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irinina [24]
4 years ago
10

The carbocation electrophile in a friedel-crafts reaction can be generated in ways other than by reaction of an alkyl chloride w

ith alcl3. For example, reaction of benzene with 2-methylpropene in the presence of h3po4 yields tert-butylbenzene. Draw a structure for the electrophile.

Chemistry
1 answer:
Firdavs [7]4 years ago
4 0
<h3>Answer:</h3>

            The structure of Electrophile (2-methylpropan-2-ylium) is shown in attached picture highlighted by green color.

<h3>Explanation:</h3>

                     Electrophiles are those species which are electron deficient and loves electron. These species often carry positive charge or contain atoms which have incomplete octet.

                     Also, we know that Benzene undergo Electrophilic Substitution reactions in which hydrogen atom is replaced by other electrophile. In given statement the electrophile to be reacted with benzene to form tert-Butylbenzene is generated from 2-methylpropene. 2-methylpropene when treated with phosphoric acid adds one proton across the double bond (electrophilic addition reaction) and generate a positively charged tertiary carbocation (stable) (highlighted green) which acts as an electrophile. Further on reaction with benzene this electrophile is substituted with proton and results in the formation of tert-Butylbenzene.

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8 0
3 years ago
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The equilibrium constant Kp for the reaction I2(g) + Br2(g) ⇀↽ 2 IBr(g) + 11.7 kJ is 280 at 150◦C. Suppose that a quantity of IB
mixas84 [53]

Answer:

\large \boxed{\text{0.0120 atm }}

Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

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p(IBr) = 0.200 atm

1. Set up an ICE table.

Let p = the initial pressure of IBr. Then

\begin{array}{ccccccc}\rm \text{I}_{2}& + & \text{Br}_{2} & \, \rightleftharpoons \, & \text{2IBr} &  &  \\0 & & 0 & & p & & \\+x &   & +x  & & -2x &   &\\x &   & x} &   & 280 & & \\\end{array}

2. Calculate p(I₂)

\begin{array}{rcl}K_{\text{p}}&=&\dfrac{p_{\text{IBr}}^{2}} {p_{\text{I}_2}^{2}}\\\\280&=&{\dfrac{0.200^{2}}{x^{2}}&&\\\\280x^{2} & = &0.0400\\x^{2} & = &\dfrac{0.0400}{280 }\\\\& = & 1.429 \times 10^{-4}\\x & = & \textbf{0.0120 atm}\\\end{array}\\\text{The partial pressure of iodine is $\large \boxed{\textbf{0.0120 atm }}$}}

Check:

\begin{array}{rcl}{\dfrac{0.200^{2}}{0.0120^{2}}}&=&280\\\\280& =& 280\\\end{array}

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