1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
BigorU [14]
3 years ago
12

Determine whether each description applies to electrophilic aromatic substitution or nucleophilic aromatic substitution.

Chemistry
1 answer:
Alborosie3 years ago
3 0

Answer:

a. electrophilic aromatic substitution

b. nucleophilic aromatic substitution

c. nucleophilic aromatic substitution

d. electrophilic aromatic substitution

e. nucleophilic aromatic substitution

f. electrophilic aromatic substitution

Explanation:

Electrophilic aromatic substitution is a type of chemical reaction where a hydrogen atom or a functional group that is attached to the aromatic ring is replaced by an electrophile. Electrophilic aromatic substitutions can be classified into five classes: 1-Halogenation: is the replacement of one or more hydrogen (H) atoms in an organic compound by a halogen such as, for example, bromine (bromination), chlorine (chlorination), etc; 2- Nitration: the replacement of H with a nitrate group (NO2); 3-Sulfonation: the replacement of H with a bisulfite (SO3H); 4-Friedel-CraftsAlkylation: the replacement of H with an alkyl group (R), and 5-Friedel-Crafts Acylation: the replacement of H with an acyl group (RCO). For example, the Benzene undergoes electrophilic substitution to produce a wide range of chemical compounds (chlorobenzene, nitrobenzene, benzene sulfonic acid, etc).

A nucleophilic aromatic substitution is a type of chemical reaction where an electron-rich nucleophile displaces a leaving group (for example, a halide on the aromatic ring). There are six types of nucleophilic substitution mechanisms: 1-the SNAr (addition-elimination) mechanism, whose name is due to the Hughes-Ingold symbol ''SN' and a unimolecular mechanism; 2-the SN1 reaction that produces diazonium salts 3-the benzyne mechanism that produce highly reactive species (including benzyne) derived from the aromatic ring by the replacement of two substituents; 4-the free radical SRN1 mechanism where a substituent on the aromatic ring is displaced by a nucleophile with the formation of intermediary free radical species; 5-the ANRORC (Addition of the Nucleophile, Ring Opening, and Ring Closure) mechanism, involved in reactions of metal amide nucleophiles and substituted pyrimidines; and 6-the Vicarious nucleophilic substitution, where a nucleophile displaces an H atom on the aromatic ring but without leaving groups (such as, for example, halogen substituents).

You might be interested in
Precision Problems:
Lorico [155]

Answer:

Precision = 2.70 \± 0.1\ cm

Explanation:

Given

The data in the table

Required

Follow the steps appended to the question;

Step 1: Calculate the Mean or Average

Mean = Summation of lengths divided by number of teams;

Mean = \frac{2.65 + 2.75 + 2.80 + 2.77 + 2.60 + 2.65 + 2.68}{7}\ cm

Mean = \frac{18.9}{7}\ cm

Mean = 2.70\ cm

Step 2: Get The Range

Range = Highest - Lowest

Range = 2.80cm - 2.60cm

Range = 0.2\ cm

Step 3: Divide Range by 2

Approximate\ Range = \frac{1}{2}Range

Approximate\ Range = \frac{1}{2} * 0.2\ cm

Approximate\ Range = 0.1\ cm

Step 4: Determine the Precision

Precision = Average \± Approximate\ Range

Substitute 2.70 for Average and 0.1 for Approximate Range

Precision = 2.70 \± 0.1\ cm

4 0
3 years ago
if i add water to 100 mL of a 0.75 M NaOH solution un til the final volume is 165mL what will the molarity of the diluted soluti
ehidna [41]
Answer is: <span>the molarity of the diluted solution 0,454 M.
</span>V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0,1 L.
c₁(NaOH) = 0,75 M = 0,75 mol/L.
n₁(NaOH) = c₁(NaOH) · V₁(NaOH).
n₁(NaOH) = 0,75 mol/L · 0,1 L.
n₁(NaOH) = 0,075 mol
n₂(NaOH) = n₁(NaOH) = 0,075 mol.
V₂(NaOH) = 165 mL ÷ 1000 mL/L = 0,165 L.
c₂(NaOH) = n₂(NaOH) ÷ V₂(NaOH).
c₂(NaOH) = 0,075 mol ÷ 0,165 L.
c₂(NaOH) = 0,454 mol/L.
7 0
3 years ago
5. 16.3 g of NaCl is dissolved in water to make 1.75 L of solution. What is the molarity of this solution? A 0.159 M B 0.278 M C
ira [324]

Answer: The molarity of this solution is 0.159 M.

Explanation:

Given: Mass of solute = 16.3 g

Volume = 1.75 L

Number of moles is defined as the mass of substance divided by its molar mass.

Hence, moles of NaCl (molar mass = 58.44 g/mol) ar calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{16.3 g}{58.44 g/mol}\\= 0.278 mol

Molarity is the number of moles of a substance present in a liter of solution.

So, molarity of the given solution is calculated as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.278 mol}{1.75}\\= 0.159 M

Thus, we can conclude that the molarity of this solution is 0.159 M.

