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Natasha2012 [34]
3 years ago
15

1-Bromopropane is treated with each of the following reagents. Draw the major substitution product if the reaction proceeds in g

ood yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
a. with H_2 O
b. with H_2 SO_4
c. with 1 equiv of KOH
d. with Csl
e. with NaCN
f. with HCl
g. with (CH_3)_2 S
h. with 1 equiv of NH_3
j. with CI_2
k. with KF

Chemistry
1 answer:
sammy [17]3 years ago
5 0

Answer:

Explanation:

If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane  a primary alkyl-halide.

Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used  otherwise it will result to a reactive mechanism if a weak electrophile is used.

However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.

The attached diagrams portraying this notions is shown in the attached file below.

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Earth is currently 150 million km away from the Sun. If Earth was 200 million km away from the Sun, why would it no longer be ab
timurjin [86]

Answer:

The temperature would be too cold

Explanation:

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3 years ago
The following calculations must be handwritten in your notebook. – Acetic Acid ■ Hydrogen ion concentration ■ Ka ■ % Error – Ace
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Answer:

Explanation:

1) Acetic acid

Concentration is given as 0.103 M

The average pH of this solution = 2.96

we know that pH = - log [H+] therefore [H+] = 10-pH

[H+] = 10-2.96

= 1.1 x 10-3 M = 0.0011 M

Consider the equilibrium

CH3COOH ⇄CH3COO- + H+

Initial 0.103 0 0

Change -x +x +x

equlibrium 0.103 -x x x

Ka = x2 / 0.103 - x

Here the initial concentration of CH3COOH = 0.103 M

the equilibrium concentration of H+ = x = 0.0011 M

Therefore the equilibrium conc of acetic acid = 0.103 - 0.0011 = 0.1019 M

Therefore Ka = 0.0011 x 0.0011 / 0.1019 = 1.187 x 10-5

2) Acetic acid + NaOH

pH measured = 4.48 , therefore [H+} = 10-4.48 = 3.3 x 10-5

Volume and conc of acetic acid = 10 mL of 0.103 M

= 10 mL x 0.103 mmol / mL

= 1.03 mmol

Volume and conc of NaOH added = 4 mL of 0.0992 M

= 4 x 0.0992 mmol

= 0.397 mmol

Consider the equation

CH3COOH + NaOH -----------> CH3COONa + H2O

Initial 1.03 0.397 0

Final 0.633 0 0.397

0.397 mmole of NaOH will convert 0.397 mmole of acetic acid to sodium acetate.

Thus the final moles of acetic acid and sodium acetate in the solution are 0.633 and 0.397

therefore [salt] / [acid] = 0.397 / 0.633 = 0.627

By Hendersen equation pH = pKa + log[salt / acid]

pH = pKa + log 0.627 = pKa - 0.203

or pKa = pH + 0.203 = 4.48 + 0.203 [ since the measured pH = 4.48]

= 4.683

Ka = 10-4.683 = 2.07 x 10-5

3) Phosphate salts:

(i) mass of NaH2PO4 taken = 0.613 g

molar mass of NaH2PO4 = 120

therefore moles = 0.613 / 120 = 0.0051 mole

= 5.1 mmol

The volume is 30 mL therefore concentration = 5.1 /30 mmol/mL

= 0.17 M

consider the equilibrium

H2PO4-⇄ HPO42- + H+

Initial 0.17 0 0

Change -x +x + x

equilibrium 0.17-x x x

Ka = x2 / 0.17-x = 6.2 x 10-8 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.17 x 6.2 x 10-8

x =  1.03 x 10-4

Thus the equilirium conc of H+ = 1.03 x 10-4 therefore pH = - log 1.03 x 10-4 = 3.99

(ii) Mass of Na2HPO4.7H2O =0.601 g

therefore no of moles = 0.601 / 268.07 = 0.00224 mole

= 2.24 mmol

The volume = 30 mL , therefore conc = 2.24 / 30 mmol/ml

= 0.075 M

consider the equilibrium

HPO42- ⇄ PO43- + H+

Initial 0.075 0 0

Change -x +x + x

equilibrium 0.075-x x x

Ka = x2 / 0.075-x = 4.8 x 10-13 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.075 x 4.8 x 10-13

x =  1.9 x 10-7

Thus the equilirium conce of H+ = 1.9 x 10-7 therefore pH = - log 1.9 x 10-7 = 6.7

(iii) Mass of Na3PO4.12H2O taken = 0.208 g

moles of trisodiumphosphate 0.208/ 380 = 0.00055 moles

= 0.55 mmol

Volume = 10 mL therefore conc = 0.55/10 = 0.055 mmol/mL

= 0.055 M

Consider the equilibrium reaction

PO43- + H2O  ⇄ HPO42- + OH-

initial 0.055 0 0

Change -x +x +x

equilibrium 0.055-x x x

Kb = x2/ 0.055 -x = 0.0208 [Kb = Kw / Ka = 10-14 / 4.8 x 10-13 = 0.0208]

x2 + 0.0208x - 0.001144 = 0 Solving this equation we get x = 0.025

That is the conce of OH- ion = 0.025M

Therefore pH = 14 - pOH = 14 - 1.6 =12.4

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3 years ago
Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine
yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

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3 years ago
Which statement best describes the reaction pathway graph for an exothermic reaction but not an endothermic reaction? It has a h
ruslelena [56]
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Explanation:

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The hills that you can see in a reaction energy diagram are not related with the final change of energy. The hills are an indication of the activation energy needed to start the reaction, but they do not measure the change of energy from the products to the reactants.

The enthalpy that is a state variable that identifies the content of heat. Then the change of enthalpy for the exothermic reactions is negative, meaning that the energy of the products is lower than the energy of the reactants. 
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If an electron fell to a lower orbit or energy level, energy would be
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It would emit energy in most of the cases in form of light
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