Answer:
Compound A and compound B are constitutional isomers with molecular formula C3H7Cl.
When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination reaction predominates.
Explanation:
Constitutional isomers are the one which differs in the structural formula.
When compound A is treated with sodium methoxide, a substitution reaction predominates.
That means sodium methoxide is a strong base and a strong nucleophile.
But when it reacts with primary alkyl halides it forms a substitution product and when it reacts with secondary alkyl halide it forms mostly elimination product.
The reaction and the structures of A and B are shown below:
Net overall dissociation:
Na2CO3 ---> 2Na(-) + CO3(2-)
*The ion charge is in parenthesis
Explanation:
because there are 4 Iodines on the left, we'll put. 4 in front of NaI to balance it. This would result in 4 Na on the left, so we'll put a 2 in front of Sodium Sulfate to balance the right side. Now we have 4 Na and I on both side, as well as 2 Sulfate on both sides. Pb is already balanced. The equation is now complete.
<span>the highest ionization energy is Be, it is higher and righter than other elements in periodic table, so it harder to remove electron from its atom</span>