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lina2011 [118]
2 years ago
6

Draw the major organic product(s) for the reaction. The starting material is a benzene ring with one substituent. The substituen

t is a nitrogen bonded to a hydrogen and a carbonyl that is bonded to a methyl group. THis reacts with tert butyl bromide and A l b r 3 to give the product.
Chemistry
1 answer:
Mice21 [21]2 years ago
3 0

Answer:

See Explanation

Explanation:

In electrophilic aromatic substitution, the benzene ring undergoes substitution when it is reacted with suitable electrophiles.

The products of electrophilic aromatic substitution depends on the substituents already present on the benzene ring. Some substituents activate the ring towards electrophilic substitution and direct the incoming electrophile to the ortho and para positions on the ring while some substituents deactivate the benzene ring towards electrophilic substitution and direct the incoming electrophlle to the meta position on the ring.

The amide substituent is moderately activating and is an ortho, para director hence the products shown in the mage attached to this answer.

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3 years ago
Read 2 more answers
a 125 g chunk of aluminum at 182 degrees Celsius was added to a bucket filled with 365 g of water at 22.0 degrees Celsius. Ignor
Diano4ka-milaya [45]
<h3>Answer:</h3>

32.98°C

<h3>Explanation:</h3>

We are given the following;

Mass of Aluminium as 125 g

Initial temperature of Aluminium as 182°C

Mass of water as 265 g

Initial temperature of water as 22°C

We are required to calculate the final temperature of the two compounds;

First, we need to know the specific heat capacity of each;

Specific heat capacity of Aluminium is 0.9 J/g°C

Specific heat capacity of water is 4.184 J/g°C

<h3>Step 1: Calculate the Quantity of heat gained by water.</h3>

Assuming the final temperature is X°C

we know, Q = mcΔT

Change in temperature, ΔT = (X-22)°C

therefore;

Q = 365 g × 4.184 J/g°C × (X-22)°C

    = (1527.16X-33,597.52) Joules

<h3>Step 2: Calculate the quantity of heat released by Aluminium </h3>

Using the final temperature, X°C

Change in temperature, ΔT = -(X°- 182°)C (negative because heat was lost)

Therefore;

Q = 125 g × 0.90 J/g°C × (182°-X°)C

  = (20,475- 112.5X) Joules

<h3>Step 3: Calculating the final temperature</h3>

We need to know that the heat released by aluminium is equal to heat absorbed by water.

Therefore;

(20,475- 112.5X) Joules = (1527.16X-33,597.52) Joules

Combining the like terms;

1639.66X = 54072.52

             X = 32.978°C

                = 32.98°C

Therefore, the final temperature of the two compounds will be 32.98°C

7 0
3 years ago
If you burn 29.4 g of hydrogen and produce 263 g of water, how much oxygen reacted?
vovikov84 [41]

The reaction between hydrogen and oxygen to form water is given as:

H_2 + O_2 \rightarrow H_2O

The balanced reaction is:

2H_2 + O_2 \rightarrow 2H_2O

According to the balanced reaction,

4 g of hydrogen (4\times 1) reacts with 32 g of oxygen (2\times 16).

So, oxygen reacted with 29.4 g of hydrogen is:

\frac{29.4\times 32}{4} = 235.2 g

Hence, the mass of oxygen that is reacted with 29.4 g of hydrogen is 235.2 g.

7 0
3 years ago
Please give me solved answer pic its really importen ​
inn [45]
Where’s the picture?
5 0
3 years ago
A sample of 3.62 moles of diphosphorous trioxide is
Sindrei [870]

H_3PO_32H_3PO_32H_3PO_3P_2O_3Answer:

B

Explanation:

This question is about stoichiometry. From the balanced equation P_2O_3 + 3H_2O⇒2H_3PO_3, we see that 3 moles of water is needed to react with 1 mole of P_2O_3.

This means that, to fully react 3.62 moles of P_2O_3, we would need 3*3.62 or  10.86 moles of water. However, we only have 6.31 moles, so water is the limiting reactant.

Since 3 moles of water react with 1 mole of P_2O_3, 6.31 moles of water can fully react with 6.31÷3 or 2.1033 moles of P_2O_3.

From the balanced equation, we see that every mole of P_2O_3 reacted gets you 2 moles of 2H_3PO_3. Therefore, 2.1033 moles of P_2O_3 would give you approximately 4.21 moles of H_3PO_3.

5 0
2 years ago
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