Draw the product of the following reaction between propanoyl chloride and p-methoxynitrobenzene. Include all formal charges.
2 answers:
Answer:
The product will be :
1-(2-methoxy-5-nitrophenyl)propan-1-one
Explanation:
This is electrophilic substitution reaction.
The electrophile is acylium ion.
The reaction will occur in presence of anhydrous aluminium chloride, which will remove chloride ion and will generate acylium ion which will act as electrophile.
The electrophile will attack on the para or other position of methoxy group.
The mechanism is shown in figure.
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Answer:
Molarity of the solution? 0,262 M.
Concentration of the potassium cation? 0,786 M.
Concentration of the phosphate anion? 0,262 M.
Explanation:
Potassium phosphate (K₃PO₄; 212,27 g/mol) dissolves in water thus:
K₃PO₄ → 3 K⁺ + PO₄³⁻ <em>(1)</em>
Molarity is an unit of chemical concentration given in moles of solute (K₃PO₄) per liters of solution.
There are 250 mL of solution≡0,25 L
The moles of K₃PO₄ are:
13,9 g of K₃PO₄ × = 0,0655 moles of K₃PO₄
The molarity of the solution is:
= 0,262 M
In (1) you can see that 1 mole of K₃PO₄ produces 3 moles of potassium cation. The moles of potassium cation are:
0,0655 moles×3 = 0,1965 moles
The concentration is:
= 0,786 M
The moles of K₃PO₄ are the same than moles of PO₄³⁻, thus, concentration of phosphate anion is the same than concentration of K₃PO₄. 0,262 M
I hope it helps!