Answer:
this is the answer
I hope it helps you although
Answer:
Enolate Alkylation
The anions from ketones, called enolates, can act as a nucleophile in SN2 type reactions. Overall an α hydrogen is replaced with an alkyl group and a new carbon-carbon bond is formed. These alkylations are affected by the same limitations as SN2 reactions previously discussed. A good leaving group, chloride, bromide, iodide or tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for preparing the enolate from the ketone. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylations occurring, and competing SN2 reactions with the base.
Explanation:
Design is illustrated in the attached document
Answer:
Each molecule contains one atom of A and one atom of B. The reaction does not use all of the atoms to form compounds.
A + B ⟶ Product
Particles: 6 8 6
If six A atoms form six product molecules, each molecule can contain only one A atom.
The formula of the product is ABₙ.
If n = 1, we need six atoms of B.
If n = 2, we need 12 atoms of B. However, we have only eight atoms of B, so the formula of the product must be AB.
Thus, 6A + 6B ⟶ 6AB, with two B atoms left over.
Explanation:
Credit goes to @znk
Hope it helps you :))
It’s the BOA not the dog or kangaroo
Answer:
1. Cu
2. Cu
3. 2 electrons.
Explanation:
Step 1:
The equation for the reaction is given below:
3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)
Step 2:
Determination of the change of oxidation number of each element present.
For Cu:
Cu = 0 (ground state)
Cu(NO3)2 = 0
Cu + 2( N + 3O) = 0
Cu + 2(5 + (3 x -2)) =0
Cu + 2 (5 - 6) = 0
Cu + 2(-1) = 0
Cu - 2 = 0
Cu = 2
The oxidation number of Cu changed from 0 to +2
For N:
HNO3 = 0
H + N + 3O = 0
1 + N + (3 x - 2) = 0
1 + N - 6 = 0
N = 6 - 1
N = 5
NO = 0
N - 2 = 0
N = 2
The oxidation number of N changed from +5 to +2
The oxidation number of oxygen and hydrogen remains the same.
Note:
1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1
2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1
Step 3:
Answers to the questions given above
From the above illustration,
1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.
2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.
3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.