ORGANIC STUDY MATERIAL, Study notes of Organic Chemistry

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MAHENDRA KALRA

NAME OF STUDENT

K.V.ITARANA

IMPORTANT REASONING BASED QUESTIONS

UNIT: 10.HALOALKANES & HALOARENES

1. Aryl halides are extremely less reactive towards Nucleophilic Substitution reactions. Answer: Aryl halides

are extremely less reactive towards Nucleophilic Substitution reactions. because C —X bond acquires a partial

double bond character due to resonance. As a result, the bond cleavage in haloarene is difficult than haloalkane

and therefore, they are less reactive towards nucleophilic substitution reaction.(draw the resonating structures of

chlorobenzene)

2. Haloarenes are less reactive than haloalkanes and haloalkenes.
3. C-X bond length in halobenzene is smaller than C-X bond length in CH 3 -X Answ er: because C —X bond in

halobenzene acquires a partial double bond character due to resonance

4. Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as main

product. Answer: Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms

isocyanides as main product.since KCN is predominantly ionic and provides cyanide ions in solution. Although

both carbon and Nitrogen can donate electron pair but carbon donates electron pair instead of Nitrogen to form

more stable C-C bond. How ever, AgCN is mainly covalent in nature and Nitrogen is free to donate electron pair

forming isocyanide as the main product.

5. Allyl chloride is more reactive than n - propyl chloride towards nucleophilic substitution reaction.

Answer: Allyl chloride is more reactive than n - propyl chloride towards nucleophilic substitution

reaction.because allyl carbocation is stabilized by resonance whereas n propyl carbocation is stabilized by +I

effect only.

6. Allyl chloride is hydrolysed more readily than n -propyl chloride.
7. Tert -Butylbromide reacts with aq. NaOH by SN1 mechanism while n-butylbromide reacts by SN

mechanism

8. Ethyl iodide undergoes S 2 undergoes reaction faster than ethyl bromide N . Answer: Since I ion is better -

leaving group than Br- ion, Ethyl iodide reacts faster than ethyl bromide in S N

9. Benzyl chloride undergoes S 1 reaction faster than cyclohexyl methyl chloride. Answer N

: because in case of

benzyl chloride the carbocation formed is stablised by resonance.

10. p - nitro chlorobenzene undergoes nucleophilic substitution faster than chlorobenzene Answer: p- nitro

chlorobenzene undergoes nucleophilic substitution faster than chlorobenzene,due to presence of electron

withdrawing groups such as nitro group in p- nitro chlorobenzene.,it forms more stable carbanion

11. The presence of nitro group ( – NO) at ortho and para postions in haloarenes increases the reactivity of 2

haloarenes towards Nucleophilic substitution reaction. Answer: The presence of nitro group ( – NO 2 ) at ortho

and para postions in haloarenes helps in stabilization of resulting carbanion by – R and – I effects and hence

increases the reactivity of haloarenes towards Nucleophilic substitution reaction.

12. Electrophilic aromatic substitution reactions in haloarenes occur slowly. Answer: Chlorine withdraws

electrons through inductive effect and releases electrons through resonance. The inductive effect is stronger than

resonance and causes net electron withdrawal as a result Electrophilic aromatic substitution reactions in

haloarenes occur slowly

13. Although chlorine is an electron withdrawing group, yet it is ortho - , para - directing in electrophilic

aromatic substitution reactions. Answer: As the weaker resonance (+R) effect of Chlorine which stabilize the

carbocation formed tends to oppose the inductive effect of Chlorine which destabilize the carbocation at ortho

and para postions and makes deactivation less for ortho and para postion.

14. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. Answer: The dipole moment

of chlorobenzene is lower than that of cyclohexyl chloride.because in chlorobenzene electron withdrawing

inductive effect is opposed by electron releasing resonance effect there fore it is relatively less polar. On the other

hand in cyclohexyl chloride there is only electron with drawing is inductive effect of -Cl atom due to which more

polar.

