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kozerog [31]
4 years ago
12

Compound W, C8H17Br, and X, C8H17Cl, are products of the radical bromination (W), and chlorination (X), of 2,4-dimethylhexane. W

and X are each chiral. W reacts readily in water to form a chiral substitution product Y, C8H17OH. Reaction of a single enantiomer of W produces a single enantiomer of Y as the only substitution product. X likewise reacts readily in water though at a slower rate than W; a chiral substitution product Z, C8H17OH, is again formed. Unlike W, however, reaction of a single enantiomer of X produces Z that is nearly racemic. Propose structures for W and X.

Chemistry
1 answer:
oksian1 [2.3K]4 years ago
5 0

Answer:

The possible structures are shown in figure.

Explanation:

The seven isomers that could be formed from the given compound due to radical halogenation are shown in the figure.

We have mentioned only chlorine as group for showing just the chiral nature of the compound.

A chiral compound has a carbon with all the four groups different from each other.

There are three structures which are chiral (as shown).

W is the brominated product

X is the chlorinated product.

1) W reacts readily in water to form a chiral substitution product Y, C8H17OH.

Reaction of a single enantiomer of W produces a single enantiomer of Y as the only substitution product.: It means that the substitution is SN2 which is possible if the compund is not tertiary.

2) X likewise reacts readily in water though at a slower rate than W; a chiral substitution product Z, C8H17OH, is again formed. Unlike W, however, reaction of a single enantiomer of X produces Z that is nearly racemic.

It means that it will undergo SN1 reaction,. So the alkyl halide must be tertiary.

The possible structures are shown in figure.

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The equilibrium constant is  K =0.02867

The equilibrium pressure for oxygen gas is  P_{O_2} =  0.09367\  atm

Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

   The Gibbs free energy forM_3 O_4_{(s)} is  \Delta G^o_{1}  = -8.80 \ kJ/mol

  The Gibbs free energy forM{(s)} is  \Delta G^o_{2}  = 0 \ kJ/mol

    The Gibbs free energy forO_2{(s)} is  \Delta G^o_{3}  = 0 \ kJ/mol

The Gibbs free energy of the reaction is mathematically represented as

         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

         \Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)

Substituting values

From the balanced equation

         \Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]

        \Delta G^o_{re} = 8.80 kJ/mol =8800J/mol

The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          K_p =[ P_{O_2}]^{\frac{3}{2} }

Where   [ P_{O_2}] is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

            K =0.02867

Now substituting this into the equation above to obtain the equilibrium of oxygen

           0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}

multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

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