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Flauer [41]
3 years ago
10

Hydrogenation (H2/Pd-C) of 3,4-dimethyl-3-hexene (3,4-dimethyl hex-3-ene) would be expected to form A) an enantiomeric pair for

both the (E) and (Z) isomers. B) an enantiomeric pair for the (E) isomer and one meso compound for the (Z) isomer. C) one meso compound for the (E) isomer and an enantiomeric pair for the (Z) isomer. D) will not form any stereoisomers since the product is a saturated hydrocarbon. E) None of the above provide a correct response.
Chemistry
1 answer:
NeTakaya3 years ago
3 0

Answer:

D) will not form any stereoisomers since the product is a saturated hydrocarbon.

Explanation:

Stereoisomerism, or spatial isomerism,refers to a type of isomerism in which  two or more molecules have the same molecular formula and atom to atom connectivity but different three-dimensional orientations of those atoms in space.

Stereoisomerism occurs in alkenes because free rotation about the C=C bond is hindered. Thus, the molecule is "locked" in a particular orientation. In a saturated alkane, there is unrestricted free rotation about the C-C bond hence straight chain alkanes do not have steroisomers.

The product formed is 3,4-dimethyl hexane which is a straight chain saturated alkane. This compound does not exhibit stereoisomerism.

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The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol..[1]

O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol..[2]

NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol..[3]

NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}

2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol

2\times \Delta H^o_{4}=470.3 kJ/mol

\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

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