Chapter 13 Organic Compounds Containing Nitrogen (Amines)

• In case of Triethylamine, the number of H’s of ammonia replaced by tert-butyl group (an alkyl group) is 3. Hence it is an example of a 3 ^◦ amine (R₃N).

• Other 3 amines, ,

1-methylcyclohexylamine tert-butylamine

N-methylaniline

are secondary (2 ^◦ R₂NH), primary (1^◦ RNH₂) and again secondary (2 ^◦ R₂NH) amines respectively.

3 2 1

CH₂ = C H C H₂NHCH₃

• In these compound main carbon chain has total of 3 carbon atoms (prop), and there is one double bond in the 2 positions (one) and 1 N-methyl amine in the 1 position. Therefore IUPAC name is N-methylprop-2-en-1-amine.

• Option (i) Allylmethylamine is actually the common name for the given compound and not IUPAC name.

• Options (ii) 2-amino-4-pentene(5 carbons in the main chain).

• and(iii) 4-aminopent-1-ene are incorrect names for the given compound.

Basicity of amines depends on its structure and the classification (electron releasing/withdrawing) other groups attached to the adjacent carbon atoms which determine ease of formation of the cation by accepting protons from the acid.

(iii) (CH₃)₂ NH is the strongest base due to the efficient +I effect(electron releasing effect ) of alkyl(here –CH₃). This effect of alkyl group pushes the unshared electron pair(lone pair) more towards the N atoms and thus make them more available for protonation by acid.

• For the same reason (i) CH₃NH₂ will be less basic than (iii) because it has less of the effect.

• In case of (ii) NCCH₂NH_2, CN(cyanide C≡N^:) is an electron-withdrawing group i.e. it pulls electrons towards itself from the N atom hence making the lone pairs less available for protonation, therefore it is not a strong base.

• In comparison between alkyl and aryl amines, the former is always more basic than the latter because in case of aryl amines:

the -NH2 (amine) group is attached directly to the benzene(aryl) ring. hence the unshared electron pair on nitrogen atom undergoes conjugation (resonance) with the benzene ring and thus it becomes almost unavailable for protonation.

thus (iv) C₆H₅NHCH₃ might be theweakestamongst the given amines.

• Here the -NH2 (amine) group is attached directly to the benzene ring. Hence, the unshared electron pair on the nitrogen atom(i.e. lone pair) will be engaged in conjugation (forming resonating structures) with the benzene ring and thus it becomes almost unavailable for protonation by the acids.

• Other 3 options (ii) (iii) and (iv) CH₃NH₂ the above scenario is not possible and therefore they are better and stronger Brönsted bases ; (iv) being the strongest.

• C₆H₅CH₂Br is the is best-suited alkyl halides for this reaction through S_N1 mechanism because it has a bulky benzene ring which will be a good leaving group in the last step of alkylation and reinforces the formation of C₆H₅CH₂NHR.

• For example,R = CH₃

+ CH₃-X →

• With other 3 (i) CH₃Br(ii) C₆H₅Br(iv) C₂H₅ Br reaction will not be as effective as (iii)

• When nitrobenzene reacts with LiAlH₄ in ether medium, it does not transform itself to amine, instead another product is formed which is azobenzene . So this reagent will not be a good choice for converting of nitrobenzene to amine.

Other 3 given reagents (i) H₂ (excess)/Pt (iii) Fe and HCl and (iv) Sn and HCl are actually efficient choice for the conversion.

• When alkyl halide is heated under reflux with a solution of potassium cyanide in ethanolic medium, the halogen is replaced by a -CN group and a nitrile is produced.; here the N required comes from KCN.

• This is a very important reaction for obtaining some compounds like amine. The nitrile formed can be converted to a primary amine by treating it with LiAlH₄ or Na/C₂H₅OH reagents.

CH₃CH₂CH₂Br + CN^-→ CH₃CH₂CH₂CN + Br^-

CH₃CH₂CH₂Br + KCN → CH₃CH₂CH₂CN + KBr

Hence the source of N is KCN and other 3 options (i) Sodium amide, NaNH₂

(ii) Sodium azide, NaN₃ (iv) Potassium phthalimide, C₆H₄(CO)₂N^- K⁺ are entirely wrong

• Gabriel phthalimide synthesis is a chemical reaction that transforms primary alkyl halides into primary amines.

• generally, potassium pthalamide is the reactant which reacts with ethanolic KOH at first and potassium salt of phthalimide forms, on heating with a certain alkyl halide (production of N-aikyl pthalimide) followed by hydrolysis in alkaline medium produces corresponding primary amine.

• Hence the source of nitrogen is C₆H₄(CO)₂N^–K⁺

• When condensation occurs between a primary amine and aldehyde or a ketone a new imine compound ( Schiff base)n is formed.

