The reaction of compound $\mathbf{P}$ with $\mathrm{CH}_{3} \mathrm{MgBr}$ (excess) in $\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}$ followed by addition of $\mathrm{H}_{2} \mathrm{O}$ gives $\mathbf{Q}$. The compound $\mathbf{Q}$ on treatment with $\mathrm{H}_{2} \mathrm{SO}_{4}$ at $0^{\circ} \mathrm{C}$ gives $\mathbf{R}$. The reaction of $\mathbf{R}$ with $\mathrm{CH}_{3} \mathrm{COCl}$ in the presence of anhydrous $\mathrm{AlCl}_{3}$ in $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ followed by treatment with $\mathrm{H}_{2} \mathrm{O}$ produces compound $\mathbf{S}$. [Et in compound $\mathbf{P}$ is ethyl group] (see structure) The product S is

Alcohols Phenols and Ethers — JEE Advanced · 2017 · 21 May 2017 · Paper 2 Question with Solution

Passage

The reaction of compound

P

with

CH_{3} MgBr

(excess) in

(C_{2} H_{5})_{2} O

followed by addition of

H_{2} O

gives

Q

. The compound

Q

on treatment with

H_{2} SO_{4}

0^{\circ} C

gives

R

. The reaction of

R

with

CH_{3} COCl

in the presence of anhydrous

AlCl_{3}

CH_{2} Cl_{2}

followed by treatment with

H_{2} O

produces compound

S

. [Et in compound

P

is ethyl group]

(see structure)

The reaction of compound

P

with

CH_{3} MgBr

(excess) in

(C_{2} H_{5})_{2} O

followed by addition of

H_{2} O

gives

Q

. The compound

Q

on treatment with

H_{2} SO_{4}

0^{\circ} C

gives

R

. The reaction of

R

with

CH_{3} COCl

in the presence of anhydrous

AlCl_{3}

CH_{2} Cl_{2}

followed by treatment with

H_{2} O

produces compound

S

. [Et in compound

P

is ethyl group]

(see structure)

The product S is

(A)(see structure)
✓ Correct
(B)(see structure)
(C)(see structure)
(D)(see structure)

Answer: (A)

Solution

P (a tetralone derivative) reacts with excess

CH_{3} MgBr

(Grignard reagent). The ketone is converted to a tertiary alcohol Q with two methyl groups added.

Treatment of Q with

H_{2} SO_{4}

at 0°C causes dehydration, forming an alkene R through intramolecular Friedel-Crafts alkylation, creating a new ring.

Friedel-Crafts acylation of R with

CH_{3} COCl

and

AlCl_{3}

introduces an acetyl group at the most activated position of the aromatic ring, giving S.

The product S is the tricyclic structure with a tert-butyl group and an acetyl substituent as shown in option [A].

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Alcohols Phenols and Ethers

Solution

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