Answer:
The correct order of increasing reactivity toward nucleophilic acyl substitution is E < D < C < A < F < B.
Explanation:
The stability of the leaving group best determines the manner of reactivity of carboxylates to nucleophilic substitution after the substitution of the nucleophile to the leaving group. The leaving group should, therefore, be protonated with hydrogen ion in the solution to form a stable molecule. From the given list: The leaving group for A, Ethyl thioacetate will be ethanethiol. For B, Acetyl chloride will be Hydrochloric acid. For C, Sodium acetate will be Sodium Hydroxide. For D, Ethyl acetate will be Ethanol. For E, Acetamide will be Ammonia, and for F, Acetic anhydride will be Ethanoic acid. The reactivity of the substitution reaction is dependent on the stability of these leaving groups. The stability of these leaving groups depends on their pKa, and the more the pKa, the lesser the acidity of the leaving group, and the lower the reactivity. Therefore, considering their pKa: A is 8.5, B is -7, C is 13.8, D is 15.9, E is 36, and F is 4.8. When we rearrange this pKa in descending order, we have E, D. C, A, F, B. Which is also the increased reactivity of the nucleophilic acyl substitution.
Answer:
37.3
263.5
Explanation:
The scale measures hundreds of units, tens of units, units, and parts of units (1 decimal place.
Scale 1
Hundreds 0 * 100 = 0
Tens: 3 * 10 = 30
Units: 7 * 1 = 7
1/10 unit = 3* 0.1 = 0.3
Total 30 + 7 + 0.3 = 37.3
Scale 2
Hundreds 2 * 100 = 200
Tens: 6 * 10 = 60
Units: 3 * 1 = 3
1/10 unit = 5* 0.1 = 0.5
Total = 200 + 60 + 3 + 0.5 = 263.5
A simple circuit consists of a battery to provide power, wires to carry the electrical power, and
load that uses the electrical power for example a light globe
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Electronegativity<span> is the measure of the ability of an atom to attract electrons to itself. Fluorine is the most </span>electronegative<span> element and francium is one of the least</span>electronegative<span>. ... The </span>molecule's polarity<span> will be determined on the negative and positive regions on the outer atoms in the </span>molecule<span>.</span>