• The reduction of this N-substituted imine on reduction with H₂/Pt or any other strong reducing agents gives secondary amines.

Among the other 3 options (i) 2° R—Br + NH₃ (ii) 2° R—Br + NaCN followed by H₂/Pt and (iv) 1° R—Br (2 mol) + potassium phthalimide followed by H₃O+/heat do not necessarily form a secondary amine.

• When 2–phenyl propanamide reacts with LiAlH₄ in ether it undergoes reduction and produces correspondent amine i.e. 2-phenyl propanamine , without changing the carbon number and remaining the main carbon chain intact as the parent compound

• Amongst the other 3 options

(i)excess H₂ (ii) Br₂ in aqueous NaOH and (iii) iodine in the presence of red phosphorus none of them brings the change of amide to amine with the carbon numbers remaining the same as the reactant compound 2-phenyl propanamine.

• (i)excess H₂ would not be able to bring the change because a catalyst would be required to activate the compound.

• (ii) Br₂ in aqueous NaOH is actually the Hoffman bromamide degradation reaction, where migration of the alkyl or aryl group takes place from carbonyl carbon to the N atom, thus by producing amine 1 carbon less than the reactant amide compound.

• And at last (iii) iodine in the presence of red phosphorus do not bring the desired change of amides.

• This is actually a Hoffman bromamide degradation reaction.

• This reaction occurs by treating an amide with bromine in the presence of the aqueous or ethanolic solution of sodium hydroxide.

• In this reaction, migration of an alkyl or aryl group happens from the carbonyl carbon of the amide to the nitrogen atom of the same amide.

• Hence the amine formed contains one carbon less than that of the reactant amide.

• With other 3 reagents (i) excess H₂/Pt (iii) NaBH₄/methanol (iv) LiAlH₄/ether the same mechanism does not applicable, hence they are wrong answers.

• Hoffmann Bromamide Degradation reaction occurs by treating an amide with bromine in the presence of the aqueous or ethanolic solution of sodium hydroxide.

• In this reaction, migration of an alkyl or aryl group happens from the carbonyl carbon of the amide to the nitrogen atom of the same amide.

• Hence the amine formed contains one carbon less than that of the reactant amide.

(i) ArNH₂ (iii) ArNO₂ and (iv) ArCH₂NH₂ do not undergo this degradation reaction.

• p-toluedene (III) is the strongest base among the 3 given compounds because the presence of −CH₃ group in the p position, which has electron-donating (+I effect) that increases the electron density on the N-atom atom and make the electron pairs more available for protonation.

• In case of I and II aniline is moe basic than III p-nitroaniline; but aniline is less basic than III p- toluedene.

• p-nitroaniline there is a nitro group in the ‘para’ position of the benzene ring which is an electron-withdrawing group and involves the lone pair on N atom in resonance thus decreasing its ability to donate unshared electron pairs to acid.

• In case of aniline the conjugation is not as effective as p-nitroaniline so it is a better base.

• Nitrous acid us unstable and methylamine produces alcohols sometimes alkenes by reacting with it . in intermediate step aliphatic diazonium salts are formed which being unstable forms alcohol and liberates quantitatively nitrogen gas.

Hence the other 3 options (i) CH₃—O—N==O (ii) CH₃—O—CH₃ and (iv) CH₃CHO are incorrect.

• When methylamine reacts with nitrous acid (prepared in situ by reaction of mineral acid and sodium nitrite) and forms an intermediate aliphatic diazonium salt which is being unstable readily forms alcohol and evolution of nitrogen gas occurs simultaneously . for this reason this reaction is used in estimation of amino acids and proteins.

• Other 3 options i), (iii) and iv) is not possible for this reaction.

at first, this step takes place because there is a requirement for stronger electrophile and for that nitric acid has to be activated for nitration of benzene.

• Sulphuric activates nitric acid and produces a strong electrophile nitronium ion NO₂⁺, its electrophilicity tends to attack the benzene ring and electrophillic substitution takes place by producing nitrobenzene.

•

The other 3 species (i) NO₂(ii) NO⁺ and (iv) NO₂^– cannot be formed during the activation of nitric acid.

• Aromatic nitro compounds undergo catalytic reduction with the reagents Fe and HCl and always form aromatic primary amines .

• Fe being the catalyst provides electrons for the reaction to occur and the acid provides reaction with protons.

Other 3 options (i), (ii) and (iv) these products do not support the reaction.

• being the strongest base among the given alkyl amines and interacts far better than towards dilute hydrochloric acid.

Reason being the presence of the 2 –CH₃ groups on the adjacent carbon atoms which is more than (i) CH₃—NH₂ as the electron releasing + I effect pushes electron more effectively to N atom in case of .

