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3 years ago

Compound A has molecular formula C7H15Br. Treatment of compound

Chemistry

1 answer:

Ira Lisetskai [31]3 years ago

3 0

Answer:

See images below.

Explanation:

The sodium ethoxide is a strong nucleophile and a strong base. When it acts as a strong nucleophile (a substance that has unpaired electrons and needs to share or donate it to be stable), an elimination reaction occurs. When it acts as a strong base, a substitution reaction occurs.

For a halide, the elimination takes place when the halogen is no steric hindrance, so the halogen must be in position 1. Then the compound A must be the must steric hindrance compound with formula C₇H₁₅Br, which is the compound showed below (3-Bromo-3-ethyl-pentane), to promote only elimination product. It happens because it's a tertiary halide.

When it reacts with sodium ethoxide (H₃C-H₂C-O-Na), the Br⁻ will leave the compound with close hydrogen, and a double bond will be formed between the carbons, so compound B must be the one showed below (3-Ethyl-2-pentene).

When the compound B is treated with dilute sulfuric acid, an addition reaction will occur, so the double bond will be broken and a hydrogen and a hydroxyl must be added to the compound. For the Markov-Nikov rule, the hydrogen will be in the most hydrated carbon. So, compound C will be the 3-Ethyl-3-pentanol showed below.

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VikaD [51]

Answer : The value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

Explanation :

The balanced cell reaction will be,

$Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)$

The half-cell reactions are:

Oxidation reaction (anode) : $Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$

Reduction reaction (cathode) : $2Ag^+(aq)+2e^-\rightarrow 2Ag(g)$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

$\Delta G^o=-2\times 96500\times 0.93$

$\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol$

Now we have to calculate the value of 'K'.

$\Delta G^o=-RT\ln K$

where,

$\Delta G_^o$ = standard Gibbs free energy = -180 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

$-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K$

$K=3.6\times 10^{31}$

Therefore, the value of $\Delta G^o$ and K is, -180 kJ/mol and $3.6\times 10^{31}$

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