Organic Compounds- Chemistry - The Uraniumtheuranium.org/content-images/organic-compounds... · 2012-04-10 · Chapter-V Chemistry of organic comounds ALKANES PREPARATION OF ALKANES

Chapter-VChemistry of organic comounds

ALKANES

PREPARATION OF ALKANES WITH SPECIAL REFERENCE TOMETHANE 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. Methaneoccurs 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 calciumoxide(CaO).

CH3 C O Na

O+

_

NaOH CaO CH4 Na2CO3+ +

Here CaO oxide does not take part in the reaction. Look to the box shown in the equation. Afterthe removal of Na

2CO

3(contained inside the box), the CH

3 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 H

2 gas is passed over finely divided nickel at 3000C, we get methane.

CO H2 CH4 H2O+ +3 Ni

3000C

3. Action of water on aluminium carbide.Al

4C

3 + 12H

2O ---------> 3CH

4 + 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 severalother 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.

CH3 C

O

O K- + +-

CH3 C

O

O + K

When dissolved in water potassium acetate dissociates to produce free acetate ion(anion) andpotassium ion(cation). When this solution is connected to a cathode and anode and electricity ispassed through the solution, ethane is evolved at the anode and hydrogen gas is produced at thecathode.

At anode:CH3 C

O

O

CH3 C

O

O

-

- CH3 CH3 CO2 e-

+ +2 2

Two acetate ions move together to the anode and undergo oxidation at anode. 2 molecues ofCO

2 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 forwhich it accepts the electron and brings about oxidation.

At cathode: There are two competing ions for the cathode. One is K+ frompotassium acetate and the other H+ ion produced from H

2O. Note that H

2O produces H+ and

OH- due to dissociation. Out of the two cations, H+ and K+ ; H+ ion has a greater tendency toundergo reduction i.e to gain electron. So it will be preferentially discharged at the cathode tofree 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 Na Cl CH3+ +2 CH3 CH3 2 NaClether

In this case, two molecules of methyl chloride reacts with two atoms of sodium to produceethane and sodium chloride. In the above scheme, you find that after the removal of 2 NaClmolecules(shown within the rectangula box), one CH

3 group joins with the other CH

3 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 groupswill join to produce an alkane containing double the number of carbon atoms as compared to thealkyl halide.

+R Cl Na Cl R+ +2 R R 2 NaClether

R stands for any alkyl group like methyl, ethyl, propyl etc. The only demerit in Wurtz reaction isthat we always get alkane containing even number of carbon atoms(as the number of carbonatoms 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 H

2 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 CH3 CH3+

Ni

200-3000C

Ethane can also be prepared from ethyne(acetylene) by the same method. In this case2 moles of hydrogen are necessary. First, acetylene is reduced by one mole of hydrogen gas toproduce ethylene. Ethylene then reacts with the second mole of hydrogen to give ethane. Thetriple bond of acetylene first changes to double bond in ethyelene and then the double bondchanges to single bond in ethane.

+HC CH H2 H2C CH2200-3000C

Ni

acetylene ethylene

H2C CH2 H2 CH3 CH3+

Ni

200-3000C

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 CO

2 and hydrogen is passed over finely

divided nickel at about 3000C. This is also a case of Sabatier and Senderens reduction.

Ni+CO H2 3000C

CH4 H2O+3

PROPERTIES OF ALKANES:Physical properties:(i) The first four alkanes namely methane, ethane, propane and butane are gases at roomtemperature. The next 13 memebers(C

5-C

17) 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 becausethey are non-polar in nature and there is very weak Vander waal's forces existing between themolecules. However, the boiling and melting point increases as the chain length of the alkaneincreases. That means as we go from methane to ethane, then to propane and so on, the boilingand metling points increases.(iii) They are not soluble in water because they are non-polar covalent compounds whilewater 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(Cl

2, Br

2 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, Br

2 and I

2 to produce haloalkanes. This is, in general, called halogenation

reaction. The chlorination of alkane is fastest among them and bromination is slower whileiodination is the slowest. Halogenation reactions are carried out in presence of sun-light andheat. Chorination is so fast that it is carried out only in diffused sunlight(not direct sunlight) whilebromination 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 givemono-, di-, tri- and tetra chloro methane. These are commonly called substitution reactions. Inthis type reaction, one H atom of alkane is substituted by one halogen atom(here Cl atom).

CH4 Cl2 CH3Cl HCl+ +monochloromethaneor methyl chloride

diff. sunlight

CH3Cl Cl2 CH2Cl2 HCl+ +diff.

sunlight dichloromethaneor methylene chloride

CH2Cl2 Cl2 CHCl3 HCl+ +diff.

sunlight trichloromethaneor chloroform

CHCl3 Cl2 CCl4 HCl+ +diff.

sunlight tetrachloromethaneor carbon tetrachloride

SAQ 4: What main product will be formed when CH4 reacts with

(i)two moles of Cl2 gas (iii)four moles of Cl

2 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 onemole 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 tetrabromoalkane is difficult and often not possible.

CH4 Br2 CH3Br HBrsunlight

heat ++

We get monobromomethane or methylbromide.In the same manner we can get methyl iodide but the reaction is very slow. We shall not discussthe iodination reaction now.2. Reaction with Oxygen:

All alkanes burn in air or oxgyen to give usually carbon dioxde and water. This is calledcombustion reaction. The gas fuel that we use in our kitchen(LPG gas) contains mostly butanealongwith small quanities of propane and other gases(ethane and methane). They burn in air

giving heat, light, CO2 and H

2O. These are highly exothermic reactions.

CH4 + 2O

2 ----------> CO

2 + 2H

2O + heat

C2H

6 + O

2 ----------> 2CO

2 + 3H

2O + 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 ofmotor vehicles that ply in our roads produce these dangerous substances besides CO

2 and has

been the major cause of air pollution.2CH

4 + 3O

2 --------->2 CO + 4H

2O, CH

4 + O

2 --------> C + 2H

2O

CH4 + O

2 --------> HCHO + H

2O, 2CH

4 + 3O

2 ----->2 CH

3COOH + 2H

2O

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 aretreated 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

(i) CH4

Cl2

diff. sunlightA

Na

ether B

(ii) CH3 CH3 Br2

sunlight heat/A

Naether

BCl2

diff. sunlightC + D

A and B are the organic products. The other inorganic product you shall have to write in completeequations.

ALKENESWITH A SPECIAL REFERECE TO ETHYLENE:

METHODS OF PREPARATION:1. Dehydrohalogenation of alkyl halide(Removal of HX):

CH2 CH2

H Cl

H2C CH2 HClalcoholic

KOH+

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 toconvert it to KCl and H

2O. This is called dehydrohalaogenation(removal of HX). In general, any

alkene can be prepare by this method.

CH CH2

Cl

R

H

HC CH2 HClRalcoholic

KOH+

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 havealready seen a sustitution reaction in the properties of alkane. In elimination reaction, twogroups leave from the adjacent carbon atoms in the form of a neutral molecule to result a doublebond. In this case, H from one carbon atom and halogen(X) from the adjacent carbon atomleave as HX molecule resulting the formation of a double bond between those two carbonatoms.Saytzeff' Rule: If the halogen atom is located in an internal position and there is hydrogen atomon either side of the halogen atom, then we get two alkenes. Out of the two, the one which ismore highly substituted is formed in greater quanity(major proudct).