• But the case of is contradictory because of steric factors. The N atom is surrounded by 3 bulky methyl groups which increases the chance for HCL protons to interact and bind to the unshared electron pairs on N atom, therefore, protonation is less effective here.

• being the least basic(conjugation of amino group with benzene ring) will be the least reactive towards HCL.

• The reaction of acid anhydrides with primary amines resembles their reaction with acyl chlorides and instead of production of HCL in case of acyl chlorides amines produce carboxylic acids in this reaction and obviously the corresponding amide as major product.

•

Hence, (ii) imide (iii) secondary amine and (iv) imine cannot be formed in the above reaction.

• In Sandmeyer reaction the diazonium groups get replaced by halide ions in the benzene ring. This is a typical nucleophilic substitution reaction which introduces new halide groups to the benzene ring in the presence of copper powder (Cu(I) ions).

All other reaction (ii) Gatterman reaction(iii) Claisen reaction(iv) Carbylamine reaction are other type of reactions .

• Gabriel phthalimide synthesis is a chemical reaction that transforms primary alkyl halides into primary amines.

• Generally, the reaction uses potassium Pthalimide as reactant which by reaction with ethanolic KOH at first produces potassium salt of phthalimide which con heating with a certain alkyl halide (production of N-aikylpthalimide) followed by hydrolysis in alkaline medium produces the corresponding primary amine, without changing in the number of carbon atoms of the chain.

• Other 2 preparation of amines (i) Hoffmann Bromamide reaction ( product amine will have 1 carbon less than the reactant alkyl halide) and (iv) Reaction with NH₃ ammonolysisof alkyl halides(the primary amine obtained can further undergo reaction with halide and forms secondary, tertiary or even quaternary amines ) are not necessarily form primary amines without changing the number of carbon atoms in the chain.

• In azo coupling reaction, the diazonium chloride plays the role of electrophillic substance which needs to be coupled with an activated aromatic compound (such as phenol, aniline or even anisole).

• But in case of Nitro benzene, nitro groups are electron-withdrawing groups, therefore they deactivate the benzene ring by forming canonical/resonating structures , after that the benzene ring cannot undergo azo coupling reaction.

Because of this situation nitrobenzene cannot perform this reaction whereas (i) Aniline ,

(ii) Phenol and (iii) Anisole can participate.

.

• According to Brönsted (and Lowry) concept, basicity depends (i) on the tendency of a substance to accept a proton and (ii) the stability of conjugate acid(like complementary partner of the base) of the corresponding base.

• Hence, in case of phenol After losing H+ phenol produces least stable conjugate acid(conjugate strong acid – weak base pair, where weak base is phenol) among the given compounds. Oxygen is more electronegative than N so, O- H bond is more polar and it has a very highly acidic character. Since phenol is more acidic than that of alcohol, therefore phenol has the least tendency to accept a proton and hence it is weak Brönsted base.

• Moreover, the conjugate base of phenol is resonance stabilised (4 resonating structures), which means phenol is a stronger acid rather than being a base (conjugate strong acid(SA) - weak base pair,(WB) where the SA is phenol and WB is the phenolate ion. )

.

• Among the other 3 options (i) (ii)and (iv) all of these are stronger base than phenol.

aniline - has less electro-negative N atom, therefore has a less polar bond and a tendency to accept proton or (more precisely to donate electrons : Lewis concept) though it is a weak base but definitely a stronger one(base) than phenol and also it produces a more stable conjugate acid after accepting proton.

•

• In case of (ii)and (iv) both are stronger base than phenol, as there is no conjugation possible for the N and O atoms attached to a cyclohexyl ring as the electrons pairs are more available for donation or proton acceptance.

• The main reason behind the basic properties of cyclopentyl amine is the availability of unshared electrons on the nitrogen atoms which allows the ease of protonation on the N atom and forming of cation by interacting with acids.

• And also the cyclopentyl ring makes the N more nucleophilic (electron-donating tendency, through the effect of hyperconjugation) it has more basic strength even more than ammonia.

• (i) might be weakest base among them due to conjugation of the –NH₂ group with the benzene ring which eventually makes the lone pair on N atoms less available, which is also partially the case for (iii) also.

• Amines are generally stronger bases than (ii) NH₃ because the n atoms in amines have more ability to form dative bonds with H s, as they have more nucleophilic N atom (more nucleophilic the N atom, more will be the basicity).

• Ammonia is more basic than water because NH₃ has less electronegative central atom N (rather than O), therefore it’s tendency of holding onto electrons is much less than oxygen and it is better electron donor i.e. better base than H₂O.

