The Deep blue CuSO₄^.5H₂O is blue vitriol. In its structure, one molecule of water is hydrogen-bonded from 4 sides and thus is released with more difficulty than the rest four water molecules.

The positive Brady's test (i.e., reaction with 2, 4-dinitrophenylhydrazine) indicates that the organic compound is carbonyl compound and Tollen's and Cannizaro reactions indicate that it is an aldehyde without $\mathrm{\alpha }$-hydrogen atom. As the end product is terephthalic acid, the compound must be p-ethyl benzaldehyde.

$\underset{\left(\mathrm{A}\right)}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\equiv \mathrm{CH}}$ $\underset{-{\mathrm{NH}}_3}{\overset{{\mathrm{NaNH}}_2}{\to }}$ CH₃CH₂C$\equiv \stackrel{-}{\mathrm{C}}\stackrel{+}{\mathrm{Na}}$ $\to \underset{\left(\mathrm{B}\right)}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\equiv {\mathrm{CCH}}_2{\mathrm{CH}}_3}$$\underset{\mathrm{hydrogenation}}{\overset{\mathrm{partial}}{\to }}$ $\underset{\left(\mathrm{C}\right)}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}\equiv {\mathrm{CHCH}}_2{\mathrm{CH}}_3}$ $\underset{\left(\mathrm{ii}\right) \mathrm{Zn} + {\mathrm{H}}_2\mathrm{O}}{\overset{\left(\mathrm{i}\right) {\mathrm{O}}_3}{\to }}$ $\underset{\left(\mathrm{D}\right)}{2{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHO}}$

As compound (D) does not have CH₃-C=O group, it does not give iodoform reaction.

Case I

When 200 mL distilled water is added to x mL solution

x X 0.3 = (x + 200) X y

Here, y = final normality or concentration.

$\therefore$ y = $\frac{0.3\mathrm{x}}{(\mathrm{x} + 200)}$

Case II

Number of equivalents of HCl = $\frac{0.3 \times \mathrm{x}}{1000}$

Number of equivalents of NaOH = $\frac{100 \times 0.1}{1000}$ = 0.01

Number of equivalents of HCl remained after addition of NaOH = $\left\{\frac{0.3 \mathrm{x}}{1000}- 0.01\right\}$

$\therefore$ Concentration = $\frac{\left\{{\displaystyle \frac{0.3 \mathrm{x}}{1000}- 0.01}\right\}}{100 + \mathrm{x}}$ x 1000

As final acid strength is same in both cases $\frac{\left\{{\displaystyle \frac{0.3 \mathrm{x}}{1000}-0.01}\right\} 1000}{100 + \mathrm{x}}$ = $\frac{0.3 \mathrm{x}}{200 + \mathrm{x}}$

or (0.3 x -10) x (200 + x) = (0.3x)(100 + x)

or 60 x -2000 + 0.3x² - l0x = 30x + 0.3x²

or 20x = 2000 or x = 100 ml

As dN/dt = kN (It is a Ist order reaction)

Integrated rate equation for Ist order reaction

k = 2.303/t log N/N_o

$\because$when t = 5 min, N = 2N_o

$\therefore$k = 2.303/5 log 2N_o/N_o

or k = 2.303/5 log 2

for N = 8N_o , t = ?

$\therefore$t = 2.303/k log 8N_o/N_o

By putting the value of k

t = $\left(\frac{2.303}{{\displaystyle \frac{2.303}{5}\log 2}}\right)$ log 8N_o/N

or t = $\frac{5\times 3 \log 2}{\log 2}$ = 15 min