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pshichka [43]
3 years ago
6

Identify the intermediate leads to the major product for the reaction of 2-methyl-2-butene with hydrogen bromide in the presence

of peroxide.
Chemistry
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:

II

Explanation:

We must have a good idea of the fact that there are two mechanisms that come into play when we are discussing about the addition of hydrogen halides to alkenes. The first is the ionic mechanism and the second is the radical mechanism.

The ionic mechanism is accounted for by the Markovnikov rule while the radical mechanism occurs in the presence of peroxides and is generally referred to as anti Markovnikov  addition.

The intermediate in anti Markovnikov  addition involves the most stable radical, in this case, it is a tertiary radical as shown in the images attached. The most stable radical is II hence it leads to the major product shown in the other image.

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10. A 20.00 mL sample of 0.150 mol/L ammonia (NH3(aq)) is titrated to the equivalence point by 20.0 mL of a solution of 0.150 mo
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Answer:

\large \boxed{\rm a)\, NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq);\,b)\,11.22;\, c)\, 5.19}

Explanation:

a) Balanced equation

The balanced chemical equation for the titration is

\large \boxed{\rm NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq)}

b) pH at start

For simplicity, let's use B as the symbol for NH₃.

The equation for the equilibrium is

\rm B + H_{2}O \, \rightleftharpoons\,BH^{+} + OH^{-}

(i) Calculate [OH]⁻

We can use an ICE table to do the calculation.

                      B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹:     0.150               0         0

C/mol·L⁻¹:       -x                 +x       +x

E/mol·L⁻¹:  0.150 - x            x          x

K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}

Check for negligibility:

\dfrac{0.150 }{1.8 \times 10^{-5}} = 8300 > 400\\\\x \ll 0.150

(ii) Solve for x

\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}

(iii) Calculate the pH

\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.64 \times 10^{-3}) = 2.78\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.78 = \mathbf{11.22}\\\\\text{The pH of the solution at equilibrium is } \large \boxed{\mathbf{11.22}}

(c) pH at equivalence point

(i) Calculate the moles of each species

\text{Moles of B} = \text{Moles of HI} = \text{20.00 mL} \times \dfrac{\text{0.0150 mmol}}{\text{1 mL}} = \text{3.00 mmol}

                 B    +    HI   ⇌   BH⁺ + I⁻

I/mol:       3.00    3.00         0

C/mol:    -3.00   -3.00     +3.00

E/mol/:       0          0          3.00

(ii) Calculate the concentration of BH⁺

At the equivalence point we have a solution containing 3.00 mmol of NH₄I

Volume = 20.00 mL + 20.00 mL = 40.00 mL

\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}

(iii) Calculate the concentration of hydronium ion

We can use an ICE table to organize the calculations.

                      BH⁺+ H₂O ⇌ H₃O⁺ +  B

I/mol·L⁻¹:     0.0750                 0        0

C/mol·L⁻¹:        -x                     +x      +x

E/mol·L⁻¹:   0.0750 - x             x         x

K_{\text{a}} = \dfrac{K_{\text{w}}} {K_{\text{b}}} = \dfrac{1.00 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\\\\\dfrac{x^{2}}{0.0750 - x} = 5.56 \times 10^{10}\\\\\text{Check for negligibility of }x\\\dfrac{0.0750}{5.56 \times 10^{-10}} = 1.3 \times 10^{6} > 400\\\\\therefore x \text{ $\ll$ 0.0750}

\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}

(iv) Calculate the pH

\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}

The titration curve below shows the pH at the beginning and at the equivalence point of the titration.

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