• When it comes to the basic strength order NH₂^– > OH ^–, conjugate acid-base pair concept must be taken into account i.e. the stronger the acid, the weaker will be the conjugate base.

• NH₂^– and OH ^– both of the species are the conjugate bases of NH₃ and H₂O respectively.

• As H₂O is a stronger acid than NH₃ (due to stronger proton donating ability of the former), H₂O autoionizes itself but NH₃ shows reluctance in donating protons.

• Therefore, conjugate base of water i.e. OH ^– must be a weaker base ( water is stronger acid)thanNH₂^– (as ammonia is weaker acid).

•

• Volatility of a compound depends upon its bonding; strong bonding or interactions between molecules lead to more bulkiness and therefore less volatility.

• In case of CH₃CH₂CH₃ there is no sufficient eletronegativity of C to form a hydrogen bond, which is likely to be the case for amines because they are associated with hydrogen bonding of N-H^..........N type; this significant intermolecular hydrogen bonding causes amines to have higher boiling points than comparable hydrocarbons.

Hence CH₃CH₂CH₃ will the most volatile of them all.

• But, (II) (CH₃)₃N and (III) will be more volatile than (I) CH₃CH₂CH₂NH₂ because in case of secondary and tertiary amines, hydrogen bonding is not as possible as the primary amines as there is less N-H bonds present in secondary amine and in the tertiary amine (CH₃)₃N there is no such N-H bond is present.

• (iii) is actually the reaction Gabriel pthalimide synthesis

• Gabriel phthalimide synthesis is a chemical reaction that transforms primary alkyl halides into primary amines.

•

generally, the reaction uses potassium pthalimide as reactant which by reaction with ethanolic KOH at first produces potassium salt of phthalimide which con heating with a certain alkyl halide(production of N-aikyl pthalimide) followed by hydrolysis in alkaline medium produces the corresponding primary amine, without changing in the number of carbon atoms of the chain.

Other options (i), (ii) and (iv) are not able to prepare of amines will give the same number of carbon atoms in the chain of amines as in the reactant.

• In case of (iii) Iodobenzene different reagents as KI used to obtain the product because it cannot be formed in presence of Cu(I) ( Iodine being less reactive than other halogens)

• And for (iv) Fluoro-benzene it cannot be formed by this reaction due to the high reactivity of Fluorine the reaction b/w F and benzene and rigorous and highly explosive.

The other 2 options (i) Chlorobenzene (ii) Bromobenzene can be formed by this reaction.

• Aromatic nitro compounds undergo catalytic reduction with the reagents Fe and HCl and always form aromatic primary amines .

•

•

• Fe being the catalyst provides electrons for the reaction to occur and the acid provides reaction with protons.

• Other 3 options (i), (ii) and (iv) these products do not support the reaction.

• In carbylamine test, only primary amines (aliphatic or aromatic) are heated with chloroform and ethanolic KOH produces isocyanides or carbylamines(foul smell).

• Other 2 options (iii) and (iv) are not involved in this test.

• In case of phosphonic acid (iii) H₃PO₂ it reduces benzene and the diazonium group gets replaced by hydrogen and itself oxidizes to H₃PO₃.

• benzenediazonium chloride participates in de-amination reaction with (ii) CH₃CH₂OH . it converts to bezene and as side products acetaldehyde , nitrogen and HCl gas is formed.

• Options (i) and (iv) are incorrect.

• When acetanilide is treated with bromine and acetic acid, mixture of (major) and (very little) is obtained. This is an example of para substituted electrophilic reaction specialised to form para products of benzene ring.

.

Other 2 products (iii) and (iv) are not formed in this reaction.

• Aniline reacts with bromine in presence water at room temperature to give a 2,4,6-tribromoaniline, bromination of aniline is an example of electrophillic substitution preferable to the ortho and para position (because these 2 positions of aniline are more electrophilic due to activated benzene ring of amino group) .

• And bromination goes through formation of intermediate arenium (a cyclohadienyl cation ) ions.

• Gabriel synthesis is only applicable on the synthesis for aliphatic primary amines as this is a nucleophillic substitution reaction which follows several steps to form primary amines (addition of ethanolic or alcoholic KOH/NaOH then heating with alkyl halide and then finally alkaline hydrolysis).

• Aryl halides like (ii) 2-Phenylethylamine(iii) N-methylbenzylamine and (iv) Aniline

do not undergo nucleophillic substitution reactons and therefore cannot be prepared by his method

• this is dehydrohalogenation of cyclohexyl bromide which involves removal of (elimination of) a hydrogen halide ..

• Alcoholic KOH forms hydroxide ions act as a strong nucleophile and replaces the halogen atom from cyclohexyl halide which later leads to formation of cyclohexene.

