(i) Reimer-Tiemann reaction

(ii) Williamson syntheses:
\(\underset { Alkyl halide\ Sodium\ alkoxide }{ R-X+\overset { + }{ Na } -\overset { - }{ O } -{ R }' } \longrightarrow R-O-{ R }'+NaX\)
\(\underset { Alkyl\ halide\ Sodium\ alkoxide }{ { CH }_{ 3 }-{ CH }_{ 2 }-Br } +\underset { sodium\ ethoxide }{ \overset { + }{ Na } -\overset { - }{ O } -{ CH }_{ 2 }-{ CH }_{ 3 } } \longrightarrow \underset { Diethyl\ ether }{ { CH }_{ 3 }-{ CH }_{ 2 }-O-{ CH }_{ 2 }-{ CH }_{ 3 } } +NaBr\)

Here, \(k=2.418\times { 10 }^{ -5 }{ s }^{ -1 },{ E }_{ a }=179.9\quad kJ\quad { mol }^{ -1 },\quad T=546K.\)
\(or \ logA=logk+\frac { { E }_{ a } }{ 2.303RT } =log(2.418\times { 10 }^{ -5 }{ s }^{ -1 })+\frac { 179.9 \ kJ \ { mol }^{ -1 } \ }{ 2.303\times 8.314\times { 10 }^{ -3 }kJ \ { K }^{ -1 } \ { mol }^{ -1 }\times 546K } \)
\(\\=(-5+0.3834){ s }^{ -1 }+17.2081=12.5924{ s }^{ -1 }\)
\(A=Antilog(12.5924){ s }^{ -1 }=3.912\times { 10 }^{ 12 }{ s }^{ -1 }\)
According to Arrhenius equation,
\(k=A{ e }^{ { -{ E }_{ a } }/{ RT } }or \ ln \ k=lnA-\frac { { E }_{ a } }{ RT } or \ logk=logA-\frac { { E }_{ a } }{ 2.303RT }\)

Each strand consist of 4 bases which are Adenine, Guanine, Thymine and Cytosine.
These complementary bases are the following:
Adenine (A) ⇔ Thymine (T)
Cytosine (C) ⇔ Guanine  (G)
⇔ This symbol represents that both the left and right hand side bases are complementary to each other.
The two strands are internally connected by hydrogen bonding between complementary bases. The two strands of DNA are not identical because their sequence of bases has to be complementary to each other. Ex- If one sequence is ATCG the other has to be TAGC.

Zn(s) ➝ Zn²⁺ (aq) + 2e^-
Cu²⁺(aq) + 2e^- ➝ Cu(s)
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Zn(s) + Cu²⁺(aq) ➝ Zn²+(aq) + Cu(s)
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\(E_{cell}=E^0_{cell}-{0.0591\over 2}log{[Zn^{2+}]\over [Cu^{2+}]}\)
\(=\left(E^0_{cu^{2+}/Cu}-E^0_{Zn^{2+}/Zn}\right)-{0.0591V\over 2}log{2\over 0.5}\)
\(=[+0.34V+0.76]-{0.0591V\over v2}log4\)
\(=+1.10V-{0 0591V\over 2}\times0.6021\)
= 1.10 V - 0.018 V = 1.082 V

The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry’s law. Thus:
\(x(\text { Nitrogen })=\frac{p \text { (nitrogen) }}{K_{\mathrm{H}}}=\frac{0.987 \mathrm{bar}}{76,480 \mathrm{bar}}=1.29 \times 10^{-5}\)
As 1 litre of water contains 55.5 mol of it, therefore if n represents number of moles of N₂ in solution,
\(x(\text { Nitrogen })=\frac{n \text { mol }}{n \text { mol }+55.5 \text { mol }}=\frac{n}{55.5}=1.29 \times 10^{-5}\)
(n in denominator is neglected as it is < < 55.5)
Thus n = 1.29 × 10^–5 \(\times\) 55.5 mol = 7.16\(\times\) 10^–4 mol
\(=\frac{7.16 \times 10^{-4} \mathrm{~mol} \times 1000 \mathrm{ \ mmol}}{1 \mathrm{~mol}}=0.716 \mathrm{ \ mmol}\)

(i)
 
(ii)
 
(iii)
 
(iv)

Given: Conductivity of CH₃COOH solution,
K= 3.905 x 10^-5 S cm^-1
Concentration of CH₃COOH solution,
C = 0.001 mol L^-I ,
Molar conductivity,
\(\lambda _{ m }=k\times \frac { 1000 }{ C } \)
\(=\frac { \left( 3.905\times 10^{ -5 }Scm^{ -1 } \right) \times \left( 1000\quad cm^{ 3 }L^{ -1 } \right) }{ 0.001molL^{ -1 } } \)
= 39.05 S cm²mol^-1
Molar conductivity at infinite dilution \(\left( \lambda ^{ 0 }_{ m } \right) \)  for CH₃COOH
\(\lambda ^{ 0 }_{ m }CH_{ 3 }COOH=\lambda ^{ 0 }_{ m }CH_{ 3 }COO^{ - }+\lambda ^{ 0 }_{ H+ }\)
= 40.9 + 349.6 S cm² mol^-1
= 390.5 S  cm² mol^-1
Degree of dissociation, \(\alpha =\frac { \lambda _{ m } }{ \lambda ^{ 0 }_{ m } } \)
\(=\frac { 39.05\quad s\quad cm^{ 2 }mol^{ -1 } }{ 39.05\quad s\quad cm^{ 2 }mol^{ -1 } } =0.1\)
(ii) A device used to convert the chemical energy produced in a spontaneous redox reaction into electrical energy is called an electrochemical cell.
When external potential applied becomes greater than E^0_cell of electrochemical cell. electrons flow from cathode to anode, i.e. electrochemical cell behaves like an I electrolytic cell.

