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Chapter 9: Q86P (page 452)

Which species in each pair is more stable?

Short Answer

Expert verified

1. It is a two step reaction.

2. In this carbocation is formed.

3. In this negative ion will attack from the side where there is less number of bulkier groups.

4. As the carbocation intermediate is formed so the stability order 3° carbocation>2° carbocation>1° carbocation

5. As the carboanion intermediate is formed so the stability order 1° carboanion>2° carboanion>3° carboanion


Step by step solution

01

How SN1 reaction proceed

1. It is a two step reaction.

2. In this carbocation is formed.

3. In this negative ion will attack from the side where there is less number of bulkier groups.

4. As the carbocation intermediate is formed so the stability order 3° carbocation>2° carbocation>1° carbocation

5. As the carboanion intermediate is formed so the stability order 1° carboanion>2° carboanion>3° carboanion

02

Final answer

Less substituted carboanion is more stable.

More substituted carbocation is more stable.

Stabilized by electron delocalization.

More substituted carbocation is more stable.

Stabilized by electron delocalization.

Less substituted carboanion is more stable.

More substituted carbocation is more stable.

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Most popular questions from this chapter

Amines are good nucleophiles, even though they are neutral. How would the rate of an SN2 reaction between an amine and an alkyl halide be affected if the polarity of the solvent is increased?

Which member in each pair in Problem 68 is a better leaving group?

Propose a mechanism for the following reaction:

a. Which reacts faster in an SN2 reaction?

b.Which reacts faster in an E1 reaction?

c.Which reacts faster in an SN1 reaction?

Two elimination products are obtained from the following E2 reaction

a. What are the elimination products?

b. Which is formed in greater yield? (Hint:See Problem 91.)
See all solutions

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