halogen derivative of alkane

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Chemistry – Paper II

[Inorganic and Organic Chemistry]

No. Name Marks without option Marks with option

1 f-Block Elements 3 5

2 Halogen Derivatives of Alkanes 10 14

3 Organic Hydroxy Compounds 5 8

4 Ethers 2 3

5 Aldehydes and Ketones 5 8

6 Acids and Esters 4 5

7 Amines 3 5

8 Biomolecules 4 6

9 Synthetic Fibres 2 3

10 Chemistry in Everyday Life 2 3

Total 40 60

01 f-block Elements

(Weightage: Marks)

Q 1. What is lanthanide contraction?

Ans 1. The steady decrease in atomic and ionic size of the lanthanide elements with increase in atomic

number is known as lanthanide contraction.

The ionic radii of tripositive ions of lanthanides are given in a table.

Ionic radii of M3+

ions of lanthanides.

Element La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

Ionic radii in

pm

10

6 103 101 100 98 96 95 94 92 91 89 88 87 86 85

On moving from La to Lu the decrease in ionic radii from La3+

(106 pm) Lu3+

(85pm) is of 21 pm.

However, this is quite small in comparison to the elements of other periods and groups.

Chapter-2_Halogen Derivative of Alkanes

Q 1. Explain, why halogenation of alkane is not good method for the preparation of alkyl haiide?

Ans. Halogenation of alkane passes through free radical mechanism, therefore reaction cannot stop at

monohalogen stage but it continues to produce dihalides, trihalides, tetrahalides etc. The mixture

of different alkyl halides is obtained. It is very difficult to separate this mixture, therefore

halogenation of alkane is not good method for preparation of alkyl halide.

e.g.



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Q 2. State and explain Markownikoff's rule and peroxide effect with suitable example.

Ans 2. i) Markownikoff's rule :

Statement : When an unsymmetrical reagent is added to unsymmetrical alkene, then the negative

part of adding reagent goes to that carbon, which carries least number of hydrogen atoms.

e.g. Propene gives 2-bromo propane with HBr.

ii) Peroxide effect : "When an unsymmetrical reagent is added to unsymmetrical alkene in

presence of peroxide then the negative part of the reagent get attached to that carbon which carries

more number of hydrogen atoms. "

This peroxide effect is also called "anti-Markownikoff's rule" or "Kharasch and Mayo's effect".

This effect takes place only in case of HBr and not in case of HCl and HI.

e.g. Propene gives 1- Bromopropane according to peroxide effect.

Note : Methyl halide can not be prepared by this method.

C) From alcohols :

Q 3. How will you prepare alkyl chlorides by the action PCl3, PClS and SOC12 on alcohols ? Give

suitable examples. Ans 3. Alkyl chlorides are prepared by heating alcohols with phosphorus trichloride, (PCl3) , phsophorus

pentachloride (PCl5) and thionyl chloride (SOCl2)

i)

e.g.

e.g.

iii)

(This reaction is known as Darzen reaction)

e.g.

Alkyl chloride can also be prepared by the action of dry HC1 gas on ethyl alcohol in presence of

anhydrous ZnCl2.

B) Action of alcoholic potassium cyanide (KCN) :

Q 4. What is the action of alc. KCN on alkyl halide ?

Ans 4. When alc. KCN (or NaCN) is heated with alkyl halide, alkyl cyanide or alkyl nitrile is obtained.

In this reaction, 'X' of alkyl halide is replaced by -CN group.



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e.g.

C) Action of alcoholic silver cyanide (AgCN) :

Q 5. What is the action of alc. AgCN on alkyl halide ?

Ans 5. Upon heating holoalkanes with alcoholic solution of silver cyanide (AgCN) alkyl isocyanides or

isonitriles are obtained. These compounds are also called carbyl amines. In this reaction 'X' of

alkyl halide is replaced by isocyanide group (-N C)-

R-X + .)alc(

AgCN

amine) carbyl (alkylisocyanide Alkyl

CN-R + AgX

e.g. C2H5-Br + .)alc(

AgCN

amine) carbyl (Ethylisocyanide Ethyl

52 CN-HC + AgBr

F) Action of silver salt of carboxylic acid (RCOOAg)

(Formation of esters):

Q 6. Explain the substitution reaction of alkyl halide with silver salt of carboxylic acid. Ans 6. When alkyl halide is heated with alcoholic solution of silver salt of carboxylic acid it produces an

ester. In this reaction halogen atom from alkyl halide is replaced by carboxylate group RCOO-

group.

e.g.

H) Action of sodium metal (Wurtz's reaction) :

Q 7. Write a note on Wurtz's reaction. Ans 7. When alkyl halide is treated with sodium metal in presence of dry ether, higher alkane is obtained.

This reaction is known as Wurtz's reaction. This reaction is convenient for the preparation of

alkanes containing even number of carbon atoms.

e.g.

i) When ethyl bromide is reacted with sodium metal in presence of dry ether, n-butane is obtained.