H2C CH CH2 CH3

ClH

H2C CH CH2 CH3 HCl+alcoholic

KOH but-1-ene

H3C CH CH CH3

Cl H

+alcoholic

KOHCH CH2 CH3 HClCH3but-2-ene

2-chlorobutane produces two alkenes i.e but-1-ene and but-2-ene. According to Saytzeff's rulebut-2-ene is formed in greater quantity than but-1-ene. This is because but-2-ene is more highlysubstituted. But-2-ene has two branches(two CH

3 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 themajor product.The branches have been shown inside boxes for better understanding.

CH CH CH3CH3 CHCH2 CH3CH2

2. Dedydration of Alcohol:We get ethylene when a mixture of ethyl alcohol and conc. H

2SO

4 is heated to a

temperature of 1700C. Ethylene gas comes out from the reacting vessel and is collected in a gasjar by the downward displacement of water. This is the laboratory method of preparation ofethylene. 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 notget the alkene, rather we get different products, which we shall see in higher classes.

CH2 CH2

H OH

H2C CH2 H2O+conc. H2SO4

1700C

This is also an elimination reaction in which H atom from one carbon and OH group from theadjacent carbon are removed as H

2O and double bond is formed in between the two carbon

atoms. Other alkene can also be prepared by this method. See the examples.

CH CH2

OH

CH3

H

HC CH2 H2OCH3 +conc. H2SO4

1700C (propene)

Actually dehydration of alcohol takes place in two steps. First when conc. H2SO

4 is mixed with

alcohol, we get alkyl hydrogen sulphate. In the second step, this alkyl hydrogen sulphate isheated to 1700C to give alkene and regenerate H

2SO

4.

Step1:H3C CH2 O H H O SO3H H3C CH2 O SO3H H2O

ethyl hydrogen sulphate+ +

Step 2: CH2 CH2

OSO3HH1700C 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. H

2SO

4 -------heat-------> >

PROPERTIES OF ALKENES:The first three members are gases(ethene, propene and but-1-ene/but-2-ene) are gases, thenext 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 forcesbetween 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 CH CH2

B A

+

When a molecule A-B is added to alkene, one atom(say A) adds to one carbon atom of thedouble bond and the other atom(B) adds to the other carbon atom of the double bond and thedouble bond is converted to a single bond. It is just the opposite of the elimination reaction bywhich an alkene is prepared. In addition reaction, the alkene which is an unsaturated compoundbecomes 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 H

2 at a higher temperature in presence of Ni catalyst, we get

alkane.

H2C CH2 H2 H3C CH3+Ni

200-3000C

Similarly propene will add with H2 to give propane and butene gives butane and so on.

(ii) Addition with halogens (Cl2, Br

2 and I

2 )

Addition of halogen with alkene is very fast. No sunlight or heat is necessary for thisreaction. The reaction can even take place even in dark. Alkene adds to halogen to give dihaloalkane. See these examples.

H2C CH2 Cl2 H2C CH2

Cl Cl

+1,2-dichloroethane

HC CH2 Br2CH3 HC CH2

BrBr

+ CH3(1,2-dibromopropane)

In the addition of bromine to alkenes, the red color of bromine is dicharged. This is used as testfor alkenes. The two halogen atoms lie on adjacent carbon atoms. Such dihalo alkanes aregenerally 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 unsymmetricalalkenes 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 lessnumber of H atoms with it. See this exmaple.

H3C CH CH2 H Cl+ -

H3C CH CH3

Cl

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 thecarbon which bears the less number of H atoms. You know that in HX, the negative part is X(asit is more electronegative) and H is the positive part(less electronegative). According toMarkonikoff's rule, X will add to the carbon which is attached to less number of H atoms. In thiscase X will add to the middle(C-2)carbon as it is attached to one H atom while H of HX will addto the terminal carbon as it is attached to two H atoms. Thus we get 2-chloropropane(orisopropyl chloride) as the major product. The other product, 1-chloropropane obtained by thereverse addition of HCl to propene is formed to a very small extent and has therefore not beenwritten.SAQ 10: (i)What product is obtaiend when but-2-ene reacts with H

2 gas in presence of

Ni and at 3000C?(ii)What product is obtained when but-1-ene reacts with Cl

2?

(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 secondproduct 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 CH2

OH Cl

CH2

(ethylene chlorohydrin or 2-chloroethanol)+

Since ethylene is a symmetrical alkene, hypchlorous acid(HOCl) gives only one product asshown above. If the the alkene is unsymmetrical, then the we get two products and the majorone 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 toform an ozonide first. In alkene ozonide, the three O atoms of O

3 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 betweenC-C vanishes to account for the tetravalency of carbon.

H2C CH2 O3 CH2 CH2

O O

O

+ (ethylene ozonide)

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.

C C

O O

O

H

H

H

H

H2O H C

O

H H C

O

H H2O2+ + +Zn

(methanal)

The ozonide bridge breaks down in this process and each carbon atom of the formerly doublebond takes up one O atom and the remaining O atom reacts with H

2O to form H

2O

2. 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 amixture of aldhehyde and ketone depending on the structure of alkenes. The formation of ozonideand 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. H2SO

4 to form the alkyl hydrogen sulphate. On diluting this with water,

hydrolysis of alkyl hydrogen sulphate takes place to form alcohol.

H2C CH2 H OSO3H H3C CH2 OSO3H(ethyl hydrogen sulphate)

+ H2O

OH H

CH3 CH2 OH H2SO4(ethyl alcohol)

+

+

Ethylene forms ethyl alcohol. For unsymmetrical alkenes(like propene), Markonikoff'srule is applicable. In H

2SO

4, H is the positive part and OSO

3H is the negative part. Accordingly

addition of H2SO

4 to alkene will take place.

(B) OXIDATION REACTIONS:(i) With Dilute Alkaline KMnO

4 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.

KMnO4

alk.+

(ethane-1,2-diolor ethylene glycol)CH2 CH2

OH OH

H2C CH2 H2O [O]+

KMnO4 supplies the nascent O atom which reacts with H

2O to form H

2O

2 and this

provides two OH groups which are added to the two carbon atoms of the double bond to form adiol. Note that all alkenes react with Baeyer's reagent in the similar manner as ethylene .

(ii) Combustion in air:Alkene burns in air or O

2 to give CO

2 and H

2O like alkane.

SAQ 14: What products are obtained when propene reacts with conc. H2SO

4 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. isacetylene ? Also acetylene gas is used for making oxyacetylene flame to provide very hightemperature 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.

CH CH

H Cl

Cl H

CH CH HClalcoholic

KOH+ 2

(acetylene)

When 1,2-dichloroethane reacts with alcoholic KOH, elimination of two HCl molecules takesplace in the manner shown above, and two new bonds are produced between the two carbonatoms and we get a triple bond. We get acetylene. This is similar to dehydrohalogenation of alkylhalides(haloalkanes) to produce alkenes. In haloalkanes, one HX molecule is eliminated to formalkene; while in dihaloalkanes, two HX molecules are eliminated to form alkyne. The halogenand 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 SAQbelow.SAQ 16: What product we get when 1,2-dichloropropane is treated with alcoholic KOH.