• For option (i) this reaction is preparation of alkyl amines from an alkyl halide treating with ammonia.

Other 2 options (i) and (iv ) are not possible.

• Direct nitration of aniline is not possible because it gives a mixture of ‘ortho’ and ‘para’ products.

• So acetic anhydride is used to produce acetanilide in order block the ‘ortho’ position activation of aniline and then it undergoes nitration and forms specifically

para-nitro aniline.

Both (i) Acetyl chloride/pyridine followed by reaction with conc. H₂SO₄ + conc. HNO₃.

• (ii) Acetic anyhdride/pyridine followed by conc. H₂SO₄ + conc. HNO₃ reagents can be used for the procedure.

But (iii) Dil. HCl followed by reaction with conc. H₂SO₄ + conc. HNO₃.

(iv) Reaction with conc. HNO₃ + conc. H₂SO₄ cannot produce acetanilide.

• Aniline reacts with bromine in presence water at room temperature to give a 2,4,6-tribromoaniline, bromination of aniline is an example of electrophillic substitution preferable to the ortho and para position (because these 2 positions of aniline are more electrophillic due to activated benzene ring of amino group)

acylation of aniline is also an example of electrophillic substitution reaction.

(ii) Coupling reaction of aryl diazonium salts (iii) Diazotisation of aniline are not electrophillic substitution reaction.

Nitrating mixture is the mixture of 1:1 solution of HNO₃ and H₂SO₄ and is used for the nitration of organic compounds.

HNO₃ in the nitrating mixture is used for nitration of benzene. This is because it acts as a base and provides the electrophile, NO₂⁺ ion for the nitration of benzene. The nitroniumion formed is the attacking species.

To obtain p-Nitroaniline as the major product.

Direct nitration of Aniline yields tarry oxidation products and nitro derivatives compounds. In acidic medium, aniline is protonated, results in the formation of anilinium ion which is meta-directing. Thus, ortho-, para- and meta-Nitroaniline compounds are formed. 51% of the total products is p-Nitroaniline, 47% m-Nitroaniline and 2% o-Nitroaniline.

To control the nitration reaction and tarry oxidation products and nitro derivatives products formation, the NH₂ group of aniline is acetylated before carrying out nitration. Here, the major product is p-Nitroaniline.

The amine group is protected by acetylation reaction with acetic anhydride.

C₆H₅CH₂OH as the major product.

C₆H₅CH₂NH₂ reacts with HNO₂ to form unstable diazonium salt, which in turn gives alcohol. Thus, when C₆H₅CH₂NH₂ reacts with HNO₂ benzyl alcohol is formed along with N₂ and H₂O. The chemical reaction involved in this case is

When C₆H₅CH₂NH₂ reacts with HNO₂ at low temperature, stable Diazonium salt is formed and no N₂ gas is evolved in this case.

The best reagents for the conversion of nitrile to primary amine are LiAlH₄ and Sodium/Alcohol

Nitriles can be converted to their corresponding primary amine by reduction with LiAlH₄ or catalytic hydrogenation.

The chemical reaction involved in this is conversion is:

The product formed in this chemical reaction is 3-Methylnitrobenzene.

The structure of 3-Methylnitrobenzene is:

2-Nitro-4-methylaniline reacts with NaNO₂ + HCl at 273-278K and form a stable 3-diazonium salt derivate. Then it reacts with H₃PO₂ and water to form 3-Methylnitrobenzene as the major product. The byproducts formed in this reaction are H₃PO₃, HCl and N₂ gas.

Benzenesulphonyl chloride or, C₆H₅SOCl is commonly known as Hinsberg’s reagent. Hinsberg’s reagent can be used to distinguish primary, secondary and tertiary amines.

Benzenesulphonyl chloride reacts with primary amine to yield a product that is soluble in alkali.

Benzenesulphonyl chloride reacts with secondary amine to yield a product that is insoluble in alkali whereas Benzenesulphonyl chloride does not react with the tertiary amine.

Benzene diazonium chloride is not stored and is used immediately after its preparation.

When aniline reacts with NaNO₂ + HCl at 273-278K, benzene diazonium chloride is formed. Thus product is highly unstable and therefore, is used immediately after its preparation. Benzene diazonium chloride is highly soluble in water at high temperature and is itself very stable at low temperature. Thus, benzene diazonium chloride cannot be stored for a long time.

The activating effect of –NH₂ group can be controlled by protecting the -NH₂ group by acetylation with acetic anhydride, and then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.

The acetylation of —NH₂ group of aniline reduce its activating effect because the lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance.

MeNH₂ is stronger base than MeOH because of the lower electronegativity and the presence of the lone pair of electrons on nitrogen atom in MeNH_2. We know that nitrogen is less electronegative than oxygen atom. Thus, nitrogen in MeNH₂ can easily donate electrons unlike oxygen atom in MeOH.