15. Grignard reagents should be prepared under anhydrous conditions. Answer: Grignard reagents should be

prepared under anhydrous conditions.because Grignard reagent reacts with water and get decomposed and form

alkanes RMgX + H R-H + Mg (OH) X. 2

O →→→→→

16. The treatment of alkyl chlorides with aq.KOH leads to the formation of alcohols but in the presence of

alc.KOH alkenes are major products. Answer: The treatment of alkyl chlorides with aq.KOH leads to the

formation of alcohols but in the presence of alc.KOH alkenes are major products.because in presence of

aq.KOH(more polar), nucleophilic substitution reaction takes place ,thus alcohols are formed while in

presence alc.KOH(less polar),elimination reaction takes place.thus alkenes are major products.

16. Alkyl halides, though polar, are immiscible with water. Answer: Alkyl halides, though polar, are immiscible

with water.because alkyl haides are unable to form H-bond with water as well as unable to break the existing H-

bonds among water molecules.

17. The solubility of haloalkanes in water very low.

para positions in the ring as these positions become electron rich due to the resonance effect caused by – OH

group.(also draw the resonating structures of phenol)

33. Phenol is more easily nitrated than benzene. Ans: As the presence of —OH group in phenol increases the

electron density at ortho and para positions in benzene ring by +R effect. The nitration, being an electrophilic

substitution reaction is more facile where the electron density is more.

34. Alcohols are comparatively more soluble in water than the corresponding hydrocarbons Ans: because

it can form intermolecular Hydrogen -bonding with water

35. Low molecular mass alcohols are soluble in water. Ans: because it can form intermolecular Hydrogen -

bonding with water

36. Propanol has higher boiling point than that of the hydrocarbon, butane. Ans : because propanol are held

together by stronger intermolecular hydrogen bonding.whereas the molecules of butane are held together by

weak yan der Waals forces of attraction

37. The boiling point of alcohols and phenols are higher in comparison to hydrocarbons, ethers,

haloalkanes. Ans: because the – OH group in alcohols and phenols is involved in intermolecular hydrogen

bonding

38. Alcohols and ethers of comparable molecular mass have different boiling points. Ans: because alcohols

can form intermolecular Hydrogen -bonding

39. The boiling point of alcohols and phenols increase with increase in number of carbon atoms. Ans:

because of increase in van der Waals forces.

40. In alcohols the boiling point decreases with increase in branching. Ans: because of decrease in van der

Waals forces with decrease in surface area

41. The acid strength of alcohols decreases in the following order:

o

>2 >

o o

. Ans: Because more is number of

electron releasing groups lesser is the acidic strength

42. The presence of electron withdrawing groups such as nitro group enhances the acidic strength of

phenol. Ans: Because withdrawing groups, such as nitro group, i ngeneral, favour the formation of phenoxide

ion resulting in increase in acid strengthThis effect is more pronounced when such a group is present at ortho

and para positions.

43. The presence of electron releasing groups such as alkyl group decreases the acidic strength of phenol

Ans: Because electron releasing groups, such as alkyl groups, i ngeneral, do not favour the formation of

phenoxide ion resultin g in decrease in acid strength. Cresols, for example, are less acidi cthan phenol

44. Ortho – nitrophenol is more acidic than Ortho – methoxyphenol. Ans: Ortho-nitrophenol is more acidic

because nitro group is electron withdrawing and will increase +Ve Charge on Oxygen to make it more acidic

whereas – OCH group is electron releasing and it will not break easily 3

45. O-nitrophenol is more acidic than o-cresol. Ans: because nitro group is electron withdrawing group

whereas – OHgroup is electron releasing

46. p -nitrophenol is more acidic than phenol. Ans: because nitro group is electron withdrawing group.
47. 2-chloroethanol is more acidic than ethanol. Ans :Due to – I effect of chlorine atom.
48. Phenol is more acidic than alcohols(Ethanol) Ans: Because Phenoxide ion is more stablised than alkoxide

since in phenoxide ion, the charge is delocalised.,The delocalisation of negative charge makes phenoxide ion

more stable and favours the ionisation of phenol.( also draw the resonating structures of phenoxide ion)

49. Phenol does not undergo protonation easily. Ans: Because in phenols, the lone pairs of electrons on the

oxygen atom are delocalized over the benzene ring due to resonance and hence are not easily available for

protonation.

50. Phenol has small dipole moment than methanol. Ans: Due to electron-withdrawing effect of the benzene

ring, the C ―O bond in phenol is less polar but in case of methanol due to electron-donating effect of CH 3

group, C ―O bond is more polar.