2. Hydrolysis of calcium carbide:

CaC2 + H

2O ---------> Ca(OH)

2 + C

2H

2

Ca++

C

C

-

-

H2O

HOH

HOH

CH CH Ca(OH)2+ +2

CaC2 is an ionic compound containing Ca2+ and C

22-. In the carbide(C

22-) there is a triple bond

between the two carbond atoms. When water reacts with calcium carbide, hydrolysis occurs attwo 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 H

2: When alkyne reacts with H

2 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 Cl2 CH CH

Cl

Cl

Cl

Cl

(1,2,3,4-tetrachloroethane)+ 2

When two molecules are added, the triple bond is converted to a single bond.(iii) With HOX (hypohalous acid):

CH CH (HO) Cl CH

Cl

Cl

CH

OH

OH

CH

Cl

Cl

C

O

HH2O-

(2,2-dichoroethanal)

+ 2

Two OH and two Cl groups are added repeated to the same carbon atoms and form an unstablecomound shown within the bracket. Two OH groups cannot be attached to one carbon atom.A molecule of H

2O 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 H

2O in presence of catalyst HgSO

4 and

H2SO

4 at 60-800C to form aldehyde or ketone. Acetylene forms aldehyde(acetaldehyde) and all

other alkynes form ketones.

CH CH H2O CH CH

H

H

OH

OH

CH3 C

O

H(acetaldehyde)

+ 2H2O-HgSO4

H2SO4

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. Acetyleneforms 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(HOOC-

COOH) when reacts with Baeyer's reagent.SAQ 17: Write the products of the following reactions.

(i) Propyne Br2 ?2+ (ii) ?Propyne H2OHgSO 4

H2SO4 heat+

(iii) ?Propyne HOBr+ 2

SAQ 18: Solve the road map problem.

HgSO4/ H2SO4

CH4 A B C D

E F G

Cl2

diff. sunlight

Na

Ether

Br2/lightheat

alcoholicKOH

Cl2

alcoholicKOH

H2O

(B) 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 calledacetylide is formed. First monosodium acetylide is formed with one mole of sodium and thendisodium acetylide is formed with the second mole of Na.

+ +CH CH Na CH C Na H21/2monosodium acetylide

1/2CH C Na Na Na C C Na H2+ +disodium acetylide

This test is used to distinguish between ethylene and acetylene. With sodium, ethylene does notproduce H

2 gas while acetylene produces H

2 gas.

HALOALKANES(ALKYL HALIDES)PREPARATION:

1. From alkanes:R-H + X

2 ---------R-X + HX (Where R=alkyl group and X=Cl,Br,I)

We know that when an alkane reacts with a halogen(Cl2,Br

2,I

2) 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 treatingalcohol with selective reagents. For chlorination we use PCl

3, PCl

5, SOCl

2(thionyl chloride) and

for bromination we use PBr3 and iodiation PI

3.

Br2/P4

I2/P4R OH PI3 R I H3PO33 3+ +

R OH PBr3 R Br H3PO33 3+ +

+++ R Cl SO2 HClR OH SOCl2+++ R Cl POCl3 HClR OH PCl5

++ 33 R Cl H3PO3R OH PCl3

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 PBr

3.

(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 thisreaction, 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 alcoholicKOH, dehydrohalogenation(-HX) takes place and we get an alkene. You noticed that merechange of the solvent from water(aqueous) to alcohol(alcoholic), the KOH performs differentfunction. Aqueous KOH brings about substitution reaction to give an alcohol while alcoholicKOH 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(NH

3):

++ R NH2 HXR X H NH2

Alkyl halide reacts with ammonia under high pressure to give amine(primary amine). Here alsothe halogen atom is substituted by amino(NH

2) 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 PREPARATIONTypes of Alcohols:There are three types of alcohols namely;

(i)Primary(10) (ii)Secondary(20) and (iii)Tertiary(30):(i) Primary alcohol(or 10 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 thatcarbon which bears 1 H atoms.

(iii) Tertirary alcohol (30 alcohol) is the alcohol in which the the OH group is attachedto that carbon which bears no hydrogen atoms.

Tertiary(30)secondary(20)primary(10)

C

R

R''

R''

OHR

CHR'

OHOHCH2R CH3OH,

1. From Alkyl halides:When alkyl halides(haloalkanes) reacts with aqueous alkali(KOH), we get an alcohol. This is asubstitution reaction in which the halogen is substituted by OH group.We have already studiedthis 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, youshall get a different product, alkene due to elimination of HX. We have done this in alkenechapter.

Examples: ++ CH3 OH KClCH3 Cl K OH(aq)(methyl alcohol)

CH2 Br K OH(aq)CH3 CH2 OH KBr+ +CH3(ethyl alchohol)

2. 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 alcoholby 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.

+ 2 R CH2 OHLiAlH4R C

O

H [H]prim. alcohol

R C

O

R' [H]LiAlH4 R CH OH

R'

2+(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'=CH

3-), 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 distillationof wood. Therefore it was called wood alcohol. The word methyl originates from the Greekword methy: wine and yle: wood i.e the wine produced from wood. When wood is distilled inthe absence of air(O

2), 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 wasobtained by separating it from the other two constituents of pyroligneous acid. Note that aceticacid(vinegar) was also manufactured from pyroligneous acid by this method. Methyl alcohol ismanufactured nowadays from CO and H

2(water gas) as follows.

CO H2

ZnO/ Cr2O3

4000C, 150 atm.CH3OH+ 2

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 issurprising to know that recently methanol is used to prepare(synthesise) biologically importantproteins. Do you know the laboratory where this synthesis is carried out? It is inside a livingbeing, a single celled species such as bacteria and years. When these microorganisms are fedwith methanol and other aqueous nuitrient salt solutions containing P, S and N, the bacteria/yeastsynthesise very useful proteins from these. This is indeed marvellous!!!MANUFACTURE OF ETHYL ALCOHOL(C

2H

5OH)

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: (C

12H

22O

11)

Molasses or cane sugar(obtained from sugar cane) and different fruit juices contain thecarbohydrate called disaccharides(mainly sucrose, maltose etc.). The chemical formula of alldisaccharides is C

12H

22O

11. When such substances are kept in the absence of air(O

2) for a long

period, microorganisms such as yeast develop in it and bring about chemical transformation ofthe sugar(disaccharides). This chemical reaction catalysed by microorganisms is calledfermentation. The catalysts produced by microorganisms are called enzymes. The enzymeproduced by yeast in this case is invertase which converts disaccharides to simplesugars(monosaccarides).

C12H22O11 C6H12O6 C6H12O 6+invertase

sucrose 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 anotherenzyme called zymase.

C6H12O6 C2H5OH CO 2

zymase

monosaccharides+2 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 (C6H

10O

5)

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 anotherenzyme called maltase converts maltose to simple sugar(glucose). Then the third enzyme, zymaseconverts simple sugar into ethyl alcohol and carbon dioxide.

+(starch)

diastase(C6H10O5)n H2O C12H22O11(maltose)

C12H22O11 H2O C6H12O6+maltase

2(glucose)(maltose)

C6H12O6 C2H5OH CO 2

zymase

monosaccharides+2 2

22 +zymase

C2H5OH CO 2C6H12O6(glucose)

USES: Commercial alcohol is a constant boiling mixture(azeotrope) containing 95% ethyl alcoholand 5% water and it cannot be further purified by distillation. To remove the remaining alocoholto obtain 100% pure alcohol(absolute alcohol), the commercial alcohol is treated with quicklime(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 largescale 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 alkenebecause of the same reason.1. Oxidation of Alcohols:

(a) Primary alcohol on oxidation gives first an aldehyde which on further oxidationforms a carboxylic acid.Look to the structure below. The H atom attached to O and another H atom attached to theadjacent carbon are removed with the help of a nascent oxgyen as H

2O. 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.

