Text Book Back Questions and Answers - Chapter 13 - Organic Nitrogen Compounds - 12th Chemistry Guide Samacheer Kalvi Solutions

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Organic Nitrogen Compounds
Text BookEvaluation
Multiple Choice Questions
Question 1.
Which of the following reagent can be used to convert nitrobenzene to aniline?
(a) $\mathrm{Sn} / \mathrm{HCl}$
(b) $\mathrm{ZnHg} / \mathrm{NaOH}$
(c) $\mathrm{LiAIH}_4$
(d) All of these
Answer:
(a) $\mathrm{Sn} / \mathrm{HCl}$
Question 2.
The method by which aniline cannot be prepared is
(a) degradation of benzamide with $\mathrm{Br}_2 / \mathrm{NaOH}$
(b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous $\mathrm{NaOH}$ solution.
(c) Hydrolysis of phenylcyanide with acidic solution
(d) reduction of nitrobenzene by $\mathrm{Sn} / \mathrm{HCI}$
Answer:
(b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous $\mathrm{NaOH}$ solution.
Question 3.
Which one of the following will not undergo Hofmann bromamide reaction?
(a) $\mathrm{CH}_3 \mathrm{CONHCH}_3$
(b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CONH}_2$
(c) $\mathrm{CH}_3 \mathrm{CONH}_2$
(d) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONH}_2$
Answer:
(a) $\mathrm{CH}_3 \mathrm{CONHCH}_3$
Only primary amides undergo hoffmann bromamide reaction
Question 4.
Assertion : Acetamide on reaction with $\mathrm{KOH}$ and bromine gives acetic acid
Reason: Bromine catalyses hydrolysis of acetamide.
(a) if both assertion and reason are true and reason is the correct explanation of assertion,
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false

Question 5.

(a) bromomethane
(b) a - bromo sodium acetate
(c) methanamine
(d) acetamide
Answer:
(c) methanamine

Question 6.
Which one of the following nitro compounds does not react with nitrous acid?
(a) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NO}_2$
(b) $\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{NO}_2$
(c) $\left(\mathrm{CH}_3\right)_3 \mathrm{C} \mathrm{NO}_2$
(d)

Answer:
(c) $\left(\mathrm{CH}_3\right)_3 \mathrm{CNO}_2-3^0$ nitroalkane
Question 7.
Aniline + benzoylchloride $\stackrel{\mathrm{NaOH}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5-\mathrm{NH}-\mathrm{COC}_6 \mathrm{H}_5$ this reaction is known as
(a) Friedel - crafts reaction
(b) HVZ reaction
(c) Schotten - Baumann reaction
(d) none of these
Answer:
(c) Schotten - Baumann reaction
Question 8.
The product formed by the reaction an aldehyde with a primary amine
(a) carboxylic acid
(b) aromatic acid
(c) schiff 's base
(d) ketone
Answer:
(c) schiff ' $s$ base

Question 9 .
Which of the following reaction is not correct.

Answer:

Question 10.
When aniline reacts with acetic anhydride the product formed is
(a) o-aminoacetophenone
(b) $\mathrm{m}$ - aminoacetophcnone
(c) $\mathrm{p}$ - aminoacetophenone
(d) acetanilide
Answer:
(d) acetanilide

Question 11.
The order of basic strength for methyl substituted amine solution is
(a) $\mathrm{N}\left(\mathrm{CH}_3\right)_3>\mathrm{N}\left(\mathrm{CH}_3\right)_2 \mathrm{H}>\mathrm{N}\left(\mathrm{CH}_3\right) \mathrm{H}_2>\mathrm{NH}_3$
(b) $\mathrm{N}\left(\mathrm{CH}_3\right) \mathrm{H}_2>\mathrm{N}\left(\mathrm{CH}_3\right)_2 \mathrm{H}>\mathrm{N}\left(\mathrm{CH}_3\right)_3>\mathrm{NH}_3$
(c) $\mathrm{NH}_3>\mathrm{N}\left(\mathrm{CH}_3\right) \mathrm{H}_2>\mathrm{N}\left(\mathrm{CH}_3\right)_2 \mathrm{H}>\mathrm{N}\left(\mathrm{CH}_3\right)_3$
(d) $\mathrm{N}\left(\mathrm{CH}_3\right)_2 \mathrm{H}>\mathrm{N}\left(\mathrm{CH}_3\right) \mathrm{H}_2>\mathrm{N}\left(\mathrm{CH}_3\right)_3>\mathrm{NH}_3$
Answer:
(d) $\mathrm{N}\left(\mathrm{CH}_3\right)_2 \mathrm{H}>\mathrm{N}\left(\mathrm{CH}_3\right) \mathrm{H}_2>\mathrm{N}\left(\mathrm{CH}_3\right)_3>\mathrm{NH}_3$
Question 12 .

