Question 12 Marks
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Question 22 Marks
Describe the basic strength of aryl amine compounds relative to ammonia.
Question 32 Marks
State the importance of diazonium salts in the synthesis of aromatic compounds.
Answer
View full question & answer→→The diazonium salts are very good intermediates for the introduction of $- F ,- Cl ,- Br ,- I ,- CN$, $OH ,- NO _2$ groups into the aromatic ring.
→Aryl fluorides and iodides can not be prepared by direct halogenation. The cyano group can not be introduced by nucleophilic substitution of chlorine in chlorobenzene but cyanobenzene can be easily obtained from diazonium salt.
→Thus, the replacement of diazo group by other groups is helpful in preparing those substituted aromatic compounds which can not be prepared by direct substitution in benzene or substituted benzene.
→Aryl fluorides and iodides can not be prepared by direct halogenation. The cyano group can not be introduced by nucleophilic substitution of chlorine in chlorobenzene but cyanobenzene can be easily obtained from diazonium salt.
→Thus, the replacement of diazo group by other groups is helpful in preparing those substituted aromatic compounds which can not be prepared by direct substitution in benzene or substituted benzene.
Question 42 Marks
Explain the displacement of diazonium group of benzene diazonium salt by hydrogen with equation.### Explain the reduction process of benzene diazonium salt by giving an equation.
Question 52 Marks
Explain the displacement equation of diazonium group of benzene diazonium salt by iodide ion and chloride ion.
Question 62 Marks
State the types of chemical reaction of diazonium salts.
Answer
View full question & answer→→The reactions of diazonium salts can be broadly divided into two categories:
(A)Reactions involving displacement of nitrogen, and
(B) Reactions involving retention of diazo group.
(A)Reactions involving displacement of nitrogen, and
(B) Reactions involving retention of diazo group.
Question 72 Marks
Describe the physical properties of benzenediazonium salts.
Answer
View full question & answer→→Benzene diazonium chloride is a colourless crystaline solid.
→It is readily soluble in water and is stable in cold but reacts with water when warmed.
→It decomposes easily in the dry state.
→Benzene diazonium fluroborate is water insoluble and stable at room temperature.
→It is readily soluble in water and is stable in cold but reacts with water when warmed.
→It decomposes easily in the dry state.
→Benzene diazonium fluroborate is water insoluble and stable at room temperature.
Question 82 Marks
Give the resonance structure of arenediazonium ion.
Answer
View full question & answer→→Resonance structure of arenediazonium ion is as follows.
→They are named by suffixing diazonium to the name of the parent hydrocarbon from which they are formed, followed by the name of anion such as chloride, hydrogensulphate, etc.
→The $N _2^{+}$group is called diazonium group.
→For example, $C _6 H _5 \stackrel{+}{ N } Cl ^{-}$is named as benzene diazonium chloride and $C _6 H _5 \stackrel{+}{ N } HSO _4^{-}$is known as benzenediazonium hydrogen sulphate.
→They are named by suffixing diazonium to the name of the parent hydrocarbon from which they are formed, followed by the name of anion such as chloride, hydrogensulphate, etc.
→The $N _2^{+}$group is called diazonium group.
→For example, $C _6 H _5 \stackrel{+}{ N } Cl ^{-}$is named as benzene diazonium chloride and $C _6 H _5 \stackrel{+}{ N } HSO _4^{-}$is known as benzenediazonium hydrogen sulphate.
Question 92 Marks
Give the general formula of diazonium salt. State the possible negative ions present in this salt and explain with an example how diazonium salts are named.
Answer
View full question & answer→→The diazonium salts have the general formula $RN _2^{+} X ^{-}$, where R stands for an aryl group and $X ^{-}$ion may be $Cl ^{-}, Br ^{-}, HSO _4^{-}, BF _4^{-}$etc.
Question 102 Marks
Short note: Sulphonation of Aniline.
Question 112 Marks
Explain the alkylation process of amine compounds with example.