- H2OCH3 CH H [O]

O H

CH3 C

O

H CH3 C

O

OH[O]

+CH3[O]

Since ethyl alcohol is a primary alcohol, we first get acetaldeyde(aldehyde) and then on furtheroxidation get acetic acid(carboxylic acid). In the first step H

2O is formed which has been shown

below the arrow mark(-H2O).

(b) Secondary alcohol on oxidation gives a ketone.

CH3

+ CH3 C

O

CH3 H2OCH3 C H [O]

O H[O]

+

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 andketones 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 aketone. Here H

2 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.

3000C

CuCH3 C

O

HCH3 CH H

O H

(ethyl alcohol) (acetaldehyde)

CH3 C H

O H

CH3

CH3 C

O

CH3Cu

3000C

(isopropyl alcohol)

(acetone)

2. Formation of Alkyl halides:

In the chapter Alkyl halides, we have studied how alocohol reacts with various reagentsto 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 theevolution of H

2 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 analcoholic person or a drunkard and experienced the smell of alcohol?). Carboxylic acids give adifferent smell(pungent).

CH3-OH + Na ----------> CH

3-O- Na+ + H

2

The other product is sodium methoxide(in general metal alkoxide). Other alcohols react in thesimilar manner. Remember that H atom attached to oxygen atom is acidic in nature and is easilydisplaced by active metals like Na, K etc.

ALDEHYDES AND KETONE

Aldehydyes and ketones are commonly called carbonyl comounds as each of them contains acarbonyl group(C=O) in it. That is why they have many similarities in their methods of preparationand 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]---------> AldehydeWe know from the chapter alcohols that primary alcohol on oxidation produces first aldehydeand subsequently carboxylic acid. To get aldehyde from primary alcohol, we shall carry out mildoxidation and arrest the reaction at the aldehyde stage.Formaldehyde (Methanal):methyl alcohol on mild oxidation gives formaldehyde.

AgCH3 OH H C

O

H H2+methyl alcohol formaldehyde3000C

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(O

2), we get water instead of H

2 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 becauseit changes to a plastic mass. Therefore it is preserved and marketted as a 37% aqueous solutioncalled formalin. In this form it is also used as a disinfectant and preservative.

Acetaldehyde(Ethanal):Ethyl alcohol on mild oxidation gives acetaldehyde.

acetaldehydeethyl alcohol+CH3 C

O

H H2CH2 OHCH3Ag

3000C

Heated silver(Ag) at 3000C is used here also as the catalyis to bring about thisoxidation(dehydrogenation). If the reaction is carried out in presence of air(O

2), H

2O is produced

in stead of H2. MnO

2 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(calciumformate) is strongly heated(distilled), aldehyde is distilled out leaving behind CaCO

3 as residue.

See the following scheme. Acetaldehyde and other higher aldhedydes are parepared by thismethod. Formaldehyde is not prepared by this method.

CH3 C

O

O

Ca

O

O

CCH3

Ca

O

O

C

C

O

H

O

H

CH3 C

O

H CaCO32 2+

+

calcium acetate calcium formate

acetaldehyde

Two molecules of CaCO3 are removed as shown in the above scheme.

SAQ 22:(i)What happens when propan-1-ol is treated with MnO

2 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 thechapter, alcohols. Acetone(propanone), the first memeber of ketone family and is prepared bythe 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.

calcium acetate

+

CH3 C

O

O

Ca

O

O

CCH3

C

O

CH3 CH3 CaCO3acetone

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 followingscheme.

+ R C O

H

Y X-+

-+R C O X

H

Y ( If H=R', it becomes a ketone)

XY is the compound which adds on the aldehyde or ketone at the C=O double bond. XY iscalled 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 partof aldehyde/ketone i.e carbon(C) atom. Thus double bond vanishes and we get the product asshown above.

(a) Addition with NH3:

+-+

-+CH3 C O H

H

NH2 CH3 C

H

NH2

OH(acetaldehyde ammonia)

1-aminoethanol

In NH3, H is the positve part and NH

2 is the negative part. A ketone adds with NH

3 in the same

manner as aldehdye.(b) With NaHSO

3(Sodium bisulphite):

CH3 C CH3 H

O

SO3Na CH3 C

OH

SO3Na

CH3 +

-

-++ (acetone bisulphite)

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 thesame manner as ketones.

(c) With HCN:

CH3 C H H

O

CN CH3 C

OH

CN

H +

-

-++ (acetaldehyde cyanohydrin)

In HCN, H is the positve part and CN is the negative part. Thus an aldehyde or ketone adds toHCN to give a compound called the cyanohydrin of the parent compound.

(B) CONDENSATION REACTIONS:Adehydes and ketones react with reagents like hydroxyl amine(NH

2OH), hydrazine(NH

2-NH

2)

or phenyl hydrazine(NH2NHC

6H

5) to give oxime, hydrazone and phenyl hydrazone of the parent

compound respectively. In these reactions, the O of the carbonyl group of aldehyde or ketonecondenses with two H atoms of the reagent to eliminate a molecule of water. Such reactions inwhich water molecule is removed by the reaction between two substances are calledcondensation reactions. C atom of aldehyde and ketone is linked with the N atom of thereagent by a double bond.

CH3C

CH3

O H2 N OH

CH3

C

CH3

N OH H2O+(acetone oxime)

+

Here acetone reacted with NH2OH(hydroxyl amine) to form the acetone oxime. Aldehyde

reacts in the same manner as ketones. Hydrazine(NH2NH

2) and phenyl hydrazine(NH

2NHC

6H

5)

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 hydrazinegives a yellow precipitate when reacts with aldehydes and ketones. This reaction iscommonly used as a test for aldehydes and ketones. More about condensation reaction shallbe taken in higher classes.(C) REDUCTION OF ALDEHYDES AND KETONES:You know that on reduction with reducing agents like LiAlH

4(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, aldehydesshow some different reactions which are not shown by ketones. These reactios are used todistinguish 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 ared precipitate of Cu

2O 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 propertieswhich 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 aldehydesand ketones react) to form hexamethylene tetramine[(CH

2)

6N

4].

6HCHO + 4NH3 ----------> (CH

2)

6N

4 + 6H

2O

SAQ 23: Solve the road map problem.

(i) 2-chloropropane A B Caq. KOH CrO3 H2NNHC6H5

(ii) BNaHSO3

AMnO2propan-1-ol

(iii) calcium propanoate A B Cheat LiAlH4 conc. H2SO4

1700C

(iv)

acetone A B CO3

H2O/ZnD ELiAlH4

PCl5 alcoholic

KOH+

SAQ 24: You are given acetone and ethyl alcohol both of which are colourless liquids. Withouttaste 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, wehave also studied that aldhehyde on oxidation gives carboxylic acid. Methyl alcohol givesformaldehyde first and then formic acid(methanoic acid). If you start from formaldehyde, wealso get formic acid.

