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AVprozaik [17]
3 years ago
12

What is the major organic product obtained from the sequence of reactions 2-phenyl-4 bomobutane and NaN3?

Chemistry
1 answer:
Anna11 [10]3 years ago
3 0

Answer:

(E)-1-phenylbut-1-ene

Explanation:

2-phenyl-4 bromobutane is an amphetamine that contains a phenyl group. It forms a major stable product with other reacting agents.

The major organic product that is obtained from the sequence of the reactions of the 2-phenyl-4 bomobutane when it reacts with NaN_3 is the (E)-1-phenylbut-1-ene.

Thus the answer is 2-phenyl-4 bromobutane is an amphetamine that contains a phenyl group. It forms a major stable product with other reacting agents.

The major organic product that is (E)-1-phenylbut-1-ene.

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4 years ago
Calculate by (a)% weight and (b) %mole each of the elements present in sugar
Musya8 [376]

Explanation:

Molecular mass of sugar = C_{12}H_{22}O_{11} : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100

Percentage of carbon by weight in C_{12}H_{22}O_{11}:

\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%

Percentage of hydrogen by weight in C_{12}H_{22}O_{11}:

\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%

Percentage of oxygen by weight in C_{12}H_{22}O_{11}:

\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%

b) Percentage of mole each of the elements present in sugar:

=\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in C_{12}H_{22}O_{11}:

\frac{12 mol}{45 mol}\times 100=26.66\%

Percentage of hydrogen by mole in C_{12}H_{22}O_{11}:

\frac{22 mol}{45 mol}\times 100=48.88\%

Percentage of oxygen by mole in C_{12}H_{22}O_{11}:

\frac{11 mol}{45 mol}\times 100=24.44\%

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4 years ago
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