Add CHCI₃ and alc. KOH,C₆H₅ - NH₂ gives foul smell of isocyanide whereas C₆H₅ - NH - CH₃ does not.  

It is because alkyl groups are electron releasing in alkyl amines, they will increase electron density on 'N', therefore, they are more basic than NH₃.

\((i) \ Rate \ =k{ \left[ A \right] }^{ 1 }{ \left[ B \right] }^{ 2 }\)
\(Rate \ =k{ \left[ A \right] }{ \left[ 3B \right] }^{ 2 }\)
\(Rate \ =9k{ \left[ A \right] }{ \left[ B \right] }^{ 2 }\)
If concentration of 'B' alone increases three times, the rate will increase nine times.
(ii) If concentration of 'A' as well as 'B' is doubled, rate \(=k{ \left[ 2A \right] }^{ 1 }{ \left[ 2B \right] }^{ 2 }\), the rate will increase eight times.
So rate \(=k{ \left[ 2A \right] }^{ 1 }{ \left[ 2B \right] }^{ 2 }\).

It is because concentration of reactants goes on decreasing with time.

1386.6 s.

-NH₂ group has lone pair of electrons, therefore, it increases electron density at o and p-positions and that is why electrophilic substitution takes place at o- and p-positions.

\(k={2.303\over t}log{[R]_0\over [R]}\)
\(={2.303\over 100}log{[R]_0\over {40\over 100}[R]_0}\)
[60% is complete, 40% is left]
\(k={2.303\over 100}(log5-log2)\)
\(k={2.303\over 100}(0.6990-0.3010)\)
\(k={2.303\over 100}\times0.3980\ min^{-1}\)
\(t_{90./.}={2.303\over k}log{[R]_0\over [R]}\)
[90% is complete, 10% is left]
\(t_{90./.}={2.303\over k}log{[R]_0\over {10\over 100}[R]_0}\)
\(={2.303\times100\over 2.303\times0.3980}log10\)
\(={100\over 0.3980}=251.26min\)

(a) (i) Ammonolysis:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{NH}_{3} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{HCl}\)
Chloroethane                           Ethanemine
(ii) Coupling reaction:

(iii) Acetylation of amines:

(b) Primary amines reacts with Hinsberg reagent (C₆H₅ SO₂Cl) to form a compound soluble in KOH or NaOH.

Secondary amines react with C₆H₅ SO₂CI to form compound insoluble in KOH.
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}+\mathrm{R}_{2} \mathrm{NH} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{NR}_{2} \stackrel{\mathrm{KOH}}{\longrightarrow} \text { Insoluble. }\)
Tertiary amines do not react with C₆H₅ SO₂Cl.

For a first order reaction
\(k=\frac{2.303}{t}\log\frac{[R_0]}{[R]},\) where \([R]_0\) = initial concentration, [R] = conc. after time t
When half of the reaction is completed, [R] = [R]⁰/2. Representing, the time taken for half of the reaction to be completed, by t_1/2' equation becomes:
\(k=\frac{2.303}{t_{1/2}}\log\frac{[R]_0}{[R]_0/2} \)
\(\Rightarrow\ \ \ \ t_{1/2}=\frac{2.303}{k}\log2\)
\(\Rightarrow\ \ \ \ t_{1/2}=\frac{2.303}{k}\times 0.3010\)
\(\Rightarrow\ \ \ \ t_{1/2}=\frac{0.693}{k}\)
The above equation shows that half life first order reaction is independent of the initial concentration of the reactant.
(ii) For a first order reaction
\(\Rightarrow\ t=\frac{2.303}{k}\log\frac{a}{a-x}\)
\(\Rightarrow\ \ \ \ t_{99\cdot /\cdot }=\frac{2.303}{k}\log\frac{100}{1}\)
\( =\frac{2.303}{k}\log 100\)
\( =\frac{2.303\times 2}{k}=\frac{4.606}{k}\)
and \(\Rightarrow\ \ t_{90\cdot /\cdot }=\frac{2.303}{k}\log\frac{100}{10}\)
\(=\frac{2.303}{k}\log{10}=\frac{2.303}{k}\)
\(\frac{t_{99\cdot /\cdot }}{t_{90\cdot /\cdot } }=2\)
\(t_{99\cdot /\cdot }=2\times t_{90\cdot /\cdot }\)

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) tert-Butyl bromide being a 3° alkyl halide on treatment with a base (i.e. NH₃) prefers to undergo elimination rather than substitution. Therefore, the product is isobutylene rather than tert-butylamine.
 
1° amines containing tert-alkyl groups can be prepared by action of suitable Grignard reagents on O-methylhydroxylarnine. For example,

(a) (i) It is sum of powers to which cone. terms are raised in rate law or rate equation.
(ii) Each step of complex reaction (which takes place in more than one step) is called elementary, step in a reaction.
\((b) \ k=0.00510 \ { min }^{ -1 }\)
\(t=\frac { 2.303 }{ k } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } }\)
\(3\times 60\times 60=\frac { 2.303 }{ 0.00510 } \log { \frac { 0.1 }{ \left[ R \right] } }\)
\(\log { \frac { 0.1 }{ \left[ R \right] } } =\frac { 10800\times 0.00510 }{ 2.303 } \)
\(=23.94\)
\(\frac { 0.1 }{ \left[ R \right] } =Antilog \ 23.94\)
\(\frac { 0.1 }{ \left[ R \right] } =8.71\times { 10 }^{ 23 }\)
\(\left[ R \right] =\frac { 0.1 }{ 8.71\times { 10 }^{ 23 } }\)
\(\left[ R \right] =0.1148\times { 10 }^{ -24 }\)
\(\left[ R \right] =1.148\times { 10 }^{ -25 }M\)

Carbylamine reaction yields isocyanides.

(c) : For every 10^oC rise in temperature, rate constant is doubled, Hence, for 20^oC rise in temperature, rate constant will become 4 times, i.e., K₁ = 4 K₂ or K₂ = 0.25 K₁

(a) Rate = \(\frac {dx}{dt} = k[A_o]^{o} = k \) or \(k = \frac {dx}{dt} = \frac {conc}{Time} =\frac {mol L^{-1}}{s}\)mol L^-1 s^-1.

(d) : Rate = k ab. When volume is reduced to 1/4 th, Concentrations will become = 4 times

New rate = k (4 a) (4 b) = 16 k ab = 16 times.