2-Methyl-4-oxo-pentanoic acid is unlikely to produce 2-Methyl-3-butanone upon strong heating.
Upon heating, the β ketoacid becomes unstable and decarboxylates, leading to the formation of the methyl ketone.
A carboxylic acid is an organic acid that contains a carboxyl group (C(=O)OH) attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO2H, with R referring to the alkyl, alkenyl, aryl, or other group.
Carboxylic acids occur widely. Important examples include the amino acids and fatty acids. Deprotonation of a carboxylic acid gives a carboxylate anion.
Full question :
Q. Which reactant is unlikely to produce the indicated product upon strong heating?
- A) 2,2-Dimethylpropanedioic acid 2-methylpropanoic acid
- B) 2-Ethylpropanedioic acid Butanoic acid
- C) 2-Methyl-3-oxo-pentanoic acid 3-Pentanone
- D) 2-Methyl-4-oxo-pentanoic acid 2-Methyl-3-butanone
- E) 4-Methyl-3-oxo-heptanoic acid 3-Methyl-2-hexanone
Hence, option (D) is correct.
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Answer:
Option b. 22 g of He will have the greatest volume at STP
Explanation:
In order to determine the volume, we apply the Ideal Gases Law equation:
P . V = n . R . T
V = n . R . T / P
R, T and P are the same in all the situation we must define n (number of moles).
The one that has the greatest number of moles will have the greatest volume at STP
22 g of Ne . 1mol / 20.1 g = 1.09 moles of Ne
22g of He . 1mol / 4 g = 5.5 moles of He
22 g of O₂ . 1mol / 32g = 0.68 moles of O₂
22 g of Cl₂ . 1mol / 70.9 g = 0.31 moles of Cl₂
Answer:
N2(g) + 3H2(g) → 2 NH3(g)
Explanation:
N2(g) + H2(g) → NH3(g)
We start equaling the number of N atoms in both sides multiplying by 2 the NH3.
N2(g) + H2(g) → 2 NH3(g)
So we equals the H atoms (there are six in products sites)
N2(g) + 3 H2(g) → 2 NH3(g)
Answer:
Key Points. Ionic bonds are formed through the exchange of valence electrons between atoms, typically a metal and a nonmetal.
Explanation:
