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3 years ago

Propose a reasonable synthesis to get benzylamine from toluene

Chemistry

1 answer:

yawa3891 [41]3 years ago

4 0

Benzyl Amine is synthesized from Toluene in two steps;

Step 1: Bromination of Methyl Group in Toluene:
Bromination is carried out on methyl group present on Benzene ring through free radical mechanism by treating toluene with N-Bromosuccinimide (NBS) and Benzyl Bromide is formed.

Step 2: Benzyl Bromide into Benzyl Amine;
This step is carried out utilizing a neme reaction "Gabriel Reaction". Benzyl bromide is treated with Potassium Phthalamide and an intermediate is formed which on treatment with Hydrazine produces Benzyl amine.

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suppose you want to treat if a sample compound is ionic. you have these materials: the solid compound, a glass container, a spoo

xz_007 [3.2K]

An ionic compound is a substance that contains atoms that are bonded together through an ionic bond, where electrons are transferred from one atom to another. For the given experiment above, the material that is missing from the list is WATER or solvent. It is necessary because conduction through water will help you identify if a sample compound is ionic through its electrolytes. Ionic compounds that are soluble are typically electrolytes

6 0

2 years ago

Read 2 more answers

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

3 0

2 years ago

What is Sulobility??

AleksAgata [21]

Amount of a substance (called the solute) that dissolves in a unit volume of a liquid substance (called the solvent) to form a saturated solution under specified conditions of temperature and pressure. Solubility is expressed usually as moles of solute per 100 grams of solvent.

4 0

3 years ago

Read 2 more answers

Unit 4.2 quiz chemistry

Maslowich

Additional oxygen

Explanation : H2O is two hydrogens and one oxygen not 2 hydrogens two oxygens

5 0

2 years ago

Which is a example of a solution

Diano4ka-milaya [45]

Answer:

Salt water