Write an equation for the reaction of vitamin E with an oxidizing radical (RO·) to give ROH and a less reactive free radical.

Short Answer

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Radical inhibitors help to protect the cells of living systems. Vitamin E is a phenol (an aromatic ring with an -OH group), and it is thought to react with radicals by losing the OH hydrogen atom.

Step by step solution

01

Role of radical inhibitor

Radical inhibitors help to protect the cells of living systems. Vitamin E is a phenol (an aromatic ring with an -OH group), and it is thought to react with radicals by losing the OH hydrogen atom.

02

Step 2:

Abstracting of a proton from Vitamin E by oxidizing radical (RO·) results in free radical. This free radical is stabilized by resonance as shown in above diagram.

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Most popular questions from this chapter

(a) When n-heptaneburns in a gasoline engine, the combustion process takes place too quickly. The explosive detonation makes a noise called knocking. When 2,2,4 trimethylpentane (isooctane) is burned, combustion takes place in a slower, more controlled manner. Combustion is a free-radical chain reaction, and its rate depends on the reactivity of the free-radical intermediates. Explain why isooctane has less tendency to knock than doesn-heptane

(b) Alkoxy radicals (R-O)are generally more stable than alkyl (R) radicals. Write an equation showing an alkyl free radical (from burning gasoline) abstracting a hydrogen atom from tert-butyl alcohol,(CH3)3COH. Explain whytert-butyl alcohol works as an antiknock additive for gasoline.

(c) Use the information in Table 4-2 (page 203) to explain why toluene (PhCH3) has a very high-octane rating of 111. Write an equation to show how toluene reacts with an alkyl free radical to give a relatively stable radical.

Question: For each reaction, estimate whether S0for the reaction is positive, negative or impossible to predict.

a.

b.The formation of diacetone alcohol:

c.

The following reaction is a common synthesis used in the organic chemistry laboratory course.

When we double the concentration of methoxide ion (CH3O-) , we find that the reaction rate doubles. When we triple the concentration of 1-bromopropane , we find the reaction rate triples.

(a) What is the order of this reaction with respect to 1-bromopropane? What is the order with respect to methoxide ion? Write the rate equation for this reaction. What is the overall order?

(b) One lab textbook recommends forming the sodium methoxide in methanol solvent, but before adding 1-bromopropane ,it first distills off enough methanol to reduce the mixture to half of its original volume. What difference in rate will we see when we run the reaction (using the same amounts of reagents) in half the volume of solvent?

Treatment oftert - butyl alcohol with concentrated HBrgives tert butyl bromide.

When the concentration of H+is doubled, the reaction rate doubles. When the concentration of tert- butyl alcohol is tripled, the reaction rate triples. When the bromide ion concentration is quadrupled, however, the reaction rate is unchanged. Write the rate equation for this reaction.

Question: (a) Write the propagation steps leading to the formation of dichloromethane (CH2Cl2).

(b) Explain why free-radical halogenation usually gives mixture of products.

(c) How could an industrial plant control the proportions of methane and chlorine to favor production of CCl4? To favor production of CH3Cl?

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