Question:

Arrange the following (i) In decreasing order of the pKb​ values: C₂H₅NH₂,C₆H₅NHCH₃,(C₂H₅)2​NH and C₆H₅NH₂ (ii) In increasing order of basic strength: C₆H₅NH₂,C₆H₅N(CH₃)2​,(C₂H₅)2​NH and CH₃NH₂ (iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine

Answer:

(i) C₆H₅NH₂, C₆H₅NHCH₃, (C₂H₅)2​NH, C₂H₅NH₂

(ii) CH₃NH₂, C₆H₅NH₂, (C₂H₅)2​NH, C₆H₅N(CH₃)2​

(iii) Aniline, p-toluidine, p-nitroaniline

Question:

Write short notes on the following: Carbylamine reaction

Answer:

Carbylamine reaction, also known as Hofmann’s isocyanide reaction, is an organic reaction in which an isocyanide is produced from an amine and an aldehyde or ketone. It is a useful synthetic tool for the synthesis of amines and isocyanides. The reaction involves the addition of an aldehyde or ketone to an amine in the presence of a strong base, such as sodium hydroxide, to form an intermediate carbylamine. This intermediate then undergoes a base-catalyzed elimination reaction to form an isocyanide. This reaction is named after the German chemist August Wilhelm von Hofmann, who first reported it in 1881.

Question:

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (ii) CH₃(CH₂​)2​NH₂ (iii) CH₃NHCH(CH₃)2​ (iv) (CH₃)3​CNH₂ (v) C₆H₅NHCH₃ (iv) (CH₃CH₂​)2​NCH₃ (vii) m−BrC₆H4​NH₂.

Answer:

(ii) CH₃(CH₂​)2​NH₂ - 2-Aminopropane - Primary amine

(iii) CH₃NHCH(CH₃)2​ - 2-Amino-2-methylpropane - Secondary amine

(iv) (CH₃)3​CNH₂ - Trimethylmethanamine - Tertiary amine

(v) C₆H₅NHCH₃ - Anilinomethane - Primary amine

(vi) (CH₃CH₂​)2​NCH₃ - N-Methylethanamine - Secondary amine

(vii) m−BrC₆H4​NH₂ - 4-Bromoaniline - Primary amine

Question:

Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Answer:

(i) Aromatic Primary Amines + Nitrous Acid → Aromatic Diazonium Salt

(ii) Aliphatic Primary Amines + Nitrous Acid → Aliphatic Hydrazines

Question:

Account for the following: (i) pKb​ of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

Answer:

(i) Aniline is a stronger base than methylamine because its pKb is higher. This is due to the presence of an additional electron-withdrawing group (the benzene ring) in aniline, which makes it a stronger base than methylamine.

(ii) Aniline is not soluble in water because it has a higher molecular weight and is non-polar, so it does not dissolve in water. Ethylamine, on the other hand, is soluble in water because it is polar and has a lower molecular weight.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide because the methylamine is a strong base and reacts with the ferric chloride to form hydrated ferric oxide.

Question:

Complete the following reactions: C₆H₅NH₂+CHCl3​+alc.KOH⟶

Answer:

_{6 6}

1. CH₅NH₂ + CHCl3 → CH₅NHCl + CH₃OH (nucleophilic substitution)

_{6 6} 2. CH₅NHCl + KOH (in alcohol) → CH₅N(KOH)Cl + H₂O (acid-base reaction)

Question:

Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline

Answer:

(i) Methylamine and dimethylamine: Bromine Water Test.

(ii) Secondary and tertiary amines: Hinsberg Test.

(iii) Ethylamine and aniline: Nitrous Acid Test.

(iv) Aniline and benzylamine: Nitrobenzene Test.

(v) Aniline and N-methylaniline: Sulfuric Acid Test.

Question:

An aromatic compound A on treatment with aqueous ammonia and heating forms compound B which on heating with Br₂​ and KOH forms a compound C of molecular formula C₆H7​N. Write the structures and IUPAC names of compounds A,B and C.

Answer:

Compound A: Structure:

IUPAC Name:

Compound B: Structure:

IUPAC Name:

Compound C: Structure:

IUPAC Name:

Question:

Give the structures of A,B and C in the following reactions: CH₃CH₂​I→NaCNAOH−Partialhydrolysis​B→NaOH+Br₂​C

Answer:

A: CH₃CH₂I

B: NaCNAOH-Partial hydrolysis

C: NaOH + Br₂

Question:

Why can aromatic primary amines not be prepared by Gabriel phthalimide synthesis?

