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-Dominant- [34]
3 years ago
6

In your discussion consider a SN2 reaction mechanism concept. Propose a modification of experimental procedure that would improv

e reaction yield. Give an example of another method of ether synthesis. Illustrate it with drawn reaction scheme; describe it in few sentences.

Chemistry
1 answer:
Hitman42 [59]3 years ago
6 0

Answer:

Here's what I get  

Explanation:

You may have done a Williamson synthesis of guaifenesin by reacting guaiacol with 3-chloropropane-1,2-diol.

A. Mechanism

Step 1

NaOH converts guaiacol into a phenoxide ion.

Step 2

The phenoxide acts as the nucleophile in an SN2 reaction to displace the Cl from the alkyl halide.

B. Improve the yield

You probably carried out the reaction in ethanol solution — a polar protic solvent.

You might try doing the reaction in a polar aprotic solvent— perhaps DMSO.

A polar aprotic solvent does not hydrogen bond to nucleophiles, so they become stronger.

C. Another method of ether synthesis —dehydration of alcohols

Sulfuric acid catalyzes the conversion of primary alcohols to ethers.

This is also a nucleophilic displacement reaction.

Protonation of the OH converts it into a better leaving group.

Attack by a second molecule of alcohol forms the protonated ether.

A molecule of water then removes the proton.

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a. i. 8.447 × 10⁻³ T ii.  27.14 cm

b. i. 2.14 cm ii. It is easily detectable.

Explanation:

a.

i. What strength of magnetic field is required?

Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have

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Bqv = mv²/r

B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m

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B = mv/rq

B = 1.99 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (12.5 × 10⁻² m × 1.602 × 10⁻¹⁹ C)

B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)

B = 0.8447 × 10⁻² kg/sC)

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Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have

F = F'

Bq'v = mv²/r'

r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and  = d/2 = 12.5 cm = 12.5 × 10⁻² m

So, r' = mv/Be

r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)

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Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm

ii. Is this distance large enough to be easily observed?

This distance of 2.14 cm easily detectable since it is in the centimeter range.

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