JEE Advanced · Chemistry · 13. GOC

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$P$ and $Q$ are isomers of dicarboxylic acid $\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}$. Both decolorize $\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}$. On heating, $\boldsymbol{P}$ forms the cyclic anhydride.

Upon treatment with dilute alkaline $\mathrm{KMnO}_{4}, \boldsymbol{P}$ as well as $\boldsymbol{Q}$ could produce one or more than one from $\boldsymbol{S}, \boldsymbol{T}$ and $\boldsymbol{U}$.

Question:
Compounds formed from $\boldsymbol{P}$ and $\boldsymbol{Q}$ are, respectively

A Optically active S and optically active pair (T,U)
B Optically inactive S and optically inactive pair (T,U)
C Optically active pair (T,U) and optically active S
D Optically inactive pair (T,U) and optically inactive S

Verified Solution

Answer & Solution

Correct Answer

(B) Optically inactive S and optically inactive pair (T,U)

Step-by-step Solution

Detailed explanation

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.

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From JEE Advanced

A charged particle is introduced at the origin

x = 0, y = 0, z = 0

with a given initial velocity

\vec{v}

. A uniform electric field

\vec{E}

and a uniform magnetic field

\vec{B}

exist everywhere. The velocity

\vec{v}

, electric field

\vec{E}

and a uniform magnetic field

\vec{B}

are given in the columns below

Column 1	Column 2	Column 3
1. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(i) $\vec{E} = E_{0} \hat{z}$	(P) $\vec{B} = - B_{0} \hat{x}$
2. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{y}$	(ii) $\vec{E} = - E_{0} \hat{y}$	(Q) $\vec{B} = B_{0} \hat{x}$
3.Proton with $\vec{v}$ =0	(iii) $\vec{E} = {- E}_{0} \hat{x}$	(R) $\vec{B} = - B_{0} \hat{y}$
4. Proton with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(iv) $\vec{E} = E_{0} \hat{x}$	(S) $\vec{B} = - B_{0} \hat{z}$

In which case will the particle describe a helical path with axis along positive z direction?

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Column I	Column II
A. \(B _2\)	p. Paramagnetic
B. \(N_2\)	q. Undergoes oxidation
C. \(O _2^{-}\)	r. Undergoes reduction
D. \(O _2\)	s. Bond order
	t. Mixing of 's' and 'p' orbitals