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ii) Methyl iodide gives ethane

But when mixture of ethyl bromide and methyl bromide is reacted with sodium metal in presence

of dry ether, the mixture of ethane, propane and n-butane is obtained.

Limitations of Wurtz's reaction :

i) Methane cannot be prepared by this method.

ii) This method is suitable for the preparation of symmetrical alkanes only.

Note ; If same reaction is carried out with Zn metal, it is termed as Frankland's reaction.

Chloroform

Formula: CHCl3 IUPAC name ; Trichloromethane

Q 8. How will you prepare chloroform from ethyl alcohol and aqueous bleaching powder?

Ans 8. Preparation of chloroform : Chloroform is prepared by distillation of paste of bleaching powder

with ethyl alcohol. This mixture is heated for three to four hours. Chloroform distilled at about

334K. This is then collected, dried over fused CaCl2 and purified.

Reactions : i) Decomposition of bleaching powder : Bleaching powder is decomposed by water

to give calcium hydroxide and chlorine gas is liberated.

CaOa2 + H2O → Ca(OH)2 + Cl2

ii) Oxidation of ethyl alcohol : The liberated chlorine oxidises ethyl alcohol to give

acetaldehyde.

iii) Chlorination of acetaldehyde : Acetaldehyde reacts with excess of chlorine to form trichloro

acetaldehyde or chloral.

iv) Formation of chloroform : Chloral finally hydrolysed by calcium hydroxide to give

chloroform

B) Oxidation (Action of air and sunlight) :

Q 9. What happens when chloroform is exposed to air and sunlight ?

Ans 9. When chloroform is exposed to air and sunlight it slowly oxidises to phosgene. (Carbonyl chloride

which is a poisonous gas).

Note: Chloroform produces highly poisonous gas i.e. phosgene hence it is always stored in airtight

amber coloured bottle in dark place.



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Also 1% ethyl alcohol is added to it. Thus added ethyl alcohol reacts with phosgene and convert it

into a non toxic ethyl carbonate.

D) Carbylamine reaction :

Q 10. Explain carbylamine reaction of chloroform.

Ans 10. Carbylamine reaction is the identification test for chloroform and also for primary amines.

When primary amines are heated with chloroform in presence of alcoholic KOH, it gives alkyl

isocyanide (carbylamine) which is having very strong disagreable smell, this reaction is called

carbylamine reaction.

e.g.

i) Ethyl amine on heating with chloroform and alc. KOH gives ethyl isocyanide

ii) Aniline gives phenyl isocyanide.

Q 11. Discuss the alkaline hydrolysis of methyl bromide with energy profile diagram.

OR

Explain in detail SN2 reaction mechanism.

Ans 11. Alkaline hydrolysis of methyl bromide is a typical example of SN2 reaction. Where,

S - stands for substitution

N - stands for nucleophilic

2 - stands for bimolecular

Methyl bromide on hydrolysis by aq. KOH, produces methyl alcohol and potassium bromide.

Ionic form of above reaction,

Kinetic expression : The rate of this reaction is directly proportional to both substrate i.e. CH3Br

and nucleophile (OH) therefore it is a bimolecular reaction or second order reaction.

Rate of reaction [CH3Br] [OH]

Rate of reaction = K [CH3Br] [OH]

Where K is called as specific rate constant.

Mechanism : This is a single step reaction and described as follows.

i) Backside attack of nucleophile : The attack of nucleophile OH- on methyl bromide

takes place from the back side of leaving nucleophile Br-. This is because Br

- ion being a

nucleophile repels the approach of other nucleophile OH- from the same side.

ii) Formation of transition state :



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a) Attack of nucleophile OH- and detachment of Br

- from carbon takes place

simultaneously.

b) As OH- approaches CH3Br the kinetic energy of the system changes into potential energy

and potential energy increases slowly resulting in the formation of transition state.

c) At this state energy is maximum and molecule is least stable. OH" group and bromine

atom are loosely bonded to the carbon atom both acquiring partial negative charges. It is

a slow step and hence rate determining step.

d) At this state the three hydrogen atoms are in a plane perpendicular to the OH-C-Br axis

and H-C-H bond angle is 120°.

iii) Formation of products : The bond between carbon and Br breaks completely and bond

between carbon and OH- forms fully to give the product methyl alcohol with 100 %

inversion of configuration.

iv) Inversion of configuration : During the attack of nucleophile OH from backside, the

three hydrogen atoms attached to carbon move on opposite side giving rise to opposite

arrangement of atoms in the product to that of methyl bromide. This opposite

arrangement is called inversion of configuration.

v) Energy profile diagram : The graph obtained by plotting the potential energy changes

against rate of reaction is called energy profile diagram.

Q 12. Discuss the alkaline hydrolysis of tertiary butyl bromide with energy profile diagram.

OR

Explain in detail SN1 reaction mechanism.

Ans 12. Alkaline hydrolysis of tertiary butyl bromide is a typical example of SN1 reaction where,

S - stands for substitution

N - stands for nucleophilic

1 - stands for unimolecular

Tertiary butyl bromide on hydrolysis by aq. KOH, produces tertiary butyl alcohol and potassium

bromide.