(a) $\mathrm{H}_3 \mathrm{PO}_2$ and $\mathrm{H}_2 \mathrm{O}$
(b) $\mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}$
(c) $\mathrm{HgSO}_4 / \mathrm{H}_2 \mathrm{SO}_4$
(d) $\mathrm{Cu}_2 \mathrm{Cl}_2$
Answer:
(a) $\mathrm{H}_3 \mathrm{PO}_2$ and $\mathrm{H}_2 \mathrm{O}$

Question 13.
$
\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2 \stackrel{\mathrm{Fe} / \mathrm{HCl}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{NaNO}_2 / \mathrm{HCl}}{273 \mathrm{~K}} \rightarrow \mathrm{B} \underset{283 \mathrm{~K}}{\longrightarrow} \mathrm{C} \text {. 'C' }
$
(a) $\mathrm{C}_6 \mathrm{H}_5-\mathrm{OH}$
(b) $\mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}_2 \mathrm{OH}$
(c) $\mathrm{C}_6 \mathrm{H}_5-\mathrm{CHO}$
(d) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$
Answer:
(a) $\mathrm{C}_6 \mathrm{H}_5-\mathrm{OH}$
$
\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2 \stackrel{\mathrm{Fe} / \mathrm{HCl}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \stackrel{\mathrm{NaNO}_2 / \mathrm{HCl}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5-\mathrm{N}_2 \mathrm{Cl} \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5-\mathrm{OH}+\mathrm{N}_2+\mathrm{HCl}
$
Question 14.
Nitrobenzene on reaction with at $80-100^{\circ} \mathrm{C}$ forms which one of the following products?
(a) 1,4 - dinitrobenzene
(b) 2, 4, 6-tirnitrobenzene
(c) 1,2-dinitrobenzene
(d) 1,3 - dinitrobenzene
Answer:
(d) 1,3-dinitrobenzene

Question 15 .
$\mathrm{C}_5 \mathrm{H}_{13} \mathrm{~N}$ reacts with $\mathrm{HNO}_2$ to give an optically active compound - The compound is
(a) pentan - 1- amine
(b) pentan -2 - amine
(c) $\mathrm{N}, \mathrm{N}$ - dimethylpropan - 2 - amine
(d) $\mathrm{N}-$ methylbutan -2 - amine
Answer:
(d) $\mathrm{N}$ - methylbutan -2 -amine

Question 16.
Secondary nitro alkanes react with nitrous acid to form
(a) red solution
(b) blue solution
(c) green solution
(d) yellow solution
Answer:
(b) blue solution
Question 17.
Which of the following amines does not undergo acetylation?
(a) $t$ - butylamine
(b) ethylamine
(c) diethylamine
(d) triethylamine
Answer:
(d) triethyl amine ( $3^{\circ}$ amine)
Question 18.
Which one of the following is most basic?
(a) 2,4-dichloroaniline
(b) 2, 4-dimethyl aniline
(c) 2,4-dinitroaniline
(d) 2, 4-dibromoaniline
Answer:
(b) 2,4-dimethyl aniline
$\mathrm{CH}_3$ is a +1 group, all other - I group. +1 group increase the electron density on $\mathrm{NH}_2$ and hence increases the basic nature.
Question 19.
When is reduced with $\mathrm{Sn}$ / $\mathrm{HCI}$ the pair of compounds formed are
(a) Ethanol, hydrozylamme hydrochloride
(b) Ethanol, ammonium hydroxide
(c) Ethanol, $\mathrm{NH}_2 \mathrm{OH}$
(d) $\mathrm{C}_3 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{H}_2 \mathrm{O}$
Answer:
(a) Ethanol, hydrozylamine hydrochloride
Question 20.
TUPAC name for the amine

(a) 3-Bimethylamino-3-methyl pentane
(b) $3(\mathrm{~N}, \mathrm{~N}$ - Triethyl) -3 -amino pentane
(c) $3-\mathrm{N}, \mathrm{N}-$ trimethyl pentanamine
(d) $3-(\mathrm{N}, \mathrm{N}-$ Dimethyl amino) - 3 - methyl pentane
Answer:
(d) 3 - (N, N-Dimethyl amino) - 3 - methyl pentane
Question 21.

