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Zina [86]
3 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]3 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.

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umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

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