Question 122 Marks
State the factors that determine the basicity of alkyl amine compounds in aqueous medium and explain the order of basicity of alkyl amine compounds based on these factors.
Answer
View full question & answer→$\rightarrow$ There is a subtle interplay of the inductive effect, solvation effect and steric hinderance of the alkyl group which decides the basic strength of alkyl amines in aqueous state.
$\rightarrow$ In the aqueous phase the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group $(+I)$ but also by solvation with water molecules.
$\rightarrow$ The greater the size of the ion, lesser will be the solvation and the less stabilized is the ion.
$\rightarrow$ The order of stability of ions are as follows:
$\rightarrow$ Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base.
$\rightarrow$ Thus, the order of basicity of aliphatic amines should be primary $>$ secondary $>$ tertiary which is opposite to the inductive effect based order.
$\rightarrow$ Secondly, when the alkyl group is small like $-CH_3$ group there is no steric hinderance to $H-$bonding.
$\rightarrow$ In case the alkyl group is bigger than $CH _3$ group there is steric hinderance to $H-$bonding.
$\rightarrow$ Secondly when the alkyl group is small like $- CH _3$ group there is no steric hinderance to $H-$donimg.
$\rightarrow$ In case the alkyl group is bigger than $CH_3$ group there is steric hinderance to $H-$bonding.
$\rightarrow$ Therefore, the change of nature of the group e.g. from $- CH _3$ to $- C _2 H _5$
results in change of the order of basic strength.
$\rightarrow$ The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution is as follows:
$\rightarrow$ $ \left( C _2 H _5\right)_2 NH >\left( C _2 H _5\right)_3 N< C _2 H _5 NH _2> NH _3$
$ \left( CH _3\right)_2 NH > CH _3 NH _2>\left( CH _3\right)_3 N> NH _3$
$\rightarrow$ In the aqueous phase the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group $(+I)$ but also by solvation with water molecules.
$\rightarrow$ The greater the size of the ion, lesser will be the solvation and the less stabilized is the ion.
$\rightarrow$ The order of stability of ions are as follows:
$\rightarrow$ Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base.
$\rightarrow$ Thus, the order of basicity of aliphatic amines should be primary $>$ secondary $>$ tertiary which is opposite to the inductive effect based order.
$\rightarrow$ Secondly, when the alkyl group is small like $-CH_3$ group there is no steric hinderance to $H-$bonding.
$\rightarrow$ In case the alkyl group is bigger than $CH _3$ group there is steric hinderance to $H-$bonding.
$\rightarrow$ Secondly when the alkyl group is small like $- CH _3$ group there is no steric hinderance to $H-$donimg.
$\rightarrow$ In case the alkyl group is bigger than $CH_3$ group there is steric hinderance to $H-$bonding.
$\rightarrow$ Therefore, the change of nature of the group e.g. from $- CH _3$ to $- C _2 H _5$
results in change of the order of basic strength.
$\rightarrow$ The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution is as follows:
$\rightarrow$ $ \left( C _2 H _5\right)_2 NH >\left( C _2 H _5\right)_3 N< C _2 H _5 NH _2> NH _3$
$ \left( CH _3\right)_2 NH > CH _3 NH _2>\left( CH _3\right)_3 N> NH _3$
Question 132 Marks
Explain the basicity of alkanamine compounds relative to ammonia in gaseous state based on inductive effect of the alkyl group.
Answer
View full question & answer→→Let us consider the reaction of an alkanamine and ammonia with a proton to compare their basicity.
→Due to the electron releasing nature of alkyl group is (R) pushes electrons towards nitrogen and thus makes the unshared electron pair more available for sharing with the proton of the acid.
→Moreover, the substituted ammonium ion formed from the amine gets stabilized due to dispersal of the positive charge by the +I effect of the alkyl group.
→Hence, alkylamines are stronger bases than ammonia.
→Thus, the basic nature of aliphatic amines should increase with increase in the number of alkyl groups.