'A' is iron chromite (FeCr2 04 ), '8' is sodium chromate (Na₂CrO₄ ), 'C' is sodium dichromate (Na₂CrO₇) and 'D' is potassium dichromate (K₂Cr₂O₇). 
\(4 \mathrm{FeCr}_{2} \mathrm{O}_{4}+8 \mathrm{Na}_{2} \mathrm{CO}_{3}+7 \mathrm{O}_{2} \longrightarrow 8 \mathrm{Na}_{2} \mathrm{CrO}_{4}\) + 2Fe₂O₃ + 8CO₂ (g)
    'A'                                                                'B'
(Chromite ore)                                            (Yellow solution)
\(2 \mathrm{Na}_{2} \mathrm{CrO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\) \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} \longrightarrow 2 \mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) 
                                                     'C'                                                      'C'                                                                   'D'
                                                                                                                                                                                (Orange crystals)

The elevation (\(\triangle\)T_b) in the boiling point = 354.11 K – 353. 23 K = 0.88 K
\(M_{2}=\frac{2.53 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1.8 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{0.88 \mathrm{~K} \times 90 \mathrm{~g}}=58 \mathrm{~g} \mathrm{~mol}^{-1}\)
Therefore, molar mass of the solute, M₂ = 58 g mol^–1.

Vitamin A is essential to us because its deficiency causes xerophthalmia (hardening of cornea of eye) and night blindness.
Sources: Fish liver oil, carrots, butter, and milk.
Vitamin C. Vitamin C is essential to us because its deficiency causes scurvy (bleeding gums) and pyorrhea (loosening and bleeding of teeth).
Sources: Citrous fruits; arnIa, green leafy vegetables.

As sucrose decomposes according to first order rate law, \(k=\frac { 2.303 }{ t } \log { \frac { { \left[ A \right] }_{ 0 } }{ { \left[ A \right] } } } \)
The aim is to find \({ \left[ A \right] }/{ { \left[ A \right] }_{ 0 } }\)
As \({ t }_{ { 1 }/{ 2 } }=3.0\quad hour,\quad \therefore k=\frac { 0.693 }{ { t }_{ { 1 }/{ 2 } } } =\frac { 0.693 }{ 3\quad hr } =0.231\quad { hr }^{ -1 }\)
Hence, \(0.231 \ { hr }^{ -1 }\)\(=\frac { 2.303 }{ 8\quad hr } \log { \frac { { \left[ A \right] }_{ 0 } }{ { \left[ A \right] } } } \) or \(\frac { { \left[ A \right] }_{ 0 } }{ { \left[ A \right] } } =0.8024\quad \)or \(\frac { { \left[ A \right] }_{ 0 } }{ { \left[ A \right] } } =Antilog\quad (0.8024)=6.345\)or \(\frac { { \left[ A \right] } }{ { { \left[ A \right] }_{ 0 } } } =\frac { 1 }{ 6.345 } =0.158\)

In the S_N2 mechanism, the reactivity of halides for same halides group increase down the group. Because increase in size increase, the halide becomes a better leaving group. Therefore, CH₃I will react faster than CH₃Br in S_N2 reactions with OH^-.

\((i)\ \left[ Co{ \left( { NH }_{ 3 } \right) }_{ 5 }{ SO }_{ 4 } \right] Cl+{ AgNO }_{ 3 }(aq)\longrightarrow AgCl(s)+\left[ Co{ \left( { NH }_{ 3 } \right) }{ SO }_{ 4 } \right] { NO }_{ 3 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 'A'\quad \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad white\ ppt\)
\( \left[ Co{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] { SO }_{ 4 }+{ BaCl }_{ 2 }\longrightarrow { BaSO }_{ 4 }(s)+\left[ Co{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] { Cl }_{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad 'B'\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad white\ ppt\)
(ii) Ionisation isomerism
(iii) 'N' is pentaamrnine chloridocobalt(III) sulphate.
'B' is pentaammine sulphatocobalt(III) chloride.

(b): Ti³⁺ has [Ar]3d¹ configuration. Thus, d-d transition is possible and thereby it shows colour.

(a) (CH₃)₃N, N, N-dimethyl methanamine.
(b) 
 a foul smelling compound.
(c) 

(d) Add NaNO₂ and cone. HCI. Cool it to 0 to 5° C. Then add alkaline solution of phenol. Aniline gives orange dye where as ethyl amino does not.
(e) \(\begin{array}{ll} \mathrm{A}^{\prime} \text { is } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} & { }^{\prime} \mathrm{B}^{\prime} \text { is } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-} \end{array}\)
(f) 'A' is C₆H₅CN, 'B' is C₆H₅COOH
(g) It is used as cationic detergents used in hair conditioners and shampoo.

(i) (c): Ligands are named in alphabetical order irrespective of their charge.
(ii) (b)
(iii) (d)
(iv) (d)