The side of each water molecule with the oxygen atom uncovered will be marginally negative.
The side of each water molecule with the hydrogen atoms uncovered will be marginally positive.
So the two Cl{-} particles will be pulled in to the biggest number of positive charges, which happen in the boxes on the upper right and lower left.
The two Na{+} particles will be pulled in to the biggest number of negative charges, which happen in the boxes on the upper left and lower right.
Answer:
satalites i think but not sure
<span>The option A is correct answer. The isotopes are X and Y. The isotopes are those which have same atomic number but different mass number. Since atomic number has same number of protons or electrons. Thus, isotopes have same number of protons or electrons. Therefore, X and Y are isotopes.</span>
Answer:
See explanation
Explanation:
The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.
On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.