Answer:

1. Gabriel phthalimide synthesis is a reaction used to prepare alkyl and aryl primary amines from aryl halides.

2. Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because the reaction involves the use of a strong base, such as sodium hydroxide, which would react with the aromatic ring and destroy it.

Question:

How will you convert: (i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane (iii) Methanol to ethanoic acid (iv) Ethanamine into methanamine (v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine (vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid?

Answer:

(i) Ethanoic acid into methanamine Step 1: Add aqueous ammonia to ethanoic acid to form a salt, ammonium ethanoate. Step 2: Heat the ammonium ethanoate to form ammonium carbamate. Step 3: Heat the ammonium carbamate to form methanamine.

(ii) Hexanenitrile into 1-aminopentane Step 1: Add hydrogen cyanide to hexanenitrile to form hexanoic acid cyanide. Step 2: Add aqueous ammonia to the hexanoic acid cyanide to form a salt, ammonium hexanoate. Step 3: Heat the ammonium hexanoate to form ammonium pentanoate. Step 4: Heat the ammonium pentanoate to form 1-aminopentane.

(iii) Methanol to ethanoic acid Step 1: Oxidize methanol using an oxidizing agent, such as chromic acid, to form formaldehyde. Step 2: Add formaldehyde to aqueous ammonia to form a salt, ammonium formate. Step 3: Heat the ammonium formate to form ethanoic acid.

(iv) Ethanamine into methanamine Step 1: Add hydrogen cyanide to ethanamine to form ethanoic acid cyanide. Step 2: Add aqueous ammonia to the ethanoic acid cyanide to form a salt, ammonium ethanoate. Step 3: Heat the ammonium ethanoate to form ammonium carbamate. Step 4: Heat the ammonium carbamate to form methanamine.

(v) Ethanoic acid into propanoic acid Step 1: Add hydrogen chloride to ethanoic acid to form a salt, ethanoic acid chloride. Step 2: Add aqueous ammonia to the ethanoic acid chloride to form a salt, ammonium ethanoate. Step 3: Heat the ammonium ethanoate to form ammonium propanoate. Step 4: Heat the ammonium propanoate to form propanoic acid.

(vi) Methanamine into ethanamine Step 1: Add hydrogen cyanide to methanamine to form methanol cyanide. Step 2: Add aqueous ammonia to the methanol cyanide to form a salt, ammonium methanoate. Step 3: Heat the ammonium methanoate to form ammonium ethanoate. Step 4: Heat the ammonium ethanoate to form ethanamine.

(vii) Nitromethane into dimethylamine Step 1: Add aqueous ammonia to nitromethane to form a salt, ammonium nitromethane. Step 2: Heat the ammonium nitromethane to form ammonium dimethylamine. Step 3: Heat the ammonium dimethylamine to form dimethylamine.

(viii) Propanoic acid into ethanoic acid Step 1: Add hydrogen chloride to propanoic acid to form a salt, propanoic acid chloride. Step 2: Add aqueous ammonia to the propanoic acid chloride to form a salt, ammonium propanoate. Step 3: Heat the ammonium propanoate to form ammonium ethanoate. Step 4: Heat the ammonium ethanoate to form ethanoic acid.

Question:

Give plausible explanation for each of the following: (i) Why are amines less acidic than alcohols of comparable molecular masses? (ii) Why do primary amines have higher boiling point than tertiary amines? (iii) Why are aliphatic amines stronger bases than aromatic amines?

Answer:

(i) Amines are less acidic than alcohols of comparable molecular masses because the nitrogen atom in amines is more electronegative than the oxygen atom in alcohols. This means that the nitrogen atom in amines attracts electrons more strongly than the oxygen atom in alcohols, making the amines less likely to donate protons and therefore less acidic.

(ii) Primary amines have higher boiling points than tertiary amines because primary amines can form hydrogen bonds with other molecules, while tertiary amines cannot. These hydrogen bonds increase the intermolecular forces between the molecules, resulting in a higher boiling point.

(iii) Aliphatic amines are stronger bases than aromatic amines because the aliphatic amines have an unshared pair of electrons on the nitrogen atom, while the aromatic amines have a delocalized electron system. This unshared pair of electrons makes the aliphatic amines more likely to accept protons from other molecules, making them stronger bases.