH3 CH2 C O
CH2 C CH2 CH3 + CaCO3
pentan-3-one (A)
OH
O
H3C
CH2 C
LiAlH4
CH2 CH3 + 2 [H]
H3C
CH2 CH CH2
pentan-3-ol
CH3
(B)
OH H
H3C
CH2 CH CH
pentan-3-ol
CH3
conc. H2S O4
H3C CH2 CH CH CH3
(pent-2-ene)
1700C
(C)
In the first step, calcium propanoate gives pentan-3-one(A) on strong heating(dry distillation)
and leaves the residue CaCO3. In the second step, a ketone is reduced to a secondary alcohol(B).
In the third step, dehydration of alcohol take place to form an alkene.
OH
O
iv)
CH3
C
LiAlH 4
CH3 + 2 [H]
CH 3
OH
CH 3
CH 3
H
PCl5
+
alcoholic
CH 3 CH CH2
KOH
CH 3 CH CH 3 + POCl3 + HCl
2-chloropropane(B)
CH 3 CH CH2 + HCl
propene(C)
O
H3 C
CH
CH 3
Cl
CH
Cl
CH
propan-2-ol (A)
CH2
O3
O3
CH3
CH
O
CH2
O
H2O/Zn
CH3
C H +
ethanal (D)
O
H C H + H2O2
methanal (E)
propene ozonide O
In the first step, reduction of ketone gives a secondary alcohol(A). In the second step, alcohol is
converted to alkyl chloride(B). Alkyl chloride on treatment with alcoholic KOH in the third step
gives alkene(C). Alkene on ozonolysis first gives an ozonide which hydrolyses to give a mixture
of ethanal(D) and methanal(E) in the fourth step.
SAQ 24: A piece of sodium metal is dropped on both the samples. The one which produces H2
gas(effervescence) is an alchohol(ethyl alcohol). Ketone does not give this test.
Alternatively: Two are allowed to react with phenyl hydrazine, the one which gives a yellow
precipitate(phenyl hydrazone) is acetone(ketone). Alcohol does not give this test.
SAQ 25: The two substances are treated with Tollen's reagent. The one which gives a silver
mirror is an aldehyde(acetaldehyde). Ketones do not give this test.
SAQ 26:
1.
CH 3
OH +
[O]
CrO3
H
O
C H
O
H C OH
formic acid (A)
[O]
In the first step a primary alcohol on oxidation gives carboxylic(formic acid)
H
O
C OH +
H O CH2 CH2 CH3
H2SO4
H
O
C O CH2 CH2 CH3 + H2O
propyl formate (B)
In the second step, carboxylic acid(formic acid) reacts with n-propyl alcohol to form an ester(propyl
formate or propyl methanoate). This is esterification reaction.
2.
CH 3 CH 2 CH 2 Cl + K CN
CH3 CH2 CH2 CN + KCl
n-propyl cyanide (A)
In the first step alkyl halide(n-propyl chloride) reacts with KCN to form its cyanide. In the
second step, alkyl cyanide is hydrolysed in the presence of acid to form caboxylic acid(butanoic
acid) and ammonia.
CH 3 CH 2 CH2 CN
+ 2 H 2O
H+
CH 3
O
CH2 CH2 C OH + NH3
butanoic acid(B)
O
CH3 CH2 CH2 C Cl
butanoyl chloride (C)
O
CH3 CH 2 CH 2 C OH + PCl3
+
H3PO 3
In the third step, carboxylic acid reacts with PCl3 to form its acid chloride(butanoyl chloride) and
phoshophours acid. The equation is not balanced.
3.
O
CH 3 CH 2 C
OH
O CH2 CH3
H
+
H2O
H+
CH3 CH2
O
C OH + CH3
propionic acid (A)
CH2 OH
ethyl alcohol (B)
This is acidic hydrolysis of ester reaction to form a carboxylic acid and an alcohol.
SAQ 27:
1.
Formic acid first forms ammonium formate which on heating gives formamide.
HCOOH + NH3 -------> HCOONH4 -------heat-------> HCONH2 + H2O
2.
Aldehyde on oxidation gives carboxylic acid.
CH3CH2CHO + [O] ----------> CH3CH2COOH(propanoic acid)
3.
This is alkaline hydrolysis(or saponification) of ester to form salt of carboxylic acid (sodium
propionate)and alcohol(methyl alcohol)
CH3CH2COOCH3 + NaOH ---------> CH3CH2COONa + CH3OH
ANSWERS TO PRACTICE QUESTIONS
1.(a) CH4 + Br2 --------> CH3Br(A) + HBr;
CH3Br + 2Na + BrCH3 --------> CH3-CH3(B) + 2NaCl
CH3-CH3 + Cl2 --------> CH3-CH2-Cl(C) + HCl
CH3-CH2-Cl + KOH(aq) -------> CH3CH2OH(D) + KCl
CH3CH2OH + [O] ---------> CH3CHO(E) + H2O
A=methyl bromide;B=ethane; C=ethyl chloride;
D=ethyl alcohol
E=acetaldehyde
(b) CH3CH2CH2OH + SOCl2 -------> CH3CH2CH2Cl(A) + SO2 + HCl
CH3CH2CH2Cl ------alc.KOH-------> CH3CH=CH2(B) + HCl
CH3CH=CH2 + HCl --------> CH3CH(Cl)CH3(C)
A: n-propyl chloride;
B=propene(elimanation reaction i.e -HCl)
C=2-chloropropane(Markonikoff's addition)
(c) CH3CH(OH)CH3 -----conc. H2SO4/heat ------> CH3CH=CH2(A) + H2O
CH3CH=CH2 + O3 ---> Propene ozonide ----H2O/Zn---> CH3CHO(B) + HCHO(C)
A= propene;
B=acetaldehyde;
C=formaldehyde
2.
