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lina2011 [118]
3 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]3 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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Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as
nikitadnepr [17]

The question in incomplete, complete question is;

Determine the theoretical yield:

Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.

Answer:

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

Moles of copper(II) nitrate = \frac{469 g}{187.5 g/mol}=2.5013 mol

Moles of sodium sulfide = \frac{156 g}{78 g/mol}=2 mol

According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.

Then 2 moles of sodium sulfide will react with:

\frac{1}{1}\times 2mol= 2 mol of copper (II) nitrate

As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.

According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:

\frac{2}{1}\times 2mol=4 mol sodium nitrate

Mass of 4 moles of sodium nitrate :

85 g/mol × 4 mol = 340 g

Theoretical yield of sodium nitrate = 340 g

The theoretical yield of sodium nitrate is 340 grams.

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