CH2 O [O]

HH

H C

O

H H C

O

OHCrO3+[O]

(formic acid)(formaldehyde)(methyl alcohol)

Ethyl alcohol on oxidation first gives acetaldehyde and then acetic acid. If we start fromacetaldehyde, we shall also get acetic acid(ethanoic acid).

CH O [O]

H

CH3

H

CH3 C

O

H CH3 C

O

OHCrO3+[O]

acetaldehyde acetic acidethyl alcohol

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.

CH3 C N H2OH+OH

OHOH

H

HH

CH3 C

O

OH NH3+ +2

In this reaction, methyl cyanide(ethanenitrile) reacts with water in presence of acid(dil. HCl, dil.H

2SO

4 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 H

2O 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 preparedby this method.3. From salt of Carboxylic acid:

Salt of carboxylic acids such as sodium acetate, potassium formate etc. react with dilutemineral acids like H

2SO

4, HCl to form carboxylic acid.

CH3 C

O

O Na H2SO4 CH3 C

O

OH Na2SO4+ +sodium acetate acetic acid

Sodium acetate(sodium ethanonate) forms acetic acid(ethanoic acid).3. Hydrolysis of Ester:

(a)Acidic hydrolysis: When ester reacts with water in presence dilute mineralacids like HCl, H

2SO

4 etc., hydrolysis occurs to form carboxylic acid and alcohol.

OH HCH3 C

O

O CH2 CH3 H2OH

+

CH3 C

O

OH CH3 CH2 OH+

ethyl acetate acetic acid ethyl alcohol+

Here C-O single bond breaks to form a mixture of carboxylic acid and alcohol. Ethyl acetate(ethylethanonate) hydrolyses to form acetic acid(ethanoic acid) and ethyl alcohol(ethanol). Hereacid(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 carboxylicacid and alcohol. The salt of carboxylic acid is then treated with mineral acid like H

2SO

4 to get

free carboxylic acid.

H C

O

O CH2 CH3 NaOHHONa

H C

O

ONa CH3 CH2 OH+ +

ethyl formate sodium formate ethyl alcohol

H C

O

ONa H2SO 4 H C

O

OH Na2SO 4+ +sodium formate formic acid

In the above example, ethyl formate(ethylmethanoate) is boiled with alkali(NaOH) to form sodiumformate(sodium methanoate) and ethyl alcohol first. Sodium formate is then treated with diluteH

2SO

4 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 hydrolysisof natural esters(fats and oils). Soaps are sodium or potssium salts of higher carboxylic acids(fattyacids). 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 6-10atmosphere and 2100C. Sodium formate is obtained first. On treatement with dilute H

2SO

4,

sodium formate forms formic acid.

CO NaOH H C

O

ONa6-10 atm

2100C+

sodium formate

H C

O

ONa H2SO4 H C

O

OH Na2SO4+ +formic acid

Manufacture of acetic acid:6-10% aqueous solution of acetic acid is called vinegar. You know that vinegar is used in foodmaterial for enhancing taste. This is industrially prepared by quick vinegar process. Ethylalcohol is oxidised by air in presence of bacteria, mycoderma aceti grown naturally on woodshavings to form acetic acid(vinegar).

CH3 CH2 OH O2 CH3 C

O

OH H2Omycoderma aceti+ +

PROPERTIES:Carboxylic acids are corrosive and are weak acids. The R-COOH hydrogen is acidic andundergoes 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.

CH3 C

O

OH Na CH3 C

O

ONa H2+ +

CH3 C

O

OH KOH CH3 C

O

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. H

2SO

4 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. H2SO

4

acts as a dehydrating agent.

CH3 C

O

O H H O CH3 CH3 C

O

O CH3 H2O+ +conc. H2SO4

heatacetic acid methyl alcohol methyl acetate

3. Formation of acid chloride:Carboxylic acids react with PCl

5 or PCl

3 or SOCl

2(thionyl chloride) to form the

corresponding acid chloride in the same way as alcohols give alkyl chloride.

CH3 C

O

OH PCl5 CH3 C

O

Cl POCl3 HCl+ + +acetyl chloride

If we use PCl3 in place of PCl

5 , we shall get H

3PO

3 in stead of POCl

3 and HCl. If we use

SOCl2, we shall get SO

2 and HCl gas alongwith the acid chloride.

4. Formation of acid amides:When NH

3 gas reacts with carboxylic acid, first the ammonium salt of carboxylic acid is

formed which on heating forms acid amide and water.

CH3 C

O

OH NH3 CH3 C

O

O(NH4)+ +CH3 C

O

NH2 H2Oacetamideammonium acetate

heat

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.

CH3 C

O

OH [H] CH3 CH2 OH H2O+ +6 2LiAlH4

Acetic acid is reduced to ethyl alcohol by the help of nascent hydrogen produced by LiAlH4.

SAQ 26: Solve the following road map problems.

1.CrO3

H2SO4

n-propyl alcoholACH3 OH

2. CH3 CH2 CH2 Cl A B CKCN H2O/H+ PCl3

3. CH3 CH2 C

O

O CH2 CH3H2O/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 KMnO

4.

3. Methyl propionate is heated in water in presence of dil NaOH.

PRACTICE QUESTIONS1. Solve the following road map problems.

(a) CH4Br2

A Na/ether BCl2/light

C Daq. KOH E

MnO2

heat/light

(b) CH3CH2CH2OH SOCl2 A alcoholic

KOHB HCl C

(c) +B CH2O/Zn

O3A

heat

conc. H2SO 4propan-2-ol

2. Make the following conversions.(i)Methyl alcohol to Ethyl alcohol (ii) Ethyl alcohol to methyl

alcohol(iii)Methane to Ethane (iv)Ethane to methane(v)Formaldehyde to acetaldehdye (vi)Acetaldehyde to

formaldehyde(vii)ethyl alcohol to acetic acid (viii)Acetic acid to methyl

alcohol

3. What happens when. Write the equations.(i)Calcium acetate is dry distilled and the product is treated with LiAlH

4.

(ii)Propene is treated with HI.(iii)Ethyl alcohol is treated with conc. H

2SO

4 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 H

2SO

4 and heated.

(ix)Propyne is treated with sodium metal(x)Propionic acid reacts with ammonia and the product formed is heated.

RESPONSE TO SAQsSAQ 1:

CH2 C O Na

O

CH3+

_

NaOH CaO+ CH3 CH3 Na2CO3+

When Na2CO

3 is removed(shown in the box), CH

3-CH

2-(ethyl) group joins with H to produce

ethane(CH3-CH

3 or C

2H

6). 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 productssuch as methane, hydrogen, ethylene besides ethane. So this method is best suitable to preparepure methane gas.SAQ 2:

+CH2 Cl Na Cl CH2H3C CH3+ +2 2 NaClether H3C CH2 CH2 CH3

butaneThe 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.

NiH2C CH CH3 H2 3000C H3C CH2 CH3+

SAQ 4: . (i)CH4 + 2 Cl

2 ---------> CH

2Cl

2 + 2HCl

When we take two moles of chlorine, two H atoms are replaced by two Cl atoms as shown bythe first two steps in the text and we get dichloromethane as the main product.

