Which of the following features is common to both acids and bases? A. ability to burn skin B. similar taste C. pH measurements g
2 answers:
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Answer:
1,4-dihydro = 113 kJ·mol⁻¹
1,2-dihydro = 101 kJ·mol⁻¹
Explanation:
In 1,4-dihydronaphthalene, the 2,3-double bond is isolated from the benzene ring.
In 1,2-dihydronaphthalene, the 3,4-double bond is conjugated with the benzene ring.
Thus, 1,2-dihydronaphthalene is partially stabilized by resonance interactions between the ring and the double bond (think, styrene).
1,2-Dihydronaphthalene is at a lower energy level because of this stabilization.
The heat of hydrogenation of 1,2-dihydronaphthalene is therefore less than that of the 1,4-isomer when each is hydrogenated to the common product, 1,2,3,4-tetrahydronaphthalene.
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> () and a <em>carboxylic acid</em> group (
). Additionally, we have an <u>alcohol </u>(
) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (
).
When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base (
) removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
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