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Hitman42 [59]
3 years ago
10

Give regents and mechanism

Chemistry
1 answer:
Kryger [21]3 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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What is the order of reaction with respect to a and b for a reaction that obeys the rate law: rate = k[a]4[b]5?
spin [16.1K]

The order of reaction with respect to a is 4 and with respect to b is 5 for a reaction that obeys the rate law.

The order of reaction is equal sum of power of concentration of the each reactant present in the reaction  that obeys the rate law . The order of individual reactant or species present in the rate law is equal to the power of the of concentration of the respective species or reactant .

Example : If rate law is given as,

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3 0
2 years ago
A sample of 8.00 g of Mg(OH)2 is added to 24.2 mL of 0.205 M HNO3. How many moles of Mg(OH)2 our present after the reaction is c
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The amount of Mg(OH)2  present after the reaction is complete is 0.136 moles of Mg(OH)2.

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

Number of moles of Mg(OH)2 = 8.00 g/58 g/mol = 0.138 moles

Number of moles of HNO3 = 0.205 M × 24.2 mL/1000 = 0.00496 moles

Given that;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

0.00496 moles of HNO3 reacts with 0.00496 moles ×  1 mole /2 moles = 0.00248 moles of Mg(OH)2

Hence, Mg(OH)2 is the reactant in excess.

The amount of Mg(OH)2 remaining = Amount present - Amount reacted

Hence; 0.138 moles - 0.00248 moles = 0.136 moles of Mg(OH)2

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