Question

Match the LIST-I (Reactant) with LIST-II (Product): [H] {|l|l|l|l|} \multicolumn{2}{|c|}{LIST-I & \multicolumn{2}{c|}{LIST-II
\multicolumn{2}{|c|}{(Reactant) & \multicolumn{2}{c|}{(Product)
A. & Benzaldehyde + ethyl alcohol (dry HCl) & I. & Diethyl acetal of benzaldehyde
B. & [c]{@{}l@{}}Phenyl magnesium bromide
with ethylene oxide followed by water & II. & 2 Phenyl ethanol
C. & Acrolein + dry HCl (g) ($-10$ degree C) & III. & Beta chloro propionaldehyde
D. & Crotonic acid + hydrogen bromide (g) (20 degree C) & IV. & Beta bromo butyric acid

• A. A-I, B-II, C-III, D-IV
• B. A-II, B-I, C-III, D-IV
• C. A-III, B-II, C-I, D-IV
• D. A-IV, B-II, C-III, D-I

Hint:

Grignard + Ethylene Oxide always adds exactly two carbons and ends in an -OH.

Answer 1

The Correct Option is A

Step 1: Concept
This involves various organic synthesis reactions including acetal formation and Grignard reactions.

Step 2: Meaning
Each reactant set leads to a specific functional group transformation.

Step 3: Analysis
A matches I: Aldehyde + Alcohol + dry HCl $\rightarrow$ Acetal. B matches II: Grignard + Ethylene oxide $\rightarrow$ Primary alcohol (with 2 extra carbons). C matches III: Acrolein + HCl $\rightarrow$ Beta-chloropropionaldehyde (Michael addition). D matches IV: Crotonic acid + HBr $\rightarrow$ Beta-bromobutyric acid (Anti-Markovnikov addition is not applicable here, addition follows electronic effects).

Step 4: Conclusion
The matches follow the standard chemical outcomes for these reactions.

Final Answer: (A)