The activating effect of –NH₂ group can be controlled by protecting the -NH₂ group by acetylation with acetic anhydride in presence of pyridine and then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine. We know that pyridine is a base due to the presence of lone pair of electrons on the nitrogen atom. Thus, this strong base is used to get rid of the side product HCl from the reaction.

The coupling reaction of aryldiazonium chloride with aniline is carried out in mild basic medium. Aryldiazonium chloride reacts with aniline to form a yellow dye of p-Aminoazobenzene. This is an example of electrophilic substitution reaction. The chemical reaction involved in this case is:

The products formed in the reaction of aniline with bromine in non-polar solvent such as CS₂ are 4-Bromoaniline and 2-Bromoaniline where 4-Bromoaniline is the major product. In non-polar solvent medium, the activating effect of –NH₂ group of aniline is reduced because of resonance and thus, mono-substitution occurs only at ortho- and para-positions.

CH₃CH₂CH₃ < CH₃CH₂NH₂ < CH₃CH₂OH

We know that nitrogen atom is less electronegative than oxygen atom. Thus, dipole moment of CH₃CH₂OH is greater than that of CH₃CH₂NH₂. CH₃CH₂CH₃ has the least dipole moment among the three given compounds because it is almost a non-polar molecule. Thus, the increasing order of dipole moment is CH₃CH₂CH₃ < CH₃CH₂NH₂ < CH₃CH₂OH.

The structure of the compound, allyl amine is:

CH₂₌CH-CH_2-NH₂

IUPAC name of allyl amine is Prop-2-en-1-amine.

The number of carbon atoms in the longest chain is three. There is a double bond between C₂ and C₃ atom and the functional group attached to this compound is amine group. Thus, IUPAC name of allyl amine is Prop-2-en-1-amine.

IUPAC name of the given compound is N,N-Dimethyl aminobenzene.

Here, the functional group attached to this compound is amine group. The nitrogen atom is attached to two methyl groups and one phenyl ring and therefore, it is a tertiary amine. Thus, the IUPAC name of the given compound is N,N-Dimethyl aminobenzene.

Z is ethylmethylamine.

According to the question, C₃H₉N reacts with C₆H₅SO₂Cl or Hinsberg’s reagent to give a solid, insoluble in alkali which means that C₃H₉N is a secondary amine. The product obtained in this reaction has no replaceable hydrogen attached to the nitrogen atom of the amine group.

Thus the structure of given amine is

The chemical reaction involved in this case is:

RNH₂ can form only 2° amine when carbylamine reaction is done followed by treatment with H₂/Pd.

When RNH₂ is heated with chloroform and KOH, isocyanides or carbylamines are formed. Isocyanides or carbylamines are then converted to 2° amine by H₂/Pd.

The chemical reaction involved in this case is:

RNH₂ + KOH + CHCl₃ + heat → RNC + H₂/Pd → R-NH-CH₃

Benzene diazonium chloride reacts with phenol under mild basic condition to form p-hydroxyazobenzene as the major product. This type of reaction is known as coupling reaction. P- hydroxyazobenzene are often coloured and are used as orange dyes.

The chemical reaction involved in this case is:

We know that aniline is a colourless liquid and is sparingly soluble in water.

Aniline reacts with aqueous HCl to form Anilinium chloride. Anilinium chloride is soluble in water. Thus, aniline is soluble in aqueous HCl.

The chemical reaction involved in this case is:

The suggested route for the above conversion is:

When the given compound is treated with Br₂/ KOH, cyclohexylamine is formed. Cyclohexylamine is then heated with KOH and CHCl₃ and isocyanides or carbylamines are formed. Isocyanides or carbylamines are then converted to 2° amine by H₂/Pd.

A=

B=

When the given compound reacts with KCN, the chloride ion is replaced by –CN group and is further converted to primary amine by H₂/Pd.

(i) toluene → p-toluidine

Toluene is allowed to react with nitrating reagent to form p-nitrotoulene and the nitro group is further reduced to p-toluidine by Fe/HCl.

(ii) p-toluidine diazonium chloride → p-toluic acid

The chloride ion in p-toluidine diazonium chloride is replaced by –CN group in the presence of Cu⁺ ion. This reaction is called Sandmeyer reaction. The –CN group in then hydrolysed to -COOH group.

(i) nitrobenzene → acetanilide

Nitrobenzene is reduced to aniline in the presence of Sn/HCl. The amino group of aniline is then acetylated with acetic anhydride in the presence of pyridine.

(ii) acetanilide → p-nitroaniline

Acetanilide is allowed to react with nirating mixture to form p-Nitroacetanilide. P-nitroacetanilide is then converted to p-nitroaniline by hydrolysis.