$\text { Product ' } \mathrm{P} \text { ' in the above reaction is }$

Answer:

Question 22 .
Ammonium salt of bcnzoic acid is heated strongly and the product so formed is reduced and then treated with $\mathrm{NaNO}_2 / \mathrm{HCl}$ at low temperature. The final compound formed is .............
(a) Benzene diazonium chloride
(b) Benzyl alcohol
(c) Phenol
(d) Nitrosobenzene
Answer:
(b) Benzyl alcohol

Question 23.
Identify $\mathrm{X}$ in the sequence give below.

Answer:

Question 24.
Among the following, the reaction that proceeds through an electrophilic substitution, is

Answer:

Question 25.
The major product of the following reaction

Answer:

Short Answer Questions
Question 1.
Write down the possible isomers of lthe $\mathrm{C}_4 \mathrm{H}_9 \mathrm{NO}_2$ give their IUPA names.
Answer:

Question 2.
There are two isomers with the formula $\mathrm{CH}_3 \mathrm{NO}_2$. How will you distinguish between them?
Answer:
$\mathrm{CH}_3 \mathrm{NO}_2$ has two isomers. They are
1. $\mathrm{CH}_3-\mathrm{NO}_2$ (Nitromethane)
2.

Question 3.
What happens when
1. 2 - Nitropropane boiled with $\mathrm{HCI}$
2. Nitrobenezen electrolytic reduction in strongly acidic medium.
3. Oxidation of tert - butylamine with $\mathrm{KMnO}_4$
4. Oxidation of acetoneoxime with triuluoroperoxy acetic acid.
Answer:
1. 2 - Nitropropane boiled with $\mathrm{HCI}$ :
2 - nitropropane upon hydrolysis with boiling $\mathrm{HCl}$ give a ketone (2 - propanone) and nitrous oxide.

2. Nitrobenezen electrolytic reduction in strongly acidic medium:
Electrolytic reduction of nitrobenzene in weakly acidic medium gives aniline but in strongly acidic medium, it gives $\mathrm{p}$ - aminophenol obviously through the acid - cataLysed rearrangement of the initially formed phenyihydroxylamine.

3. Oxidation of tert - butylamine with $\mathrm{XMnO}_4$ :
In general, primary amines, in which the $-\mathrm{NH}_2$ group is attached to a tertiary carbon can be oxidised with $\mathrm{KMnO}_4$ to the corresponding nitro compound in excellent yield. Therefore $3^{\circ}-$ butylamine oxidised to give 2 - methyl - $2-$ nitropropane.

4. Oxidation ofacetoneoxime with trifluoroperoxy acetic acid: Oxidation ofacetoneoxime with trifluoroperoxy acetic acid gives 2 - nitropropane.

Question 4.
How will you convert nitrobenzene into
1. 1,3,5-trinitrobenzene
2. $o$ and $\mathrm{p}$ - nitrophenol
3. $\mathrm{m}$ - nitro aniline
4. azoxybenzene
5. hydrozabenzene
6. $\mathrm{N}$ - phenylhydroxylamine
7. aniline
Answer:
1. Conversion of nitrobenzene into $1,3,5$-trinitrobenzene:

2. Conversion of nitrobenzene into $\mathrm{o}$ and $\mathrm{p}$ - nitrophenol:
(a) Method I:
Nitrobenzene heated with solid $\mathrm{KOH}$ at $340 \mathrm{~K}$ gives a low yield of a mixture of $0-$ and $\mathrm{P}$ - nitrophenols.

