Answer:
See explanation
Explanation:
In order to do this, we need to use 3 reagents to get the final product.
The first one, and logic is the halogenation of the alkene. Doing this, with Br2/CCl4, we'll get an alkane with two bromines, one in carbon 2 and the other in carbon 3.
Then, the next step is to eliminate one bromine of the reactant. The best way to do this, is using sodium ethoxide in ethanol. This is because sodium ethoxide is a relatively strong base, and it will promove the product of elimination in major proportions rather than the sustitution product. If we use NaOH is a really strong base, and it will form another product.
When the sodium ethoxide react, it will form a double bond between carbon 1 and 2 (The carbon where one bromine was with the methyl, changes priority and it's now carbon 3).
The final step, is now use acid medium, such H3O+/H2O or H2SO4/H2O. You can use any of them. This will form an carbocation in carbon 2 (it's a secondary carbocation, so it's more stable that in carbon 1), and then, the water molecule will add to this carbon to form the alcohol.
See the attached picture for the mechanism of this.
Affects populations,
Occurs when there is genetic variation
Selects organisms with certain traits to survive.
Answer:
The calculated concentration of sodium thiosulphate solution will be less than the actual value.
Explanation:
When IO3^2- solution is added to KI solution, I2 gas is released ,then sulphuric acid is now added to facilitate reduction. In order to prevent the escape of iodine (I2) gas ,the solution must immediately be titrated with thiosulphate.
If the solution is not immediately titrated with thiosulphate, the concentration of iodine available in the system decreases. When this occurs, it will also cause a decrease in the amount of iodine available to react with thiosulphate thus decreasing the concentration of thiosulphate obtained from calculation
Answer:
there is no d electron that can be promoted via the absorption of visible light
Explanation:
One of the properties of transition elements is the possession of incompletely filled d orbitals. This property accounts for their unique colours.
The colours of transition metal compounds stem from d-d transition of electrons due to the presence of vacant d orbitals of appropriate energy to which electrons could be promoted.
For elements whose atoms have a d10 configuration, such vacant orbitals does not exist hence their compounds are not colored.
Sometimes, the colour of transition metal compounds stem from ligand to metal charge transfer(LMCT) for instance in KMnO4.