Ionic form of the above reaction,



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Kinetic expression : The rate of this reaction is directly proportional to concentration of only one

reactant i.e. tert-butyl bromide, therefore it is unimolecular reaction or first order reaction.

Rate of reaction [(CH3)3 C-Br] Rate of reaction = K [(CH3)3C-Br] Where, K is specific

rate constant.

Mechanism The mechanism can be described in two steps.

Step-I : i) Formation of tert-butyl carbonium ion :

a) In the first step the tert-butyl bromide undergo heterolysis and forms tert-butyl

carbonium ion ( C-Br bond is more polar and it breaks heterolytically due to +ve as well

as -ve Inductive effect).

This step is a very slow step hence rate determining step of the reaction.

b) Thus formed carbonium ion has a planar structure and carbon atom is sp2 hybridised. The

bond angle between CH3-C-CH3 is 120°.

c) Before formation of carbonium ion first transition state is formed.

Step-II: ii) Formation of products : In the second step the nucleophile OH

- attack on the planer

positively charged tert. butyl carbonium ion from backside as well as front side. This

results in the formation of the product (tertbutyl alcohol) with 50 % retention and 50 %

inversion of configuration. This is a very fast step and formation of C-OH bond passes

through transition state - II.

Energy profile diagram :

The graph obtained by plotting potential energy against rate of reaction is called energy profile

diagram'.



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Energy profile diagram of SN

1 reaction

Chapter 3_Organic Hydroxy Compounds Q 1. What is the action of moist silver oxide on -i) Bromomethane (Methyl bromide) ii) n-propyl

iodide (1-iodopropane) ?

Ans 1. i) When moist silver oxide is heated with methyl bromide it gives methyl alcohol.

(Ag2O + H2O → 2AgOH)

ii) When moist silver oxide is heated with n-propyl iodide it gives n-propyl alcohol.

B) By hydration of alkenes :

Addition of molecule of water across the double bond of alkene is known as hydration of alkene.

Q 2. How will you prepare alcohol from alkene (olefin) ?

Ans 2. When alkene is passed into cold and conc. sulphuric acid, it undergoes addition reaction to form

an alkyl hydrogen sulphate which on heating with water get hydrolysed to form corresponding

alcohol.

In case of unsymmetrical alkene, addition takes place according to Markownikoff's rule (-OSO3H

group gets attached to carbon atom of alkene carrying least number of hydrogen

Note : This is a commercial method mainly useful for the preparation of secondary alcohols.

Q 3. How will you obtain primary alcohols and secondary alcohols by reduction of carbonyl

compounds ?

OR

How will you convert carbonyl group )OC( into

i) -CH2-OH group ii) >CH-OH group ?



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Ans 3. Aldehydes on reduction gives primary alcohols while ketones on reduction gives secondary

alcohols. Reduction can be carried out by two ways.

Catalytic reduction (Hydrogenation) : Aldehydes and ketones are reduced by passing hydrogen

gas in presence of Ni or Pt or Pd at 413 K. Aldehydes gives primary alcohols on reduction.

Here, -CH=O group is converted into -CH2-OH group.

Ketones gives secondary alcohols on reduction.

Here, >C=O group is converted into >CH-OH group.

ii) Reduction by sodium amalgam and water : Aldehydes and ketones are reduced by using

sodium -amalgam and water, aldehyde gives primary alcohol and ketone gives secondary alcohol.

The reduction is due to nascent hydrogen which is liberated by the action of water on sodium-

amalgam.

In these reactions, OC group is converted to -CH2-OH group and OHCH group.

Q 4. How will you prepare following from carbonyl compounds

i) Ethanol ii) Propan 2-ol iii) 2-methyl propan-2-ol?

OR

What is the action of methyl megnesium iodide(Grignards reagent) on-

i) Formaldehyde (Methanal)

ii) Acetaldehyde (Ethanal)

iii) Acetone (Propanone) ?

OR

How will you convert:

i) Methanal into ethanol.

ii) Ethanal into propan-2-ol.

iii) Propanone into 2-methyl propan-2-ol ?

Ans 4. i) When formaldehyde is reacted with methyl magnesium iodide in presence of dry ether, it gives

addition product, which on acid hydrolysis gives ethyl alcohol (primary alcohol).

Here, >C=O group is converted to -CH2-OH group.



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ii) When acetaldehyde is reacted with methyl magnesium iodide in presence of dry ether it gives

addition product, which on acid hydrolysis gives iso-propyl alcohol (sec. alcohol).

Here, > C=O group is converted to >CH-OH group.

iii) When acetone is reacted with methyl magnesium iodide in presence of dry ether, it give

addition product, which on acid hydrolysis gives tertiary butyl alcohol. 1

Here, OC group is converted to group.

I) Action of HCl :

Q 5. Explain the action of conc. HCl on -

i) 1° alcohol ii) 2° alcohol iii) 3° alcohol.