(b) Method II:

3. Conversion of nitrobenzene into $\mathrm{m}$ - nitro aniline:

4. Conversion of nitrobenzene into azoxybenzene:

5. Conversion of nitrobenzene into hydrazobenzene:

6. Conversion of nitrobenzene into $\mathrm{N}$ - phenylhydrozylamine:

7. Conversin of nitrobenzene into aniline:

Question 5.
Identify compounds $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in the following sequence of reactions.
i.
ii. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2 \mathrm{Cl} \stackrel{\mathrm{CuCN}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{NH}_3}{\longrightarrow} \mathrm{C}$
iii.
iv. $\mathrm{CH}_3 \mathrm{NH}_2 \stackrel{\mathrm{CH}_3 \mathrm{Br}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{CH}_3 \mathrm{COCl}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{B}_2 \mathrm{H}_6}{\longrightarrow} \mathrm{C}$

(v)

(vi)

(vii)

Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Question 6.
Write short flotes on the following
1. Hoffmann's bromide reaction
2. Ammonolysis
3. Gabriel phthalimide synthesis
4. Schotten - Baumann reaction
5. Carbylamine reaction
6. Mustard oil reaction
7. Coupling reaction
8. Diazotisation
9. Gomberg reaction
Answer:
1. Hoffmann's bromide reaction:
When Amides are treated with bromine in the presence of aqueous or ethanolic solution of $\mathrm{KOH}$, primary amines with one carbon atom less than the parent amides are obtained.

2. Ammonolysis:
When Alkyl halides (or) benzylhalides are heated with alcoholic ammonia in a sealed tube, mixtures of $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ amines and quaternary ammonium salts are obtained.

3. Gabriel phthalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic $\mathrm{KOH}$ forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine.

4. Schotten - Baumann reaction:
Aniline reacts with benzoylchloride $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl}\right)$ in the presence of $\mathrm{NaOH}$ to give $\mathrm{N}$ - phenyl benzamide. This reaction is known as Schotten Baumann reaction. The acylation and benzoylation are nucleophilic substitutions.

5. Carbylamine reaction:
Aliphatic (or) aromatic primary amines react with chloroform and alcoholic $\mathrm{KOH}$ to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test used to identify the primary amines.

6. Mustard oil reaction:
When primary amines are treated with carbon disuiphide $\left(\mathrm{CS}_2\right), \mathrm{N}-$ alkyldithio carbonic acid is formed which on
subsequent treatment with $\mathrm{HgCI}_2$, give an alkyl isothiocyanate.

7. Coupling reaction:
Benzene diazonium chloride reacts with electron-rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds. Coupling generally occurs at the para position. If para position is occupied then coupling occurs at the ortho position. Coupling tendency is enhanced if an electron-donating group is present at the para position to $-\mathrm{N}_2^{+} \mathrm{Cl}^{-}$group. This is an electrophilic substitution.

8. Diazotisation:
Aniline reacts with nitrous acid at low temperature $(273-278 \mathrm{~K})$ to give benzene diazonium chloride which is stable for a short time and slowly decompose seven at low temperatures. This reaction is known as diazotization.

9. Gomberg reaction
Benzene diazonium chloride reacts with benzene in the presence of sodium hydroxide to give biphenyl. This reaction in known as the Gomberg reaction.

Question 7.
How will you distinguish between primary secondary and tertiary alphatic amines.
Answer:

Question 8.
Account for the following
1. Aniline does not undergo Friedel - Crafts reaction
2. Diazonium salts of aromatic amines are more stable than those of aliphatic amines
3. $\mathrm{pk}_{\mathrm{b}}$ of aniline is more than that of methy lamine
4. Gabriel phthalimide synthesis is preferred for synthesising primary amines.
5. Ethylamine is soluble in water whereas aniline is not
6. Amines are more basic than amides
7. Although amino group is o-and $\mathrm{p}$ - directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of $m$ - nitroaniline.
Answer:
1. Aniline does not undergo Friedel - Crafts reaction:
Aniline being a Lewis base reacts with Lewis acid $\mathrm{AiCl}_3$ to form a salt.

Due to the presence of a positive charge on $\mathrm{N}$ - atom in the salt the group $-\mathrm{NH}_2 \mathrm{AlCl}_3^{-}$acts as a strongly deactivating group. As a result, it reduces the electron density in the benzene ring and which inhibits the electrophilic substitution reaction. Therefore aniline does not under go Friedel - Crafts reaction.
2. Diazonium salts of aromatic amines are more stable than those of aliphatic amines:
The diazonium salts of aromatic amines are more stable than those of aliphatic amines due to dispersal of the
positive charge on the benzene ring as shown below.