→The order of basicity of amines in gaseous phase follows the expected order:
tertiary amine > secondary amine > primary amine > ammonia
→Due to the electron releasing nature of alkyl group is (R) pushes electrons towards nitrogen and thus makes the unshared electron pair more available for sharing with the proton of the acid.
→Moreover, the substituted ammonium ion formed from the amine gets stabilized due to dispersal of the positive charge by the +I effect of the alkyl group.
→Hence, alkylamines are stronger bases than ammonia.
→Thus, the basic nature of aliphatic amines should increase with increase in the number of alkyl groups.
→The order of basicity of amines in gaseous phase follows the expected order:
tertiary amine > secondary amine > primary amine > ammonia
Question 142 Marks
Explain the basic characterization of amine compounds with the process equation and state the importance of the process.
Answer
View full question & answer→→Amines, being basic nature react with acids to form salts.
→Amine salts on treatment with a base like NaOH, regenerate the parent amine.
$RN _{ N }^{+} H _3 X ^{-}+{ }^{-} OH \rightarrow R _{ N } H _2+ H _2 O + X ^{-}$
→Amine salts are soluble in water but insoluble in organic solvents like ether. This reaction is the basis for the seperation of amines from the non basic organic compounds insoluble in water.
→Amine salts on treatment with a base like NaOH, regenerate the parent amine.
$RN _{ N }^{+} H _3 X ^{-}+{ }^{-} OH \rightarrow R _{ N } H _2+ H _2 O + X ^{-}$
→Amine salts are soluble in water but insoluble in organic solvents like ether. This reaction is the basis for the seperation of amines from the non basic organic compounds insoluble in water.
Question 152 Marks
Discuss the effect of hydrogen bonding on the boiling point of isomeric amine compounds.
Question 162 Marks
Write note on solubility of amine compounds. Explain which of the amine and alcohol compounds has higher solubility ? Why ?
Answer
View full question & answer→→Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecule.
→However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part.
→Higher amines are essentially insoluble in water.
→Considering the electro negativity of nitrogen of amine and oxygen of alcohol as 3.0 and 3.5 respectively, this means that since the oxygen atom is more electrogative than the nitrogen atom, alcohol compounds can form strong hydrogen bonds with water. Hence, amine compounds have lower water solubility than alcohol compounds.
→Amines are soluble in organic solvents like alcohol, ether and benzene.
→However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part.
→Higher amines are essentially insoluble in water.
→Considering the electro negativity of nitrogen of amine and oxygen of alcohol as 3.0 and 3.5 respectively, this means that since the oxygen atom is more electrogative than the nitrogen atom, alcohol compounds can form strong hydrogen bonds with water. Hence, amine compounds have lower water solubility than alcohol compounds.
→Amines are soluble in organic solvents like alcohol, ether and benzene.
Question 172 Marks
State the general physical properties of amine compounds.
Answer
View full question & answer→→The lower aliphatic amines are gases with fishy odour
→Primary amines with three or more carbon atoms are liquid and still higher ones are solid.
→Aniline and other arylamines are usually colourless but get coloured on storage due to atmospheric oxidation.
→Primary amines with three or more carbon atoms are liquid and still higher ones are solid.
→Aniline and other arylamines are usually colourless but get coloured on storage due to atmospheric oxidation.
Question 182 Marks
Write a note оп formation of amine compounds by reduction of nitro compounds.
Answer
View full question & answer→→Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals in acidic medium.
→Nitroalkanes can also be similarly reduced to the corresponding alkanamines.
Reduction with iron scrap and hydrochloric acid is preferred because FeCl, formed gets hydrolysed to release hydrochloric acid during the reaction.
→$\begin{aligned} & Fe +2 HCl \rightarrow H _2+ FeCl _2 \\ & FeCl _2+ H _2 O \rightarrow Fe ( OH )_2+2 HCl \end{aligned}$
→Thus, only a small amount of hydrochloric acid is required to initiate the reaction.
→Nitroalkanes can also be similarly reduced to the corresponding alkanamines.
Reduction with iron scrap and hydrochloric acid is preferred because FeCl, formed gets hydrolysed to release hydrochloric acid during the reaction.
