Answer:
Explanation:
In this case, we have a <u>Friedel-Craft reaction</u> (see figure 1). Usually in this type of reaction, the benzene ring is <u>bonded directly to the carbon that has the halide atom</u>. But in this specific case, this is not happening.
To understand why we have to remember the reaction mechanism of this reaction. As first, step the Cl-C bond attacks the and we form a new bond between the alkyl halide and the , finally the C-Cl bond is broken and a <u>carbocation is formed</u>. In this case, we will have a primary carbocation, a very <u>unstable species</u>. Therefore we will have a <u>methyl shift</u> to obtain a tertiary carbocation.
This tertiary carbocation will react with and continues the reaction. (See figure 2)
I hope it helps!
Mole percent is the percentage of the total moles that is of a particular component.
Answer: science is a systematic enterprise that builds and organizes knowledge in the form of testable explanations and predictions about the universe
Explanation:
Answer:
The order of reactivity of metals is as follows, Potassium > Sodium > Lithium > Calcium > Magnesium > Aluminium > Zinc > Iron > Copper > Silver > Gold.
Explanation:
The reactivity of elements (metals) towards water decreases towards the right in a period. It also increases down the group. But zinc is more reactive towards water than iron. Hence the correct order is:
Iron<Zinc<Magnesium<Sodium
Answer: CuI₂ + Br₂
Explanation:
1) The activity series F > Cl > Br > I means that F is the most active and I is the least active of those four elements (the halogens, group 17 in the periodic table).
The activity is a measure of how eager is an element to react compared to other elements in the series in a single replacement reaction.
2) Choice 1: CuI₂ + Br₂
Since the activity of Br is higher than that of I, Br will react with CuI₂, displacing I, which will be left alone, as per this chemical equation:
CuI₂ + Br₂ → CuBr₂ + I₂
Being I less active than Br, it cannot displace Br in CuBr₂.
3) Choice 2: Cl₂ + AlF₃
Being Cl less active than F, the former will not displace the latter, and the reaction will not proceed.
4) Choice 3: Br₂ + NaCl
Again, being Br less active than Cl, the former will not displace the latter, and the reaction will not proceed.
5) Choice 4: CuF₂ + I₂
Once more, being I less active than F, the former will not displace the latter, and the reaction will not proceed.