5 0
3 years ago
How many kilojoules is 1,500,000 calories
pochemuha

Answer:

1 cal = 0.004187 kJ

1,500,000 cal = 6280.5 kJ

5 0
3 years ago
A certain first-order reaction has a half-life of 25.2 s at 20°C. What is the value of the rate constant k at 60°C if the activa
DochEvi [55]

Answer:

t

(

2

)

1/2

=

85.25 s

Notice how you're given the half-life (for one temperature), a second temperature, and the activation energy. The key to doing this problem is recognizing that:

the half-life for a first-order reaction is related to its rate constant.

the rate constant changes at different temperatures.

Go here for a derivation of the half-life of a first-order reaction. You should find that:

t

1/2

=

ln

2

k

Therefore, if we label each rate constant, we have:

k

1

=

ln

2

t

(

1

)

1/2

k

2

=

ln

2

t

(

2

)

1/2

Recall that the activation energy can be found in the Arrhenius equation:

k

=

A

e

−

E

a

/

R

T

where:

A

is the frequency factor, i.e. it is proportional to the number of collisions occurring over time.

E

a

is the activation energy in

kJ/mol

.

R

=

0.008314472 kJ/mol

⋅

K

is the universal gas constant. Make sure you get the units correct on this!

T

is the temperature in

K

(not

∘

C

).

Now, we can derive the Arrhenius equation in its two-point form. Given:

k

2

=

A

e

−

E

a

/

R

T

2

k

1

=

A

e

−

E

a

/

R

T

1

we can divide these:

k

2

k

1

=

e

−

E

a

/

R

T

2

e

−

E

a

/

R

T

1

Take the

ln

of both sides:

ln

(

k

2

k

1

)

=

ln

(

e

−

E

a

/

R

T

2

e

−

E

a

/

R

T

1

)

=

ln

(

e

−

E

a

/

R

T

2

)

−

ln

(

e

−

E

a

/

R

T

1

)

=

−

E

a

R

T

2

−

(

−

E

a

R

T

1

)

=

−

E

a

R

[

1

T

2

−

1

T

1

]

Now if we plug in the rate constants in terms of the half-lives, we have:

ln

⎛

⎜

⎝

ln

2

/

t

(

2

)

1/2

ln

2

/

t

(

1

)

1/2

⎞

⎟

⎠

=

−

E

a

R

[

1

T

2

−

1

T

1

]

This gives us a new expression relating the half-lives to the temperature:

⇒

ln

⎛

⎜

⎝

t

(

1

)

1/2

t

(

2

)

1/2

⎞

⎟

⎠

=

−

E

a

R

[

1

T

2

−

1

T

1

]

Now, we can solve for the new half-life,

t

(

2

)

1/2

, at the new temperature,

40

∘

C

. First, convert the temperatures to

K

:

T

1

=

25

+

273.15

=

298.15 K

T

2

=

40

+

273.15

=

313.15 K

Finally, plug in and solve. We should recall that

ln

(

a

b

)

=

−

ln

(

b

a

)

, so the negative cancels out if we flip the

ln

argument.

⇒

ln

⎛

⎜

⎝

t

(

2

)

1/2

t

(

1

)

1/2

⎞

⎟

⎠

=

E

a

R

[

1

T

2

−

1

T

1

]

⇒

ln

⎛

⎜

⎝

t

(

2

)

1/2

400 s

⎞

⎟

⎠

=

80 kJ/mol

0.008314472 kJ/mol

⋅

K

[

1

313.15 K

−

1

298.15 K

]

=

(

9621.78 K

)

(

−

1.607

×

10

−

4

K

−

1

)

=

−

1.546

Now, exponentiate both sides to get:

t

(

2

)

1/2

400 s

=

e

−

1.546

⇒

t

(

2

)

1/2

=

(

400 s

)

(

e

−

1.546

)

=

85.25 s

This should make sense, physically. From the Arrhenius equation, the higher

T

2

is, the more negative the

[

1

T

2

−

1

T

1

]

term, which means the larger the right hand side of the equation is.

The larger the right hand side gets, the larger

k

2

is, relative to

k

1

(i.e. if

ln

(

k

2

k

1

)

is very large,

k

2

>>

k

1

). Therefore, higher temperatures means larger rate constants.

Furthermore, the rate constant is proportional to the rate of reaction

r

(

t

)

in the rate law. Therefore...

The higher the rate constant, the faster the reaction, and thus the shorter its half-life should be.

Explanation:

Sorry just go here https://socratic.org/questions/588d14f211ef6b4912374c92#370588

3 0
2 years ago
Other questions:
  • How can we find the volume of this fish tank? Please I needed but Good
    10·1 answer
  • Americium-241 is used in smoke detectors. it has a first order rate constant for radioactive decay of k=1.6×10−3yr−1. by contras
    10·1 answer
  • Show the difference in the reactivity of Na,Ca,Mg,Al,Fe,Cu with H2O
    5·1 answer
  • A thermodynamically unfavorable reaction can become favorable when coupled to a highly ________ reaction. A thermodynamically un
    8·1 answer
  • Which equation represents the combined gas law?
    14·2 answers
  • When you turn on a radio, what type of energy transformation takes place?
    11·1 answer
  • In a simple model of a neutral magnesium atom with the elements most common mass number. What must happen to an atom of magnesiu
    6·1 answer
  • When a relationship is specified, related records will be displayed in a form's
    11·2 answers
  • What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed as
    8·1 answer
  • Which of the following pairs of compounds will react to form a precipitate? Why?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!