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Georgia [21]
3 years ago
14

Show the alkyl bromide and alcohol used to make methyl t-butyl ether using the Williamson ether synthesis to the right of the re

trosynthetic arrow. alcohol alkyl bromide click to edit CHa H3 H3 b) Complete the general mechanism by adding curved arrows and drawing the final organic product. (Note that R is an abbreviation for any alkyl group and can be found on the bottom row in the pull down periodic table in the drawing tools menu.) product Na +RX
Chemistry
1 answer:
Korolek [52]3 years ago
4 0

Answer:

Hi

Williamson's ether reactions imply that an alkoxide reacts with a primary haloalkane. Alkoxides consisting of the conjugate base of an alcohol and are formed by a group R attached to an oxygen atom. They are often written as RO–, where R is the organic substituent (Step 1).

Sn2 reactions are characterized by the reversal of stereochemistry at the site of the leaving group. Williamson's synthesis is no exception and the reaction is initiated by the subsequent attack of the nucleophile. This requires that the nucleophile and electrophile be in anti-configuration (Step 2).

As an example (figure 3).

In the attached file are each of the steps of Williamson's synthesis.

Explanation:

Download docx
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The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
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Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

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3 years ago
What is the maximum number of grams of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl
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<u>Answer:</u> The maximum amount of N-acetyl-p-toluidine that can be prepared is 1.7 grams.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Molarity of p-toluidine hydrochloride solution = 0.167 M

Volume of solution = 70. mL

Putting values in above equation, we get:

0.167M=\frac{\text{Moles of p-toluidine hydrochloride}\times 1000}{70}\\\\\text{Moles of p-toluidine hydrochloride}=\frac{0.167\times 70}{1000}=0.0117mol

The chemical equation for the reaction of p-toluidine hydrochloride and acetic anhydride follows:

\text{p-toluidine hydrochloride}+\text{Acetic anhydride}\rightarrow \text{N-acetyl-p-toluidine}

By Stoichiometry of the reaction:

1 mole of p-toluidine hydrochloride produces 1 mole of N-acetyl-p-toluidine

So, 0.0117 moles of p-toluidine hydrochloride will produce = \frac{1}{1}\times 0.0117=0.0117mol of N-acetyl-p-toluidine

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of N-acetyl-p-toluidine = 149.2 g/mol

Moles of N-acetyl-p-toluidine = 0.0117 moles

Putting values in equation 1, we get:

0.0117mol=\frac{\text{Mass of N-acetyl-p-toluidine}}{149.2g/mol}\\\\\text{Mass of N-acetyl-p-toluidine}=(0.0117g/mol\times 149.2)=1.7g

Hence, the maximum amount of N-acetyl-p-toluidine that can be prepared is 1.7 grams.

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