While you convert one organic compound to another compound, you may have to proceed
through several steps. In all the steps, you can use any inorganic reagent, but you are not allowed
to use any other organic compound from outside. For example, in the first bit (i), methyl alcohol
is first converted to methyl chloride with the help of PCl5. Then with the help of Wurtz reaction,
methyl chloride is converted to ethane. Here we do not use a different organic compound. The
compound(CH3Cl) derived from the starting compound(CH3OH)is used. Ethane is then converted
to ethyl chloride and finally ethyl alcohol is obtained from ethyl chloride.
(i)
CH3OH + PCl5 -------> CH3Cl + POCl3 + HCl
CH3Cl + 2Na + ClCH3 -------> CH3-CH3 + 2NaCl
CH3-CH3 + Cl2 -----light -------> CH3CH2Cl + HCl
CH3CH2Cl + KOH(aq) --------> CH3CH2OH + KCl
(ii) CH3CH2OH + [O] ----CrO3----> CH3COOH + H2O
CH3COOH + NaOH ---------> CH3COONa + H2O
CH3COONa + NaOH(CaO) -----heat ------> CH4 + Na2CO3
CH4 + Cl2 -------light -------> CH3Cl + HCl
CH3Cl + KOH(aq) ------> CH3OH + KCl
Alternatively:
CH3CH2OH ------conc. H2SO4(heat) ------> CH2=CH2 + H2O
CH2=CH2 + O3 ------> ethylene ozonide ----- H2O/Zn ------> 2HCHO + H2O2
HCHO + [H] ------LiAlH4 --------> CH3OH
(iii) CH4 + Cl2 ------light-----> CH3Cl + HCl
CH3Cl + 2Na + ClCH3 ------ether-------> CH3-CH3 + 2NaCl
(iv) CH3-CH3 + Cl2 ------light-----> CH3-CH2-Cl + HCl
CH3-CH2-Cl -------alc. KOH ------> CH2=CH2 + HCl
CH2=CH2 + O3 -------> ozonide ------H2O/Zn ------> 2HCHO + H2O2
HCHO + 4[H} -------Zn(Hg)/HCl --------> CH4 + H2O
(v) HCHO + [H] -----LiAlH4 ------> CH3OH
CH3OH + PCl3 ---------> CH3Cl + H3PO3
CH3Cl + 2Na + ClCH3 ---------> CH3-CH3 + 2NaCl
CH3-CH3 + Cl2 ------light ------> CH3-CH2-Cl + HCl
CH3-CH2-Cl + KOH(aq) -------> CH3-CH2-OH + KCl
CH3-CH2-OH + [O] ----MnO2 -----> CH3-CHO + H2O
(vi) CH3-CHO + 2[H] -------> CH3-CH2-OH
CH3-CH2-OH ------conc. H2SO4/heat -------> CH2=CH2 + H2O
CH2=CH2 + O3 ------> ozonide -----H2O/Zn -------> 2 HCHO + H2O2
(vii) CH3-CH2-OH + [O] ------> CH3-CHO ------[O]-------> CH3COOH
(viii) CH3-COOH + NaOH -------> CH3COONa + H2O
CH3-COONa + NaOH(CaO) ------> CH4 + Na2CO3
CH4 + Cl2 -----light-------> CH3Cl + HCl
CH3Cl + KOH(aq) -------> CH3OH + KCl
3.
(i)
(CH3COO)2Ca -----heat------> CH3COCH3(acetone)+ CaCO3
CH3COCH3 + 2[H] ------LiAlH4 ------> CH3CH(OH)CH3 (propan-2-ol)
(ii) CH2=CH-CH3 + HI ------Marknokoff's addition-----> CH3CH(I)CH3 (2-iodopropane)
(iii) CH3-CH2-OH -------conc. H2SO4/heat -------> CH2=CH2 (ethylene)+ H2O
(iv) CH3COCH3 + H2NOH ------> (CH3)2 C= N-OH(acetone oxime) + H2O
(Refer the reaction of aldehyde/ketone)
(v) CH3CHO forms a silver mirror with Tollen's reagent.
(vi) Cl-CH2CH(Cl)CH3 ------alc. KOH ------> CH≡C-CH3 (propyne)+ 2HCl
(Refer the chapter alkyne)
(vii) CH2=CH-CH2-CH3 + O3 -----> ozonide
-------H2O/Zn-------> HCHO(methanal) + CH3-CH2-CHO(propanal)
(viii) HCOOCH2CH3 + H2O ------H2SO4 ------->
HCOOH(formic acid) + CH3-CH2-OH(acetic acid)
(ix) HC≡C-CH3 + Na ----------> Na+ -C≡C-CH3 (sodium propynide) + ½ H2
(x) CH3-CH2-COOH + NH3 ------> CH3-CH2-COONH4(ammonium propanoate)
------heat----> CH3-CH2-CONH2 + H2O
(propanamide)