Chemistry, 02.09.2019 21:20 amuijakobp78deg

What is the best method for the preparation of p-chlorotoluene in high yield?

Answers: 2

Show answers

Answers

Answer from: adenn3693

The answer is: p-aminotoluene; NaNO₂/HCl, 0⁰C; CuCl

Explanation:

To prepare p-chlorotoluene from p-aminotoluene:

Diazotization reaction: This reaction is used form the preparation aryldiazonium salt from primary aromatic amines.

Therefore, p-aminotoluene is treated with NaNO₂/HCl at 0-5⁰C to prepare the aryl diazonium salt.

Sandmeyer reaction: This reaction is used for the preparation of aryl halides from aryl diazonium salts.

Therefore, the aryl diazonium salt formed by the diazotization reaction of p-aminotoluene, is heated with cuprous chloride to give p-chlorotoluene.

Answer from: Quest

4d^55s^1 ( third choice)

Answer from: Quest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

Another question on Chemistry

Chemistry, 21.06.2019 18:00

Temperature and kinetic energy are proportional. a) adirectly b) directly c) indirectly

Answers: 2

Answer

Chemistry, 22.06.2019 04:00

The rules of engagement (roe) working group is often used to (select all that apply.)

Answers: 2

Answer

Chemistry, 22.06.2019 11:00

As air becomes more dense, (select all that apply) o. air weighs less o. gas molecules are closer together o. air is colder o. air weighs more o. gas molecules are further apart o. air is hotter

Answers: 3

Answer

Chemistry, 22.06.2019 19:30

Chlorine and water react to form hydrogen chloride and oxygen, like this: 2cl2 (g) + 2h2o (g) → 4hcl (g) + o2 (g) also, a chemist finds that at a certain temperature the equilibrium mixture of chlorine, water, hydrogen chloride, and oxygen has the following composition: compound concentration at equilibrium cl2 0.55m h2o 0.57m hcl 0.53m o2 0.34m calculate the value of the equilibrium constant kc for this reaction. round your answer to 2 significant digits.