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Leokris [45]
3 years ago
13

What reaction conditions most effectively convert a carboxylic acid to a methyl ester

Chemistry
2 answers:
ycow [4]3 years ago
4 0

Answer:

Fischer Esterification

Explanation:

Fischer esterification or Fischer-Speier esterification is a carboxylic esterification reaction that uses an inorganic acid as a catalyst, mainly sulfuric acid (H2SO4).

The carboxylic acids are usually too weak acids, so they have little tendency to yield the proton (H +) that they lose in the esterification.

A stronger acid, such as sulfuric acid, behaves like a proton donor and facilitates the esterification reaction.

First, sulfuric acid injects a proton into the carboxyl group

The alcohol molecule, R’-OH, has an electron-rich oxygen atom, is attracted to the protonated carboxylic structure

This complex is not very stable and stabilizes towards the most favorable energy molecule, which is the ester, releasing a water molecule and a proton (H +). The proton is then used to regenerate sulfuric acid

dem82 [27]3 years ago
4 0

Answer: The convenient and efficient sonochemical method for methyl esterification of carboxylic acids reaction conditions are ;

• catalyzation by polymer-supported triphenylphosphine (0.1 eq. PS-PPh3)

• in the presence of 2,4,6-trichloro-1,3,5-triazine (1.0 eq. TCT) and Na2CO3 (2.0 eq. )

• Temperature 50°C for 10 - 30 mins

Explanation:

Methyl esters of various carboxylic acids bearing reactive hydroxyl groups as well as acid- or base-labile functionalities could be rapidly prepared within short times in good to excellent yields in high purities without column chromatography.

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Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o
jekas [21]

Answer:

The % yield is 74.45 %

Explanation:

<u>Step 1:</u> The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

<u>Step 2</u>: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

<u>Step 3:</u> Calculate moles of glucose

Moles glucose = Mass glucose  / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

<u>Step 4:</u> Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

<u>Step 5:</u> Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

Mass ethanol = 0.682 moles * 46.08 g/mol

Mass ethanol = 31.43 grams = theoretical mass

<u>Step 6:</u> Calculate % yield

% yield = actual mass / theoretical mass

% yield = (23.4 grams / 31.43 grams) * 100%

% yield = 74.45 %

The % yield is 74.45 %

7 0
3 years ago
1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th
vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy,  , for a homogeneous substance to be a function of specific volume  and temperature  .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T)  dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

<u>This last equation is the Clapeyron equation.</u>

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0
3 years ago
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