(ii) CH4 + 4Cl

2 ----------> CCl

4 + 4HCl

In this case we get tetrachloromethane as the main product by the substitution of four H atomsby 4 chlorine atoms. All the four steps as shown in the test will be taking place.SAQ 5: CH

3-CH

3 + Cl

2 ---------> CH

3-CH

2-Cl + HCl

We get monochloroethane or ethyl chloride as the product. For substituting all the six hydrogenatoms of ethane, we have to take 6 moles of Cl

2 so that we get hexachloroethane(C

2Cl

6) and six

molecules of HCl.SAQ 6: CH

3-CH

2-CH

3 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 CH

3-CH

3 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 ofeach other. See this.

+

isopropyl chloride2-chloropropane

H3C CH CH3 HCl

Cl

or n-propyl chloride1-chloropropane

+H3C CH2 CH2 Cl HCl

H3C CH2 CH3 Cl2+

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)

(iii)+

H3CCH

H3C

Cl Na Cl CHCH3

CH3

2 +

H3CCH

H3C

CHCH3

CH3

2,3-dimethylbutane

ether

The joining of two isopropyl groups in the Wurtz reaction gives 2,3-dimethylbutane.(iv) CH

4 + 3Cl

2 -------> CHCl

3 + 3HCl (Trichloromethane or chloroform will be formed)

(v) H3C CH2 CH2 CH3 Cl2

H3C CH2 CH2 CH2 Cl HCl

H3C CH2 CH CH3 HCl

Cl

+

+

1-chlorobutane

2-chlorobutane

+

Butane has two types of carbon atoms, namely CH3- and -CH

2- and will give two

monochloro butanes(1-chlorobutane and 2-chlorobutane).

(vi) (a)

H3C C

CH3

CH3

CH3 Cl2 H3C C

CH3

CH3

CH2 Cl HCl+ +

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-2,2-

dimethylpropane or neopentyl chloride.

(b)

+ HClH3C CH CH2 CH2

CH3

Cl

+ HClH3C CH CH CH3

CH3 Cl

+ HClH3C C CH2 CH3

CH3

Cl

+ HClCl H2C CH CH2 CH3

CH3

H3C CH CH2 CH3 Cl2

CH3

+

In 2-methylbutane there are four different types of carbon atoms hence we get four differentmonochloro compounds. When a H atoms from the left terminal CH

3 group is replaced we get

1-chloro-2-methylbutane(shown in the top). When a H is replaced from the second carbonfrom left(CH), we get 2-chloro-2-methylbutane(second from top), and when one H is replacedfrom the 3rd carbon atom(CH

2) we get 3-chloro-2-methylbutane(second from bottom) and

when H is replaced from the fourth carbon(right terminal CH3), we get 1-chloro-3-

methylbutane(shown at the bottom). Thus all the four products are obtained in this reaction.

SAQ 8: (i) CH4 Cl2 CH3Cl HCl+ +diff. sunlight

(A)

CH3 Cl 2Na Cl CH3+ +ether

CH3 CH3 2NaCl+(Wurtz reaction) (B)

In the first step, monochlorination occurs to give methyl choride(A) which reacts with Nain presence of ether(Wurtz reaction) to produce ethane(B).

(ii)(A)light/heat

Br2++ HBrH3C CH2 BrH3C CH3 Br2

H3C CH2 Br Br CH2 CH32Na+ + H3C CH2 CH2 CH3 HBr+(B)

ether(Wurtz reaction)

H3C CH2 CH2 CH3 Cl2

H3C CH2 CH2 CH2 Cl HCl

H3C CH2 CH CH3 HCl

Cl

+

+

+

(1-chlorobutane)

(2-chlorobutane)

(C)

(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, undergoesmonochlorination(substitution reaction) to give two monochloroproducts, 1-chlorobutane and2-chlorobutane(C and D).

SAQ 9: (i) H2C CH CH3

BrH

H2C CH CH3 HBr+alcoholic

KOH (propene)

(ii)but-1-ene

H2C CH CH2 CH3

OHH

H2C CH CH2 CH3 H2O+conc. H2SO4

heat

H3C CH CH CH3

OH H

+conc. H2SO4

heat CH CH CH3 H2OCH3but-2-ene

In this case we get two alkenes: but-1-ene and but-2-ene but Saytzeff's rule holds goodhere also. But-2-ene, therefore,is the major product.SAQ 10:

(i)Ni

200-3000CH3C CH CH CH3 H2 H3C CH2 CH2 CH3+

(butane)

(ii) +(1,2-dichlorobutane)

H2C CH CH2 CH3

Cl Cl

H2C CH CH2 CH3 Cl2

(iii) H2C CH CH2 CH3 H I H3C CH CH2 CH3

I

++ -

(2-iodobutane)(Markonikoff' rule)

This is an unsymmetrical alkene and the negative part of HI i.e I adds on the C-2 whichbears one H atom according to Markonikoff's rule. So the mojor product is 2-iodobutane.

(iv) H3C CH CH CH3 HCl H3C CH CH2 CH3

Cl

(2-chlorobutane)+

But-2-ene is a symmetrical alkene. The additon of Cl can take place on any carbon atom of thedouble bond and we get the same compound 2-chlorobutane. Note that there is nothing called3-chlorobutane. This is a wrong name. If you number the carbon chain in the reverse way, it isactually 2-chlorobutane.SAQ 11:

H3C CH CH2 (HO) Br+-

H3C CH CH2

OH Br(propene bromhydrinor 1-bromopropan-2-ol)+

Note that the positve part in HOX is X and negatvie part OH although X is written to the right ofHO. The negative part, OH adds on to the middle carbon which bears less number of H atomsto give 1-bromopropan-2-ol or called propene bromohydrin.SAQ 12:

CH3 CH CH2 O3 H3C CH CH2

O O

O

+

H2O/Zn

CH3 C

O

H H C

O

H H2O2+ +(ethanal) (methanal)

First propene ozonide will be formed which will by hydrolysed to formethanal(acetaldehyde), methanal(formaldehyde) and H

2O

2. Thus two different aldehydes are

formed in this case.

SAQ 13: (i)

CH3 CH CH CH3 H3C CH CH

O O

O

CH3+

CH3 C

O

H CH3 C

O

H H2O2+ +(ethanal)

O3

(ethanal)

In this case we get two same aldehyde(ethanal) molecules.(ii)

CH3 C CH CH3

CH3

H3CC

OH3C

CH3

O

O

CH+

CH3 C

O

CH3 CH3 C

O

H H2O2+ +

O3

(ethanal)(propanone)

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 toMarknikoff's rule. We get propan-2-ol as the main product.

H3C CH CH2 H OHH2SO4/H2O-+

CH3 CH CH3

OH

+propan-2-ol

SAQ 15: We get propane-1,2-diol by the addition of one OH group on either side of the doublebond.SAQ 16: We get propyne by the removal of 2HCl molecules from 1,2-dichloropropane.

C CH

H Cl

Cl H

CH3 C CH HClCH3alcoholic

KOH+ 2

(propyne)

SAQ 17:

(i) CH3 C CH Br2 CH3 C

Br

Br

CH

Br

Br

+ 2 (1,1,2,2-tetrabromoproane)

(ii)CH3 C CH H2O CH3 C

OH

OH

CH

H

H+ -

H OH

CH3 C

O

CH32+H2O-HgSO4

H2SO4

Propyne is an unsymmetrical alkyne. Markonikoff's rule will be applied here. The negativepart of H

2O 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 whenreacts with water in presence of warm HgSO

4/H

2SO

4.

