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

Paragraph:
A tertiary alcohol $H$ upon acid catalysed dehydration gives a product $I$. Ozonolysis of I leads to compounds $\mathrm{J}$ and $\mathrm{K}$. Compound $\mathrm{J}$ upon reaction with
$\mathrm{KOH}$ gives benzyl alcohol and a compound $\mathrm{L}$, where $\mathrm{K}$ on reaction with $\mathrm{KOH}$ gives only $M$.

Question:
The structures of compounds $J, K$ and $L$, respectively, are

A $\mathrm{PhCOCH}_3, \mathrm{PhCH}_2 \mathrm{COCH}_3$ and $\mathrm{PhCH}_2 \mathrm{COO}^{-} \mathrm{K}^{+}$
B $\mathrm{PhCHO}, \mathrm{PhCH}_2 \mathrm{CHO}$ and $\mathrm{PhCH}_2 \mathrm{COO}^{-} \mathrm{K}^{+}$
C $\mathrm{PhCOCH}_3, \mathrm{PhCH}_2 \mathrm{CHO}$ and $\mathrm{CH}_3 \mathrm{COO}^{-} \mathrm{K}^{+}$
D $\mathrm{PhCHO}, \mathrm{PhCOCH}_3$ and $\mathrm{PhCOO}^{-} \mathrm{K}^{+}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\mathrm{PhCHO}, \mathrm{PhCOCH}_3$ and $\mathrm{PhCOO}^{-} \mathrm{K}^{+}$

Step-by-step Solution

Detailed explanation

$
J=\mathrm{PhCH}=\mathrm{O} K=\mathrm{PhCOCH}_3 \quad$$L=\mathrm{PhCOO}^{-} \mathrm{K}^{+}
$

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 correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions, is

List - I	List - II
(P) Passing $H _2 S$ in the presence of $NH _4 OH$	(1) $Cu ^{2+}$
(Q) $\left( NH _4\right)_2 CO _3$ in the presence of $NH _4 OH$	(2) $Al ^{3+}$
(R) $NH _4 OH$ in the presence of $NH _4 Cl$	(3) $Mn ^{2+}$
(S) Passing $H _2 S$ in the presence of dilute HCl	(4) $Ba ^{2+}$
	(5) $Mg ^{2+}$

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Let

p, q, r

be non-zero real numbers that are, respectively, the

10^{th}, 100^{th}

and

1000^{th}

terms of a harmonic progression. Consider the system of linear equations

x + y + z = 1

10 x + 100 y + 1000 z = 0

q r x + p r y + p q z = 0

	List- $I$		List- $II$
$(I)$	If $\frac{q}{r} = 10$ , then the system of linear equations has	$(P)$	$x = 0, y = \frac{10}{9}, z = - \frac{1}{9}$ as a solution
$(II)$	If $\frac{p}{r} \neq 100$ , then the system of linear equations has	$(Q)$	$x = \frac{10}{9}, y = - \frac{1}{9}, z = 0$ as a solution
$(III)$	If $\frac{p}{q} \neq 10$ , then the system of linear equations has	$(R)$	infinitely many solutions
$(IV)$	If $\frac{p}{q} = 10$ , then the system of linear equations has	$(S)$	no solution
		$(T)$	at least one solution

The correct option is:

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List - I	List - II
(P) Passing \(H _2 S\) in the presence of \(NH _4 OH\)	(1) \(Cu ^{2+}\)
(Q) \(\left( NH _4\right)_2 CO _3\) in the presence of \(NH _4 OH\)	(2) \(Al ^{3+}\)
(R) \(NH _4 OH\) in the presence of \(NH _4 Cl\)	(3) \(Mn ^{2+}\)
(S) Passing \(H _2 S\) in the presence of dilute HCl	(4) \(Ba ^{2+}\)
	(5) \(Mg ^{2+}\)