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julsineya [31]
3 years ago
11

When ethylbenzene is treated with NBS and irradiated with UV light, two stereoisomeric compounds are obtained in equal amounts.

Draw the products and explain why they are obtained in equal amounts:

Chemistry
1 answer:
Stells [14]3 years ago
5 0

Answer: 1-Phenyl ethyl radical is formed as an intermediate in the reaction and since Phenyl ethyl radical has  a trigonal planar geometry so it is a planar molecule having two faces.  So Br radical radical can recombine  with the  two faces with equal probability  leading to a racemic mixture in 50:50ratio of products.Hence two products are formed which are known as enatiomers.

Explanation:

When we irradiate the ethylbenzene with UV light , it leads to homolytic cleavage and 1- Phenyl ethyl free radical is generated.

Phenyl ethyl free radical is generated because it is very stable as it is on a  secondary carbon center as well as on a benzylic position so it can be stabilized by the resonance as well as inductive effect at the secondary carbon center.

NBS(N-bromosuccinimide) is a source of bromine radical and provides bromine free radical.

Once the  1- Phenyl ethyl free radical is generated then bromine free radical can recombine with benzyl free radical leading to product formation.

Since  1- Phenyl ethyl free radical  has a trigonal planar geometry so it is a planar molecule which has two faces  and hence the radical recombination with bromine free radical can occur with either of the two faces available.

Kindly refer the attachments for structure as well as the mechanism of the reaction.

So two isomers which are enantiomers are produced  are obtained.

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Sulfurous acid, H2SO3, breaks down into water (H20) and sulfur dioxide (SO2).
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Explanation:

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3 years ago
Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
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The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

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