When a solution containing p-Toluene diazonium chloride and p-nitrophenyl diazonium chloride is allowed to react with alkaline solution of phenol, an azo dye is formed. This is an example of electrophilic substitution reaction. In alkaline medium, phenol is converted to phenoxide which is more electron rich than phenol and therefore, phenoxide ion is more reactive for electrophilic attack. p-Nitrophenyl diazonium cation is stronger electrophile than p-Toluene Diazonium cation because of –NO₂ group. Thus, p-Nitrophenyl diazonium cation couples with phenol to form an azo dye.

The structure of the product formed is:

The p-nitroaniline is first treated with Br₂/CH₃COOH and is converted to corresponding Diazonium salt. Finally the intermediate compound is treated with Cu₂Br₂/HCl to obtain the desired product.

Benzene is first nitrated using nitrating reagent followed by reduction of nitro group. The amine group of aniline is then protected by acetylation and a nitro group is attached at para position by nitration. The protected group is then deprotected to form p-nitoaniline.

The above conversion can be done in the following way:

Here, the aniline is first converted to diazonium salt and finally to nitro-benzene. Then bromine ion is attached to meta position with respect to nitro-group in nitro-benzene to form 3-bromo nitrobenzene.

i) This conversion involves several chemical reactions like protection of amine group of aniline, nitration, deprotection of amine group, bromination, diazonium salt formation. The –N₂Cl group is replaced by proton.

The following conversion can take place in following ways:

ii) The following conversion can take place in presence of KI. The –N₂Cl group is replaced by Iodine ion.

(i) Ammonolysis - (d) Reaction of alkylhalides with NH₃

(ii) Gabriel phthalimide synthesis - (c) Reaction of phthalimide with KOH and R—X

(iii) Hoffmann Bromamide reaction - (a) Amine with lesser number of carbon atoms

(iv) Carbylamine reaction - (b) Detection test for primary amines.

(i) Ammonolysis - (d) Reaction of alkylhalides with NH₃

Alkylhalides are treated with NH₃ to form amines. The C-halogen bond is cleaved by NH₃ molecules.

(ii) Gabriel phthalimide synthesis - (c) Reaction of phthalimide with KOH and R—X

Gabriel synthesis is used for the preparation of primary amines. Phthalimide is treated with KOH to form potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.

(iii) Hoffmann Bromamide reaction - (a) Amine with lesser number of carbon atoms

Hoffmann Bromamide reaction is used for the preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide.

(iv) Carbylamine reaction - (b) Detection test for primary amines.

Aliphatic and aromatic primary amines on heating with chloroform and KOH form isocyanides or carbylamines which are foul smelling substances but secondary and tertiary amines do not show this reaction. Thus, Carbylamine reaction can be used to detect primary amines.

(i) Benzene sulphonyl chloride - (b) Hinsberg reagent

(ii) Sulphanilic acid - (a) Zwitter ion

(iii) Alkyl diazonium salts - (d) Conversion to alcohols

(iv) Aryl diazonium salts - (c) Dyes

(i) Benzene sulphonyl chloride - (b) Hinsberg reagent

Benzenesulphonyl chloride or, C₆H₅SOCl is commonly known as Hinsberg’s reagent. Hinsberg’s reagent can be used to distinguish primary, secondary and tertiary amines.

(ii) Sulphanilic acid - (a) Zwitter ion

The net charge on sulphanilic acid is zero. Therefore, Sulphanilic acid is a zwitter ion or dipolar ion.

(iii) Alkyl diazonium salts - (d) Conversion to alcohols

When alkyl diazonium salts are treated with water molecules, they are converted to alcohols.

(iv) Aryl diazonium salts - (c) Dyes

Acylation is when primary and secondary amines react with acid chlorides, anhydrides and esters to give amides as products.

Alkylation is when primary and secondary amines react with alkyl halides to form a higher substituted amine. Acylation is carried out in the presence of a base stronger than the amine like pyridine which causes the shifts the equilibrium to the right side.

Hoffmann’s bromamide reaction is given by amides for the preparation of primary amines, not by primary amines themselves. Primary amines are more basic than secondary amines.

Hinsberg reagent/benzenesulphonyl chloride on reaction with primary amine ethylamine yields N-Ethylbenzene sulphonamide. The hydrogen attached to N-Ethylbenzene sulphonamide is acidic in nature. This is due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali.

Hinsberg reagent/benzenesulphonyl chloride on reaction with secondary amine N,N-diethylbenzenesulphonamide. This compound is insoluble in alkali as it does not contain any reactive acidic hydrogen attached to the nitrogen atom.

