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Alenkinab [10]
2 years ago
11

identify the reagents necessary to achieve each of the following transformations o3 dms, pcc, ch2cl, h2so4,h2o,hgso4

Chemistry
1 answer:
hoa [83]2 years ago
3 0

the reagents necessary to convert alcohol to ketoneNa_2Cr_2O_7 , H_2O which involves oxidation of alcohols.

<h3>What is oxidation of alcohols?</h3>
  • Alcohol oxidation is a significant organic chemistry process. Secondary alcohols can be oxidized to produce ketones, while primary alcohols can be oxidized to produce aldehydes and carboxylic acids.
  • In contrast, tertiary alcohols cannot be oxidized without the C-C bonds in the molecule being broken.
  • In order to cause primary alcohols to oxidize into aldehydes
  1. Cr_2O_7 ^-^2 (dichromate)
  2. CrO_3/pyridine (Collins reagent)
  3. Chromium pyridinium compound (PCC)
  4. Dichromate of pyridinium (PDC, Cornforth reagent)
  5. Periodinane by Dess-Martin
  6. Oxalyl chloride with dimethylsulfoxide (DMSO) for Swern
  • oxidation of secondary alcohols to ketones
  1. Cr_2O_7 ^-^2 (dichromate)
  2. CrO_3/pyridine (Collins reagent)
  3. Chromium pyridinium compound (PCC)
  4. Dichromate of pyridinium (PDC, Cornforth reagent)
  5. Periodinane by Dess-Martin
  6. Oxalyl chloride and dimethyl sulfoxide (DMSO) (Swern oxidation)
  7. CrO_3, H_2SO_4/acetone (Jones oxidation)
  8. Acetone with aluminum isopropoxide (Oppenauer oxidation)

To learn more about oxidation of alcohols with the given link

brainly.com/question/7207863

#SPJ4

<u>Question:</u>

Identify the reagents necessary to achieve each of the following transformations

a. O_3 , DMS

b.H_2SO_4 , H_2O , HgSO_4

c.Na_2Cr_2O_7 , H_2O

d.Fe ^ {2+}, NaOH

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3 0
2 years ago
Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

6 0
2 years ago
If the freezing point of an aqueous 0.10 m glucose solution is −x°c, what is the approximate freezing point of a 0.10 m nacl sol
maks197457 [2]
Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m -  molality, moles of solute per kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
8 0
3 years ago
A sample of Ca(OH)2 is considered to be an Arrhenius base because it dissolves in water to yield
postnew [5]
The answer is C. Arrhenius bases increase the concentration of OH- in solution.
8 0
3 years ago
Read 2 more answers
A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon
ikadub [295]

<u>Answer:</u> The enthalpy of the reaction is 269.4 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

q_1=c\Delta T

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_1=675J/^oC\times 29.62^oC=19993.5J

To calculate the heat absorbed by water, we use the equation:

q_2=mc\Delta T

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J

Total heat absorbed = q_1+q_2

Total heat absorbed = [19993.5+114690.12]J=134683.62J=134.7kJ

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol

Hence, the enthalpy of the reaction is 269.4 kJ/mol

6 0
3 years ago
Read 2 more answers
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