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pentagon [3]
2 years ago
11

Acetylene is an important synthetic building block that can be used to produce organic compounds with a wide variety of other fu

nctional groups. Consider the transformation of 3-phenylpropanal from acetylene shown below, which requires several steps.

Chemistry
1 answer:
Neporo4naja [7]2 years ago
4 0

In organic synthesis, we are trying to produce a compound by series of manipulations.

<h3>What is organic synthesis?</h3>

In organic synthesis, we are trying to produce a compound by series of manipulations. Note that organic synthesis is akin to an architect who is building a house.

In this case, we are trying to construct the molecule  3-phenylpropanal from acetylene. The first step involves the reaction of the alkene with RX. This is a nucleophilic reaction. Next we reduce the alkyne using Na/NH3. Lastly, the reagent 9-BBN and H2O2/OH- are used to obtain the product as shown.

Learn more about organic synthesis:brainly.com/question/3633466

#SPJ1

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AlladinOne [14]
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Yes.

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6 0
3 years ago
How many grams of phosphorus are required to produce 4.21x10^22 molecules of phosphorus trifluoride?
kolezko [41]

2.1618 grams of P4 is required to produce 4.21x10^22 molecules of phosphorus trifluoride.

Explanation:

The balanced chemical reaction for formation of PF3 from yellow phosphorus is given by:

P4 + 6F2 ⇒ 4PF3

1 mole of P4 reacts to give 4 moles of PF3

It is given that 4.21x10^22 molecules of PF3 are produced

so the number of moles can be calculated by the relation

number of molecules = number of moles × Avagadro number

number of moles = \frac{number of molecules}{avagadro number}

n = \frac{4.21x10^22}{6.023. 10^23}

n= 0.06989 moles of PF3 is formed.

Applying stoichiometry,

1 mole of P4 gives 4 moles PF3

x mole will produce 0.06989 moles of PF3

\frac{4}{1} = \frac{0.0698}{x}

4x= 0.0698 × 1

 x =  \frac{0.0698}{4}

x= 0.01745 moles of P4 will be required.

The weight of the phosphorus can be obtained by number of moles × atomic mass of one mole of P4

= 0.01745 × 123.89

= 2.1618 grams of P4.

     

3 0
4 years ago
When light is shown on a mixture of chlorine and chloromethane, carbon tetrachloride is one of the components of the final react
velikii [3]

A free-radical substitution reaction is likely to be responsible for the observations. The reaction mechanism of a reaction like this can be grouped into three phases:

  • Initiation; the "light" on the mixture deliver sufficient amount of energy such that the halogen molecules undergo homologous fission. It typically takes ultraviolet radiation to initiate fissions of the bonds.  
  • Propagation; free radicals react with molecules to produce new free radicals and molecules.
  • Termination; two free radicals combine and form covalent bonds to produce stable molecules. Note that it is possible for two carbon-containing free-radicals to combine, leading to the production of trace amounts of long carbon chains in the product.

Initiation

\text{Cl}-\text{Cl} \stackrel{\text{UV}}{\to} \text{Cl}\bullet + \bullet\text{Cl}

where the big black dot indicates unpaired electrons attached to the atom.

Propagation

\text{CH}_3\text{Cl}+ \text{Cl}\bullet \to \bullet\text{CH}_2\text{Cl} + \text{HCl}

\bullet\text{CH}_2\text{Cl} + \text{Cl}_2 \to \text{CH}_2\text{Cl}_2 + \text{Cl}\bullet

\text{CH}_2\text{Cl}_2 + \text{Cl}\bullet \to \bullet\text{CHCl}_2 + \text{HCl}

\bullet\text{CHCl}_2+ \text{Cl}_2 \to \text{CHCl}_3 + \bullet \text{Cl}

\text{CHCl}_3 + \text{Cl}\bullet \to \bullet\text{CCl}_3 + \text{HCl}

\bullet\text{CCl}_3 + \text{Cl}_2 \to \text{CCl}_4 + \text{Cl}\bullet

Termination

\text{Cl}\bullet + \bullet\text{Cl} \to \text{Cl}-\text{Cl}

8 0
3 years ago
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