51. The C —O—H b ond angle in alcohols slightly less than the tetrahedral angle whereas the C —O—C bond

angle in ether is slightly greater.

52. Out of ethanol and propanol, ethanol gives iodoform test whereas propanol does not do so. Ans:

because ethanol has CH CH(OH)- linkage while propanol has not. 3

53. Thionyl chloride preferred over PCl in nucleophilic substitution reaction to convert alcohol to alkyl 5

halide. Ans : because all by – product are gases and escape easily and pure alkyl halide is Obtained CH 3 CH 2 OH

• SOCl 2

→ CH

3

CH

2 2

Cl + SO (g) + HCl (g)

54. Among HI, HBr, HCl, HI is most reactive towards alcohol. Ans :.Due to low bond dissociation energy of HI
55. In aryl alkyl ethers (i) the alkoxy group activates thebenzene ring towards electrophilic substitution

and (ii) it directs the incoming substituents to ortho and para positions in benzene ring... Ans: In aryl

alkyl ethers (i) the alkoxy group activates thebenzene ring towards electrophilic substitution and (ii) it directs

the incoming substituents to ortho and para positions in benzene ring. because the – OR(alkoxy group) group

activates the benzene ring towards elecrophilic substitution and directs the substituents to Ortho and para

positions in benzene ring. Also, it directs the incoming group to ortho and para positions in the ring as these

positions become electron rich due to the resonance effect caused by – OR group.(also draw the resonating

structures of aryl alkyl ethers).

56. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method.

Ans: because dehydration of secondary and tertiary alcohol will take place and alkene will be product.

57. O=C=O is nonpolar while R —O—R is polar. Ans :Due to lone pair of electrons on Oxygen atom.

Unit:12.ALDEHYDES,KETONES & CARBOXYLIC ACIDS

58. Aldehydes and Ketones have lower boiling point than alcohols.

• It is because alcohol molecules are associated with intermolecular H-bonding whereas aldehydes and ketones

are not.

59. Aldehydes are more reactive than Ketones towards Nucleophilic addition reaction

• Because aldehydes are more polar than ketons due to lesser number of electron releasing alkyl groups which

reduce +Ve charge in aldehyde than ketones.

60. Cyclohexanone forms cyanohydrin in good yield but 2,2,6 – tri methylcyclohexanone does not.

• because there is Steric hinderance of three methyl groups in 2,2,6 – tri methylcyclo hexanone.

61. There are two – NHgroup in semi carbazide however only one is involved in the formation of semi 2

carbazones.

• Since the NH group attached to carbonyl group is stabilized by resonance and has double bond character. 2

62. During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid

catalyst, the water or the ester should be removed as fast as it is formed.

• .It is done so that ester formed does not get hydrolysed.

63. Melting Point of an acid with even number of carbon atoms higher than those of its neighbours with

odd number of carbon atoms

• .It is because acids having even number of carbon atoms can fit into crystal lattice more easily than odd ones,

therefore have higher melting point.

64. It is necessary to control the pH during the reaction of aldehydes and ketones with ammonia

derivatives.

• The addition of ammonia derivative to aldehyde and ketones is done in weakly acidic medium in case the

medium is strongly acidic, then the ammonia derivative will be also protonated and will not be able to act as a

nucleophile.

65. Formaldehyde does not take part in Aldol condensation.

• Due to absence of hydrogen atom.

66. Benzaldehyde gives a positive test with Tollens reagent but not with Fehling and Benedict’s solution.

• Benzaldehyde gives a positive test with Tollens reagent but not with Fehling and Benedict’s solution. Because

The electron donating resonance effect of the benzene ring increases the electron density in the carbonyl

group of benzaldehyde. Hence only stronger oxidizing agent can oxidise C-H to COOH to form carboxylic acid

but weaker reagent (like fehling, Benedict) fail to oxidise.

67. Carboxylic acids do not give the characteristic reactions of carbonyl group.

• Carbonyl carbon of carboxyl group is less electrophilic than carbon of carbonyl group in aldehyde.

68. Chloroacetic acid is stronger than acetic acid.

• because Cl is electron withdrawing, therefore chloroactate ion is more stable than acetate ion that is why

chloroacetic acid is stronger than acetic acid.