(iii) CH3 C CH (HO) Br CH3 C

OH

OH

CH

Br

Br

+-CH3 C

O

CH

Br

Br

2+H2O

(1,1-dibromopropanone)

-

Markonikoff's rule is applied here. The negative part of HOBr i.e OH group repeatedlyadds to the middle carbon which does not bear any H atom. After the elimination of H

2O, we get

1,1-dibromopropanone.

SAQ 18: (i) CH4 Cl2 CH3Cl HCl+ +(A)

(ii) CH3 Cl Na Cl CH3 CH3 CH3 NaCl+ + +(B)

(iii) CH3 CH3 Br2 CH3 CH2 Br HBr+ +

(C)

(iv) CH2

H

CH2

Br

CH2 CH2 HBralc. KOH

+

(D)

(v) CH2 CH2 Cl2 CH2 CH2

Cl Cl

+

(E)

(vi) CH

H

Cl

CH

Cl

H

CH CH HClAlc. KO H

+ 2(F)

(vii)CH CH H2O CH

H

H

CH

OH

OH

CH3 C

O

H+ 2 - H2O

(G)

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 undergoesmonobromination to give monobromoethane or ethyl bromide(C) which undergoesdehydrobrimination(-HBr) by alcoholic KOH to give ethene or ethylene(D). Ethylene undergoesaddition reaction with Cl

2 to form 1,2-dichloroethane(E) which undergoes dehydrochorination

(-2HCl) to form ethyne or acetylene(F). Acetylene reacts with water in presence of HgSO4/

H2SO

4 to give ethanal or acetaldehyde(G).

SAQ 19:

(i) CH3 CH2 OH PCl5 CH3 CH2 Cl POCl3 HCl+ + +(ethyl chloride)

(ii) H3C CH CH3 PBr3

OH

H3C CH CH3 H3PO3

Br

33 + +(2-bromopropane)

(iii) CH3 OH PI3 CH3 I H3PO3+3 3I2/P4

+

SAQ 20:

(i) CH3 Cl KOH CH3 OH KCl+ + (We get methyl alcohol)

(ii) CH3 CH CH2

Br H

CH3 CH CH2 HBralc. KOH

+(propene)

Here a molecule of HBr is eliminated and we get propene.

(iii) ++ CH2 CN KICH3CH2 I K CNCH3(ethyl cyanide)

We get ethyl cyanide(propanenitrile) in this reaction.

(iv) ++H3C CH2 CH2 Cl H NH2 H3C CH2 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 thecarbon atom to which OH group is attached. If it is 2 or 3 it is a primary alcohol, if it is 1, asecondary alcohol and if none it is tertiary.

(a)prim. (b)prim. (c)prim. (d)sec. (e)tert.

(ii) (a) CH3 CH CH3 KOH(aq)

I

CH3 CH CH3

OH

(propan-2-ol)+

(b) CH3 C

O

H [H]LiAlH4 CH3 CH2 OH2+

(ethyl alcohol)

SAQ 22: (i) CH3 CH2 CH

O H

H CH3 CH2 C

O

H H2OMnO 2

[O]+

propan-1-ol propanal

Propanal-1-ol is a primary alcohol and it is oxidised by MnO2 in acetone to form propanal and

water.

(ii)

calcium formate

++ 2CH2 C

O

H CaCO3CH3Ca

O

O

C

C

O

H

O

HCH2 C

O

O

Ca

O

O

CCH2CH3

CH3

heat

propanal

calcium propanoate

2

When calcium propanoate and calcium formate mixture is distilled, we get propanal.SAQ 23:

(i) +CH3 CH Cl K OH(aq.)

CH3

CH3 CH OH KCl

CH3

+2-propanol(A)

acetone(B)+

CrO3+ CH3 C

O

CH3 H2OCH3 C O H [O]

CH3

H

CH3 C

O

CH3 H2 NNHC6H5 CH3 C

N NHC6H5

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 secondaryalcohol, on oxidation with chormic oxide, in the second step, we get a ketone(B). Ketone reactswith phenyl hydrazine in the third step to give a yellow precipiate(phenyl hydrazone). This is acondensation reaction.

(ii) CH3 CH2 CH O H [O]

HMnO2 CH3 CH2 C

H

O H2O+ +propanal(A)

CH3 CH2 C O H SO3Na

H

+

+ -

-CH3 CH2 C

H

SO3Na

OH propanal bisulphite(B)+

In the first step, propan-1-ol, which is a primary alcohol, on mild oxidation gives analdehyde(propanal,B). Propanal reacts with sodium bisulphite to give a white precipitate ofpropanal bisulphite. This is an addition reaction of aldehyde.

(iii)pentan-3-one

+heat

H3C CH2 C CH2 CH3 CaCO3

O

CH3 CH2 C

O

O

Ca

O

O

CCH2CH3

(A)

H3C CH2 C CH2 CH3 [H]

O

H3C CH2 CH CH2

OHLiAlH4

2+ CH3pentan-3-ol

(B)

1700C

conc. H2SO4H3C CH2 CH CH

HOH

CH3pentan-3-ol

H3C CH2 CH CH CH3(pent-2-ene)

(C)

In the first step, calcium propanoate gives pentan-3-one(A) on strong heating(dry distillation)and leaves the residue CaCO

3. 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.

iv)LiAlH4CH3 C CH3 [H]

O

CH3 CH CH3

OH

2+propan-2-ol (A)

CH3 CH CH3 PCl5

OH

CH3 CH CH3 POCl3 HCl

Cl

+ ++

2-chloropropane(B)

KOH

alcoholic+CH3 CH CH2

HCl

CH3 CH CH2 HClpropene(C)

H2O/ZnO3H3C CH CH2 O3 CH3 CH CH2

O O

O

CH3 C

O

H H C

O

H H2O2

propene ozonide

ethanal methanal+

(D) (E)+

In the first step, reduction of ketone gives a secondary alcohol(A). In the second step, alcohol isconverted to alkyl chloride(B). Alkyl chloride on treatment with alcoholic KOH in the third stepgives alkene(C). Alkene on ozonolysis first gives an ozonide which hydrolyses to give a mixtureof 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 H

2

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 yellowprecipitate(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 silvermirror is an aldehyde(acetaldehyde). Ketones do not give this test.SAQ 26:

1.formic acid

+

(A)H C

OOHH C

OHCH3 OH [O]

CrO3 [O]

In the first step a primary alcohol on oxidation gives carboxylic(formic acid)

H2SO4H C

O

OH H O CH2 CH2 CH3+ H C

O

O CH2 CH2 CH3 H2Opropyl formate

+

(B)

In the second step, carboxylic acid(formic acid) reacts with n-propyl alcohol to form an ester(propylformate or propyl methanoate). This is esterification reaction.

2. CH3 CH2 CH2 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 thesecond step, alkyl cyanide is hydrolysed in the presence of acid to form caboxylic acid(butanoicacid) and ammonia.

CH3 CH2 CH2 CN H2OH+

CH3 CH2 CH2 C

O

OH NH3butanoic acid(B)

+ +2

CH3 CH2 CH2 C

O

OH PCl3 CH3 CH2 CH2 C

O

Cl H3PO3++butanoyl chloride (C)

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.

