the main product of the following is ch3-ch2-ch(ch3)-ch2-o-ch2-ch3 in presence of 1mole of HI is
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Answer:
CH
3
−CH
2
−O−CH
2
−CH
3
+
excess
2HI
→
ethyl iodide
2CH
3
CH
2
I
+H
2
O
when HI is present in limited quantity CH
3
CH
2
I and CH
3
CH
2
OH both are formed.
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