69. The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of comparable

molecular masses.

• .It is due toweak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions.

70. The lower members of aldehydes and ketones such as methanal,ethanal and propanone are miscible

with water in all proportions.

• because they form hydrogen bond with water

71. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition

reactions.with reasons (i) Ethanal, Propanal, Propanone, Butanone.(ii) Benzaldehyde, p -Tolualdehyde,

p -Nitrobenzaldehyde, Acetophenone.(iii) Acetaldehyde, Acetone, Di- tert -butyl keto ne, Methyl tert -

butyl ketone

• (i) Butanone <Propanone <Propanal <Ethanal.

(ii), Acetophenone< p -Tolualdehyde < Benzaldehyde,. p -Nitrobenzaldehyde

(iii) Di- tert -butyl ketone ,<Methyl tert -butyl ketone< Acetone< Acetaldehyde

72. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions

than propanal.

94. Gabriel phthalimide synthesis is preferred for synthesising primary amines. Ans because Gabriel

phthalimide synthesis gives pure primary amines without any contamination of secondary and tertiary

amines therefore it is preferred for synthesising primary amines.

95. Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis. Ans because in

preparation by Gabriel phthalimide synthesis,Ar-X is needed and aryl halides do not undergo nuleophilic

substitution esily due to presence of partial double character.

96. Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline

on nitration gives a substantial amount of m -nitroaniline. Ans because Nitration is usually carried out

with a mixture of conc HNO 3 and conc H 2 SO 4. In presence of these acids aniline gets protonated to form the

anilinium ion which is meta directing.

97. Direct nitration of aniline is not carried out. Ans Because by doing this it gets oxidized to protonated anilne

which gives 47% m-nitroaniline.

98. Aniline does not undergo Friedel-Crafts reaction. because aniline being a lewis base reacts with lewis acid

AlCl 3

to form a salt

99. Amines are less acidic than alcohols of comparable molecular masses. Ans because oxygen is more

electronegative than nitrogen, therefore O-H bond breaks early than N – H bond.

100. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. because

Methylamine being more basic than water ,accepts a proton from from water liberate OH ions ,these OH ions

combine with Fe to form a browm ppt of hydrated ferric oxide.

3+

101. Aniline is soluble in aqueous HCl. Ans because Aniline forms the salt anilinium chloride which is

water soluble.

102. MeNH 2 is stronger base than MeOH. Ans because Nitrogen is less electronegative than oxygen

therefore lone pair of electrons on nitrogen is readily available for donation. Hence, MeNH2 is more basic than

MeOH.

103. NH 2 group of aniline acetylated is before carrying out nitration. Ans Because the acetyl group

reduces the reactivity of of the ring thus its oxidation does not occur easily with HNO 3

104. Acetylation of —NH 2

group of aniline reduce its activating effect Ans Because with acetylation of

aniline result in decrease of electron density on Nitrogen.

105. Arrange the following in increasing order of:

a. C 2 H 5 NH2, C 2 H 5 OH, (CH 3 ) 3 N (boiling point)

b. C 2

H ) H

5

OH, (CH

3 2

NH, C

2 5

NH

2

(boiling point)

c. C 6

H

5

NH

2 2 5 2 2 5 2

, (C H ) NH, C H NH. (solubility in water)

d. CH 3 NH 2 (CH 3 ) 2 NH (CH 3 ) 3 N ( basic strength in aqueous phase )

e. (C (C 2

H ) H

5 3

N, C

2 5

NH

2, 2

H )

5 2

NH ( basic strength in aqueous phase )

f. C 2

H H NHCH

5

NH

2

, C

6 5 3

, (C

2

H ) H

5 2

NH and C 6 5

NH

2

(p Kb values)

g. Aniline, p -nitroaniline and p -toluidine( basic strength )

h. C 6 H 5 NH 2 , C 2 H 5 NH2, (C 2 H 5 ) 2 NH, NH 3 ( basic strength in aqueous phase )

i. C 2 H 5 NH2, (C 2 H 5 ) 2 NH, (C 2 H 5 ) 3 N,NH 3 ( basic strength )

j. C 2

H

5

NH

2,

C

6 5 3 2 5 6 5

H NHCH , (C H )

2

NH, C H NH

2

( basic strength )

k. C 6

H H N(CH )

5

NH

2,

C

6 5 3 2

, (C

2

H ) NH,CH (

5 2 3

NH

2

basic strength )

MECHANISM BASED QUESTIONS

UNIT: 10 HALOALKANES AND HALOARENES.