OR

What happens when -

i) Ethyl alcohol reacted with conc. HCl in presence of ZnCl2

ii) Iso-propyl alcohol reacted with conc. HC1 in presence of ZnCl2 iii) Tert-butyl alcohol reacted with conc. HCl ?

Ans 5. HCl is least reactive halogen acid therefore it require the catalyst anhydrous ZnCl2 for reaction

with 1° and 2° alcohols.

i) When ethyl alcohol (1° alcohol) reacts with conc. HCl in presence of anhydrous zinc chloride as

catalyst it gives ethyl chloride.



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ii) Iso-propyl alcohol.(2° alcohol) gives iso-propyl chloride.

iii) But, tertiary alcohol reacts with conc. HCl even in absence of catalyst at room temperature.

Note: The mixture of conc. HCl and anhydrous ZnCl2 is called Lucas reagent.

D) Dehydration of alcohols :

(Reaction involving both alkyl group and -OH group) :

Q 6. Explain dehydration of alcohols. Ans 6. Elimination of water molecule from alcohol is called dehydration of alcohol.

Dehydration of alcohols can be carried out by using conc. H2SO4 or heated alumina (Al2O3). The

ease of dehydration of alcohols is Tertiary > Secondary > Primary

i) Primary Alcohols : Primary alcohols are dehydrated by 95% H2SO4 at 443 K or in presence of

Al2O3 at 623K. They give corresponding alkenes.

e.g.

ii) Secondary alcohols : Secondary alcohols are dehydrated by 60 % H2SO4 at 373 K or in

presence of Al2O3 at 523K to form corresponding alkenes.

e.g.

iii) Tertiary alcohols : Tertiary alcohols are dehydrated by 20 % H2SO4 at 363 K or in presence of

Al2O3 at 423K.

e.g.

Note : Dehydration of alcohols by conc. H2SO4 is known as liquid phase dehydration.

Dehydration of alcohols by using Al2O3 is known as vapour phase dehydration.

Q 7. How will you distinguish between primary, secondary and tertiary alcohols?

OR How will you distinguish between ethanol, propan-2-ol and 2-methyl propan-2-ol ?



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Ans 7. Different classes of alcohols behave differently towards oxidising agents. Hence this reaction can

be used to distinguish between primary, secondary and tertiary alcohols.

i) Oxidation of primary alcohols : Primary alcohols on oxidation give aldehydes which on

further oxidation give carboxylic acids. Both aldehyde and acid contain the same number

of carbon atoms as the original primary alcohol.

e.g.

ii) Oxidation of secondary alcohols : Secondary alcohols on oxidation give ketones with

same number of carbon atoms as the original alcohol. Ketone on further oxidation gives

a carboxylic acid with one carbon atom less than original alcohol.

e.g.

ii) Oxidation of tertiary alcohols : A tertiary alcohol resists oxidation in neutral or alkaline

medium. First, the tert. alcohol gets dehydrated into alkene. Alkene on further oxidation

gives ketone with one less number of carbon atom as the original alcohol. Ketone on

further oxidation gives an acid with still one less number of carbon atom.

e.g.

Q 8. How is phenol prepared from chlorobenzene by Dow's process ?

Ans 8. In Dow's process chlorobenzene is heated at 623K with 6-8% NaOH solution under 300 atm.

pressure, sodium phenoxide is obtained which on acidification with dil.HCl gives phenol.



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Q 9. How will you prepare phenol from cumene (iso-propyl benzene or 2-phenyl propane) ?

Ans 9. This is a commercial method of preparation of phenol. Preparation of phenol from cumene

involves two steps.

Step I : Oxygen or air is passed through the suspension of cumene in aqueous Na2CO3 solution in

presence of cobalt naphthenate as catalyst at 403 K, cumene gets oxidised to cumene

hydroperoxide.

Step II : Cumene hydroperoxide is decomposed by hot dil. H2SO4 and gives phenol and acetone.

Acetone is valuable biproduct which is separated by distillation.

D) From aniline by diazotization :

Q 10. What is diazotization ? How will you prepare phenol by diazotization reaction? Ans 10. The replacement of primary amino group ( -NH2) from aromatic amine by a diazonium chloride

)NClN( group is called diazotization. In this reaction aniline is treated with nitrous acid

(HNO2) in cold i.e. a mixture of NaNO2 and HCl at 273 - 278 K (below 5°C), benzene diazonium

chloride is formed. (NaNO2 + HCl → NaCl + HNO2)

The aqueous solution of this benzene diazonium chloride (salt) on boiling with dil. H2SO4 gives

phenol.



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Q 11. What is the action of following on phenol -

i) Br2 in water (bromine water) ii) Br2 in carbon disulphide (CS2) ?

Ans 11. i) Action of Br2 in water (bromine water) : Phenol reacts with bromine water and gives

yellowish-white ppt. of 2, 4, 6- tribromophenol.

ii) Action of Bromine in CS2 : Phenol reacts with liquid bromine in carbon disulphide at 273 K it

gives a mixture of O-bromophenol and p-bromophenol (p - bromophenol is a major product).

