Use a sheet of paper to answer the following question. Take a picture of your answers and attach to this assignment. Predict the
product formed in the nucleophilic aromatic substitution reaction between 1-chloro-2,4-dinitrobenzene and sodium methoxide (NaOCH3). Draw the mechanism of the reaction in (a), showing why the product you have selected is formed.
2 answers:
Answer:1-methoxy-2,4-dinitrobenzene
Explanation:
The nitro groups are strongly electron withdrawing and promote nucleophilic substitution reactions where one of the original substituents is removed and replaced by a strong nucleophile such as the methoxy group. The mechanism of the reaction is attached below. The electron withdrawing nitrogroup assists the formation of the intermediate in the reaction as shown.
Answer: 1-Methoxy-2,4-dinitrobenzene is formed which is solublw in the solution of hot methanol giving it almost a 95% transperency.
Explanation:
Its not SN1 or SN2 infact it is "Nucleophilic aromatic substitution and a SNAr reaction"
At first initial step is the attach of the nucleophile (CH₃O⁻) as the methanol and Sodium Ethoxide are in equiliberium with each other where the Pi bonds are broken and then where the chlorine is attached and then this pi bond is travels to the Electron withdrawing group that is ortho or para to the leaving group and the double bond forms between the carbon and the Electron withdrawing group. The last step follows the elimination of chlorine and the methoxide group cannot bond with electron withdrawing group as they both negative in charge.
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Answer: 178.9 g
Explanation:
Density =
find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³
1.0 cm³ = 1.0 mL
so 195.112 cm³ = 195.112 mL
plug value into density equation:
0.917 g/mL = (mass) / (195.112 mL)
and solve for mass!
Answer 1:
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.
In present case, following reaction is involved:
2NO2 ↔ 2NO + O2
Here, K =![\frac{[PNO]^2[O2]}{[PNO2]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BPNO%5D%5E2%5BO2%5D%7D%7B%5BPNO2%5D%5E2%7D%20)
Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence, K =![\frac{[0.0022]^2[0.0011]}{[0.247]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5B0.0022%5D%5E2%5B0.0011%5D%7D%7B%5B0.247%5D%5E2%7D%20)
= 8.727 X 10^-8
Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................
Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.
Therefore, Reaction quotient = </span>
=![\frac{[0.021]^2[0.037]}{[0.192]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5B0.021%5D%5E2%5B0.037%5D%7D%7B%5B0.192%5D%5E2%7D%20)
= 4.426 X 10^-4.
Here, Reaction quotient > Equilibrium constant.
Hence, <span>the reaction need to go to reverse direction to reattain equilibrium </span>
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.
In present case, following reaction is involved:
2NO2 ↔ 2NO + O2
Here, K =
Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence, K =
= 8.727 X 10^-8
Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................
Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.
Therefore, Reaction quotient = </span>
=
= 4.426 X 10^-4.
Here, Reaction quotient > Equilibrium constant.
Hence, <span>the reaction need to go to reverse direction to reattain equilibrium </span>