Explain the S N1 & S N2 mechanism with suitable examples.

Answer:

1. S

N

1 mechanism

• Substitution Nucleophilic Unimoecular (S N)

1

• In this reaction, the rate of reaction depends only on the concentration of alkyl halide ie rate=K [ R X ]
• It is mainly shown by tertiary alkyl halids eg. Tertiary butyl halide.
• Example : The reaction between tert -butyl bromide and hydroxide ion yields tert -butyl alcohol and

follows the first order kinetics

• It is Two step reactions.
• Step I : In the first step slow dissociation of alkyl halide takes place by reversible reaction forming a

carbocation.

• Step II : The carbocation at once combines with the nucleophile to form final product.
• Order of reactivity : 3

0

2 > 1 alkyl halide

0 0

• Greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will

be the rate of reaction. In case of alkyl halides, 3 alkyl halides undergo S 1 reaction very fast because of

0

N

the high stability of 3 carbocations.

0

• For a given alkyl group, the reactivity of the halide, R-X, follows the order R–I> R– Br>R– Cl>>R– F .( As

iodine is a better leaving group because of its large size, it will be released at a faster rate in the presence

of incoming nucleophile.)

2. SN2 mechanism - Substitution Nucleophilic bimolecular (S N

• In this reaction, the rate of reaction depend on the concentration of both the alkyl halide and Nucleophile ie

rate = [ R X ] [ Nu ]

• It is mainly given by primary alkyl halides e.g. n-alkyl halide
• Example: The reaction between CH Cl and hydroxide ion to yield methanol and chloride ion 3
• CH 3 Cl + OH

CH 3 OH + Cl

• It is One step reaction
• It results in complete inversion of configuration
• Order of reactivity of alkyl halide : Primary halide1 > Secondary halide2 > Tertiary halide 3

0 0 0

• Of the simple alkyl halides, methyl halides react most rapidly in SN2 reactions because there are only three

small hydrogen atoms. Tertiary halides are the least reactive because bulky groups hinder the approaching

nucleophiles.

In the following pairs of halogen compounds, which would undergo S N

2 reaction faster?

Answer: ( 1 one is faster It is primary halide and therefore undergoes S 2 reaction faster.

st

N )

(1 one is faster As iodine is a better leaving group because of its large size, it will be released at a faster rate in

st

th presence of incoming nucleophile)

Mechanism

The mechanism of the reaction involves the following three steps:

Step 1 : Protonation of alkene to form carbocation by electrophilic attack of H 3 O.

Step 2: Nucleophilic attack of water on carbocation.

Step 3: Deprotonation to form an alcohol.

2. Mechanism for the acidic Dehydration of alcohols to give alkenes

Mechanism

Step 1 : Formation of protonated alcohol.

Step 2: Formation of carbocation: It is the slowest step and hence, the rate determining step of the reaction.

Step 3: Formation of ethene by elimination of a proton.

The acid used in step 1 is released in step 3. To drive the equilibrium to the right, ethene is removed as it is formed.

3. Mechanism for the acidic Dehydration of alcohols to give ethers

Mechanism

4. Mechanism for the addition Grignard reagent on carbonyl compounds.

The first step of the reaction is the nucleophilic addition of Grignard reagent to the carbonyl group to form an adduct.

Hydrolysis of the adduct yields an alcohol.

5. Mechanism for the reaction of HI on Ethoxymethane

When HI is in excess and the reaction is carried out at high temperature, ethanol reacts with another molecule of

HI and is converted to ethyl iodide.

1. Give mechanism for the Hydration of alkenes
2. Give mechanism for the acidic Dehydration of alcohols to give alkenes.. OrWrite the mechanism of acid

dehydration of ethanol to yield ethene.

3. Give mechanism for the acidic Dehydration of alcohols to give ethers or Give mechanism of preparation of

ethoxyethane from ethanol.

