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NEET CHEMISTRYHydrocarbonsMedium

Question

CH3−CH2−CH=CH2→HBr,peroxideY→C2H5ONaZCH_3-CH_2-CH=CH_2 \xrightarrow{HBr, \text{peroxide}} Y \xrightarrow{C_2H_5ONa} Z

The product Z in the above-mentioned reaction is:

A

CH3−(CH2)3−O−CH2CH3CH_3-(CH_2)_3-O-CH_2CH_3

B

(CH3)2CH−O−CH2CH3(CH_3)_2CH-O-CH_2CH_3

C

CH3(CH2)4−O−CH3CH_3(CH_2)_4-O-CH_3

D

CH3CH2−CH(CH3)−O−CH2CH3CH_3CH_2-CH(CH_3)-O-CH_2CH_3

Step-by-Step Solution

  1. Step 1: Formation of Y (Peroxide Effect): The reaction of But-1-ene (CH3−CH2−CH=CH2CH_3-CH_2-CH=CH_2) with HBr in the presence of peroxide proceeds via a free radical mechanism (Anti-Markovnikov addition). The bromine radical adds to the terminal carbon to form a more stable secondary free radical, which then abstracts a hydrogen atom. This yields the primary alkyl halide: 1-Bromobutane (CH3−CH2−CH2−CH2−BrCH_3-CH_2-CH_2-CH_2-Br) . CH3−CH2−CH=CH2+HBr→peroxideCH3−CH2−CH2−CH2−Br (Y)CH_3-CH_2-CH=CH_2 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2-CH_2-Br \text{ (Y)}
  2. Step 2: Formation of Z (Williamson Ether Synthesis): 1-Bromobutane (a primary alkyl halide) reacts with sodium ethoxide (C2H5ONaC_2H_5ONa) via an SN2S_N2 nucleophilic substitution reaction. The ethoxide ion (C2H5O−C_2H_5O^-) displaces the bromide ion to form an ether . CH3(CH2)3Br+C2H5ONa→CH3(CH2)3−O−CH2CH3+NaBrCH_3(CH_2)_3Br + C_2H_5ONa \rightarrow CH_3(CH_2)_3-O-CH_2CH_3 + NaBr
  3. Conclusion: The final product Z is 1-ethoxybutane (CH3−(CH2)3−O−CH2CH3CH_3-(CH_2)_3-O-CH_2CH_3). Option 4 would be the product if the reaction followed Markovnikov addition (absence of peroxide).

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Hydrocarbons. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYHydrocarbonschchchchxrightarrowhbrtextperoxidexrightarrowchonaproduct

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