CH3 CH2 C

O

O CH2 CH3 H2OOH H

H+

+ CH3 CH2 C

O

OH CH3 CH2 OHpropionic acid +

ethyl alcohol(A) (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 -------> HCOONH

4 -------heat-------> HCONH

2 + H

2O

2. Aldehyde on oxidation gives carboxylic acid.CH

3CH

2CHO + [O] ----------> CH

3CH

2COOH(propanoic acid)

3. This is alkaline hydrolysis(or saponification) of ester to form salt of carboxylic acid (sodiumpropionate)and alcohol(methyl alcohol)CH

3CH

2COOCH

3 + NaOH ---------> CH

3CH

2COONa + CH

3OH

ANSWERS TO PRACTICE QUESTIONS1.(a) CH

4 + Br

2 --------> CH

3Br(A) + HBr;

CH3Br + 2Na + BrCH

3 --------> CH

3-CH

3(B) + 2NaCl

CH3-CH

3 + Cl

2 --------> CH

3-CH

2-Cl(C) + HCl

CH3-CH

2-Cl + KOH(aq) -------> CH

3CH

2OH(D) + KCl

CH3CH

2OH + [O] ---------> CH

3CHO(E) + H

2O

A=methyl bromide;B=ethane; C=ethyl chloride; D=ethyl alcoholE=acetaldehyde

(b) CH3CH

2CH

2OH + SOCl

2 -------> CH

3CH

2CH

2Cl(A) + SO

2 + HCl

CH3CH

2CH

2Cl ------alc.KOH-------> CH

3CH=CH

2(B) + HCl

CH3CH=CH

2 + HCl --------> CH

3CH(Cl)CH

3(C)

A: n-propyl chloride; B=propene(elimanation reaction i.e -HCl)C=2-chloropropane(Markonikoff's addition)

(c) CH3CH(OH)CH

3 -----conc. H

2SO

4/heat ------> CH

3CH=CH

2(A) + H

2O

CH3CH=CH

2 + O3 ---> Propene ozonide ----H

2O/Zn---> CH

3CHO(B) + HCHO(C)

A= propene; B=acetaldehyde; C=formaldehyde2. While you convert one organic compound to another compound, you may have to proceedthrough several steps. In all the steps, you can use any inorganic reagent, but you are not allowedto use any other organic compound from outside. For example, in the first bit (i), methyl alcoholis first converted to methyl chloride with the help of PCl

5. Then with the help of Wurtz reaction,

methyl chloride is converted to ethane. Here we do not use a different organic compound. Thecompound(CH

3Cl) derived from the starting compound(CH

3OH)is used. Ethane is then converted

to ethyl chloride and finally ethyl alcohol is obtained from ethyl chloride.(i) CH

3OH + PCl

5 -------> CH

3Cl + POCl

3 + HCl

CH3Cl + 2Na + ClCH

3 -------> CH

3-CH

3 + 2NaCl

CH3-CH

3 + Cl

2 -----light -------> CH

3CH

2Cl + HCl

CH3CH

2Cl + KOH(aq) --------> CH

3CH

2OH + KCl

(ii) CH3CH

2OH + [O] ----CrO

3----> CH

3COOH + H

2O

CH3COOH + NaOH ---------> CH

3COONa + H

2O

CH3COONa + NaOH(CaO) -----heat ------> CH

4 + Na

2CO

3

CH4 + Cl

2 -------light -------> CH

3Cl + HCl

CH3Cl + KOH(aq) ------> CH

3OH + KCl

Alternatively:CH

3CH

2OH ------conc. H

2SO

4(heat) ------> CH

2=CH

2 + H

2O

CH2=CH

2 + O

3 ------> ethylene ozonide ----- H

2O/Zn ------> 2HCHO + H

2O

2

HCHO + [H] ------LiAlH4 --------> CH

3OH

(iii) CH4 + Cl

2 ------light-----> CH

3Cl + HCl

CH3Cl + 2Na + ClCH

3 ------ether-------> CH

3-CH

3 + 2NaCl

(iv) CH3-CH

3 + Cl

2 ------light-----> CH

3-CH

2-Cl + HCl

CH3-CH

2-Cl -------alc. KOH ------> CH

2=CH

2 + HCl

CH2=CH

2 + O

3 -------> ozonide ------H

2O/Zn ------> 2HCHO + H

2O

2

HCHO + 4[H} -------Zn(Hg)/HCl --------> CH4 + H

2O

(v) HCHO + [H] -----LiAlH4 ------> CH

3OH

CH3OH + PCl

3 ---------> CH

3Cl + H

3PO

3

CH3Cl + 2Na + ClCH

3 ---------> CH

3-CH

3 + 2NaCl

CH3-CH

3 + Cl

2 ------light ------> CH

3-CH

2-Cl + HCl

CH3-CH

2-Cl + KOH(aq) -------> CH

3-CH

2-OH + KCl

CH3-CH

2-OH + [O] ----MnO

2 -----> CH

3-CHO + H

2O

(vi) CH3-CHO + 2[H] -------> CH

3-CH

2-OH

CH3-CH

2-OH ------conc. H

2SO

4/heat -------> CH

2=CH

2 + H

2O

CH2=CH

2 + O

3 ------> ozonide -----H

2O/Zn -------> 2 HCHO + H

2O

2

(vii) CH3-CH

2-OH + [O] ------> CH

3-CHO ------[O]-------> CH

3COOH

(viii) CH3-COOH + NaOH -------> CH

3COONa + H

2O

CH3-COONa + NaOH(CaO) ------> CH

4 + Na

2CO

3

CH4 + Cl

2 -----light-------> CH

3Cl + HCl

CH3Cl + KOH(aq) -------> CH

3OH + KCl

3.(i) (CH

3COO)

2Ca -----heat------> CH

3COCH

3(acetone)+ CaCO

3

CH3COCH

3 + 2[H] ------LiAlH

4 ------> CH

3CH(OH)CH

3 (propan-2-ol)

(ii) CH2=CH-CH

3 + HI ------Marknokoff's addition-----> CH

3CH(I)CH

3 (2-iodopropane)

(iii) CH3-CH

2-OH -------conc. H

2SO

4/heat -------> CH

2=CH

2 (ethylene)+ H

2O

(iv) CH3COCH

3 + H

2NOH ------> (CH

3)

2 C= N-OH(acetone oxime) + H

2O

(Refer the reaction of aldehyde/ketone)(v) CH

3CHO forms a silver mirror with Tollen's reagent.

(vi) Cl-CH2CH(Cl)CH

3 ------alc. KOH ------> CH≡C-CH

3 (propyne)+ 2HCl

(Refer the chapter alkyne)(vii) CH

2=CH-CH

2-CH

3 + O

3 -----> ozonide

-------H2O/Zn-------> HCHO(methanal) + CH

3-CH

2-CHO(propanal)

(viii) HCOOCH2CH

3 + H

2O ------H

2SO

4 ------->

HCOOH(formic acid) + CH3-CH

2-OH(acetic acid)

(ix) HC≡C-CH3 + Na ----------> Na+ -C≡C-CH

3 (sodium propynide) + ½ H

2

(x) CH3-CH

2-COOH + NH

3 ------> CH

3-CH

2-COONH

4(ammonium propanoate)

------heat----> CH3-CH

2-CONH

2 + H

2O

(propanamide)