Hinsberg's test is used for the identification of primary, secondary, and tertiary amines.
Hinsberg's reagent is benzenesulphonyl chloride (C_6​H₅​SO_2​Cl).
It reacts differently with primary, secondary, and tertiary amines.
(i) Hinsberg's reagent reacts with primary amines to form N− alkylbenzenesulphonyl amide which is acidic in nature and soluble in alkali.
Note: N− alkylbenzenesulphonyl amide contains a strong electron-withdrawing sulphonyl group. Due to this, the H− atom attached to nitrogen can be removed easily. Hence, it is acidic.
(ii) Hinsberg's reagent reacts with secondary amines to form a sulphonamide which is insoluble in alkali.
Note: As there is no hydrogen atom attached to the N atom in the sulphonamide, it is not acidic and insoluble in alkali.
(iii) Hinsberg's reagent does not react with tertiary amines.

(i) and (ii)
Eight isomers of C₄H₁₁N are
(a) \(\stackrel{4}{\mathrm{C}} \mathrm{H}_3-\stackrel{3}{\mathrm{C}} \mathrm{H}_2-\stackrel{2}{\mathrm{C}} \mathrm{H}_2-\stackrel{1}{\mathrm{C}} \mathrm{H}_2-\mathrm{NH}_2\)
                     Butan -1 -amine
                               (Primary)
(b)           CH₃
                    |
\(\stackrel{3}{CH}_3-{ }^2 \mathrm{CH}-\stackrel{1}{C} \mathrm{H}_2-\mathrm{NH}_2\)
            2-methyl propan -1-amine
                     (Primary)

(iii) Isomerism exhibited by different amines are:
(a) Chain isomers, i.e. have different carbon chains, (a) and (b), (c) and (d) (as discussed in part (i) and (ii)]
(b) Position isomers, i.e. functional group occupy different positions, (a) and (c), (b) and (d). 
(c) Metamers, i.e. different alkyl groups are attached to the same functional group, (e) and (f), (e) and (g).
(d) Functional isomers, i.e. they have different functional groups. All the three categories (1°, 2° and 3°) of amines are the functional isomers of each other.

\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right)\)
\(\log { \frac { 2.39\times { 10 }^{ -7 } }{ 2.15\times { 10 }^{ -8 } } } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \left( \frac { 1 }{ 650 } -\frac { 1 }{ 700 } \right)\)
\({ E }_{ a }=\frac { 19.147\times 650\times 700\times \left( \log { 23.9 } -\log { 2.15 } \right) }{ 50 } \)
\(=\frac { 19.147\times 650\times 700\times \left( 1.3783-0.3324 \right) }{ 50 } \)
\(=\frac { 19.147\times 650\times 700\times 1.0459 }{ 50\times 1000 } \)
\({ E }_{ a }=\frac { 19.147\times 13\times 7\times 1.0459 }{ 10 }\)
\(=182.23\ kJ/mol\ \)

\({ T }_{ 1 }=295K, \ { T }_{ 2 }=305K, \ { k }_{ 2 }=2{ k }_{ 1 } \ (given)\)
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
\(\log { 2 } = \ \frac { { E }_{ a } }{ 2.303\times 8.314 } \left( \frac { 305-295 }{ 305\times 295 } \right) \)
\({ E }_{ a } \ = \ \frac { 19.147\times 305\times 295\times 0.3010 }{ 10 } \)
\( { E }_{ a } \ = \ 51854.8 \ J \ { mol }^{ -1 }\)
\(=51.85 \ KJ \ { mol }^{ -1 }\)

For first order reaction, \(t=\frac { 2.303 }{ k } \log { \frac { a }{ a-x } } \)
99% completion means that x = 99% of a = 0.99 a
\({ t }_{ 99 }\%\)\(=\frac { 2.303 }{ k } \log { \frac { a }{ a-0.99a } } =\frac { 2.303 }{ k } \log { { 10 }^{ 2 } } =2\times \frac { 2.303 }{ k } \)
90% completion means that x = 90% of a = 0.90 a
\(\therefore \quad { t }_{ 90 }\%\)\(=\frac { 2.303 }{ k } \log { \frac { a }{ a-0.99a } } =\frac { 2.303 }{ k } \log { { 10 } } =2\times \frac { 2.303 }{ k } \)
\(\therefore \quad \frac { { t }_{ 99 } }{ { t }_{ 90 } } ={ \left( \frac { 2\times 2.303 }{ k } \right) }/{ \left( \frac { 2.303 }{ k } \right) }=2\quad \)\({ t }_{ 99 }\%\)\(=2\times { t }_{ 90 }\%\)