Q 12. Why phenols are acidic in nature?

Ans 12. i) Any substance which gives proton is called Bronsted acid. Phenol gives proton hence phenol

acts as acid.

ii) Phenols dissociates to give proton.

iii) Since phenoxide ion is more stable than phenol due to resonance stabilization. Hence the

forward reaction (dissociation) is favoured. Therefore phenol gives H+ ions and behaves

as acid.

Note : The resonance stabilisation of phenoxide ion can be explained with the help of the

following contributing structures.

Chapter 4_Ethers



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Q 1. Explain metamerism in ethers with suitable example. Ans 1. Definition : Two or more compounds may have same molecular formula and belonging to the

same homologous series but differ from one another in the nature of alkyl radicals linked to the

polyvalent functional group are called metamers and this phenomenon is called metamerism.

e.g. Molecular formula C4H10O represents following metamers.

i) ii)

iii)

Ethers containing at least four carbon atoms show metamerism.

A) By Williamson's synthesis :

Q 2. How will you prepare simple and mixed ethers by Williamson's synthesis ?

OR

Write a note on Williamson's synthesis. Ans 2. When sodium alkoxide is heated with alkyl halide in alcohol ethers are formed. This method is

called Williamson's synthesis. By this method simple as well as mixed ethers can be prepared.

i) Simple ether : When sodium methoxide is heated with methyl bromide in alcohol, simple ether

i.e. dimethyl ether is obtained.

ii) Mixed ether : When sodium methoxide is heated with ethyl bromide in alcohol mixed ether i.e.

ethyl methyl ether is obtained.

B) By continuous etherification process :

Q 3. Write a note on continuous etherification process.

OR

How will you prepare simple ethers by continuous etherification process? Ans 3. When excess of alcohol is heated with conc. H2SO4 at 413K, it gives alkyl hydrogen sulphate as

an intermediate which reacts with second molecule of alcohol to give ether. It is two step reaction.

Step-I: Formation of intermediate :

Step-II: Formation of ether :

In this method, H2SO4 is regenerated and the same amount of H2SO4 reacts with more alcohol to

form ether. Hence, this reaction is called continuous etherification. e.g. Ethyl alcohol with conc.

H2SO4 at 413K gives diethyl ether.



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i)

ii)

Note : This method is suitable only for the preparation of simple ethers. Mixed ethers can not be

prepared by this method.

B) By alkaline hydrolysis of geminal dihalides :

Chapter 5_Aldehydes and Ketones

Q 1. How will you prepare aldehydes and ketones by hydrolysis of geminal dihalides ?

Ans 1. a) Preparation of aldehydes (Hydrolysis of terminal geminal dihalides): When geminal

dihalide (containing two halogen atoms attached to the same terminal arbon atom) is boiled with

aq. alkali (KOH or NaOH). Then on hydrolysis unstable dihydroxy compound is formed which is

converted into aldehyde by elimination of water molecule.

b) Preparation of ketones (Hydrolysis of non-terminal geminal dihalides): When geminal

dihalide (containing two halogen atoms attached to non-terminal carbon atom) is boiled with aq.

alkali (NaOH or KOH), then on hydrolysis unstable dihydroxy compound is formed which is

converted to ketone by elimination of water molecule.

ii) When iso-propylidene dichloride boiled with aq. KOH, it gives acetone.

iii) When 3,3-dichloropentane boiled with aq. KOH, gives diethyl ketone.



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ii) Preparation of acetaldehyde : A mixture of calcium formate and calcium acetate on dry

distillation gives acetaldehyde (ethanal).

Addition Reactions

A) Addition of HCN :

Q 2. What is the action of H-CN on -

i) Formaldehyde ii) Acetaldehyde iii) Acetone ? Ans 2. i) Action of H-CN on formaldehyde : Formaldehyde reacts with HCN

(a mixture of NaCN and HCl) to form formaldehyde cyanohydrin.

ii) Action of H-CN on acetaldehyde : Acetaldehyde reacts with HCN to form acetaldehyde,

cyanohydrin.

iii) Action of HCN on acetone : Acetone reacts with HCN to form acetone cyanohydijin

B) Addition of NaHSO3 (Sodium bisulphite) :

Q 3. What is the action of NaHSO3 on

i) Acetaldehyde ii) Acetone ?



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Ans 3. i) Action of NaHSO3 on acetaldehyde : When acetaldehyde reacts with saturated aq. solution of

NaHSO3 it forms acetaldehyde sodium bisulphite compound.

ii) Action of NaHSO3 on acetone : When acetone reacts with saturated aq. solution of NaHSO3, it

forms acetone sodium bisulphite.

Q 4. What is the action of ammonia on

i) Formaldehyde (Methanal)

ii) Acetaldehyde (Ethanal)


Ans 4. i) Action of ammonia on formaldehyde :

When formaldehyde reacts with ammonia, it gives hexamethylene tetramine (Urotropine) which is

used as urinary antiseptic.

(Formaldehyde is most reactive and the reaction does not stop at the first step).

