Compound A has molecular formula C7H15Br. Treatment of compound
1 answer:
Answer:
See images below.
Explanation:
The sodium ethoxide is a strong nucleophile and a strong base. When it acts as a strong nucleophile (a substance that has unpaired electrons and needs to share or donate it to be stable), an elimination reaction occurs. When it acts as a strong base, a substitution reaction occurs.
For a halide, the elimination takes place when the halogen is no steric hindrance, so the halogen must be in position 1. Then the compound A must be the must steric hindrance compound with formula C₇H₁₅Br, which is the compound showed below (3-Bromo-3-ethyl-pentane), to promote only elimination product. It happens because it's a tertiary halide.
When it reacts with sodium ethoxide (H₃C-H₂C-O-Na), the Br⁻ will leave the compound with close hydrogen, and a double bond will be formed between the carbons, so compound B must be the one showed below (3-Ethyl-2-pentene).
When the compound B is treated with dilute sulfuric acid, an addition reaction will occur, so the double bond will be broken and a hydrogen and a hydroxyl must be added to the compound. For the Markov-Nikov rule, the hydrogen will be in the most hydrated carbon. So, compound C will be the 3-Ethyl-3-pentanol showed below.
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Answer:
See explanation
Explanation:
We define the formal charge on an atom in a molecule as the charge it carries assuming that electrons in all chemical bonds of the molecule were shared equally between atoms irrespective of the electronegativity of each atom.
The formula for calculating the formal charge on an atom in a molecule is;
Formal Charge = [number of valence electrons on neutral atom] – [(number of lone electron pairs) + (½ number of bonding electrons)] ·
The formal charge on the two nitrogen atoms in diazomethane is obtained as follows;
Middle nitrogen atom = 5 – 8/2 – 0 = +1
Last nitrogen atom = 5 – 4/2 – 4 = –1
The Lewis structure of the molecule is shown in the image attached.
