The stability of \(\alpha \) -helix structure is due to intramolecular H-bonding between -NH- and -CO- groups of the same polypeptide chain. The \(\alpha \) -helix is termed as \({ 3.6 }_{ 13 }\) helix since each turn of the helix has approximately 3·6 amino acids and the hydrogen bonding leads to the formation of a 13-membered ring.

The (-CO-NH-) in nylon gets hydrolysed

(i) Since compound A (C₄H₁₀O) does not react with Na metal or KMnO₄, it cannot be an alcohol.
(ii) Since compound A dissolves in cone, H₂SO₄, it may be an ether.
(iii) Since ether A on heating with excess of HI gives a single alkyl halide, therefore, ether (A) must be symmetrical. Now the only symmetrical ether having M.F. C₄HIOO is diethyl ether (CH₃CH₂OCH₂CH₃).

(i) Since compound A on oxidation gives compound 8 with M.F. C₂H₄O₂, therefore, compound 8 may be acetic acid, CH₃COOH and A may be acetaldehyde, CH₃CHO.
(ii) Since compound A, i.e., acetaldehyde reacts with dil. NaOH, therefore, it undergoes aldol condensation to afford an aldol.Further since this aldol on heating gives compound (C), therefore, (C) must be an α, β- unsaturated aldehyde, i.e., but-2-en-I-al (crotonaldehyde).
(iii) Since compound (C) on catalytic hydrogenation gives compound D, therefore, D may be either I-butanal or I-butanol depending upon the extent of hydrogenation.
AIl the reactions involved in this question are explained below:

Si (E.N. = 8) is more electropositive than C (E.N. = 2.5), therefore, Me₃Si (trimethylsilyl group) has greater +l-effect than that of Me₃C (r-butyl group). As a result, Me₃Si intesifies the -ve charge on the carboxylate ion relative to r-butyl group and hence Me₃CCH₂COOH is a stronger acid that Me₃SiCH₂COOH.

(i) HOCI in presence of an acid generates the reactive electrophile, chloronium ion «r, which attacks benzene to give chlorobenzene (A).
(ii) Chlorobenzene (A) on treatment with NaNH2/liq. NH₃ undergaes-dehydrohalogenation via benzyne to afford aniline (B).
(iii) Aniline (B) on treatment with HBF4 forms the corresponding salt anilinium tetrafluoroborate (C) which on heating with NaNO₂ undergoes Balz-Schiemann reaction through the intermedium formation of benzenediazonium tetrafluoroborate to afford fluorobenzene (D).

According to the question,
The relative ratios of these two isomeric chlorobutanes can be easily calculated by knowing: (i) the number and type of hydrogens (i.e. 1°, 2° or 3°) to be substituted and (ii) their relative rates of substitution (i.e. I : 3·8 : 5·0 for CI₂ at 298 K). Thus
\({1-Chlorobutane\over 2-Chlorobutane}={No.\ of\ 1^0H\over No.\ of\ 2^0H}\times{Reactivity\ o\ 1^0H\over Reaativity\ of\ 2^0H}={6\over 4}\times{1\over 3.8}={6\over 15.2}={28./.\over 72./.}\)

Comparing oxidation potential of iron with those of metals A, B and C, iron has higher oxidation potential than C only. Hence, when coating of C is broken, rusting will become faster.

At pH = 5, as half-life period is found to be independent of initial concentration of sucrose, this means with respect to sucrose, it is a reaction of first order, i.e., Rate = k [Sucrose].
If n is the order with respect to H⁺ion, t_1/2 ∝ [H⁺]^I-n,
i.e., 500 ∝ (10^-5)^I-n [PH = 5 means [H⁺] = 10^-5 M]                 .......(I)
and 50 ∝ (l0^-6)I-n [pH = 6 means [H⁺] = 10^-6 M]                    .......(ii)
Dividing (i) by (ii), 10 = (lo)l-n i.e. 1 - n = 1 or n = 0, i.e., order with respect to H+ion = O.Hence, overall rate law is Rate = k [Sucrose] [H⁺]^o.