Question

Monocyclic compounds $ P, Q, R $ and $ S $ are the major products formed in the reaction sequences given below.
The product having the highest number of unsaturated carbon atom(s) is:

• A. \( P \)
• B. \( Q \)
• C. \( R \)
• D. \( S \)

Hint:

Always count each C involved in multiple bonding (C=C, C≡C, C=O) and the 3 C from the aromatic ring to evaluate total unsaturation.

Answer 1

The Correct Option is C

Let us examine each reaction and determine the structure of the products \( P, Q, R, S \) and count their unsaturated carbon atoms. Step 1: Compound \( P \) Reaction: \[ \text{Phenylpropanoic acid} \xrightarrow{\text{Br}_2/\text{Red P}, \text{H}_2O} \text{Phenylethylamine (Gabriel phthalimide reaction)} \] Actually, this is the Hunsdiecker reaction: \[ \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{COOH} \xrightarrow{\text{Br}_2/\text{Red P}} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \] So product \( P \) = ethylbenzene (C₆H₅–CH₂–CH₃) \[ \Rightarrow \text{Aromatic ring (3 unsaturated C)} + \text{no other double/triple bond} \Rightarrow \text{Total unsaturated carbon atoms} = \boxed{3} \]
Step 2: Compound \( Q \) Reaction: Crossed aldol between benzaldehyde and formaldehyde in base \[ \text{C}_6\text{H}_5\text{CHO} + \text{HCHO} \xrightarrow{\text{NaOH}, 293\,K} \text{C}_6\text{H}_5\text{CH}=\text{CH}\text{CHO} \Rightarrow \text{Cinnamaldehyde} \] So \( Q = \text{C}_6\text{H}_5\text{CH}=\text{CH}\text{CHO} \) Unsaturated C = 3 (benzene) + 2 (C=C, C=O) = \boxed{5}
Step 3: Compound \( R \) Reaction: Phenylacetylene reacts with NaNH₂ and then 1-bromopropene (an alkyne coupling), followed by Hg²⁺/H₃O⁺ gives enol → keto Intermediate: \[ \text{Phenyl–C≡C–CH=CH}_2 \xrightarrow{\text{Hg}^{2+}, H_3O^+} \text{Phenyl–C(=O)–CH₂–CH₃} \] But actually after hydration of triple bond, we get: \[ \text{Phenyl–C≡C–CH=CH}_2 \Rightarrow \text{Phenyl–CO–CH=CH}_2 \Rightarrow \text{Aromatic + C=C + C=O} \] So: - 3 from benzene - 2 from alkene + ketone = \boxed{5} unsaturated C However, more precisely, the product is conjugated: \[ \text{C}_6\text{H}_5–C≡C–CH=CH_2 \Rightarrow \text{no hydration} = directly conjugated enyne \Rightarrow triple bond (2 unsaturated C), alkene (1 C), benzene (3 C) \Rightarrow \boxed{6} unsaturated carbon atoms
Step 4: Compound \( S \) Reaction: 1. Ozonolysis of styrene → benzaldehyde 2. 2 eq. \( \text{CH}_3\text{MgBr} \) adds to two carbonyls → alcohols 3. Acidic dehydration → formation of alkenes Final product: \[ \text{C}_6\text{H}_5–C(CH_3)=CH_2 \Rightarrow \text{Aromatic + C=C} \Rightarrow 3 (benzene) + 1 (alkene) = \boxed{4} unsaturated C % Comparison of unsaturated carbon atoms:

• \( P \): 3
• \( Q \): 5
• \( R \): 6
• \( S \): 4