→$\begin{aligned} & Fe +2 HCl \rightarrow H _2+ FeCl _2 \\ & FeCl _2+ H _2 O \rightarrow Fe ( OH )_2+2 HCl \end{aligned}$
→Thus, only a small amount of hydrochloric acid is required to initiate the reaction.
Question 192 Marks
Explain the classification of amine compounds.
Answer
View full question & answer→$\rightarrow$ Amine compounds are classified as primary $(1^0)$ secondary $(2^0)$ and tertiary $(3^0)$ depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecule.
$\rightarrow$ If one hydrogen atom of ammonia is replaced by $R$ or $Ar,$ we get $RNH_2$ or $ArNH_2$ a primary amine $(10)$
$\rightarrow$ If two hydrogen atoms of ammonia or one hydrogen atom of $R - NH_2$ are replaced by another alkyl/aryl $(R \ ')$ group we get secondary amine $\left(2^0\right) R - NH - R ^{ \ \prime}$
$\rightarrow$ The second alkyl/aryl group may be same or different
$\rightarrow$ Replacement of another hydrogen atom by alkyl/aryl group leads to the formation of tertiary amine $\left(3^0\right)$.
Primary $\left(1^{\circ}\right)$ secondary $\left(2^{\circ}\right)$ tertiary $\left(3^{\circ}\right)$
$\rightarrow$ Amines are said to be 'simple' when all the alkyl or aryl groups are the same, and 'mixed' when they are different.
$\rightarrow$ If one hydrogen atom of ammonia is replaced by $R$ or $Ar,$ we get $RNH_2$ or $ArNH_2$ a primary amine $(10)$
$\rightarrow$ If two hydrogen atoms of ammonia or one hydrogen atom of $R - NH_2$ are replaced by another alkyl/aryl $(R \ ')$ group we get secondary amine $\left(2^0\right) R - NH - R ^{ \ \prime}$
$\rightarrow$ The second alkyl/aryl group may be same or different
$\rightarrow$ Replacement of another hydrogen atom by alkyl/aryl group leads to the formation of tertiary amine $\left(3^0\right)$.
Primary $\left(1^{\circ}\right)$ secondary $\left(2^{\circ}\right)$ tertiary $\left(3^{\circ}\right)$
$\rightarrow$ Amines are said to be 'simple' when all the alkyl or aryl groups are the same, and 'mixed' when they are different.
Question 202 Marks
Explain the structure of amine compounds by drawing a diagram.
Answer
View full question & answer→$\rightarrow$ Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons.
$\rightarrow$ Nitrogen orbitals in amines are therefore, $SP^3$ hybridized and the geometry of amines is pyramidal.
$\rightarrow$ Each of the three $SP^3$ hybridized orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines.
$\rightarrow$ The fourth orbital of nitrogen in all amines contains an unshared pair of electrons.
$\rightarrow$ The angle $C-N-E ($Where $E$ is Cor $H)$ is less than $109.5^\circ$
$\rightarrow$ For instance, it is $108^\circ$ in case of trimethylamine as in the figure below:
$\rightarrow$ Nitrogen orbitals in amines are therefore, $SP^3$ hybridized and the geometry of amines is pyramidal.
$\rightarrow$ Each of the three $SP^3$ hybridized orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines.
$\rightarrow$ The fourth orbital of nitrogen in all amines contains an unshared pair of electrons.
$\rightarrow$ The angle $C-N-E ($Where $E$ is Cor $H)$ is less than $109.5^\circ$
$\rightarrow$ For instance, it is $108^\circ$ in case of trimethylamine as in the figure below:
Question 212 Marks
What are amine compounds? Explain with an example ?
Answer
View full question & answer→→Amines can be considered as derivatives of ammonia.
→The compounds obtained when one, two or three hydrogen atoms in ammonia are displaced by an alkyl or aryl group are called amine compounds.
→The compounds obtained when one, two or three hydrogen atoms in ammonia are displaced by an alkyl or aryl group are called amine compounds.