nigrogen containing compounds - einstein …einsteinclasses.com/nitrogen_co_c.pdf · ... nitrogen...

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Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road

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NIGROGEN CONTAINING COMPOUNDS

C1A Structure : of the organic compounds that show appreciable basicity (e.g. those strong enough to

turn litmus blue), by for the most important are the amines. An amine has the general formulaeRNH

2, R

2NH or R

3N where R is an alkyl or aryl group. For e.g.

Nomenclature : Aliphatic amines are named by naming the alkyl group or groups attached tonitrogen and following these by the word-amine e.g.

Salts of amine are generally named by replacing amine by ammonium (or aniline by anilinium), andadding the name of the anion (chloride, nitrate, sulfate etc.) e.g.

C6H

5NH

3+Cl– (C

2H

5NH

3+)

2SO

42–

Anilininum EthylammoniumChloride Sulfate

C1B Physical Properties of amines : Like ammonia, amines are polar compounds and except for tertiaryamines, can form intermolecular hydrogen bonds. Amines have higher boiling points thannon-polar compounds of same molecular weight, but lower boiling points that alcohols or carboxylicacids.

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Amines of all three classes are capable of forming hydrogen bonds with water. As a result smalleramines are quite soluble in water, with borderline solubility is reached with six carbon atoms.

Amines are soluble in less polar solvents like ether, alcohol, benzene etc.

C1C Stereochemistry of Nitrogen : Consider quaternary ammonium salts, compounds in which fouralkyl groups are attached to nitrogen. Here all four sp3 orbitals are used to form bonds and quatenarynitrogen is tetrahedral. Thus quaternary ammonium salts in which nitrogen holds four differentgroups have been found to exist as configurational enantiomers, capable of showing optical activity.

C2 Methods of Preparation :

1. Reduction of Nitro Compounds :

2. Reaction of halides with ammonia or amines :

XRNRNRHNRNHRNH

)4(salts

ammoniumQuaternary

|R

|

RX

Amine3

|R

R|

RX

Amine2

|R

XR

Amine12

RX3

0

R

0

00

RX must be alkyl or aryl with electron withdrawing substituents.

The presence of large excess of ammonia lessens the importance of these last reactions and increasesthe yield of primary amine.

3. Reductive Amination :

C = O + NH3

CNNaBHor

Ni,H

3

2 CH – NH2

10 Amine

C = O + RNH2

CNNaBH

Ni,H

3

2 CH – NHR 20 Amine

C = O + R2NH

CNNaBH

Ni,H

3

2 CH – NR2

30 Amine

4. Reduction of nitriles (Higher carbon number is obtained)

22Catalyst,H2

NHCHRNCR 2

)(1diamineeneHexamethyl

224222Ni,H

leAdiponitri42

NaCN2222

0

2 NHCH)(CHNCHHCN)CH(NCClCHCHCHClCH

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5. Hoffman degradation of amides :

ePentylaminn2423

KOBr

e)(HexanamidCapromide

2423

2322

OBr22

NH)(CHCHCONH)(CHCH

COArNHorNHRArCONHorRCONH

Discussion : From the above reaction it is clear that in this reaction, the rearrangement occurs, sincethe group joined to carbonyl carbon in the amide is found joined to nitrogen in the product.

The reaction is believed to proceed by the following steps :

1.

2.

5. 232

OHCORNHOH2OCNR 2

Steps (3) and (4) are generally takes place simultaneously. The attachment of R to nitrogen helps topushout halide ion.

(3,4)

Step (5) is the hydrolysis of an isocyanate (R – N = C = O) to form amine and carbonate ion.

If the Hoffman degradation is carried in absence of water an isocyanate is actually isolated.

