JEE Main Chemistry question
2-Chlorobutane →(Alc. KOH, Δ) X (major) →(i. O₃; ii. Zn/H₂O) Y. Methyl formate + ethylmagnesium bromide →(i. Dry ether; ii. H⁺, H₂O) P. Product Y and P differ in molecular mass by u.
Answer: 14
Solution
Zaitsev elimination of 2-chlorobutane with alc. KOH gives but-2-ene as the major product (X); reductive ozonolysis of this symmetrical internal alkene gives acetaldehyde, CH₃CHO (Y, MW = 44). Methyl formate reacts with two equivalents of ethylmagnesium bromide (the first displaces methoxide to give propionaldehyde in
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