3. $\mathrm{pK}_{\mathrm{b}}$ of aniline is more than that of methylamine:
In aniline, the lone pair of electrons on the $\mathrm{N}$ - atom is delocalized over the benzene ring. As a result electron density on the nitrogen decreases. In contrast in $\mathrm{CH}_3 \mathrm{NH}_2$, $+\mathrm{I}$ effect of $\mathrm{CH}_3$ increases the electron density on the $\mathrm{N}$ atom. Therefore, aniline is a weaker base than methylamine and hence its $\mathrm{pK}$ value is more than that of methyl amine.
4. Gabriel phthalimide synthesis is preferred for synthesising primary amines:
Gabriel phthalimide reaction gives pure 10 amine without any contamination of $2^{\circ}$ and $3^{\circ}$-amines. Therefore it is preferred for synthesising primary amines.

5. Ethylamine is soluble in water whereas aniline is not:
Ethylamine when added to water forms intermolecular $\mathrm{H}$ - bonds with water. And therefore it is soluble in water. But aniline does not form $\mathrm{H}$ - bond with water to a very large extent due to the presence of a large hydrophobic $\mathrm{C}_6 \mathrm{H}_5$ group. Hence, aniline is insoluble in water.

6. Amines are more basic than amides:
In simple amines, the lone pair of electrons is on nitrogen and hence available for protonation. In amides on the other hand, the electron pair on nitrogen is delocalised to the carboxyl oxygen through resonance and thus it is not available for protonation. So amines are more basic than amides.

7. Although amino group is $o-$ and $\mathrm{p}$ - directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of $\mathrm{m}$ - nitroaniline:
Nitration is usually carried out with a mixture of conc $\mathrm{HNO}_3$ and conc $\mathrm{H}_2 \mathrm{SO}_4$. In the presence of these acids, most
of aniline gets protonated to form anilinium ion. Therefore, in the presence of acids, the reaction mixture consists of aniline and anilinium ion.
Now - $\mathrm{NH}_2$, group in aniline is $\mathrm{O}, \mathrm{P}$ - directing and activating while the $-\mathrm{NH}_3$ group is anilinium ion is meta directing and deactivating. Whereas nitration of aniline (due to steric hindrance at $o-$ position) mainly gives $\mathrm{p}$ nitroaniline, the nitration of anilinium ion gives $\mathrm{m}$ - nitro aniline. In actual practice, approximately a $1: 1$ mixture of $\mathrm{P}$ and $\mathrm{m}$ - nitroaniline is formed.

Question 9.
Arrange the following
1. In increasing order of solubility in water, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2,\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$
2. In increasing order of basic strength
- aniline, $\mathrm{p}$ - toludine and $\mathrm{p}$ - nitroaniline
- $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{p}-\mathrm{Cl}-\mathrm{C}_6 \mathrm{H}_4-\mathrm{NH}_2$
3. In decreasing order of basic strength in gas phase. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2,\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{NH},\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}$ and $\mathrm{NH}_3$
4. In Increasing order of boiling point $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH},\left(\mathrm{CH}_3\right)_2 \mathrm{NH}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$
5. In decreasing order of the $\mathrm{pK}_{\mathrm{b}}$ values $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3,\left(\mathrm{C}_2 \mathrm{H}\right)_2 \mathrm{NH}_4$ and $\mathrm{CH}_3 \mathrm{NH}_2$
6. Increasing order of basic strength $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right)_2,\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{NH}$ and $\mathrm{CH}_3 \mathrm{NH}_2$
7. In decreasing order of basic strength

Answer:
1. Solubility decreases with increase in molecular mass of amines due to increase in the size of a hydrophobic hydrocarbon part and with decrease in the number of $\mathrm{H}$ - atoms on the $\mathrm{N}$ - atom which undergo $\mathrm{H}$ - bonding.
Now among the given compounds $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$ has the highest molecular mass of 93 followed by $\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}_{\text {with }}$ molecular mass of 73 with $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$ has the lowest molecular mass of 45 . Thus the solubility increases in the order in which molecular mass decreases.
2. (a) The electron - donating groups increases the basic strength of amines while the electron - withdrawing groups decrease the basic strength of amines. Therefore $\mathrm{p}$ - nitroaniline is the weakest base followed by aniline while $\mathrm{p}$ -
toluidine, which has methyl group and therefore it is the strongest base. Basic strength increases in the order. $\mathrm{P}$ nitro aniline $<$ aniline $<\mathrm{p}$ - toluidine
(b) Chlorine atom has both $-\mathrm{I}$ effect and $+\mathrm{R}$ effect since - I effect out weights the $+\mathrm{R}$ effect, therefore $\mathrm{p}-$ chloro aniline is weak base than aniline. Aikyl groups are electron - donating groups."