* When the migrating group is aryl the rate of degradation is increased by the presence of electronreleasing substituents in the aromatic ring. Thus substituted benzamide show the following order ofreactivity :

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G : –OCH3 > –CH

3 > –H > –Cl > – NO

2

6. Gabriel Phthalimide Synthesis :

Practice Problems :

1. Boiling of C2H

5NCO + NaOH leads to the formation of

(a) C2H

5COOH + NH

3(b) C

2H

5NH

2 + Na

2CO

3

(c) CH3NH

2 + CH

3COONa (d) None

2. Intermediates of this reaction are except :

(a) R – N = C = O (b)

(c) (d) a, c

3. Which of the following would you predict as incorrect

(a) can be hydrolysed to C2H

5COOH

(b) (CN)2 can be hydrolysed to

(c) can be hydrolysed to C2H

5NH

2 and NO

3—

(d) can be hydrolysed to C2H

5NH

2 and CH

3COOH

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4. Which of the following reactions does not yield an amine

(a) ......OHNCRH

2

(b) ......NHXR 3

(c) ......]H[NOHCHRNa

OHHC 52 (d) ......]H[4RCONH 4LiAlH

2

[Answers : (1) b (2) b (3) c (4) a]

C3 Chemical properties of Amines : The tendency of nitrogen to share the unpaired electronsunderlines the entire chemical behaviour of amines : their basicity, their action as nucleophiles - inboth aliphatic and acyl substitution - and the usually high reactivity of aromatic rings bearing aminoor substituted amino groups.

1. Basicity of Amines : Salt formation :

R – NH2 + H+ R+NH

3

R2NH + H+ R

2NH

2+

R3N + H+ R

3NH+

Example :

Structure and Basicity :

Let us see how basicities are related to the structure.

We shall compare the stabilities of amines with the stabilities of their ions; the more stable the ionrelative to the amine from which it is formed, the more basic the amine.

First of all, amines are more basic than alcohols, ethers, esters etc. for the same reason thatammonia is more basic than water, Nitrogen is less electronegative than oxygen and can betteraccomodate the positive charge of the ion.

An aliphatic amine is more basic than ammonia : because the electron-releasing alkyl groups tend todisperse the positive charge of the substituted ammonium ion;

How can be account for the fact that aromatic amines are weaker bases than ammonia ?

Let us compare the structure of aniline with anilinium ion with the structures of ammonia and theammonium ion.

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Aniline i.e. Aromatic amines are less basic due to the fact that amine is stablized by resonance to thegreater extent than its ion.

From another point of view we can say that its electron pair is partly shared by ring and is lessavailable for sharing with a hydrogen ion.

Effect of substituents on basicity of aromatic amines :

Electron releasing substituents like CH3, increases the basicity of aniline, and electron withdrawing

substituents like –X, –NO2 decreases the basicity.

The electron releasing substituents tends to disperse the positive charge of the anilinium ion, andthus stablizes the ion relative to amine. The electron withdrawing tends to intensify the positivecharge of the anilinium ion, and thus destablizes the ion relative to the amine.

We notice that base strengthening substituents are the ones that activate an aromatic ring towardselectrophilic substitution; the base-weakining substituents are the ones that deactivate an aromaticring towards electrophilic substitution.

2. Alkylation :

XArNRArNRArNHRArNH

XNRNRNHRRNH

3RX

2RXRX

2

4RX

3RX

2RX

2

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Example :

)4(iodideammonium

propylnTrimethyl3373

ICH3

|CH

73ICH

3

|H

73ICH

273

0

3

3

33 I)CH(NHCnCHNHCnCHNHCnNHHCn

3. Hoffmann elimination from quatenary Ammonium Salts :

Example :

This reaction is called Hoffmann elimination, is quite analogous to the dehydrohalogenation of analkyl halide. Most commonly reaction is E

2 :

Hydroxide ion abstracts a proton from carbon; a molecule of tertiary amine is expelled.

4. The Cope Elimination :

Tertiary amine oxide are prepared easily by treating tertiary amines with H2O

2.

5. Reactions of Amines with Nitrous acid :

Nitrous acid is a weak acid & is unstable also. It is prepared by treating sodium nitrite (NaNO2) with

an aqueous solution of a strong acid :

HCl (aq) + NaNO2(aq) HONO(aq) + NaCl(aq)

H2SO

4(aq) + NaNO

2(aq) HONO(aq) + Na

2SO

4(aq)

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Nitrous acid reacts with all kinds of amines. The products that we obtain from these reactionsdepends on whether the amine is primary, secondary or tertiary and whether the amine is aliphaticor aromatic.

