Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have
as nucleophile. Also, this compound is also in excess. So, we will have as solvent
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For N₂,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference
0.00 (Non Polar Covalent)
For Na₂O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic)
For CO₂,
E.N of Oxygen = 3.44
E.N of Carbon = 2.55
________
E.N Difference 0.89 (Polar Covalent)
Answer:
Radio waves were first developed in WWII
Explanation:
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A bimolecular reaction can be kinetically first order in behaviour provided one of the reactants is taken in such a large excess that its concentration may hardly change. Such a reactant will not contribute to the order. Thus, a bimolecular reaction will be of first order. These reactions are known as pseudo chemical reactions.