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Chapter 4: Q32P (page 233)

Propionitrile()is deprotonated by very strong bases. Write resonance forms to show the stabilization of the carbanion that results.

Short Answer

Expert verified

Effect of a strong base

Strong bases abstract protons from the alpha carbon attached to the electron-withdrawing group.

Step by step solution

01

Step 1:

Effect of a strong base

Strong bases abstract protons from the alpha carbon attached to the electron-withdrawing group.

02

Step 2:

Proton abstraction by strong base in propionitrile

Here in propionitrile, the nitrile group is a strong electron-withdrawing group, and base abstract proton from the adjacent carbon results in the formation of carbanion. The carbanion formed is stabilized by the nitrile group as the negative charge moves from carbon to nitrogen of the nitrile group (shown above). In conclusion, we say that the resulting carbanion is stabilized by resonance.

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Most popular questions from this chapter

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?

2,3-Dimethylbutanereacts with bromine in the presence of light to give a mono brominate product. Further reaction gives a good yield of a dibrominated product. Predict the structures of these products, and propose a mechanism for the formation of the monobrominated product.

Question: (a) Propose a mechanism for the free-radical chlorination of ethane.

(b) Calculatefor each step in this reaction

(c) Calculate the overall value offor this reaction.

For each compound, predict the major product of free-radical bromination. Remember that bromination is highly selective, and only the most stable radical will be formed.

(a) cyclohexane

(b) methylcyclopentane

(c) decalin

(d) hexane

(e)

(f)

The following reaction has a value ofΔG0= -2.1 kJ/mol  (-0.50  kcal/mol)

CH3Br  +  H2S  ⇄  CH3SH  +  HBr

(a) Calculate Keqat room temperature(250C) for this reaction as written.

(b)Starting with a1M solution of CH3Brand H2S, calculate the final concentrations of all four species at equilibrium.

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