Reactions of primary aliphatic amines with nitrous acid :

Primary aliphatic amines react with nitrous acid through diazotisation reaction giving high yield ofunstable diazonium salts.

Even at low temperature they decompose to form nitrogen (N2) and carbocation (R+)

R+ reacts with H2O to form ROH, or alkene or R – X. It means mixture of products produced.

Reactions of Primary Arylamines with Nitrous acid :

Primary arylamine react with HONO acid to give arenediazonium salts. These salts are althoughunstable but are more stable than the diazonium salt of aliphatic primary amine.

X:NNArArNH

C50Temp.

OHHONO,

arylaminePrimary2 0

2

Reaction of Secondary amine with Nitrous acid :

Secondary amines both aliphatic and aromic react with nitrous acid to yeild N-nitroamines usuallyseparate from reaction mixture as oily yellow liquid. Specific Examples :

oil)yellow(aamineethylNitrosodimN

23OH

HONO2

..

23 NON)(CHNaNOHClHN)(CH2

Reaction of Tertiary amine with nitrous acid :

When tertiary aliphatic amine is mixed with nitrous acid an equilibrium is established among thetertiary amine, its salt and an N-nitrosoammonium ion compound

23 NaNOHX:NR

OXNNRHXNR..

3saltAmine

3

Tertiary aryl amine react with nitrous acid to form p-nitroso aromatic compound.

Nitrosation exclusivery takes place at para position.

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Replacement Reactions of Arene Diazonium Salts :

[H3PO

2 is known as hypophosphorous acid]

6. Ring Substitution in Aromatic Amines :

Example :

(a)

(b)

(c)

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Nitric acid not only nitrates but also oxidizes the highly reactive ring as well.

In the strongly acidic condition aniline is converted into anilinium ion (–NH3+) because of its positive

charge it directs substitution to the meta position.

To overcome this difficulty, we protect the amino group : we acetylate the amine, then carry out thesubstitution and finally hydrolyze the amide to the desired substituted amine

For Example :

(a)

(b)

7. Sulfonation of Aromatic Amines :

Sulphanilamide. The Sulfa drug :

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8. Analysis of Amines : Hinsberg Test :

reactionNoClSOHCNR

.rxnnoRNSOHCClSOHCNHR

NHRSOHCKNRSOHCNHRSOHCClSOHCNHR

25633

KOH

solutionKOHinInsoluble

|R

2562562

2

256H

SolutionClear256

KOH256256

12

0

0

0

9. Carbyl amine test : It is given by 10 alkyl amine and aryl amine.

OHKCl3RNCKOH3CHClRNH 232

unpleasent smell of alkyl isocyanide is obtained. 20 amine and 30 amine does not give this test.

Practice Problems :

1. Which compound is obtained at the end of the following reaction,

)C()B()A(Ethylamine 352 NHPClHNO

(a) ethyl cyanide (b) ethyl amine (c) methyl amine (d) acetamide

2. When ethyl amine is heated with chloroform and alcoholic KOH, a bad odour compound is formed.The compound is

(a) a secondary amine (b) an acid

(c) a cyanide (d) an isocyanide

3. CH3NH

2 + CHCl

3 + 3KOH X + Y + 3H

2O; compounds X and Y are

(a) CH3CN + 3KCl (b) CH

3NC + 2KCl

(c) CH3CONH

2 + 3KCl (d) CH

3NC + K

2CO

3

4. In the following series of reaction, A is

OHHC)B()A( 52HNOductionRe 2

(a) CH3CN (b) CH

3NC (c) C

2H

5CN (d) CH

3NO

2

5. Which one of the following is least basic

(a) C2H

5NH

2(b) NH

3(c) (C

2H

5)

2NH (d) C

6H

5NH

2

6. Which of the following is least basic

(a) aniline (b) p-methylaniline

(c) diphenylamine (d) triphenylamine

7. A positive carbylamine test is given by

(a) N, N-dimethyl aniline (b) 2, 4-dimethyl aniline

(c) N-methyl-o-methylaniline (d) p-methyl benzylamine

8. A nitrogenous substance X is treated with HNO2 and the product so formed is further treated with

NaOH solution, which produces blue colouration. X can be

(a) CH3CH

2NH

2(b) CH

3CH

2NO

2

(c) CH3CH

2ONO (d) (CH

3)

2CHNO

2

9.

