Identify Types of Reactions in Conversion of Intermediate Q to S

To determine the sequential chemical reactions converting compound Q to R and R to S, let us analyze each transformation step-by-step:

1. Structure of Compound Q:

Compound P is an ester containing a $p\text{-tert-butylphenyl}$ moiety: $\text{4-(tert-butyl)phenyl}-\text{CH}_2\text{CH}_2\text{CO}_2\text{Et}$

When compound P is treated with excess methylmagnesium bromide ($\text{CH}_3\text{MgBr}$) followed by aqueous workup ($\text{H}_2\text{O}$), the ester group undergoes nucleophilic addition twice to form a tertiary alcohol, compound Q: $\text{Q} = \text{4-(tert-butyl)phenyl}-\text{CH}_2\text{CH}_2\text{C(OH)(CH}_3)_2$

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2. Conversion of Q to R:

Treatment of tertiary alcohol Q with $\text{H}_2\text{SO}_4$ at $0^\circ\text{C}$ leads to protonation of the hydroxyl group followed by loss of a water molecule, generating a stable tertiary carbocation: $\text{4-(tert-butyl)phenyl}-\text{CH}_2\text{CH}_2\text{C}^+(\text{CH}_3)_2$

This tertiary carbocation undergoes an intramolecular electrophilic aromatic substitution attack on the benzene ring, forming a five-membered indane ring (compound R). This process of carbocation attack on an aromatic ring to form a carbon-carbon bond is classified as an intramolecular Friedel-Crafts alkylation.

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3. Conversion of R to S:

Reacting compound R with acetyl chloride ($\text{CH}_3\text{COCl}$) in the presence of anhydrous aluminum chloride ($\text{AlCl}_3$) generates an acylium ion ($\text{CH}_3\text{C}^+=\text{O}$).

The acylium ion acts as an electrophile and attacks the aromatic ring of compound R to introduce an acetyl group ($-\text{COCH}_3$). This classic electrophilic aromatic substitution reaction is known as a Friedel-Crafts acylation.

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Conclusion:

• Q to R: Friedel-Crafts alkylation
• R to S: Friedel-Crafts acylation

Hence, the correct sequential chemical reactions are Friedel-Crafts alkylation and Friedel-Crafts acylation.

Correct Answer: D