Arrange the following compounds in decreasing order of their boiling points: Butanol, Butanamine, Butane

Molecules containing hydrogen bonds have higher boiling temperatures than molecules of the same size that don't have a - OH or - NH group. Due to the presence of hydrogen bonding, the molecules are held together more tightly, requiring more heat (energy) to separate them. Hence, butanamine and butanol have higher boiling points compared to butane. Further, in butanol, the hydrogen is bonded to oxygen which is more electronegative compared to the nitrogen bonded to hydrogen in butanamine. Therefore, the decreasing order of boiling points is: Butanol > Butanamine > Butane.

How will you distinguish between the following pairs of compounds: Aniline and N - methylaniline

Aniline is a primary amine, so it gives carbylamine test whereas N-methyl aniline is a secondary amine, so it does not give this test. Carbylamine test- When aniline is heated with an alcoholic solution of KOH and CHCl 3 , so it gives the foul smell of phenyl carbylamine. C 6 H 5 - NH 2 Aniline + CHCl 3 + 3 KOH → C 6 H 5 - NC Phenyl carbylamine + 3 KCl + 3 H 2 O

How will you distinguish between the following pairs of compounds: Aniline and Ethanamine

Azo-dye test can be used to distinguish between ethylamine and aniline. When aromatic amines react with H N O 2 ( N a N O 2 + d i l . H C l ) at 0 - 5 o C followed by a reaction with an alkaline solution of 2 -naphthol, a dye is formed. The dye formed is usually yellow, red or orange coloured. In the above reaction, diazotization takes place. At pH 9 - 10 in the presence of phenol and diazonium salt, the coupling reaction takes place and forms an azo dye. Aliphatic amines give a brisk effervescence under similar conditions due to the evolution of N 2 gas. C H 3 C H 2 - N H 2 + H O N O → 0-5 °C C 2 H 5 O H + N 2 ↑ + H 2 O

Give the structures of A and B in the following sequence of reactions: C 6 H 5 N 2 + Cl - → ∆ CuCN A → H 2 O / H + B

Diazonium group being a very good leaving group is substituted by other groups such as Cl - , Br - , I - , CN - and OH – which displace nitrogen from the aromatic ring. The nitrogen formed escapes from the reaction mixture as a gas. The CN – nucleophiles can easily be introduced in the benzene ring in the presence of Cu ( I ) ion. The reaction is as follows: C 6 H 5 N 2 + Cl - → ∆ CuCN C 6 H 5 CN benzonitrile A Nitriles are hydrolysed to amides and then to acids in the presence of H + or OH − as catalyst. Similarly, benzonitrile undergoes hydrolysis to form benzamide and finally gives benzoic acid. C 6 H 5 CN → H 2 O H + / OH - C 6 H 5 CONH 2 Benzamide → ∆ H 3 O + C 6 H 5 COOH Benzoic acid B + NH 3

Give the structures of A and B in the following sequence of reactions: C 6 H 5 NO 2 → Fe / HCl A → 0 o - 5 o C NaNO 2 + HCl B

Nitro compounds are reduced to amines with iron scrap and hydrochloric acid. In this reaction, FeCl 2 is formed which gets hydrolysed to release hydrochloric acid during the reaction. SImilarly, nitrobenzene gives aniline on reaction with iron scrap and hydrochloric acid. C 6 H 5 NO 2 → Fe / HCl C 6 H 5 NH 2 Aniline A Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at 273 - 278 K . Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid. The conversion of primary aromatic amines into diazonium salts is known as diazotisation. The reaction is as follows: C 6 H 5 NH 2 + NaNO 2 + HCl → 273 - 278 K C 6 H 5 N 2 + Cl - + NaCl + 2 H 2 O

Arrange the following in increasing order of boiling points : ( CH 3 ) 2 NH , C 2 H 5 OH , C 2 H 5 NH 2

The boiling point of compounds depend on the hydrogen bonding present in the compound. The more extensive the H -bonding in the compound, the higher is the boiling point. C H 3 2 N H contains only one H -atom, whereas C 2 H 5 N H 2 contains two H -bonding than C H 3 2 N H . Hence, the boiling point of C 2 H 5 N H 2 is higher than that of C H 3 2 N H . Boiling point ∝ force of attraction ∝ H - bond Further, O is more electronegative tha n N . Thus, ethyl alcohol forms stronger H-bonds than ethyl amine. As a result, the boiling point of C 2 H 5 O H is stronger than C 2 H 5 N H 2 and C H 3 2 N H . Hence, increasing order of their boiling points as follows: C H 3 2 N H < C 2 H 5 N H 2 < C 2 H 5 O H

Arrange the following in the increasing order of their pKb values : C 6 H 5 NH 2 , C 2 H 5 NH 2 , C 6 H 5 NHCH 3

The increasing order for pKb values is C 2 H 5 NH 2 ( 3 . 29 ) < C 6 H 5 NHCH 3 ( 9 . 30 ) < C 6 H 5 NH 2 ( 9 . 38 ) . The numbers in bracket indicates pKb. Higher is the pKb value, lower is the basic strength of amine. Thus, the increasing order for pKb values indicates decreasing order of the basic strength of amine. Aniline is least basic and ethanamine is most basic. Aniline is least basic as unshared pair of electrons on N atom is involved in resonance with benzene ring and cannot be easily donated. When an alkyl group such as methyl or ethyl is present, the electron releasing inductive effect ( + I effect ) of alkyl group increases the electron density on N atom and unshared pair of electrons on N atom can be easily donated.

Write the structures of the main products of the following reaction :

Benzene diazonium chloride can be converted to benzene by using ethanol or phosphinic acid. In this reaction, benzene diazonium ion is reduced. The diazonium group is replaced by − H .

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Give a simple chemical test to distinguish between Aniline and N,N-dimethylaniline.

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