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X is :

(a) (b) (CH3)

3N + CH

3 – CH = CH

2 + H

2O

(c) (CH3)

3N + CH

3CH

2CH

2OH (d)

10. OHCHNONHC 1042)x(114 (30 alcohol) hence X will give :

(a) carbyl amine reaction

(b) Hofmann mustard oil reaction

(c) diazonium salt (as the intermediate) with HNO2

(d) all are correct

11. In the following compounds,

the order of basicity is

(a) IV > I > III > II (b) III > I > IV > II

(c) II > I > III > IV (d) I > III > II > IV

12. NH2 A

OH

MgBrCH2POCl

3

33

(a) (b)

(c) (d)

[Answers : (1) b (2) d (3) b (4) a (5) d (6) d (7) b (8) d (9) b (10) d (11) d (12) a]

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SINGLE CORRECT CHOICE TYPE

1. The product (D) in the following sequence ofreaction is

)B()A(COOHCHheatNH

33

)D()C(OHHCNaOP 5252

(a) ester (b) amine

(c) acid (d) alcohol

2. Ethyl amine on oxidation with acidified KMnO4

gives

(a) an acid

(b) an alcohol

(c) an aldehyde

(d) a nitro compound

3. The correct increasing order of basic strength in,

CH3CH

2CN, CH

3CH

2NH

2, CH

3N = CHCH

3 is

(a) CH3N = CHCH

3, CH

3CH

2NH

2,

CH3CH

2CN

(b) CH3CH

2NH

2, CH

3N = CHCH

3,

CH3CH

2CN

(c) CH3CH

2CN, CH

3N = CHCH

3,

CH3CH

2NH

2

(d) CH3CH

2CN, CH

3CH

2NH

2, CH

3N =

CHCH3

4. Chlorobenzene can be prepared by reacting anilinewith

(a) HCl

(b) Cu2Cl

2

(c) Chlorine in presence of anhydrous AlCl3

(d) nitrous acid followed by heating withCu

2Cl

2

5. Nitrobenzene on reduction with Zn and NH4Cl

forms

(a) aniline

(b) nitrobenzene

(c) hydrazobenzene

(d) phenyl hydroxylamine

6. In the reaction,

)B()A(NHHCKCN

CuCN

C50

HClNaNO256 0

2

)C(OH/H 2

the product (C) is

(a) C6H

5CH

2NH

2(b) C

6H

5COOH

(c) C6H

5OH (d) none of these

7. Which of the following is least basic

(a) aniline

(b) p-methylaniline

(c) diphenylamine

(d) triphenylamine

8. An organic compound X having molecular formulaC

6H

7O

2N has 6 carbon atoms in a ring system, two

double bonds and also a nitro group as substitu-tion, X is

(a) homocyclic but not aromatic

(b) aromatic but not homocyclic

(c) homocyclic and aromatic

(d) heterocyclic

9. By the action of bromine and alkali on benzamideit gives

(a) benzene (b) aniline

(c) bromo benzene (d) acetanilide

10. Nitroso amines (R2N – N = O) are water

soluble. On heating with conc. HCl, they givesecondary amines. The reaction is called

(a) Perkin reaction

(b) Fries reaction

(c) Liebermann nitroso reaction

(d) Etard reaction

11. Which nitro compound will show tautomerism

(a) C6H

5NO

2(b) (CH

3)

3CNO

2

(c) CH3CH

2NO

2(d) o-nitrotoluene

12. HCONHR 3POCl

Pyridine ? is

(a) RCN (b) RNC

(c) RCNO (d) RNCO

13. R — N C + HgO A + Hg2O; What is AA

(a) RNH2

(b) RCONH2

(c) R — NCO (d) RCOOH

14. Identify X in the series,

42

3

SOH

HNO intermediate

OH2 X

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(a) (b)

(c) (d)

15. The reactions,

RI

2OH2 RNH2

illustrate :