Structure of Hexamethylene tetramine is :

ii) Action of ammonia on acetaldehyde : Acetaldehyde reacts with ammonia to form

acetaldehyde ammonia.

iii) Action of ammonia on acetone : Ketones do not give simple addition product with ammonia.

When acetone reacts with ammonia it gives diacetone amine. Two molecules of acetone reacts

with each other. First, it gives mesityl oxide. Further addition of ammonia in mesityl oxide gives

diacetone amine.



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Q 5. What is the action of Grignard's reagent on -

i) Formaldehyde (Methanal) ii) Acetaldehyde (Ethanal)


OR

How will you prepare following from carbonyl compounds

i) Ethanol ii) propan-2-ol

iii) 2-methyl propan-2-ol ?

OR

How will you convert -

i) Methanal into Ethanol ii) Ethanal into propan-2-ol iii) Propanone into 2-methyl propan-2-ol ?

Ans 5. i) When formaldehyde is reacted with methyl magnesium iodide in presence of dry ether, it gives

addition product which on acid hydrolysis gives ethyl alcohol (primary alcohol).

Here, OC group is converted to -CH2-OH group.

ii) When acetaldehyde is reacted with methyl magnesium iodide in presence of dry ether, it gives

addition product which on acid hydrolysis gives iso-propyl alcohol (sec. alcohol).

Here, OC group is converted to OHCH|

group.

iii) When acetone is reacted with methyl magnesium iodide in presence of dry ether, it gives

addition product which on acid hydrolysis gives tertiary butyl alcohol (tert-alcohol).

Here, OC is converted to group.



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Q 6. What is the action of hydroxyl amine on aldehydes and ketones? Ans. Aldehydes and ketones when reacted with hydroxyl amine, it forms corresponding oximes.

Aldehyde gives aldoxime and ketone gives ketoxime. Reactions are carried out in weakly acidic

media.

Here, OC group is converted into NOHC group.

Q 7. Explain in detail aldol condensation .

Ans 7. Definition : When two molecules of aldehyde or ketone, containing active -hydrogen atom are

warmed with alkali gives -hydroxy aldehyde (Aldol) or -hydroxy ketone (ketol). This

condensation reaction is known as aldol condensation.

The product contains, aldehyde (-CHO) as well as alcohol group hence it is called Aldol. This is

self addition reaction.

Conditions :

i) Aldehyde or ketone which undergoes aldol condensation must possess -hydrogen atom.

ii) The reaction is carried out in presence of dil. alkali like, dil. NaOH, dil KOH, Na2CO3 or

Ba(OH)2. Mechanism : In this reaction the a*-carbon atom of one molecule gets attached to the carbonyl

carbon atom of another molecule along with migration of -hydrogen towards carbonyl oxygen to

form -OH group.

e.g.

i) Acetaldehyde undergo aldol condensation as,



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Acetaldol when heated with acid looses water molecule and gives crotonaldehyde ( , -

unsaturated aldehyde).

ii) Acetone undergo aldol condensation as ,

Diacetone alcohol when heated with dilute acid, it looses water molecule and forms mesityl oxide.

Note; Formaldehyde (HCHO) and benzaldehyde (C6H5CHO) do not undergo aldol condensation

as they do not contain -hydrogen atom.

Q 8. Write a note on Cannizzaro's reaction.

OR

Write a note on self-redox reaction in aldehydes.

Ans 8. Aldehydes which do not contains -hydrogen atom when heated with concentrated alkali (50%),

undergo self oxidation-reduction (redox) reaction to form mixture of primary alcohol and salt of

acid, this reaction is called Cannizzaro's reaction.

In this reaction, one molecule of aldehyde oxidises to acid and another molecule reduces to

alcohol. The acid formed reacts with alkali and forms salt of acid.

This is base catalysed auto-redox reaction.

When formaldehyde is heated with strong alkali like NaOH(50%), it produces methyl alcohol and

sodium formate.

e.g.

This reaction is also called disproportionation reaction.

Note : Acetaldehyde and acetone do not show Cannizzaro's reaction because they contain -

hydrogen atoms.

c) Clemmensen's Reduction

Q 9. Write a note on Clenumensen's reduction. Ans 9. When aldehydes or ketones are heated with zinc-amalgam and conc. HCl, they are reduced to

corresponding alkane. This reaction is known as Clemmensen's reduction.

In this reaction, OC group is converted to -CH2- group (methylene group).

i)



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ii)

e.g.

i)

a)

Note : In this reaction aldehydes do not give satisfactory results.

Chapter 6_Acids and Esters

Q 1. Explain acid chloride formation reaction of carboxylic acids. Ans 1. Carboxylic acids react with thionyl chloride or phosphorus pentachloride or phosphorus trichloride

to form acid chloride by replacement of -OH group by -Cl atom.

e.g.

i)

ii)

iii)

Q 2. Explain amide formation reaction of carboxylic acids. Ans 2. When carboxylic acid reacts with ammonia, it gives ammonium salt of acid which on heating

converted to corresponding acid amide.

e.g.

i) Acetic acid reacts with ammonia, it gives ammonium acetate which on further heating gives

acetamide.

ii) Propionic acid reacts with ammonia, it gives ammonium propionate which on heating gives

propionamide.



