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Question:

Identify Z in the sequence of reactions: CH3CH2CH=CH2 →HBr/H2O2→ Y →C2H5ONa→ Z

CH3−(CH2)3−O−CH2CH3

(CH3)2CH−O−CH2CH3

CH3CH2−CH(CH3)−O−CH2CH3

CH3(CH2)4−O−CH3

Solution:

In the given sequence, the compound Z is CH3−(CH2)3−O−CH2CH3. HBr in presence of peroxide gives anti-Markovnikov addition product. A molecule of HBr is added across C=C double bond. Br is added to C atom having more number of H atoms. 1° alkyl halide on reaction with C2H5ONa gives SN2 reaction. Bromide ion is replaced with ethoxide ion.