(a) Gabriel’s phthalimide reaction

(b) A good method to prepare pure secondary amine

(c) A reaction which can be extended topreparation of -amino acid

(d) None

16. Boiling of C2H

5NCO + NaOH leads to the forma-

tion of

(a) C2H

5COOH + NH

3

(b) C2H

5NH

2 + Na

2CO

3

(c) CH3NH

2 + CH

3COONa

(d) None

17. Hinsberg reagent is.................and reactswith....................amine to form a product soluble inalkali :

(a) — SO2Cl, 10

(b) —SO2NH

2, 20

(c) HO— —CH3, 30

(d) —SO2Cl, 20

18. OHCHNONHC 1042)x(114 (30 alcohol)

hence X will give :

(a) carbyl amine reaction

(b) Hofmann mustard oil reaction

(c) diazonium salt (as the intermediate) withHNO

2

(d) all are correct

19.

Which is correct alternate ?

(a)

(b)

(c)

(d) none is correct

20.

Intermediates of this reaction are except :

(a) R – N = C = O

(b)

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(c)

(d) a, c

21.

X is :

(a)

(b) (CH3)

3N + CH

3 – CH = CH

2 + H

2O

(c) (CH3)

3N + CH

3CH

2CH

2OH

(d)

22. A compound A when reacted with PCl5 and then

with ammonia gave B, B when treated withbromine and cuastic potash produced C. C ontreatment with NaNO

2 and HCl at 00C and then

boiling produced orthocresol. Compound A is

(a) o-toluic acid

(b) o-chlorotoluene

(c) o-bromotoluene

(d) m-toluic acid

23. Which of the following would you predict asincorrect

(a) can be hydrolysed to

C2H

5COOH

(b) (CN)2 can be hydrolysed to

(c) can be hydrolysed

to C2H

5NH

2 and NO

3—

(d) can be

hydrolysed to C2H

5NH

2 and CH

3COOH

24. An aromatic amine (A) was treated with alcoholicpotash and another compound (Y) when a foulsmelling gas was formed with formula C

6H

5NC. (Y)

was formed by reacting a compound (Z) with Cl2

in presence of slaked lime. Compound (Z) is

(a) C6H

5NH

2(b) CH

3OH

(c) CH3COCH

3(d) CHCl

3

25. Deamination of n-BuNH2 with NaNO

2/HCl gives

two butanols, two butenes and two butyl chlorides.The possible explanation of these products is

(a) Intermediate formed is RN2+ which is

very stable

(b) As intermediate is carbocation

(c) Both (a) and (b)

(d) None

26. What products would you expect to get by theapplication of the Hofmann exhaustive methylationto Me

2CHCH

2CH

2NH

2

(a) Me2C = CH

2

(b) Me2HCCH = CH

2

(c) CH2 = CH

2

(d) All

27. Which of the following reactions does not yield anamine

(a) ......OHNCRH

2

(b) ......NHXR 3

(c)

......]H[NOHCHRNa

OHHC 52

(d) ......]H[4RCONH 4LiAlH2

28. An organic compound (A) on reduction gave acompound (B). Upon treatment with HNO

2, (B)

gave ethyl alcohol and on warming with CHCl3 and

alcoholic KOH, (A) gave offensive smell. Thecompound (A) is

(a) CH3CN (b) C

2H

5CN

(c) CH3NH

2(d) CH

3NC

29. Examine the following two structures for theanilinium ion and choose the correct statementfrom the ones give below :

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(a) II is not an acceptable canonicalstructure because carbonium ions are lessstable than ammonium ions

(b) II is not an accepted canonical structurebecause it is non-aromatic

(c) II is not an acceptable canonicalstructure because the nitrogen has 10valence electrons

(d) II is an acceptable canonical structure

30. What major products would you expect to get bythe application of the Hofmann exhaustivemethylation respectively

Me2CHCH

2NHCH

2CH

2Me; EtNHCH

2CH

2Cl

(A) (B)

(a) Me2C = CH

2, CH

2 = CH

2

(b) MeCH = CH2, CH

2 = CH

2

(c) MeCH = CH2, CH

2 = CHCl

(d) Me2C = CH

2, CH

2 = CHCl

31. In the following compounds,

the order of basicity is

(a) IV > I > III > II (b) III > I > IV > II

(c) II > I > III > IV (d) I > III > II > IV

32. In the reaction p-chloro toluene with KNH2 in

liquid NH3 the major product is

(a) o-toluidine

(b) m-toluidine

(c) p-toluidine

(d) p-chloro aniline

33. A compound X has the molecular formula C7H

7NO.

