g in the following three compounds(1,2,3) arrange their relative reactivity towards the reagent CH3Cl / AlCl3. Justify your orde
1 answer:
Answer:
3 > 2> 1
Explanation:
Aromatic compounds undergo electrophilic substitution reaction with several electrophiles.
Some substituted benzenes are more reactive towards electrophilic aromatic substitution than unsubstituted benzene.
Certain groups of substituents increase the ease with which an aromatic compound undergoes aromatic substitution.
If we look at the compounds closely, we will notice that only toluene leads to easy reaction with CH3Cl / AlCl3. Thus is due to the +I inductive effect of -CH3 which stabilizes the negatively charged intermediate produced in the reaction.
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<u>Answer:</u> The concentration of required will be 0.285 M.
<u>Explanation:</u>
To calculate the molarity of , we use the equation:
Moles of = 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:
For the given chemical equations:
Net equation:
To calculate the equilibrium constant, K for above equation, we get:
The expression for equilibrium constant of above equation is:
As, is a solid, so its activity is taken as 1 and so for
We are given:
Putting values in above equations, we get:
Hence, the concentration of required will be 0.285 M.
Answer:
Reagent O₂ will be consumed first.
Explanation:
The balanced reaction between O₂ and C₄H₁₀ is:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles
- O₂: 13 moles
- CO₂: 8 moles
- H₂O: 10 moles
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of the compounds that participate in the reaction is:
- C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
- H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles* 58 g/mole= 116 g
- O₂: 13 moles* 32 g/mole= 416 g
- CO₂: 8 moles* 44 g/mole= 352 g
- H₂O: 10 moles* 18 g/mole= 180 g
If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?
mass of O₂= 223.78 grams
But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>