4. Give mechanism for the addition Grignard reagent on carbonyl compounds.
5. Give mechanism for the reaction of HI on methoxymethane.
6. Describe the mechanism of hydration of propene to yield propan-2-ol.

Write the mechanism of the following reaction :

7. Write the mechanism of the following reaction:

CH CH

3 2

OH + HBr→ CH Br + H 3

CH

2 2

O

UNIT: 12 Aldehydes,Ketones & Carboxylic acids

**1. Give mechanism for Nucleophilic addition reaction in carbonyl compounds.

2. Write the mechanism of reaction between actetaldehyde and HCN.
3. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition**

reactions.with reasons

• (i) Ethanal, Propanal, Propanone, Butanone.
• (ii) Benzaldehyde, p -Tolualdehyde, p - Nitrobenzaldehyde, Acetophenone.
• (iii) Acetaldehyde, Acetone, Di- tert -butyl ketone, Methyl tert -butyl ketone

Answers:1 & 2

• Mechanism - Nucleophile attacks electrophilic sp hybridised carbon. the hybridization changes to sp and a

2 3

tetrahedral alkoxide intermediate is formed which reacts with proton to form addition product.

• (i),Butanone. Propanone Propanal, Ethanal
• (ii) Acetophenone , p -Tolual dehyde ,Benzaldehyde, p -Nitrobenzaldehyde ,.
• (iii)Di- tert -butyl ketone, Methyl tert -butyl ketone, Acetone, Acetaldehyde

• •••• Fehling’s reagent: Fehling solution A is aqueous copper sulphate and Fehling solution B is alkaline sodium

potassium tartarate (Rochelle salt)

• ••• Test: On heating an aldehyde with Fehling’s reagent,(equal amount of Fehling solution A & Fehling solution B )a

reddish brown precipitate of [Cu O] is obtained. 2

• ••• Reaction :R CHO + 2 Cu + 5 OH¯ RCOO¯ + Cu O + 3 H

2+

→ 2 2 O

Aldehyde Red ppt

O

R – C – R + Fehling solution →No red ppt.

Important Note : Ketones do not give this test & also Aromatic aldehyde do not reduce fehling solution

6. Iodoform Test (Aldehydes & Ketones)

• •••• [Aldehydes & Ketones containing linkage.e.g.Ethanal(Acetaldehyde) Propanone(Acetone) etc.]
• •••• Reagent: I / NaOH 2
• •••• Test: Aldehydes & Ketones containing – COCHlinkage on reaction I / NaOH gives of 3 2

Yellow Ppt CHI 3

• •••• Reaction

O

R – C – CH

3

• I + NaOH R COONa + CHI + NaI + H O 2

3 2

Yellow Ppt

7. Sodium bicarbonate test

• •••• Aliphatic & Aromatic Carboxylic acids give this test.
• •••• Reagent: NaHCO 3 Sodium Hydrogencarbonate
• •••• Test: Carboxylic acids on reaction with NaHCO (Sodium Hydrogencarbonate ) gives 3

effervescence due to

evolution of CO gas 2

• •••• Reaction: R COOH + NaH CO R COONa + H O + CO bubbled 3

2 2 (g)

8. Test for Methanoic acid (Formic acid) - Methanoic acid (Formic acid) give Tollen’s test & Fehling test. - Reaction: H COOH + Tollen’s reagent → 2 Ag + CO 2 + H 2 O - ( Silver Mirror)

H COOH + Fehling sol → Cu 2 O + CO 2 + H 2 O

(Red ppt)

9. Isocyanide test (Carbylamine reaction) - Primary Aliphatic (e.g.Ethanamine) & Aromatic amines(e.g.Aniline) give this test. - Reagent: Chloroform(CHCl ) + Alcoholic KOH 3 - Test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium

hydroxide form isocyanides or carbylamines which are foul smelling substances. (Unpleasent odur)

• Reaction:

RNH

2

• CHCl + 3 KOH(alc) Warm R NC + 3 KCl + 3H O 3 2

Alkyl isocyanide)( (offensive smel

• Important note: Secondary and tertiary amines do not give this test. 10. Heinsberg test

• ••• To distinguish between Primary (1 ), Secondary (2 ), & Tertiary (

0 0 0

) Amines.