On treatment with Br2 and KOH, X gives an amine

Y. The latter gives carbylamine test. Y upondiazotisation and coupling with phenol gives an azodye. Thus X is

(a) C6H

5CONH

2(b) C

6H

5NO

2

(c) C6H

5COONH

4(d) None

34. Methyl ethyl propylamine formsnon-super-imposable mirror images but it does notshow optical activity because

(a) Of rapid flipping

(b) Amines are basic in nature

(c) Nitrogen has a lone pair of electrons

(d) Of absence of asymmetric nitrogen

35. Which is maximum basic in nature ?

(a)

(b)

(c)

(d)

36. End product of following sequence of reaction :

— Br CBABaOOH/ONH 233

(a) = O

(b) —OH

(c) = O

(d)

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37. NH2 A

OH

MgBrCH2POCl

3

33

(a)

(b)

(c)

(d)

ANSWERS (SINGLE CORRECTCHOICE TYPE)

10. c

11. c

12. b

13. c

14. b

15. a

16. b

17. a

18. d

19. a

20. b

21. b

22. a

23. c

1. b

2. c

3. c

4. d

5. d

6. b

7. d

8. a

9. b

24. c

25. b

26. b

27. a

28. a

29. c

30. c

31. d

32. b

33. a

34. a

35. b

36. c

37. a

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EXCERCISE BASED ON NEW PATTERN

COMPREHENSION TYPE

Comprehension-1

Mechanism of Hofmann degradation of amide goesas follows.

1. The reagent used to convert I into II is

(a) NaBr + NaOH

(b) Br2 + NaOH

(c) NaBr + NaHCO3

(d) NBS

2. The rate determining step is conversion of

(a) I II (b) II III

(c) III IV (d) IV V

3. The Hofmann degradation of

produces

(a)

(b)

(c)

(d)

MATRIX-MATCH TYPE

Matching-1

Column - A Column - B

(A) 3SOH

3 RCONHNRCOOH 42

(P) Hofmanndegradation

(B) KOH.2

Br.1 2

(Q) Carbylaminetest

(C) C2H

5NH

2 + CHCl

3 + KOH C

2H

5NC

(R) Hofmannrearrangement

(D) CH3CONH

2 + Br

2 + KOH CH

3NH

2

(S) Schmidtreaction

Matching-2

Column - A Column - B

(A) (P) o-Phenetidine

(B) (Q) o-toluidine

(C) (R) aziridine

(D) (S) anthranilic

acid

MULTIPLE CORRECT CHOICE TYPE

1. p-chloro aniline and anilinium hydrochloride canbe distinguished by

(a) Sandmayar’s reaction

(b) NaHCO3

(c) AgNO3

(d) Carbylamine test

2. Which of the following pairs do not containsisomers ?

(a) Nitromethane and Ethylnitrite

(b) Ethaninitrile and Ethyl nitrite

(c) Ethanenitrile and Ethyl isocyanide

(d) n-propylnitrite and 2-Nitro propane

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3. The reduction of which of the following compoundswill not give primary amine ?

(a) Acetaldoxime (b) Acetone

(c) Ethanenitrile (d) Bromoethane

4. Which of the following process (es) will notproduce 20 amine ?

(a) Gabriel’s synthesis

(b) Hofmann’s bromide reaction

(c) reduction of carbylamine

(d) Isopropylnitrite + Sn/HCl

5. Mixture of 10, 20, 30 amines can be seperated by

(a) Hinsberg’s method

(b) Hofmann’s method

(c) distillation

(d) chromatography

Assertion-Reason Type

Each question contains STATEMENT-1 (Assertion)and STATEMENT-2 (Reason). Each question has4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.

