Answer:
Y is a 3-chloro-3-methylpentane.
The structure is shown in the figure attached.
Explanation:
The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).
The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.
Answer:
the stabilization of the negative charge in orbitals with higher s character
Explanation:
Acetylide anion is a carbon anion compound or popularly called carbanion. Now Acetylide anion is sp hybridized. However acetylide anion tends to be more acidic as we move from sp³ to sp, hence acidicity increases, which makes sp to have the highest acidity and become the most stable.
So, we can conclude that the acetylide anion is more acidic due to the stabilization of the negative charge in orbitals with higher s character and as the s character increases, acidic nature of acetylide anion also increases.
The reaction between mercury (Hg) and sulfur (S) to form HgS is:
Hg + S ------------- HgS
Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS
The given mass of Hg = 246 g
Atomic mass of Hg = 200.59 g/mol
# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles
Based on the reaction stoichiometry,
# moles of S that would react = 1.226 moles
Atomic mass of S = 32 g/mol
Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g
39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS