All categories

3 years ago

Draw the product of the following reaction between propanoyl chloride and p-methoxynitrobenzene. Include all formal charges.

Chemistry

2 answers:

Lera25 [3.4K]3 years ago

8 0

Please find the attached document

GREYUIT [131]3 years ago

4 0

Answer:

The product will be :

1-(2-methoxy-5-nitrophenyl)propan-1-one

Explanation:

This is electrophilic substitution reaction.

The electrophile is acylium ion.

The reaction will occur in presence of anhydrous aluminium chloride, which will remove chloride ion and will generate acylium ion which will act as electrophile.

The electrophile will attack on the para or other position of methoxy group.

The mechanism is shown in figure.

You might be interested in

A gas has a volume of 20.0 L at a pressure of 950 mmHg with a temperature of 125.0 o If the conditions are changed to STP, what

ehidna [41]

Use formula: Initial Pressure x Initial Volume/Initial temperature = Final pressure x Final Volume/Final Temperature => 17.15L

6 0

3 years ago

Chemistry Lab Question: Filtration

tatyana61 [14]

For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.

For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.

6 0

3 years ago

19. I accidentally mixed bleach and ammonia when cleaning one day. It released 55 liters of deadly chlorine (CI) gas. How many g

user100 [1]

55,000 grams
Hope this helps you out :)

3 0

3 years ago

The density of gold is 19.32 g/mL. What is the volume, in mL, of a nugget of gold with a mass of 55.07 g?

Vika [28.1K]

Answer: 2.850

Explanation:

$19.32=\frac{55.07}{v}\\19.32v=55.07\\v=\frac{55.07}{19.32}=\boxed{2.850 \text{ mL}}$

5 0

2 years ago

What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa

almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of stomach acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL$

Putting values in above equation, we get:

$1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL$

Hence, the volume of HCl neutralized is 1.25 mL

3 0

3 years ago