Question

An alkene X ($C_5H_{10}$) on ozonolysis gave formaldehyde and butyraldehyde. The products from the reaction of X with (i) $H_2O/H^{\oplus}$ and (ii) $HBr((C_6H_5COO)_2$ are respectively

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

{Markovnikov:} H adds to C with more H (Rich get richer). {Anti-Markovnikov (Peroxide Effect):} Only with HBr. H adds to C with fewer H.

Answer 1

The Correct Option is A

Step 1: Identify Alkene X:
Ozonolysis Products: Formaldehyde ($HCHO$) + Butyraldehyde ($CH_3CH_2CH_2CHO$). Reconstruct the alkene by removing oxygen and joining the carbonyl carbons with a double bond. $CH_3-CH_2-CH_2-CH=O + O=CH_2 \rightarrow CH_3-CH_2-CH_2-CH=CH_2$ So, X is 1-Pentene.
Step 2: Reaction (i) - Acid Catalyzed Hydration:
Reagent: $H_2O / H^+$. Mechanism: Electrophilic addition following Markovnikov's Rule. The $H^+$ adds to the terminal carbon ($CH_2$) to form the more stable secondary carbocation. Then $OH^-$ attacks the secondary carbon. Product: Pentan-2-ol ($CH_3-CH_2-CH_2-CH(OH)-CH_3$).
Step 3: Reaction (ii) - HBr with Peroxide:
Reagent: $HBr / (C_6H_5COO)_2$ (Benzoyl Peroxide). Mechanism: Free Radical Addition following Anti-Markovnikov's Rule (Kharasch effect). The $Br^\bullet$ radical adds to the terminal carbon to form a stable secondary radical intermediate. Product: 1-Bromopentane ($CH_3-CH_2-CH_2-CH_2-CH_2Br$).
Step 4: Match with Options:
The first structure should be Pentan-2-ol. The second structure should be 1-Bromopentane. Option 1 shows exactly these structures (2-hydroxy pentane and 1-bromo pentane).