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NEET CHEMISTRYMedium

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).
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