JEE MAIN - Chemistry (2021 - 24th February Morning Shift - No. 13)
The product formed in the first step of the reaction of
_24th_February_Morning_Shift_en_13_1.png)
with excess $$\mathrm{Mg/Et_2O}$$ ($$\mathrm{Et=C_2H_5}$$) is
with excess $$\mathrm{Mg/Et_2O}$$ ($$\mathrm{Et=C_2H_5}$$) is
_24th_February_Morning_Shift_en_13_2.png)
_24th_February_Morning_Shift_en_13_3.png)
_24th_February_Morning_Shift_en_13_4.png)
_24th_February_Morning_Shift_en_13_5.png)
Explanation
Here, in first step only one mole of
Mg/ Et2O
attacks on bromine and form two 2MgBr in the first step.
On further moving in the reaction, two MgBr are eliminated to form alkene in respective positions.
_24th_February_Morning_Shift_en_13_6.png)
On further moving in the reaction, two MgBr are eliminated to form alkene in respective positions.
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