Chemistry, 19.07.2019 01:30, JakerGamer

7.50 draw all constitutional isomers with the molecular formula c4h9br, and then arrange them in order of: (a) increasing reactivity toward an sn2 reaction. (b) increasing reactivity toward an e2 reaction

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answered: zoeycrew

Isomers are:

2-bromobutane (I)

2-bromo-2-methylpropane (II)

1-bromo-2-methylpropane (III)

1-bromobutane (IV)

for SN2 reaction :

IV > III > I > II

for E2 reaction is:

IV < III < I < II

Explanation:

The constitutional isomers are (structures are shown if figure):

2-bromobutane (I)

2-bromo-2-methylpropane (II)

1-bromo-2-methylpropane (III)

1-bromobutane (IV)

For SN2 reaction : the order of reactivity of alkyl halides is

tertiary < secondary < primary

For E2 reaction: the order of reactivity of alkyl halides is

tertiary > secondary > primary

Thus the order of reactivity for SN2 reaction is:

IV > III > I > II

The order of reactivity for E2 reaction is:

IV < III < I < II

7.50 draw all constitutional isomers with the molecular formula c4h9br, and then arrange them in ord

answered: Guest

The answer is b because c don't explain why the moths are on them and it isn't a because as we know only one species of moth does this

answered: Guest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.