When the phenol shown below is treated with KOH, it forms a product whose IR spectrum does not show an absorption in the 3200-36
00 cm-1 region. Propose a structure for the product.?
1 answer:
Answer:
Explanation:
The first attached diagram shows the proposed phenol that is being treated with KOH to form Oxonine (CₐHₐO).
Reaction of Arylhalide after treatment with KOH lead to the displacement of Bromine atom with the hydrogen from the hydroxyl (OH) substituent , then forming HBr , leaving the oxy (-O) atom alone to form an Oxonine compound.
Therefore the propose structure for the product (Oxonine) CₐHₐO is shown in the second diagram attached below.
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At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.
The melting point is the temperature at which a substance passes from solid to liquid. Below the melting point, a substance is in the solid state. Above the melting point, a substance is in the liquid or gas state.
The boiling point is the temperature at which a substance passes from liquid to gas. Below the boiling point, a substance is solid or liquid. Above the boiling point, a substance is in the gas state.
At -25 °C, methanol is above the melting point (-97.6 °C) and below the boiling point (64.7 °C). Thus, it is in the liquid state.
At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.
You can learn more about the melting and boiling points here: brainly.com/question/5753603?referrer=searchResults
Answer:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in
<u>= 4.75 × 10²³ molecules of Propane</u>.
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No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =
<u>= 34.72 g</u>