As a result the electron density on the nitrogen atom increases in the ethylamine and thus they can donate lone pair of electrons niore easily. Therefore Ethylamine is more base than aromatic amines.

Due to delocalization of lone pair of electrons of the $\mathrm{N}$ - atom over the benzene ring, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}$, and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3$ are far less basic than $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$. Further due to +1 effect of the $\mathrm{CH}_3$ group, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3$ is little more basic than $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$. Therefore increasing order basic strength is
3. In the gas phase, solvent effects i.e., stabilization of the conjugate acids due to $\mathrm{H}$ - bonding, are absent. Therefore, in the gas phase, basic strength mainly depends upon the +1 effect of the alkyl groups. Since the +1 effect increases with the number of ailcyl groups,
therefore the basic strength of the amines decreases as the number of ethyl groups decreases from three in $\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}$ to two in $\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}$ to one in $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$ and zero in $\mathrm{NH}_3$. Basic strength in the gas phase decreases in the order is, $\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}>\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{~N}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{NH}_3$
4. Since the electro negativity of $\mathrm{O}$ is higher than that of $\mathrm{N}$, therefore, alcohols form stronger $\mathrm{H}-0$ bonds than amines. In other words, the boiling points of alcohols are higher than those of amines of comparable molecular masses. Therefore the boiling point of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(46)$ is higher than those of $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}(45)$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2(45)$.
Further since the extent of $\mathrm{H}$ - bonding depends upon the number of $\mathrm{H}$-atoms on the $\mathrm{N}$-atom. Therefore $1^{\circ}-$ amines with two $\mathrm{H}$ - atoms on the $\mathrm{N}$ - atom have higher boiling points than $2^{\circ}$ - amines having only one $\mathrm{H}$ - atom. Therefore the boiling point of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$ is higher than that of $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$. Increasing order of boiling point is, $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}<\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}<\mathrm{CH}_5 \mathrm{OH}$

5. Due to delocalization of lone pair of electrons of the $\mathrm{N}$ - atom over the benzene ring, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3$ is far less basic than $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2,\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{NH}$ and $\mathrm{CH}_3 \mathrm{NH}_2$. Among $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$ and $\left(\mathrm{C}_2 \mathrm{H}_5\right), \mathrm{NH},\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}$ is more basic than $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$ due to greater +1 effect of the two $\mathrm{C}_2 \mathrm{H} 5$ groups and stabilization of its conjugate acid by $\mathrm{H}-$ bonding.
Compare to Ethyl and methyl group, $\mathrm{C}_2 \mathrm{H}_5$ - group has more +1 effect than $\mathrm{CH}_3$ - group. Therefore methylamine is weak base than ethylamine. Combining all these facts the relative basic strength of these four amines decreases in the order.
$\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\mathrm{CH}_3 \mathrm{NH}_2>\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3$. Since a stronger base has a lower $\mathrm{pK}_{\mathrm{b}}$ value therefore, $\mathrm{pK}_{\mathrm{b}}$ values decrease in the reverse order. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCH}_3>\mathrm{CH}_3 \mathrm{NH}_2>\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2>\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}$
6. Due to delocalization of lone pair of electrons of the $\mathrm{N}$-atom over the benzene ring, all aromatic amines are less basic than alkylamines i.e., $\mathrm{CH}_3 \mathrm{NH}_2$. Presence of electron - donating groups $\left(-\mathrm{CH}_3\right)$ on the $\mathrm{N}-$ atom increases the basicity of substituted aniline with respect to $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$.
In $\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{NH}$, the lone pair of electrons on the $\mathrm{N}$ - atom is delocalized over two benzene rings instead of one in $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$, therefore $\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{NH}$ is much less basic than $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$. Combining all the three trends together, the basic strength of the four amines increasing in the order.
$
\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{NH}<\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2<\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}\left(\mathrm{CH}_3\right)_2<\mathrm{CH}_3 \mathrm{NH}
$
7. Aliphatic amines are more basic than aromatic amines. Therefore $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2$ and $\mathrm{CH}_3 \mathrm{NH}_2$ are more basic. Among the ethylamine and methylamine. ethylamine was experienced more +1 effect than methylamine and hence ethylamine is more basic than methylamine.
Nitrogroup has a powerful electron withdrawing group and they have both $-\mathrm{R}$ effect as well as - I effect. As a result, all the nitro anilines are weaker bases than aniline. In $\mathrm{P}$ - nitroaniline

