JEE Advanced · Chemistry · 25. Alcohols, Phenols & Ethers

Reaction of iso-propylbenzene with \(\mathrm{O}_2\) followed by the treatment with \(\mathrm{H}_3 \mathrm{O}^{+}\)forms phenol and a by-product \(\mathbf{P}\). Reaction of \(\mathbf{P}\) with 3 equivalents of \(\mathrm{Cl}_2\) gives compound \(\mathbf{Q}\). Treatment of \(\mathbf{Q}\) with \(\mathrm{Ca}(\mathrm{OH})_2\) produces compound \(\mathbf{R}\) and calcium salt \(\mathbf{S}\).
The correct statement(s) regarding \(\mathbf{P}, \mathbf{Q}, \mathbf{R}\) and \(\mathbf{S}\) is(are)

A Reaction of \(\mathbf{P}\) with \(\mathbf{R}\) in the presence of \(\mathrm{KOH}\) followed by acidification gives
B Reaction of \(\mathbf{R}\) with \(\mathrm{O}_2\) in the presence of light gives phosgene gas
C \(\mathbf{Q}\) reacts with aqueous \(\mathrm{NaOH}\) to produce \(\mathrm{Cl}_3 \mathrm{CCH}_2 \mathrm{OH}\) and \(\mathrm{Cl}_3 \mathrm{CCOONa}\)
D \(\mathbf{S}\) on heating gives \(\mathbf{P}\)

Verified Solution

Answer & Solution

Correct Answer

(A) Reaction of \(\mathbf{P}\) with \(\mathbf{R}\) in the presence of \(\mathrm{KOH}\) followed by acidification gives

Step-by-step Solution

Detailed explanation

(1)

(2) \(\underset{(\mathrm{R})}{\mathrm{CHCl}_3} \xrightarrow{\mathrm{O}_2} \underset{\text { Phosgene gas }}{\mathrm{COCl}_2}+\mathrm{HCl}\)
(3) \(Q\) does not undergo Cannizaro reaction
(4)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

The wave function, \(\psi_{ n , 1, m_{ l }}\) is a mathematical function whose value depends upon spherical polar coordinates \(( r , \theta, \phi)\) of the electron and characterized by the quantum numbers \(n , 1\) and \(m _{ l }\). Here \(r\) is the distance from the nucleus, \(\theta\) is colatitude and \(\phi\) is azimuth. In the mathematical functions given in the Table, Z is the atomic number and \(a_0\) is Bohr radius

Column 1	Column 2	Column 3
(I) 1s orbital	\(\text {(i) } \psi_{ n , 1, m_1}\) \(\propto\left(\frac{ Z }{ a _{ o }}\right)^{\frac{3}{2}} e ^{-\left(\frac{ Zr }{ e _{ o }}\right)}\)	(P)
(II) 2s orbital	(ii) One radial node	(Q) Probability density at the nucleus \(\propto \frac{1}{ a _{ o }^3}\).
(III) \(2 p _{ z }\) orbital	\(\text {(iii) } \psi_{ n }, 1, m_{ l }\) \(\propto\left(\frac{ Z }{ a _{ o }}\right)^{\frac{5}{2}} re ^{-\left(\frac{ Zr }{2 a _{ o }}\right)} \cos \theta\)	(R) Probability density is maximum at the nucleus.
(IV) \(3 d_{ z }^2\) orbital	(iv) \(x y\)-plane is a nodal plane	(S) Energy needed to excite an electron from \(n =2\) state to \(n =4\) state is \(\frac{27}{32}\) times the energy needed to excite are electron from \(n =2\) state to \(n =6\) state.

For \(\text{He} ^{+}\)ion, the only incorrect combination

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