(A) Statement-1 is True, Statement-2 is True;Statement-2 is a correct explanationfor Statement-1

(B) Statement-1 is True, Statement-2 is True;Statement-2 is NOT a correctexplanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

1. STATEMENT-1 : An aqueous solution of Me3N is

less basic as compare to same concentration ofMe

4N+OH–.

STATEMENT-2 : Me3N have small value of K

b

2. STATEMENT-1 :

2RNHNRCHR:NRC are in increasing

order of basicity.

STATEMENT-2 : The more s character in thehybrid orbital of the N with the unshared pair ofe–s, the less basic is the molecule.

3. STATEMENT-1 : In strong acid, diazo couplingcan be done easily.

STATEMENT-2 : Strong acid converts ArNH2 to

ArNH3+.

4. STATEMENT-1 : Succinic anhydride can notreact with R

3N.

STATEMENT-2 : Succinic anhydride formR

2NCOCH

2CH

2COO–R

2NH

2+ a salt when react

with R2NH.

5. STATEMENT-1 : When benzene react with(HNO

3 + H

2SO

4) it forms nitrobenzene.

STATEMENT-2 : When benzene react with dil.HNO

3 it form Nitrobenzene.

6. STATEMENT-1 :

are in decreasing order of acidity.

STATEMENT-2 : This is because of the presenceof electron donating –NO

2 group in phenol at

different positions.

7. STATEMENT-1 : RCH2CH

2NH

2 does not undergo

E2 elimination.

STATEMENT-2 : NH2

– is a vary poor leavinggroup.

8. STATEMENT-1 : Most alkylamines are morebasic than ammonia in aqueous solution.

STATEMENT-2 : pKb of CH

3NH

2 is higher than

that of NH3.

9. STATEMENT-1 : Amines are polar compounds.

STATEMENT-2 : There is difference inelectronegativity between nitrogen, carbon andhydrogen.

10. STATEMENT-1 : . In this Imino

nitrogen is more protonated than amino nitrogen.

STATEMENT-2 : This is because of notformation of symmetrical resonance stabilisedcarbon.

(Answers) EXCERCISE BASED ON NEW PATTERN

COMPREHENSION TYPE1. b 2. c 3. aMATRIX-MATCH TYPE1. [A-S ; B-R ; C-Q ; D-P] 2. [A-Q ; B-S ; C-P ; D-R]MULTIPLE CORRECT CHOICE TYPE1. a, b, c 2. a, b 3. b, d 4. a, b, d 5. a, b, c, dASSERTION-REASON TYPE1. A 2. A 3. D 4. B 5. C 6.7. A 8. C 9. A 10. C

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INITIAL STEP EXERCISE

(SUBJECTIVE)

1. Complete the following equation :

(i) heat

323 KOHCHClNHCH

(ii)heat

O)COCH(223

23NHCHCH

2. What happens when :

(i) Acetamide is heated with bromine andpotassium hydroxide.

(ii) Ethylamine is treated with aceticanhydride.

3. How would you bring the following conversions :

Ethyl chloride to n-propylamine (in 2 steps).

4. Explain the following :

Dimethylamine is a stronger base than trimethylamine.

5. Compare the basicities of :

Methylamine, Dimethylamine, Aniline,N-Methylaniline.

6. Write out the structures and names of the isomericamines of formula C

4H

11N.

7. Of a group of isomeric amines, the t-amine has thelowest b.p. Explain.

8. Deamination of n-BuNH2 with NaNO

2/HCl gives

two butanols, two butenes and two butyl chlorides.Give a possible mechanism for the formation ofthese products.

9. What products would you expect to get by theapplication of the Hofmann exhaustive methylationto :

(i) Me2CHCH

2CH

2NH

2;

(ii) Me2CHCH

2NHMe;

(iii) Me2CHCH

2NHCH

2CH

2Me;

(iv) EtNHCH2CH

2Cl ?

10. RCONHMe does not undergo the Hofmann aminepreparation reaction. Offer an explanation.

11. t-BuNH2 and neoPenNH

2 cannot be prepared by

the action of NH3 on the corresponding alkyl

bromide. Why not ? Each can be prepared from acarboxylic acid. How ?

