Answer:
In an electrophilic aromatic substitution (Friedel Crafts alkylation) first in the monoalkylation of the 1,4-dimethoxybenzenethe the methoxy groups redirects the substitution for ortho-para positions with respect to the electrophile that is going to enter (alkyl group) this is due to the increase in electron density in that position, that is , to the inductive effect.
According to the second incoming alkyl group there would be 3 positions available, from which it will choose the meta position in relation to the second methoxy group, since the alkyl group is a weak activator of the ortho meta positions and coincides with the position to which it redirects the second methoxy group.
Answer:
i do not know but i hope you do not find the answer for incorrectly answering mine.
Explanation:
Answer:
Explanation:
Answer
It's a very strong base. The closer you are to column 1 you are, the stronger the base.
Answer:
When the following redox equation is balanced with smallest whole number coefficients, the coefficient for the iodide ion will be __6__.
Explanation:
From the redox equation, we can see that NO₃⁻ is reduced to NO (from oxidation state +5 to +2), whereas I⁻ is oxidized to I₂ (from oxidation state -1 to 0). The half reactions are balanced with H⁺ (acidic solution), as follows:
Reduction : 2 x (NO₃⁻(aq) + 3 e- + 4 H⁺ → NO(g) + 2 H₂O)
Oxidation : 3 x (2 I⁻(aq) → I₂(s) + 2 e-)
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Total equation: 6 I⁻(aq) + 2 NO₃⁻(aq)+ 8 H⁺ → 3 I₂(s) + 2 NO(g) + 4 H₂O
That is the redox equation with the smallest whole number coefficients.
Accordin to this, the coefficient for the iodide ion (I⁻) is: 6.
