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
Zina [86]
4 years ago
13

Which statement is incorrect regarding the reaction of benzene with an electrophile? View Available Hint(s) A) The carbocation i

ntermediate loses a proton from the carbon bonded to the electrophile. B) Carbocation formation is the rate-determining step. C) Benzene functions as a nucleophile. D) The carbocation intermediate reacts with a nucleophile to form the addition product.

Chemistry
1 answer:
Leno4ka [110]4 years ago
3 0

Answer:

The carbocation intermediate reacts with a nucleophile to form the addition product.

Explanation:

The reaction of benzene with an electrophile is an electrophillic substitution reaction. Here the electrophile replaces hydrogen. There is no formation of carbocation as intermediate in the reaction. Infact there is transition state where the electorphile attacks on benzene ring and at the same time the hydrogen gets removed from the benzene. So a transition carbocation is formed.

The general mechanism is shown in the figure.

i) Attack of the electrophile on the benzene (which is the nucleophile)

ii) The carbocation intermediate loses a proton from the carbon bonded to the electrophile.

iii) the carbocation formation is the rate determining step.

iv) There is no formation of addition product.

Thus the wrong statement is

The carbocation intermediate reacts with a nucleophile to form the addition product.

You might be interested in
Calculate the pH and pOH of 500.0 mL of a phosphate solution that is 0.285 M HPO42– and 0.285 M PO43–. (Ka for HPO42- = 4.2x10-1
jekas [21]

Answer: when concentrations of acid and base are same, pH = pKa

PH = 12.38 pOH = 1.62

Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00

8 0
3 years ago
1. How many ATOMS of carbon are present in 7.48 grams of carbon monoxide ?
Salsk061 [2.6K]

Answer:

The answer is 1.61 × 10²³ atoms

Explanation:

To determine number of atoms, we will use the formula below

Number of atoms = number of moles (n) × avogadro's constant (6.02 x 10²³)

n was not provided, hence we will solve for n

n = mass/ molar mass

molar mass of carbon monoxide, CO (where C is 12 and O is 16) is 12 + 16 = 28

mass was provided in the question as 7.48

n = 7.48/28

n = 0.267

Hence,

number of atoms = 0.267 × 6.02 x 10²³

= 1.61 × 10²³ atoms

3 0
4 years ago
The chemical name of copper ii tetra sulphate pentahydrate​
Anastasy [175]

hope it's helpful ❤❤❤

THANK YOU.

3 0
3 years ago
CAN SOMEONE PLSSS HELP ME LIKE PLS IM GONNA CRY PLSSSSS BRAINLY15 PONITS PLS
Kitty [74]
If there is chloroplasts it’s a plant
5 0
3 years ago
Which one has a greatest mass, one atom of carbon, one atom of hydrogen, or one atom of litium
SpyIntel [72]
Lithium, it has an atomic number of about 3
5 0
3 years ago
Other questions:
  • What term describes isotopes with an unbalanced number of neutrons?
    13·1 answer
  • CH3COOH mc015-1.jpg CH3COO– (aq) + H+(aq) What will happen to the chemical equilibrium of the solution if CH3COONa is added?
    8·1 answer
  • What the branch of a zoologist
    8·2 answers
  • 224/88Ra+4/2He+? find the missing isotope
    5·1 answer
  • Can you mix copper with rubbing alcohol
    7·1 answer
  • Analysis of an unknown compound reveals that the compound is 27.29% carbon and 72.71% oxygen. The molar mass of the compound is
    15·1 answer
  • we have not visited jupiter to collect samples, so how do we know what the atmosphere is composed of?
    14·1 answer
  • Iron-59 which is used to diagnose anaemia has a half life of 45days, what fraction of it is left in 90 days
    15·1 answer
  • 6. Which change to a population would cause
    9·1 answer
  • Mention three sources of light​
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!