12. Arrange the following compound or anions in eachgroup in order of decreasing basicity :

(a)

(b) Acetanilide, aniline, N-methylaniline

13. Arrange the following in order of increasing basicnature.

(a) diphenyl amine, aniline, cyclohexyl amine

(b) pyridine, pyrolle,aniline

FINAL STEP EXERCISE

(SUBJECTIVE)

1. Complete the following equations :

(i) ??CHRCH 262 ClNHHB2

(ii) ?BrKCNEtNH KOH22

2. An acidic compound (A), C4H

8O

3 loses its optical

activity on strong heating yielding (B), C4H

6O

2

which reacts readily with KMnO4. (B) forms a

derivative (C) with SOCl2, which on reaction with

(CH3)

2NH gives (D). The compound (A) on

oxidation with dilute chromic acid gives an unstablecompound (E) which decarboxylates readily to give(F), C

3H

6O. The compound (F) gives a

hydrocarbon (G) on treatment with amalgameatedZn and HCl. Give structures of (A) to (G) withproper reasoning.

3. A phenolic compound (A), C7H

8O

2 on mild oxida-

tion gives a highly volatile oil (B). (A) forms (C) onreaction with dimethyl sulphate in alkali. Oxidation of(C) with hot KMnO

4 gives (D) which then reacts with

bromine water to give (E) containing about 72% bro-mine. Give structures of (A) to (E) with proper reason-ing.

4. Show with equations how the following compounds areprepared (equations need not to be balanced)

(i) n-Propylamine from ethyl chloride(in two steps)

(ii) Chlorobenzene from aniline (in two steps)

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(iii) benzaldehyde to cyanobenzene (not morethan three steps).

(iv) Toluene to p-nitro benzyl ethyl ether.

5. An organic compound (A), C4H

9Cl on reacting with

aqueous KOH gives (B) and on reaction alc. KOHgives (C) which is also formed on passing thevapours of (B) over heated copper. The compound(C) readily decolourizes bromine water. Ozonoly-sis of (C) gives two compounds (D) and (E). Com-pound (D) reacts with NH

2OH to give (F) and the

compound (E) reacts with NaOH to give an alcohol(G) and sod. salt (H) of an acid. (D) can also beprepared from propyne on treatment with water inpresence of Hg++ and H

2SO

4. Identify (A) to (H) with

proper reasoning.

6. A basic, volatile nitrogen compound gave a foulsmelling gas with CHCl

3 and alcoholic KOH. A

0.295 gm sample of the substance dissolved in aq.HCl and treated with NaNO

2 solution at 00C liber-

ated a colourless, odourless gas whole volume cor-responds to 112 mL at STP. After the evolution ofthe gas was complete, the aqueous solution was dis-tilled to give an organic liquid which did not con-tain nitrogen and which on warming with alkali andiodine gave a yellow precipitate. Identify the origi-nal substance and explain the reactions (Assumethat it contains one N atom per molecule).

7. (a) Identify A, B, C and D

)B(OHC)A(NHC OH/NaOH/I84

HNO94

322

HCHODC Zn/OH/O 23

(b)

,CuHNOBr/KOH DCBA 22

3HgSO,SOH CH—CCH442

Identify A, B, C and D and explain the reactions.

8. Show by writing the appropriate sequence ofequations how you could carry out each of thefollowing transformations :

(a) Cyclohexanol toN-methylcyclohexylamine

(b) to

9. Consider this two-step reaction sequence thatproceeds through intermediate E (which ordinarilyis not isolated) to final product F :

Show the structure of E and F.

10. An organic compound (A), C18

H20

O on ozonolysisgives (B), C

10H

12O and (C), C

8H

8O

2. Compound (B)

gives iodoform reaction and produces an oxime (D),C

10H

13ON on treatment with NH

2OH. Compound

(D) reacts with PCl5 in dry ether to give (E) which

on hydrolysis gives (F) C8H

11N and acetic acid. (F)

on treatment with HNO2 followed by oxidation gives

phthalic acid. Compound (C) on mild oxidationgives (G) which gives effervescence with NaHCO

3.

(G) on treatment with HI producesp-hydroxybenzoic acid and CH

3I.

Give structures of (A) to (G) with properreasoning.

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