• •••• Reagent: Benzenesulphonyl chloride C 6 H 5 SO 2 Cl Heinsberg reagent

Test:(i) Primary amine reacts with benzenesulphonyl chloride (Heinsberg reagent) to give N-ulphonyl amide

which is soluble in alkali because the hydrogen attached to nitrogen in sulphonamide is strongly acidic

(ii) Secondary amine reacts with benzenesulphonyl chloride (Heinsberg reagent) to give N,N-

diethylbenzenesulphonamide which is insoluble in alkali Since N, N-diethylbenzene sulphonamide does not

contain any hydrogen atom attached to nitrogen atom.

(iii)Tertiary amines do not react with benzenesulphonyl chloride.

Important note: Now day’s benzenesulphonyl chloride is replaced by p-toluenesulphonyl chloride****.

11. Azo dye test - Aniline & its derivative give this test. - Reagent: (NaNO 2 + HCl )[Nitrous acid] followed by

 -napthol

• Test: Aniline on reaction with NaNO + HCl at 273-278 K gives BDC which forms a brilliant orange dye with 2

 -napthol in sodium hydroxide.

• Reaction:

C + NaNO + HCl 6

H

5

- NH

C Cl¯ 6

H -N

5

2 ⎯⎯ ⎯ ⎯→

orange Azo dye

Summary of chemical Test

S.No Test Reagent Inference

1. Lucas test :To distinguish

between Primary ( 1 ),

0

Secondary (2 ), & Tertiary

0

0 ) Alcohols)

ZnCl 2

/HCl (

0 ) Alcohols gives Turbidity

(immediately), 2 Turbidity after some

0

time (5-10 min) 1 does not give

0

Turbidity at room temperture

2. Iodoform test ( Alcohols

containing CH 3

-CH(OH)-

linkage )

I 2 / NaOH Yellow Ppt of CHI 3 .is formed

3. Neutral ferric chloride

test (Phenol)

Neutral FeCl 3

Violet colouration

4. Tollens test [Aliphatic

Aldehydes(e.g.Ethanal,Propa

nal etc) & Aromatic

Aldehydes (Benzaldehyde

etc.)]

Ammoniacal.

AgNO 3

Bright silver mirror [ Ag] is produced due

to the formation of silver metal.

5. Fehling’s test [Only

Aliphatic Aldehydes]

Fehling solution A

(aqueous copper

sulphate & Fehling

solution B alkaline

sodium potassium

tartarate(Rochelle

salt)

Reddish brown precipitate of [Cu O] is 2

obtained.

6. Iodoform test ( Aldehydes &

Ketones containing – COCH 3

linkage )

I

2

/ NaOH Yellow Ppt of CHI .is form 3

ed

7. Sodium bicarbonate test

(Aliphatic & Aromatic

Carboxylic acids)

NaHCO 3

Sodium

Hydrogencarbonate

Effervescence due to evolution of CO gas. 2

8. Isocyanide test Primary

Aliphatic & Aromatic

amines.

Chloroform(CHCl 3 ) +

Alcoholic KOH

Unpleasent odur (foul smelling) of

isocyanides or carbylamines.

Secondary and tertiary amines

0 ,

0 ,& 3 Amines

0

Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical

equations of the reactions involved

NAME REACTIONS BASED QUESTIONS

UNIT: 10 HALOALKANES AND HALOARENES

Wurtz reaction : Alkyl halides react with sodium in dry ether to give hydrocarbons

(Alkanes) containing double the number of carbon atoms present in the halide

2R – X +2Na ⎯⎯ ⎯ ⎯→ R – R + 2NaX

2CH

3

- Br +2Na ⎯⎯ ⎯ ⎯→ CH 3

– CH

3

+ 2Na Br

2. Fittig reaction: Aryl halides(Haloarenes) when treated with sodium in dry ether gives in which two aryl

groups are joined together. 2Ar – X +2Na ⎯⎯ ⎯ ⎯→

Ar – Ar + 2NaX

Wurtz-Fittig reaction : A mixture of an alkyl halid e(Haloalk anes) and aryl halide(Haloarenes.) gives an

alkylarene when treated with sodium in dry ether.

Ar – X + Na + R – X ⎯⎯ ⎯ ⎯→ Ar – R + 2NaX

Finkelstein reaction: Alkyl chlorides/ bromides on reaction with NaI in dry acetone to give Alkyl iodides.