$\text { both - } \mathrm{R} \text { effect and - I effect of the } \mathrm{NO}_2 \text { group decrease the basicity. Therefore decreasing order of basic strength is, }$

Ethylamine $>$ Methylamine $>$ Aniline $>\mathrm{p}-$ nitro aniline
Question 10.
How will you prepare propan -1 - amine from
1. butane nitrile
2. propanamide
3. 1 - nitropropane
Answer:
1. Preparation of propan -1- amine from butane nitrile.
Butane nitrile treated with acid hydrolysis followed by Hoffmann's bromamide degradation. gives propan - 1 amine.

2. Preparation of propan $-1-$ amine from propanamide.
When propanamide is treated with $\mathrm{LiAIH}_4$ in the presence of water gives propan -1 - amine.

3. Preparation of propan - 1 - amine from 1 - nitropropane.

Reduction of 1 - Nitropropane using $\mathrm{H}_2 / \mathrm{Ni}$ or $\mathrm{Fe} / \mathrm{HCl}$ gives propan -1 - amine.
Question 11.
Identify $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D

Answer:

Question 12 .
How will you convert dlethylamine into
1. N, N-dlethylacetamide
2. $\mathrm{N}$ - nitrosodiethylamine
Answer:
1. Conversion of diethylamine into $\mathrm{N}, \mathrm{N}$-diethylacetamide.

Diethylamine react with acetyichioride in the presence of pyridine to form $\mathrm{N}, \mathrm{N}$ - diethyl acetamide.
2. Conversion of diethylamine into $\mathrm{N}$ - nitrosodiethylamine.

Question 13
Identify $A, B$ and $C$

Answer:

Question 14 .
Identify $A, B, C$ and $D$

Answer::

Question 15 .
Complete the following reaction

Answer:

Question 16 .
Predict $A, B, C$ and $D$ for the follwing reaction

Answer:

Question 17.
A dibromo derivative $(\mathrm{A})$ on treatment with $\mathrm{KCN}$ followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of $\mathrm{CO}_2$. (B) on heating with liquid ammonia followed by treating with $\mathrm{Br}_2 / \mathrm{KOH}$ gives (C) which on treating with $\mathrm{NaNO}_2$ and $\mathrm{HCI}$ at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74 . Identify A to D.
Answer:

Question 18.
Identify $\mathrm{A}$ to $\mathrm{E}$ in the following frequncy of reactions.

Answer:

Question 1.
Write all possible isomers for the following compounds.
1. $\mathrm{C}_2 \mathrm{H}_5-\mathrm{NO}_2$
2. $\mathrm{C}_3 \mathrm{H}_7-\mathrm{NO}_2$
Answer:
1. Possible isomers for $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NO}_2$ as following
(a) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{NO}_2-$ Nitroethane
(b) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{O}-\mathrm{N}=\mathrm{O}-$ Ethyl nitrite
(c)

(d) $\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH}-$ Glycine (amino acid)

(e)

(f)

(g)

2. Possible isomers for $\mathrm{C}_3 \mathrm{H}_7 \mathrm{NO}_2$ as follows.
(a) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NO}_2-1-$ Nitropropane
(b) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{O}-\mathrm{N}=\mathrm{O}-$ propane -1 - nitrite

(c)

(d)

(e)

(f) $\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{COOH}-$ Alanine
Question 2 .
Find out the product of the following reactions

(i)

(ii)

Answer:

(i)

(ii)

Question 3.
Predict the major product that would be obtained on nitration of the following compounds.

(i)

(ii)

(iii)

Answer:

(i)

(ii)

(iii)

Question 4.
Draw the structure of the following compounds
1. Neopentylamine
2. Tert - butylamine
3. $\alpha$-amino propionaldehyde
4. tribenzylamine
5. $\mathrm{N}-$ ethyl $-\mathrm{N}-$ methylhexan -3 - amine
Answer:
1. Neopentylamine:

2. Tert - butylamine

3. $\alpha$-amino propionaldehyde

4. tribenzylamine

5. $\mathrm{N}$ - ethyl $-\mathrm{N}-$ methylhexan -3 - amine

Question 5.
Give the correct IUPAC names for the following amines.

(i)

(ii)

(iii)

(iv)

(v)

Answer:

(i)

(ii)

(iii)

(iv)

(v)