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:
Explanation:
Figure 6. An instant cold pack consists of a bag containing solid ammonium nitrate and a second bag of water. When the bag of water is broken, the pack becomes cold because the dissolution of ammonium nitrate is an endothermic process that removes thermal energy from the water.
Answer:
1-(tert-butoxy)-2-methylpropane
Note: there is a mistake in formula, the correct formula is (CH₃)₂-CH-CH₂-O-C(CH₃)₃ not (CH₃)₂-CH-CH₂-O(CH₃)₃, because oxygen is a divalent compound.
Explanation:
<em>Structural formula is attached</em>
IUPAC naming rules
1. start numbering the chain from the functional group. In this compound we start from oxygen side.
2. Here we can see that at position 1 there is an oxy group along with a tertiary carbon having three methyl groups. So we write it as 1-tert-butoxy. Which means that there is a methoxy group at position 1 along with a tertiary carbon.
3. At position 2 we can see that there is a methyl group attached to the main chain, so we write it as 2-methyl.
4. Now we count the total number of carbons in the main chain. As we can see that there are 3 carbons in the remaining or parent chain, so we write it as propane
5. So the IUPAC name of the compound will be 1-(tert-butoxy)-2-methylpropane
I believe the answer is orbital hybridization theory
