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Hitman42 [59]
4 years ago
10

Give regents and mechanism

Chemistry
1 answer:
Kryger [21]4 years ago
4 0

Answer:

Reagents: 1) BH_3 2) H_2_O2, OH^-

Mechanism: Hydroboration

Explanation:

In this case, we have a <u>hydration of alkene</u>s reaction. But, in this example, we have an <u>anti-Markovnikov reaction</u>. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: <u>"hydroboration"</u>.

The <u>first step</u> of this reaction is the addition of borane (BH_3) to the double bond. Then in the <u>second step</u>, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the <u>third step</u>, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond "BH_2-O-OH". In <u>step number 4</u> we have the migration of the C-B bond to oxygen. Then in <u>step number 5</u>, we have the attack of OH^- on the O-BH_2 to produce an alkoxide. Finally, the water molecule produce in step 2 will <u>protonate</u> the molecule to produce the alcohol.

See figure 1

I hope it helps!

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A food substance kept at 0°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6
ZanzabumX [31]

Answer:

1.   67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

Given that:

time t_1 = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time t_2 = 10.6 hours

Temperature T_1 = 0° C = (0+273 )K = 273 K

Temperature T_2 = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since (\frac{k_2}{k_1}=\frac{t_2}{t_1})

Then we can rewrite the above expression as:

In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})

2.934 = \frac{E_a}{8.314}(\frac{30}{82719})

2.934 = \frac{30E_a}{687725.766}

30E_a = 2.934 *687725.766

E_a = \frac{2.934 *687725.766}{30}

E_a =67255.58 \ J/mol

E_a =67.2 \ kJ/mol

7 0
3 years ago
What is minimum volume of oxygen required to react with 42.5 g of aluminum in the synthesis of aluminum oxide at stp?
natka813 [3]
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Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.

Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.

Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815

Finally, you need to calculate the volume of </span>1.1815 <span>moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,

1.1815 * 22.7 = </span>26.8 liters <span>of oxygen gas.
</span>
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