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Q 3. Explain acid anhydride formation reaction of carboxylic acids. Ans 3. When carboxylic acid is heated with a strong dehydrating agent like P2O5, two molecules of acid

combines to form corresponding acid anhydride with elimination of water molecule.

[P2OS + H2O → 2HPO3 (Metaphosphoric acid)]

Excess P2O5 reacts with H2O to form metaphosphoric acid which prevents the backward reaction.

e.g. Weightage

i)

(P2O5 + H2O → 2HPO3)

ii)

(P2O5 + H2O → 2HPO3)

Q 4. Explain the ester formation property of carboxylic acid.

OR

Write a note on esterificatton. Ans 4. When carboxylic acid is refluxed with alcohol in presence of dehydrating agent like conc. H2SO4

or dry HC1 or anhydrous ZnCl2 esters is formed.

This condensation of alcohol with acid is called esterification.

e-g.

i) Acetic acid heated with ethyl alcohol in presence of conc. H2SO4 gives ethyl acetate.

ii) Acetic acid heated with methyl alcohol in presence of con. H2SO4 gives methyl acetate



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Conc. H2SO4 acts as dehydrating agent as well as catalyst.

Chapter 7_Amines A) By Ammonolysis of alkyl halide:

Q 1. Explain the ammonolysis of alkyl halide.

OR

How will you prepare amines by Hoffmann's ammonolysis method ?

Ans 1. When alkyl halide is heated with alcoholic solution of ammonia in a sealed tube at 373 K.

Mixture of primary, secondary and tertiary amines along with quaternary ammonium salt are

obtained. This reaction is called Hoffmann's ammonolysis reaction.

However, primary amines can be obtained in good yield by using ammonia in excess .

Q 2. How will you prepare primary amines by reduction of aldoxime and ketoxime.

OR

Explain the preparation of primary amines from oximes.

Ans 2. a) From aldoximes : When oximes of aldehydes are reduced in presence of sodium and ethanol as

reducing agent gives corresponding primary amines.

e.g.

i) Formaldoxime reduces to give methyl amine in presence of sodium ethanol

ii) Acetaldoxime gives ethyl amine.

b) From ketoximes : When oximes of ketones are reduced in presence of sodium and ethanol as

reducing agent gives corresponding primary amines.



25

e.g. When acetoxime is reduced in presence of sodium and alcohol as reducing agent gives iso-

propyl amine.

C) Acetylation :

Q 3. What is acetylation of amines ? Explain with suitable examples1.

OR

How will you distinguish between primary, secondary and tertiary amine by acetylation ? Ans 3. Acetylation : The replacement of hydrogen atom from the functional group of amines by acetyl

group (-COCH3) is called acetylation.

Acetylation of amines is carried out with the help of acetyl chloride or acetic anhydride in

presence of a weak base like pyridine.

Amines reacts differently with acetyl chloride or acetic anhydride hence this reaction is used to

distinguish between amines.

i) Primary amines : When primary amines are heated with acetyl chloride or acetic anhydride, on

acetylation it gives monoacetyl and diacetyl alkyl amines.

OR

ii) Secondary amine : When secondary amines are heated with acetyl chloride or acetic anhydride

on acetylation, it gives monoacetyl dialkyl amine.

iii) Tertiary Amines : Tertiary amines do not react with acetyl chloride or acetic anhydride

because they do not contain any replacable hydrogen atoms in their functional group.

D) Action of Nitrous acid (HNO2) :



26

Q 4. How will you distinguish between primary, secondary and tertiary amines by the reaction

with nitrous acid ?

Ans 4. Primary, secondary and tertiary amines behaves differently with nitrous acid and hence this

reaction is used to distinguish between them.

Nitrous acid is a unstable acid hence it is prepared immediately by reacting NaNO2 with HCl.

i) Primary amines : Primary amines reacts with nitrous acid in cold to give corresponding

primary alcohols with liberation of nitrogen.

ii) Secondary amines : Secondary amine reacts with nitrous acid in cold, to give nitroso

compound. Which is yellow coloured oily compound insoluble in water.

iii) Tertiary amine : As tertiary amines does not contain any replacable hydrogen atom in its

functional group. It does not reacts with nitrous acid. It gives only unstable addition product and

reaction is reversible.

Chapter 9_Synthetic Fibres

Q 1. What are Nylons ?

Ans 1. The synthetic poly amide fibres are called nylons. They contain -CONH- functional group.

These fibres are prepared first time in Newyork and London hence the name Nylon. Depending

upon the number of carbon atoms present in starting material used for preparation, they are of two

types. i) Nylon-6 ii) Nylon-66

Q 2. Q.4 How is Nylon-6 prepared ?

Ans 2. Nylon-6 is also called Perlon-L.

It is a polyamide fibre obtained by polymerisation of -caprolactum. Since the raw material -

caprolactum contain six carbon atom, the nylon is called Nylon-6.

