Which of the following is required to change the state of matter?

Kryger [21]

Answer:

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

Mechanism: Hydroboration

Explanation:

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

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

See figure 1

I hope it helps!

4 0

2 years ago

The last intermediate in citric acid cycle is:_________

GenaCL600 [577]

The last intermediate in citric acid cycle is Oxaloacetic acid.

<h3>What is Citric Acid Cycle?</h3>

Organic molecule HOC(CO2H)(CH2CO2H)2 is the chemical formula for citric acid. It is a weak organic acid that is colorless. Citrus fruits naturally contain it. It is a biochemical intermediary in the citric acid cycle, which is a component of all aerobic organisms' metabolism.

Every year, more than two million tons of citric acid are produced. It is frequently used as a flavoring, an acidifier, and a chelating agent.

Citrates, which include salts, esters, and the polyatomic anion present in solution, are derivatives of citric acid. Trisodium citrate is an example of the former; triethyl citrate is an example of an ester.

Learn more about citric acid with the help of the given link:

brainly.com/question/15582668

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7 0

1 month ago

The bond that join the 3'-hydroxyl group of the deoxypentose of one nucleotide to the 5'-hydroxyl group of the deoxypentose of a

lara [203]

Answer:

phosphodiester bond

Explanation:

Phosphodiester linkage/bond is found in deoxyribonucleic and ribonucleic acids. It is formed from a reaction involving the elimination of water from a reaction involving the hydroxyl groups of two different 5-carbon (pentose) sugars and a phosphate group.

The elimination of water, also known as condensation reaction occur twice, resulting in the formation of two ester bonds which then bind the phosphate group to the pentose sugars to become a phosphodiester bond.

The bond links the 3'-hydroxyl group of one of the pentose sugars and the 5'-hydroxyl group of the other pentose sugar in the nucleotides that make up nucleic acids.

8 0

2 years ago

Which is the correct formula for phosphorus pentachloride? PCl4 because a subscript of 4 indicates four Cl atoms PCl5 because a

Arlecino [84]

Answer: The correct formula for phosphorous pentachloride is $PCl_5$ because a subscript 5 indicates five chlorine (Cl) atoms.

Explanation: For the given molecule, phosphorous pentachloride, there are 2 atoms present which are phosphorous and chlorine atoms.

Number of phosphorous atoms = 1

Number of chlorine atoms = 5

So, the correct formula for phosphorous pentachloride will be $PCl_5$ because the subscript 5 represents the 5 chlorine atoms.

7 0

2 years ago

Read 2 more answers

Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.

WARRIOR [948]

Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer

5 0

2 years ago