R – Cl +NaI ⎯⎯ ⎯ → R – I + NaCl

R – Br +NaI ⎯⎯ ⎯ → R – I + NaBr

Swarts reaction Heating of alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg 2 F 2 , CoF 2

or SbF give alkyl fluorides 3

to R – X ⎯⎯ ⎯⎯⎯⎯⎯→

R – F example :

CH 3 – Br + AgF ⎯⎯ → CH 3 – F + AgBr

Markovnikov’s rule : During addition of a polar molecule to unsymmetrical alkene, the negative part of

addendum attaches to that double bonded carbon which has least no. of hydrogen atom.

CH 3 CH = CH 2 + HBr ⎯⎯ → CH 3 CHBrCH 3 (2-Bromo Propane)

Peroxide effect (kharasch effect) when a polar molecule is added to unsymmetrical alkene in the presence of

organic peroxide then addition occurs against the Markovnikov’s rule (only for addition of H Br)

CH

3

CH = CH + HBr 2 ⎯⎯ ⎯⎯→

CH CH

3

2

- CH

2

-Br (1-Bromo Propane)

Grignard Reagent: Alkyl or aryl magnesium halide is called grignard reagent. It is obtained by treating alkyl or

aryl bromide with magnesium in the presence of ether.

R – X + Mg ⎯⎯ ⎯⎯→ RMg X

CH CH

3 2

Br + Mg ⎯⎯ ⎯⎯→

CH

3

CH

2

Mg Br

Saytzeff rule : If in an elimination reaction there is availability of more than one  -hydrogen atom then that

alkene is the major product which is highly alkylated ie containing greater number of alkyl groups attached to

the doubly bonded carbon atoms.

UNIT: 11 ALCOHOLS,PHENOLS & ETHERS

Hydroboration – o xidation reaction : The alcohol obtained corresponds to anti- Markownikov ’s addition of water

on alkenes. CH 3

-CH=CH

−

CH

3

- CH

2

- CH

2

- OH

Reimer-Tiemann reaction

Cannizzaro Reaction: Aldehydes, which do not have an α-hydrogen atom, undergo self oxidation and reduction

reaction on treatment with concentrated alkali.In this reaction one molecule of the aldehyde is reduced to

alcohol and another is oxidized to carboxylic acid salt.

Clemmensen Reaction: The carbonyl group of aldehydes and Ketones are reduced to CH group on treatment 2

with zinc amalgam(Zn/Hg) and concentrated hydrochloric acid.(Conc.HCl)

Example: CH 3 CH =O ⎯⎯ ⎯ ⎯→

−

CH 3 CH 3

CH 3 CO CH 3 ⎯⎯ ⎯ ⎯→

−

CH 3 CH 2 CH 3

Wolf Kishner Reaction

Example: CH CH =O 3 ⎯⎯ ⎯⎯⎯⎯⎯→

CH

3

CH

3

CH 3 CO CH 3 ⎯⎯ ⎯ ⎯⎯⎯⎯→ CH 3 CH 2 CH 3

Rosenmund Reaction: Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on barium

sulphate. RCOCl + H 2

⎯⎯ → RCHO + HCl

Etard Reaction

Stephen Reaction

Gatterman-Koch reaction:

Ozonolysis ofalkenes involves the addition of ozone molecule to alkene to form ozonide, and then cleavage of

the ozonide by Zn-H 2 O to Aldehyde and/or Ketones.

Esterification: Esters are generally prepared by heating carboxylic acids with alcohols in the presence of a

mineral acid such as conc. H 2 SO 4 or HCl gas.

Decarboxylation : Sodium salts of acids when heated with soda lime, alkanes are formed.

Hell Volhard Zelinsky: Carboxylic acids having an α-hydrogen are ha logenated at the α-position on treatment

with chlorine or bromine in the presence of small amount of red phosphorus to give α-halocarboxylic acids

Example : CH 3 COOH + Cl 2 ⎯⎯ ⎯ → CH 2 (Cl)COOH + Cl 2

Acetylation: The introduction of acetyl (CH CO) group in alcohols or phenols is known as acetylation. 3

UNIT: 13 AMINES