Styrene on reaction with reagent X gave Y, which on hydrolysis followed by oxidation gave Z. Z gives positive 2,4-DNP test but does not give iodoform test. What are X and Z respectively?
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- \( {HBr} : {C}_6{H}_5{COCH}_3 \)
- \( {HBr} : {C}_6{H}_5{CHO} \)
- \( {HBr} : ({C}_6{H}_5{CO})_2{O}_2 : {C}_6{H}_5{CH}_2{CHO} \)
- \( {HBr} : ({C}_6{H}_5{CO})_2{O}_2 : {C}_6{H}_5{COCH}_3 \)
The Correct Option is C
Solution and Explanation
The reaction describes the conversion of styrene into products via two steps:
- Step 1: The reaction of styrene with reagent X leads to the formation of compound Y. Styrene (\( {C}_6{H}_5{CH=CH}_2 \)) reacts with HBr (hydrobromic acid) under suitable conditions. This results in the addition of HBr across the double bond, forming phenyl ethanol (\( {C}_6{H}_5{CH}_2{OH} \)).
- Step 2: Upon hydrolysis and oxidation of Y, the product Z is formed. The oxidation step converts phenyl ethanol to benzaldehyde (\( {C}_6{H}_5{CHO} \)). Z gives a positive 2,4-DNP test, indicating the presence of an aldehyde group, but it does not give the iodoform test, suggesting that Z does not contain a methyl ketone group.
Thus, the correct sequence is:
- X = HBr (which adds across the double bond)
- Z = benzaldehyde (\( {C}_6{H}_5{CHO} \)). The correct reaction follows the sequence: \[ {HBr} : {C}_6{H}_5{CHO}. \] Thus, the correct answer is (3).
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