Preparation : Step I : Polymerisation of monomer :

When -caprolactam is heated at 533 K in an inert atmosphere of nitrogen for about 4-5 hours it

undergoes polymerisation and give viscous molten mass of Nylon-6.

Step II : Spinning of polymer :

This viscous molten mass is then forced through a tiny holes of spinnerets to get nylon threads

which are then cooled by passing stream of dry air and cut finally into pieces of required length.

Q 3. How is Nylon-66 prepared Ans 3. Nylon-66 is a poly amide fibre obtained by the condensation of adipic acid and hexamethylene

diamine, as both the raw materials contain six carbon atoms the nylon is called Nylon-66.

Step-I : Preparation of nylon salt (monomer) :



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When aqueous solution of adipic acid and hexamethylene diamine are allowed to react together,

hexamethylene diammonium adipate (Nylon salt) is formed. Then water is evaporated to get dry

nylon salt, this nylon salt is called monomer.

Q 4. How is terylene (Dacron) prepared? Ans 4. Terylene (Dacron) is a polyester fibre, obtained by condensation polymerisation of ethylene glycol

and dimethyl terephthalate (D.M.T.).

Step-I : Preparation of monomer :

When ethylene glycol and dimethyl terephthalate (2:1) are heated together at about 503K in

presence of a mixed catalyst zinc acetate and Sb2O3. Trans-esteriflcation takes place to give

dihydroxy, diethyl terephthalate (monomer) by loss of methanol molecule.

Step-II: Polymerisation of monomer :

The monomer is heated at about 553-573 K in vacuum, it polymerises to form molten viscous

terylene polymer, with loss of ethylene glycol molecule.

Step-III : Spinning of polymer :

The molten viscous polymer is forced through spinneret having very fine holes. The fibre so

formed is cooled by stream of dry air.

09 - Synthetic Fibres (Weightage : Marks)



28

Q 1. What is aspirin? How will you prepare aspirin from salicylic acid ?

Ans. Aspirin is a commonly used analgesic drug. It also has antipyretic property (lowering body

temperature). The name aspirin was coined by adding ‘a’ for acetyl to spirin for spiraea, the plant

species from which salicylic acid was once prepared.

Chemically, it is acetyl salicylic acid or 2-acetoxy benzoic acid.

Preparation of aspirin from salicyclic acid : Aspirin is obtained by acetylation of salicylic acid by

acetic anhydride in presence of conc. H2SO2.

Q 2. What are tranquilizers ? Give two examplex.

Ans. Tranquillizers are the chemical substances used for the treatment of mental diseases.

Or

The drug used for the treatment of stress as well as well as mild and severe mental illness is called

tranquilizer.

These are used to relieve stress, fatigue, anxiety and induces sleep. They are essential component of

sleeping pills. These are also called psychotherapeutic drugs.

They act on CNS and help in reduceing strees, fatigue and anxiety and induces sleep. The most

commnly used tranquilizers are derivatives of barbituric acid.

For example :

i) Hypnotic : Barbituric acid derivatives -

E.g. veronal, luminal, seconal etc.

These are sleep producing agents, i.e. hypnotics.

ii) Non-hypnotic : e.g. equanil and valium.

These are used in controlling depression and hypertension.

Q 3. Distinguish between antisepetics and disinfectants.

Ans.

Antiseptics Disinfectants

1. They can prevent the growth of micro-organism

or kill them.

They kill micro-organisms.

2. They are not harmful to living tissues therefore

they can be applied to the skin.

They are harmful to the living tissues therefore they

can not be applied to the skin.

3.Antiseptic action is milder but more prolonged. Action of disinfectant is immediate but for short

duration.

4. They are always used in low concentration. They are used in high concentration.

5. They are used for dressing of wounds, ulcers and

in the treatment of diseased skin.

They are used for disinfecting floors, toilets, drains

etc.

E.g. Dettol, iodine boric acid etc. E.g. 1% solution of phenol, chlorine etc.

Q 4. What are the types of antibiotics? Give suitable examples.



29

Ans. Antibiotics are of two types :

i) Bactericidal antibiotics : The antibiotics which are used to kill micro-organisms or bacteria are

called bactericidal antibiotics. E.g. Penicillin, Aminoglycosiders, oflaxacin, cephalosporin etc.

ii) Bacteriostatic antiboitics : The antibiotics which are used to prevent thegrowth of micro-

organisms or bacteria are called bacteriostatics antibiotics.

Q 5. Give the structure of saccharin and write its applications.

Ans. The first popular artificial sweetner is sacchain. It is approximately 300 times sweeter than cane

sugar. It is non-biodegradable and does not have calorific value. It is excreted as such through

urine.

Applications :

i) It is used by diabetic patient as a substitute of sugar.

ii) Sodium salt of saccharin is used in the preparation of soft drinks, baked goods, chewing gums,

candy, jams and confectionary etc.

iii) It is palatable, therefore, it is used in supari, pan masala, pan flavouring materials.

halogen derivative of alkane

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