Amines Solutions of Practice Problems

Solution 1: - Amines containing replaceable hydrogen atom react with acid chloride or acid anhydride to form substituted amides. This reaction is called acylation reaction. In case of tertiary amines there is no replaceable H atom and therefore there do not react with acetyl chloride or acetic anhydride and hence do not undergo acylation.

Solution 2: -

i. C6H5NH2 < NH3 < (CH3)2N < CH3NH2 < (CH3)2NH

ii. p-nitroaniline < Aniline < p-methylaniline

Solution 3: - Amine has a lone pair of electron on N atom and therefore, react as necleophiles.

Solution 4: - Aniline does not undergo Friedel crafts alkylation because of the formation of salt with AlCl3 which is used as a catalyst. Due to this nitrogen of aniline acquire +ve charge and hence acts as a strong deactivating group for further reaction.

Solution 5: -Under strong acidic conditions of nitration most of the aniline is converted in to anilinium ion having NH3

+ group. This group is a m – directing group therefore, m nitro aniline is also obtained along with o- and p- products.

Amines Solutions of Practice Problems

Solution 6: - The ammononolysis of alkyl halides gives quaternary ammonium salt. The free amine can be obtained from ammonium salt by using strong base

RNH3 + NaOH RNH2 + H2O + Na+X−

Solution 7: - Aromatic amines cannot be prepared by Gabriel pthalamide synthesis because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

Solution 8: -

Solution 9: - (CH3)3N < C2H5NH2 < C2H5OH

Solution 10: -

Sn + HCl NaNO2 + HCl H2O/H+ C6H5NO2 C6H5NH2 C6H5N2

+Cl− C6H5OH 273 K

Amines Solutions of Practice Problems

Solution 11: -

i. Since the compound C of molecular formula C6H7N is formed B on treatment with Br2 and KOH, therefore the compound B must be an amide and C must be an amine. The only aromatic amine having molecular formula C6H7N is C6H5NH2 (aniline)

ii. Since C is aniline the amide from which is formed must be benzamide Br2, KOH C6H5NH2 C6H5CONH2 Thus B is benzamide.

iii. Since B is formed from A with aqueous ammonia and heating therefore compound A must be benzoic acid. aq. NH3 C6H5COOH C6H5CONH2 Heat Thus A = C6H5COOH , B = C6H5CONH2 , C = C6H5NH2

Solution 12: -

I. C6H5NH2 + CHCl3 + 3KOH C6H5NC + 3KCl + 3H2O

II. C6H5N2Cl + H3PO2 + H2O C6H6 + N2 + H3PO3 + HCl

III. C6H5NH2 + H2SO4(conc. ) C6H5NH3+HSO4

− Reduction

IV. C6H5N2Cl + C6H5OH C6H6 + CH3CHO + N2 + HCl

Amines Solutions of Practice Problems

V.

CH3COOH

VI. C6H5NH2 + (CH3CO)2O C6H5CONHCH3 + CH3COOH

HBF4 NaNO2

VII. C6H5N2Cl C6H5N2+BF4

− C6H5NO2 + BF3 + NaF Cu Δ

Solution 13: -

a) Aromatic primary amines react with HNO3 at 273 - 278 K to form aromatic diazonium salts.

Aliphatic primary amines also react with HNO3 at 273 - 278 K to form aliphatic diazonium salts. However, these are unstable even at low temperature and therefore decompose readily to form alcohols and N2 is evolved. 273 – 278 K H2O CH3CH2NH2 + HNO2 + HCl CH3CH2 − N+ = NCl− C2H5OH + N2 + HCl

b)

Amines Solutions of Practice Problems

I. Loss of proton from amines give amide ion whereas loss of a proton

from alcohol gives an alkoxide ion RNH2 RNH− + H+ ROH RO− + H+ Since O is more electronegative than N therefore RO− can accommodate the –ve charge more easily than RNH− consequently, RO− is more stable than RNH−. Thus alcohols are more acidic than amines.

II. Primary amine (RNH2) have two hydrogen atoms on the N atom and therefore form intermolecular hydrogen bonding. Tertiary amines (R3N) do not have hydrogen atom on the N atom and therefore these do not form hydrogen bonds. As a result of hydrogen bonding is primary amine, they have higher boiling point than tertiary amines of comparable molecular mass.

III. Both arylamine and alkylamines are basic in nature due to the presence of lone pair on N – atom. But arylamines are less basic than alkylamines for example aniline is less basic than ethylamine as shown by Kb values: Ethylamine: Kb = 4.7 × 10−4 Aniline: Kb = 4.2 × 10−10 The less basic character of aniline can be explained on the basis of aromatic ring present in aniline. Aniline can have the following resonating structures.

Amines Solutions of Practice Problems

Solution 14: -

i. Carbylamine reaction: - Aliphatic and aromatic primary amines when warmed with CHCl3 and an alcoholic solution of KOH form isocyanide or carbylamine which have very unpleasant or foul smell. Warm R − NH2 + CHCl3 + 3KOH R − NC + 3KCl + 3H2O Warm CH3CH2NH2 + CHCl3 + 3KOH CH3CH2NC + 3KCl + 3H2O

ii. Diazotisation: - The reaction of converting aromatic primary amines to diazonium salt is called diazotisation. NaNO2 + HCl NaCl + HONO

iii. Hoffmann’s bromide reaction: - Primary amines can be prepared from amides by treatment with Br2 and KOH or NaOH. The amine formed contains one carbon atom less than the parent amide. Therefore this method is used for stepping down the series in organic conversions. RCONH2 + Br2 + 4KOH RNH2 + K2CO3 + 2KBr + 2H2O

Amines Solutions of Practice Problems

iv. Coupling reaction: - Aromatic amines react with diazonium salt to form azo compounds in acidic medium called dyes. The reaction is known as coupling or diazo reaction. For example aniline couples with benzene diazonium salt to form diazoamino benzene which ultimetly change to p – amino azo benzene on warming with a small quantity of HCl.

v. Ammonolysis: - When an aqueous or alcoholic solution of ammonia is treated with an alkyl halide at 373 K in a sealed tube, all the three types of amines are obtained. In this reactions the alkyl halides undergo nucleophilic substitution reaction in which the nucleophile ammonia displaces the halogen atom. The ammonia molecule first attacks an alkyl halide molecules to form a primary amine salt. This salt then reacts with ammonia to give the corresponding primary amine and ammonium halide. RX + NH3 RNH3X RNH3X + NH3 RNH2 + NH4

+X− The free amine can also be obtained from the ammonium salt by treatment with a strong base. RNH3X + NaOH RNH2 + H2O + Na+X−

Amines Solutions of Practice Problems

The primary amine also acts as a nucleophile and reacts further with alkyl halide and this sequence of reaction can lead to the formation of secondary tertiary and finally quaternary ammonia salts RX 373 K RX 373 K RX 373 K RNH2 R2NH R3N R4N+X−

vi. Gabriel phthalimide synthesis: - This method is used only for preparing only primary amines. In this method phthalimide which is treated with alkyl halide or benzyl halide to form N-alkyl or aryl phthalimide. The refluxing with NaOH gives primary amine.

Solution 15: -

I.

Amines Solutions of Practice Problems

II.

III.

Amines Solutions of Practice Problems

IV.

V.