The sulphonyl group attached to nitrogen atom is a strong electron withdrawing group but that is not the reason for insolubility of the sulphonamide, the absence of acidic hydrogen.

Nitro compounds on reduction give amines as products.

The reaction can use hydrogen gas passed over finely divided nickel, palladium or platinum, but iron scrap metal and small amount of HCl is used in the presence of steam preferably. This is because FeCl₂ formed gets hydrolysed to release hydrochloric acid during the reaction.

Thus, only a small amount of hydrochloric acid is required to start the reaction.

Primary amines can be prepared by Gabriel Phthalimide synthesis but only aliphatic primary amines, aromatic primary amines cannot be prepared by this technique. When phthalimide is heated with ethanolic KOH, potassium salt of phthalimide is formed which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide, hence the formation of aromatic primary amines do not occur.

Acetanilide is formed by the acetylation of aniline under the presence of a strong base like pyridine. The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance, resulting in less availability of the electron pair on nitrogen for reaction.

This way, activating effect of –NHCOCH₃ group is less than that of amino group and acetanilide is less basic than aniline.

In case of substituted aniline, electron releasing groups increase basic strength whereas electron withdrawing groups like decrease it.

The delocalization of electrons in the Acetanilide is show in below figure;

When hydrocarbon ‘A’ reacts with HCl to give a compound ‘B’ with a chloride moiety, it means that A is an alkene and –Cl from HCl is added across the double bond to form ‘B’.

When ‘B’ reacts with NH₃, it forms an amine with the –Cl being substituted with –NH₂. Compound ‘C’ is thus C₄H₁₁N or C₄H₉NH₂.

When amine ‘C’ reacts with NaNO₂ and HCl followed by treatment with water forms a diazonium salt and later gives an alcohol. The product ‘D’ is an optically active alcohol, so the amine ‘C’ is aliphatic in nature.

Ozonolysis of ‘A’ gives 2 moles of CH₃CHO or acetaldehyde. Ozonolysis of alkenes gives aldehydes or ketones or a mixture depending on the substitution pattern of the alkene. The products – two moles of acetaldehyde mean the compound ‘A’ is But-2-ene or CH₃—CH=CH—CH₃.

The reactions are given as follows.

Firstly, when compound ‘A’ reacts with benzenesulphonyl chloride, it gives a compound which is soluble in alkali. That means the compound is a primary amine. ‘A’ is given as C₆H₇N or C₆H₅NH₂.

Compound ‘A’ is C₆H₅NH₂ or aniline if it is a primary amine, which can be proven in the following reactions. Aniline is sparingly soluble in water.

Since aniline is basic, on reaction with mineral acid like HCl, it gives compound ‘B’ which is anilium chloride which is a salt and is soluble in water.

When aniline reacts with chloroform, CHCl₃ and alcoholic potash, KOH, it forms benzene isonitrile or compound ‘C’ which has an obnoxious smell.

As mentioned above, aniline reacts with benzenesulphonyl chloride to give compound ‘D’ or N-phenylbenzenesulphonamide which is soluble in alkali.

When aniline reacts with NaNO₂ and HCl, it forms diazonium salt ie. benzene diazonium chloride, which on reaction with phenol in alkaline conditions gives an orange dye ‘F’ which is p-hydroxyazobenzone.

When p-Nitrotoluene undergoes reduction, it forms p-methyl benzyl amine. The reagents for ‘1’ are Sn + HCl or Fe + HCl which cause the reduction due to release of hydrogen. Even hydrogen gas passed over finely divided nickel, platinum or palladium in the presence of steam works to drive the reaction.

N-Methylaniline undergoes acetylation in the presence of a strong base such as pyridine to form N-methylacetanilide. The next step is nitration, with the reagents given which are HNO₃ and H₂SO₄. Due the presence of the acetyl group, nitration leads to the formation of 2-Nitro-4-methylacetanilide. Hence, ‘2’ is 2-Nitro-4-methylacetanilide.

This molecule undergoes a reaction with reagent ‘3’ to give 2-nitro-4-methylaniline. This means ‘2’ undergoes hydrolysis to give 2-nitro-4-methylaniline. ‘3’ is H₂O/H⁺.

2-nitro-4-methylaniline further reacts with NaNO₂/HCl which forms a diazonium salt, hence ‘4’ is a diazonium salt 2-nitro-4-diazonium chloride.

After reaction with reagent ‘5’, the product formed does not have the N₂⁺Cl^- group at 4C. This means the diazonium chloride has been displaced by H. Hence the reagent ‘5’ is a mild reducing agent like hypophosphorous acid H₃PO₂ (phosphinic acid) or ethanol. The reagents themselves get oxidised to phosphorous acid and ethanal, respectively.