Why would dry cleaners use non polar tetrachloroethylene,C2cl4 to get hamburger grease out of your shirt

Why do you require an acid catalyst to make an ester? Why not just mix acid and alcohol? Describe an alternate method of making

djverab [1.8K]

Answer:Acid catalyst is needed to increase the electrophilicity of Carbonyl group of Carboxylic acid as alcohol is a weak nucleophile.

Alternatively esters can be synthesised by converting carboxylic acid into acyl chloride using thionyl chloride(SOCl_{2} and then further treating acyl chloride with alcohol.

Carboxylic acid and esters can be easily distinguished on the basis of IR as carboxylic acid would contain a broad intense peak in 2500-3200cm_{-1} corresponding to OH stretching frequency whereas esters would not contain any such broad intense peak.

Alcohol and esters can also be distinguished using IR as alcohols would contain a broad intense peak at around 3200-3600cm_{-1}

Explanation: For the synthesis of esters using alcohol and carboxylic acid we need to add a little amount of acid in the reaction . The acid used here increases the electrophilicity of carbonyl carbon and hence makes it easier for a weaker nucleophile like alcohol to attack the carbonyl carbon of acid.

The oxygen of the carbonyl group is protonated using the acidic proton which leads to the generation of positive charge on the oxygen. The positive charge generated is delocalised over the whole acid molecule and hence the electrophilicity of carbonyl group is increased. Kindly refer attachment for the structures.

If we simply mix the acid and alcohol then no appreciable reaction would take place between them and ester formation would not take place because the carboxylic acid in that case is not a good electrophile whereas alcohol is also not a very strong nucleophile which can attack the carbonyl group.

Alternatively we can use thionyl chloride or any other reagent which can convert the carboxylic acid into acyl chloride. Acyl chloride is very elctrophilic and alcohol can very easily attack the acyl chloride and esters could be synthesized.

The carboxylic acid and ester can very easily be distinguished on the basis of broad intense OH stretching frequency peak at around 2500-3200cm_{-1} . The broad intense OH stretching frequency peak is present in carboxylic acids as they contain OH groups and absent in case of esters .

Likewise esters and alcohols can also be distinguished on the basis IR spectra as alcohols will have broad intense spectra at around 3200-3600cm_{-1}corresponding to OH stretching frequency whereas esters will not have any such peak. Rather esters would be having a Carbonyl stretching frequency at around 1720-1760

4 0

3 years ago

11.0 L of hydrogen and 5.52 L of oxygen are exploded together in a reaction tube. What volume of water vapor was formed, at STP?

Marysya12 [62]

Answer:

11.0 L

Explanation:

The equation for this reaction is given as;

2H2 + O2 --> 2H2O

2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O

At STP;

1 mol = 22.4 L

This means;

44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O

In this reaction, the limiting reactant is H2 as O2 is in excess.

The relationship between H2 and H2O;

44.8 L = 44.8 L

11.0 L would produce x

Solving for x;

x = 11 * 44.8 / 44.8

x = 11.0 L

4 0

3 years ago

What is the connection between Oobleck and the mantle?

lapo4ka [179]

Answer:

Oobleck is a mixture of cornstarch & water, a substance that behaves like the Mantle.

Explanation:

3 0

3 years ago

I need the measurements to the appropriate amount of significant figures on this pleasee

klio [65]

Answer:

the blue paper, 10.9

Explanation:

3 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago