Substitution Reactions are those reactions in which one nucleophile replaces another nucleophile present on a substrate. These reactions can take place via two different mechanism i.e SN¹ or SN². In SN¹ substitution reactions the leaving group leaves first forming a carbocation and nucleophile attacks carbocation in the second step. While in SN² reactions the addition of Nucleophile and leaving of leaving group take place simultaneously.
Example:
OH⁻ + CH₃-Br → CH₃-OH + Br⁻
In above reaction,
OH⁻ = Incoming Nucleophile
CH₃-Br = Substrate
CH₃-OH = Product
Br⁻ = Leaving group
Organic reactions are typically slower than ionic reactions because in organic compounds the covalent bonds are first broken, this breaking of bonds is a slower step, while, in ionic compounds no bond breakage is required as it consists of ions, so only bond formation takes place which is a quicker and fast step.
Answer:
Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.
Explanation:
The benzene ring in itself does not easily undergo electrophilic substitution reaction. Some groups activate or deactivate the benzene ring towards electrophilic substitution reactions.
-NO2 ia a highly deactivating substituent therefore, Friedel-Crafts alkylation of nitrobenzene does not take place under any conditions.
This reaction scheme is therefore flawed because Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.
Answer:
Explanation:
The work function of the sodium= 495.0 kJ/mol
It means that
1 mole of electrons can be removed by applying of 495.0 kJ of energy.
Also,
1 mole = 
So,
electrons can be removed by applying of 495.0 kJ of energy.
1 electron can be removed by applying of 
of energy.
Energy required = 
Also,
1 kJ = 1000 J
So,
Energy required = 
Also, 
Where,
h is Plank's constant having value 
c is the speed of light having value 
So,
Also,
1 m = 10⁻⁹ nm
So,
Data: molar mass 470 g/mol
Percent composition:
Hg = 85.0%
Cl = 15.0%
Solution:
1) Convert % to molar ratios
A. Base: 100 g
=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol
Cl = 15.0 g / 35.45 g/mol = 0.4231 mol
B. divide by the higher number and round to whole number
Hg = 0.4325 / 0.4231 = 1.00
Cl = 0.4231 / 0.4231 = 1.00
=> Empirical formula = Hg Cl
2) Find the mass of the empirical formula:
HgCl: 200.59 g/mol + 35.45 g/mol = 236.04
3) Determine how many times is the empirical mass contained in the molecular mass:
470 g/mol / 236.04 = 1.99 ≈ 2
=> Molecular formula = Hg2 Cl2.
Answers:
Empirical formula HgCl
Molecular Formula Hg2Cl2
<span>Answer is: Van't Hoff factor
(i) for this solution is 1.051 .
Change in boiling point from pure solvent to solution: ΔT
=i · Kb · b.
Kb - </span><span>molal boiling point elevation constant</span><span> is 0.512°C/m.
b - molality, moles of solute per kilogram of solvent.
b = 1.26 m.
ΔT = 101.63°C - 100</span>°C = 1.63°C.
i = 1.63°C ÷ (0.512°C/m · 1.26 m).
i = 1.051.
