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 allyl carbocation is resonance stabilized and hence is as stable as a 30 carbocation
Therefore, on treatment with Lucas reagent, allyl alcohol gives white turbidity in less than 1 min.l-Propanol, on the other hand, does not react with Lucas reagent.

During acetylation, one H-atom (at. mass = Iamu) of the OH group is replaced by an acetyl group, i.e., CH₃CO (mol. mass = 43 amu). .
-OH + (CH₃CO)₂O ⇾ -O-COCH₃ + CH₃COOH
In other words, acetylation of each OH group increases the mass by 43 - I = 42 amu. Now the mol. mass of C₄H₁₀O = 106 amu while that of the acetylated product is 190 amu, therefore, the number of OH groups III the compound\(={190-106\over 42}=2\)

(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).

For the cell as represented, \({ E }_{ cell }^{ 0 }={ E }_{ { Ce }^{ 4+ },{ Ce }^{ 3+ } }^{ 0 }-{ E }_{ { Fe }^{ 3+ },{ Fe }^{ 2+ } }^{ 0 }=1.61-0.77=0.84V\)
As \({ E }_{ cell }^{ 0 }>0\) , left hand electrode is anode while right hand electrode is cathode. Hence, current will flow from right to left. The current will decrease with time ultimately the reaction  stops.

(i) As pressure of X is changing with time, it cannot be a zero order reaction. Let us now check it for 1st order.
At t = 100 min, \(k={2.303\over 100}log{P_0\over P_t}={2.303\over 100}log {800\over 400}=6.932\times 10^{-3}min^{-1}\)
At t = 200 min, \(k={2.303\over 200}log{800\over 200}={2.303\over 800}log4=6.932\times10^{-3}min^{-1}\)
As k comes out to be constant, hence it is a reaction of 1st order
(ii)\(t_{75./.}={2.303\over k}log{100\over 100-75}={2.303\over 6.932\times10-3min}log4=200min\)
(iii) 2x (g)⟶ 3 y (g) + 2z (g)
Initial Pressure             800mm      0             0
Pressure after time t     800 - 2p    3 p           2 p
When pressure of X is 700 mm, 800 - 2 p = 700 or p = 50 mm
Total pressure = (800 - 2 p) + 3 p + 2 p = 800 + 3 p = 800 + 3 x 50 = 950 mm.

V₂ = 59.6 cm³