In this reaction, the analyte is heated with alcoholic potassium hydroxide and chloroform. If a primary amine is present, the isocyanide (carbylamine) is formed, as indicated by a foul odour. The carbylamine test does not give a positive reaction with secondary and tertiary amines.

(i) C₆H₅NH₂ < CH₃CH₂NH₂ < CH₃NHCH₃
(ii) (CH₃ )₃N < (CH₃)₂NH < CH₃NH₂

(i) Cannizzaro reaction Aldehydes (not having alpha hydrogen atoms) undergo self oxidation-reduction (disproportionation) reaction. One molecule is oxidized to carboxylic acid and another molecule is reduced to alcohol. Concentrated alkali is the reagent.
(iii) Cross aldol condensation between different carbonyl compounds (aldehydes and ketones). If both reactants contain alpha hydrogen atoms, four different products can be obtained.

a. Tertiary butyl ketone does not give precipitate with sodium bisulphate whereas acetone does.
b. Dialkyl cadmium is considered superior to grignard reagent for the preparation of a ketone from an acid chloride.

\({ t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ k } \)
\(=\frac { 0.693 }{ 5.5\times { 10 }^{ -14 }{ s }^{ -1 } }\)
\(=1.26\times { 10 }^{ 13 }s\)

\(\overset { 3 }{ C } { H }_{ 2 }=\overset { 2 }{ C } H\overset { 1 }{ C } { H }_{ 2 }NHCH_{ 3 }\quad N-Methylprop-2-en-1\quad amine\)

Only \({ 1 }^{ \circ }\) aromic amines undergo diazotisation. Benzylamine is a \({ 1 }^{ \circ }\) aliphatic amine and hence does not undergo diazotisation.

Carbylamine reaction yields isocyanides.

Activated complex has higher energy. When it decomposes. energy is always realeased and it may give products or reactants back.

(a) : The rate of constant of a reaction depends only on temperature and does not depend upon concentrations of the reactants.

(a) (i) Hofmann Bromamide Degradation Reaction This reaction is used for preparing amine contaning one carbon less than the starting amide. This method was developed for the preparation of primary amines by reacting an amide with Br₂/ Cl₂ in NaOH/KOH.
In this reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the N-atom.

\(\begin{equation} R-\mathrm{NH}_{2}+\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{NaBr}+2 \mathrm{H}_{2} \mathrm{O} \end{equation}\)
\(\begin{equation} \text { e.g. } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2}+\mathrm{Br}_{2}+4 \mathrm{NaOH}\\ \text { Amide } \end{equation}\)⟶\(\begin{equation} \begin{aligned} &3 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}+\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{NaBr}+2 \mathrm{H}_{2} \mathrm{O}\\ &\text { Amine } \end{aligned} \end{equation}\)
(ii) Diazotisation The conversion of primary aromatic amines into their diazonium salts is called diazotisation. Benzene diazonium chloride is prepared by the reaction of aniline with nitrous acid (which is produced by the reaction of NaNO₂ and HCI) at 273-278K or 0-5^oC as shown below:
\(\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{NaNO}_{2}+2 \mathrm{HCl} \stackrel{273-278 \mathrm{~K}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N^+}_{2}{\mathrm{Cl^-}}+ \mathrm{NaCl}+2 \mathrm{H}_{2} \mathrm{O} \end{array}\)
Due to its unstability, the diazonium salt is not generally stored and is used immediately after its preparation. 
(iii) Gabriel Phthalimide Synthesis When a phthalimide is treated with ethanolic KOH, it forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis forms corresponding primary amines as shown below.

