(i) Amino acids have zwitterionic structure. Therefore. in presence of strong acids the} exist as cations' (I) and thus on electrolysis, these migrate towards cathode.
\(\overset { + }{ { NH }_{ 3 } } -CHR-{ COO }^{ - }+{ H }^{ + }\longrightarrow \quad \overset { + }{ { NH }_{ 3 } } -CHR-{ COO }H\\ \quad \quad Zwitterion\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad Cation(I)\)
In contrast, in presence of alkalies, the amino acids exist as anions (II) and thus on electrolysis, these migrate towards anode.
\(\overset { + }{ { NH }_{ 3 } } -CHR-{ COO }^{ - }+{ OH }^{ - }\longrightarrow \quad \overset { + }{ { NH }_{ 3 } } -CHR-{ COO }H\\ \quad \quad Zwitterion\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad Anion(II)\\ \)
(ii) A monoamino monocarboxylic acid such as glycine exists as a dipolar ion \((\overset { + }{ { NH }_{ 3 } } -{ CH }_{ 2 }-{ COO }^{ - })\) .In this structure, \(\overset { + }{ { NH }_{ 3 } } \) group acts as the acid and \({ COO }^{ - }\) group acts as the base. Thus, monoamino monocarboxylic acids can act both as an acid as well as a base. Therefore, these acids have two pK values one as an acid (when titrated with a base) and other as a base (when titrated with an acid).

(i) Since compound (A) on treatment with CHCI3 and KOH (i.e., Reimer-Tiemann reaction), gives two products Band C, therefore, A must be phenol and Band C must be o-hydroxybenzaldehyde and P: hydroxybenzaldehyde respectively or vice-versa.
(ii) Since both Band C on distillation with Zn dust give the same compound (D), therefore, D must be benzaldehyde.
(iii) Since oxidation of D gives E with M.F. C7R602, therefore, E must be benzoic acid.
(iv) Since sodium salt of E, i.e.benzoic acid upon heating with soda-lime gives compound (F), therefore.
(F) must be benzene. Zn dust distillation of A would also give benzene (F).

In the acidic medium, the oxygen atom of the oxirane ring gets protonated. The presence of positive charge on the oxygen atom weakens the C-O bond more on the side of C² since the partial positive charge created on C₂ will be stabilized by the + I- effect of the two CH₃ groups in the transition state (T.S.). Consequently, the attack of the nucleophile, i.e., H₂O preferentially ocurson C₂ yielding 2-methyl propane- 1, 2-diol.

(i) HCN (PK_a 10) and NH3 (PK_a 11·25) are very weak: acids. Therefore, their conjugate bases, i.e., CN^- ion and NH₂ ^- are very strong bases. Since strong bases are bad leaving groups, their displacement in nucleophilic substitution reactions is never observed,
(ii) Iodide ion is a powerful nucleophile and hence reacts rapidly with RCI to form RI,
Further because 1- ion is a better leaving group than Cl- ion, therefore, RI is more rapidly hydrolysed than RCI to form ROH,
The I^- ion thus regenerated recycles in the above reaction thereby explaining its catalytic effect.

Tarnish will be removed if the following reaction takes place :
\(Al+{ { Ag }_{ 2 } }S\longrightarrow { Al }^{ 3+ }+2Ag+{ S }^{ 2- }\)
EMF of this cell reaction = \({ E }_{ { Ag }_{ 2 }S/{ 2Ag },{ S }^{ 2- } }^{ 0 }-{ E }_{ { Al }^{ 3+ }/{ Al } }^{ 0 }=-0.71-(-1.66)V=+0.95V\)
As EMF is positive, the reaction will take place and tarnish will be removed.

Suppose initial pressure of N₂O₅ is P mm and decrease is pressure of N₂O₅ in time t is p mm.
Then N₂O₅ (g) = 2 N02 (g) + 0·5 02 (g)
Initial   P mm
After time t,   (P-p)  2p   0·5 p Total = P + 1·5p
P ∝ 600 mm and (P + 1·5p) ∝ 960 mm or 1·5p ∝ 360 mm or p o∝ 240 mm
Mole fraction of N₂O₅ decomposed\(={p\over P}={240\over 600}=0.4\)

V₂ = 59.6 cm³