Test: 35 Year JEE Previous Year Questions: Compounds Containing Nitrogen - JEE MCQ

The reagent used in Hofmann bromamide reaction is alkaline halogen (NaOH or KOH + X₂)

Conversion of (iii) to (iv) involving rearrangement is the slowest step. Species (iii) is electron deficient (N has only 6 electrons), hence it has a tendency to get its octet completed by migration of alkyl group.

Since the reaction is intramolecular, no cross product will be formed.

In o-nitrophenol intramolecular H-bonding is possible because OH and NO₂ groups are close to each other.
This makes the ortho isomer less acidic as its capacity to donate a proton (H-atom) decreases. There is no such intramolecular H–bonding in the p-isomer.

Chlorobenzene is resonance stabilized.
Thus aryl halides (chlorobenzene) do not undergo nucleophilic substitution. Reason is correct.

TIPS/FORMULAE : Electron donating tendency to a double bond is called +M effect and the transfer of electrons take place towards the attacking reagent due to +E effect.
In strongly acidic conditions, aniline becomes protonated with the result lone pair of electrons is not available to produce +E and +M effects. Thus here aniline becomes less reactive towards electrophilic substitution. On the other hand, the   group exerts strong –I effect causing deactivation of the

The colour of the azo dye formed will be orange red but not blue. However, the colour of dye can said to be due to extended conjugation due to presence of azo group.

The alkyl groups are electron releasing group (+ I), thus increases the electron density around the nitrogen thereby increasing the availability of the lone pair of electrons to proton or lewis acid and making the amine more basic. Hence more  the no. of alkyl group more basic is the amine. Therefore the correct order is

NH₃ < CH₃NH₂ < (CH₃)₂ N

Ethyl isocyanide on hydrolysis form primary amines.

Therefore it gives only one mono chloroalkane.

Wurtz reaction is for the preparation of hydrocarbons from alkyl halide

RX + 2Na + XR → R – R + 2NaX

is most basic. Inothers the basic character is suppressed due to Resonance (see applications of resonance).

The compound is CH₄N₂O
Empirical weight = 60; Mol. wt. = 60; 
Molecular formula

On heating urea loses ammonia to give Biuret

2NH₂CONH₂ → H₂ NCO.NH.CONH₂ + NH₃

Biuret with alkaline CuSO₄ gives violet colour. Test for –CONH^– group.

NOTE : Aromatic amines are less basic than aliphatic amines. Among aliphatic amines the order of basicity is 2° > 1° > 3°. The electron density is decreased in 3° amine due to crowding of alkyl group over N atom which makes the approach and bonding by a proton relatively difficult. Therefore the basicity decreases. Further Phenyl group show – I effect, thus decreases the electron density on nitrogen atom and hence the basicity.
∴ dimethylamine (2° aliphatic amine) is strongest base among given choices.
∴ The correct order of basic strength is Dimethylamine > Methyl amine > Trimethyl amine > Aniline.

This is carbylamine reaction.

CH₃CH₂NH₂ + CHCl₃ + 3KOH → C₂H₅NC + 3KCl  + 3H₂O

Primary aromatic amines react with nitrous acid to yield arene diazonium salts.

The diazonium group can be replaced by fluorine by treating the diazonium salt with fluoroboric acid (HBF₄). The precipitated diazonium fluoroborate is isolated, dried and heated until decomposition occurs to yield the aryl fluoride. This reaction is known as Balz-Schiemann reaction.

Now since the molecular mass increases by 42 unit as a result of the reaction of one mole of CH₃COCl with one-NH₂ group and the given increase in mass is 210. hence the number of –NH₂ group is = 210/42 = 5

Reaction (III) is a Hofmann bromamide reaction formation of CH₃CH₂NH₂ is possible only from a compound CH₃CH₂CONH₂ which can be obtained from the compound CH₃CH₂COO^– NH⁺₄ (B) in (II) reaction further propanic acid (CH₃CH₂COOH) on reaction with NH3 produce CH₃CH₂COO^–NH₄^– (reaction I) hence the reaction will be

Methyl isocyanate CH₃ – N = C = O

Arylamines are less basic than alkyl amines and even ammonia. This is due to resonance. In aryl amines the lone pair of electrons on N is partly shared with the ring and is thus less available for sharing with a proton.
In alkylamines, the electron releasing alkyl group increases the electron density on nitrogen atom and thus also increases the ability of amine for protonation.
Hence more the no. of alkyl groups higher should be the basicity of amine. But a slight discrepancy occurs in case of trimethyl amines due to steric effect. Hence the correct  order is

(CH₃)₂ NH > CH₃NH₂> (CH₃)₃ N> C₆H₅ NH₂

4 moles of NaOH and one mole of Br₂ is required during production of one mole of  amine during Hoffmann's bromamide degradation  reaction.