$CH_3-CH_2-CH=CH_2 \xrightarrow{HBr, ext{peroxide}} Y \xrightarrow{C_2H_5ONa} Z$

The product Z in the above

Question

$CH_3-CH_2-CH=CH_2 \xrightarrow{HBr, 	ext{peroxide}} Y \xrightarrow{C_2H_5ONa} Z$

The product Z in the above-mentioned reaction is:

Accepted Answer

1. **Step 1: Formation of Y (Peroxide Effect):** The reaction of But-1-ene ($CH_3-CH_2-CH=CH_2$) 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** ($CH_3-CH_2-CH_2-CH_2-Br$) .
   $$ CH_3-CH_2-CH=CH_2 + HBr \xrightarrow{	ext{peroxide}} CH_3-CH_2-CH_2-CH_2-Br 	ext{ (Y)} $$
2. **Step 2: Formation of Z (Williamson Ether Synthesis):** 1-Bromobutane (a primary alkyl halide) reacts with sodium ethoxide ($C_2H_5ONa$) via an $S_N2$ nucleophilic substitution reaction. The ethoxide ion ($C_2H_5O^-$) displaces the bromide ion to form an ether .
   $$ CH_3(CH_2)_3Br + C_2H_5ONa ightarrow CH_3(CH_2)_3-O-CH_2CH_3 + NaBr $$
3. **Conclusion:** The final product Z is 1-ethoxybutane ($CH_3-(CH_2)_3-O-CH_2CH_3$). Option 4 would be the product if the reaction followed Markovnikov addition (absence of peroxide).

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