Primary amines are produce through this method without the traces of secondary or tertiary amines. So, this method is preferred for the synthesis of primary amines.

According to the given information compound A (mol formula CH₅H₁₀) is an alkene.
A on ozonolysis gives two carbonyl compound. (i.e. have group) B and C.

B gives Fehling's test, so it is an aldehyde.
B also gives Tollen's test, so it has group. That means, B is acetaldehyde.
C does not give Fehling test, so it is a ketone. Also, it gives positive iodoform test so it contain group. That means, B is acetaldehyde.
C does not give Fehling test, so it is a ketone. Also, it gives positive iodoform test so it contain group.
Now, structure of alkene A can be obtained by writing the products of ozonolysis side by side with their groups facing each other. On removing the oxygen atoms and joining the remaining fragments by a double bond, the structure of alkene 'A' can be obtained which is 2-methylbut-2-ene.

Arrhenius equation. To deduce a quantitative relationship between rate constant and temperature, Arrhenius gave the following equation:
\(k=A{ e }^{ { { -E }_{ a } }/{ RT } }\quad \quad ..(i)\)
where A is a constant of proportionality, Ea is the activation energy which represents the minimum energy that the reacting molecules must possess before undergoing a reaction, T is the absolute temperature and R is the gas constant. This equation is called Arrhenius equation.              
Taking logarithm, eq. (i) may be written as
\(ln\ \ \ k=ln\ A-\frac { { E }_{ a } }{ R } \times \frac { 1 }{ T } \quad \)
Converting to logarithm to the base 10 (InX = 2.303 log X), we get 
\(2.303\log { k } =2.303\log { A } -\frac { { E }_{ a } }{ RT } \quad \quad ...(ii)\)
When log k is plotted against \(\frac { 1 }{ T } \) , we get a straight line as shown in diagram. 
The intercept of this line is equal to log A and slope is equal to \(-\frac { { E }_{ a } }{ 2.303R } \quad \)
Therefore,
\(Slope=-\frac { { E }_{ a } }{ 2.303R } \)                                              
Knowing the value of slope and gas constant R, activation energy can be calculated as
\({ E }_{ a }=-2.303R\times Slope\)
Alternatively, \({ E }_{ a }\) and A can be calculated by determining the values of rate constant at two different temperatures. Le,t \({ k }_{ 1 }\) and \({ k }_{ 2 }\) are the rate constants for the reaction at two different temperatures \({ T }_{ 1 }\) and \(T_{ 2 }\) respectively. Then,
\(\log { { k }_{ 1 } } =\log { A- } \frac { { E }_{ a } }{ 2.303\quad R{ T }_{ 1 } } \)
\(and\ \ \log { { k }_{ 2 } } =\log { A- } \frac { { E }_{ a } }{ 2.303\quad R{ T }_{ 1 } } \)
Subtracting eq. (iv) from eq. (iii), we get
\(\log { { k }_{ 2 } } -\log { { k }_{ 1 } } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ T_{ 2 } } \right] \)
\(or\ log=\frac { { E }_{ a } }{ 2.303R } \left[ \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ T_{ 2 } } \right] \)
By substituting the values of \({ k }_{ 1 }\) and\({ k }_{ 2 }\) and temperatures \({ T }_{ 1 }\) and \(T_{ 2 }\), \({ E }_{ a }\)can be calculated.

(i) (c)
(ii) (a): A primary amine forms N-alkylbenzene sulphonamidewhich because ofthe presence of an acidic hydrogen on the N-atom dissolves in aqueous KOH.
(iii) (b)
(iv) (a): Tertiary amine does not contain a replaceable hydrogen on the nitrogen atom. So, 3^o amine does not react with Hinsberg's reagent.

(i) (a): It is an example of cross Cannizzaro reaction where aromatic aldehyde gets reduced to alcohol and aliphatic aldehyde gets oxidised to its sodium salt (both aldehydes must not contain any \(\alpha\)-hydrogen).

(ii) (c)
(iii) (a): The Cannizzaro product of given reaction yields 2, 2, 2-trichloroethanol.

(iv) (a): C-C bond is not formed in Cannizzaro reaction while other reactions result in the formation of C-C bond.