Organic Chemistry Substitution And Elimination Reactions Practice Problems

Organic chemistry is a fascinating field that deals with the structure, properties, composition, reactions, and preparation of carbon-containing compounds. Two fundamental types of reactions in organic chemistry are substitution and elimination reactions. Mastering these reactions is crucial for understanding more complex organic transformations. This comprehensive guide provides numerous practice problems to help you solidify your understanding of substitution (SN1, SN2) and elimination (E1, E2) reactions, complete with detailed explanations.

Introduction to Substitution and Elimination Reactions

Substitution and elimination reactions are two of the most common types of reactions in organic chemistry. In a substitution reaction, one atom or group is replaced by another. In contrast, an elimination reaction involves the removal of atoms or groups from a molecule, resulting in the formation of a double or triple bond.

• Substitution Reactions: These reactions involve the replacement of a leaving group with a nucleophile. The two main types are SN1 and SN2 reactions.
• Elimination Reactions: These reactions involve the removal of atoms or groups from a molecule, leading to the formation of an alkene. The two main types are E1 and E2 reactions.

Understanding the mechanisms, stereochemistry, and factors affecting these reactions is essential for predicting the outcomes of organic reactions.

Key Concepts to Review

Before diving into the practice problems, let’s briefly review some key concepts:

SN1 Reactions

• Mechanism: Two-step process involving the formation of a carbocation intermediate.
• Rate Law: Unimolecular, rate = k[substrate].
• Stereochemistry: Racemization at the chiral center due to the formation of a planar carbocation.
• Substrate Preference: Tertiary > Secondary > Primary.
• Nucleophile: Weak nucleophiles, often the solvent.
• Leaving Group: Good leaving groups stabilize the negative charge.
• Solvent: Polar protic solvents stabilize the carbocation intermediate.

SN2 Reactions

• Mechanism: One-step process with simultaneous bond breaking and bond forming.
• Rate Law: Bimolecular, rate = k[substrate][nucleophile].
• Stereochemistry: Inversion of configuration (Walden inversion) at the chiral center.
• Substrate Preference: Primary > Secondary > Tertiary (Tertiary substrates do not undergo SN2 reactions).
• Nucleophile: Strong nucleophiles, often with a negative charge.
• Leaving Group: Good leaving groups stabilize the negative charge.
• Solvent: Polar aprotic solvents favor SN2 reactions by not solvating the nucleophile.

E1 Reactions

• Mechanism: Two-step process involving the formation of a carbocation intermediate.
• Rate Law: Unimolecular, rate = k[substrate].
• Stereochemistry: No specific stereochemistry, Zaitsev’s rule often applies (more substituted alkene is favored).
• Substrate Preference: Tertiary > Secondary > Primary.
• Base: Weak bases.
• Leaving Group: Good leaving groups.
• Solvent: Polar protic solvents stabilize the carbocation intermediate.

E2 Reactions

• Mechanism: One-step process with simultaneous bond breaking and bond forming.
• Rate Law: Bimolecular, rate = k[substrate][base].
• Stereochemistry: Anti-periplanar geometry is required. Zaitsev’s rule often applies.
• Substrate Preference: Tertiary > Secondary > Primary.
• Base: Strong bases.
• Leaving Group: Good leaving groups.
• Solvent: Polar aprotic solvents favor E2 reactions.

Factors Affecting SN1, SN2, E1, and E2 Reactions

• Substrate Structure: The degree of substitution at the carbon atom attached to the leaving group.
• Nucleophile/Base Strength: Strong nucleophiles favor SN2 reactions, while strong bases favor E2 reactions.
• Leaving Group Ability: Good leaving groups stabilize the negative charge and facilitate the reaction.
• Solvent Effects: Polar protic solvents favor SN1 and E1 reactions, while polar aprotic solvents favor SN2 and E2 reactions.
• Temperature: Higher temperatures generally favor elimination reactions (E1 and E2) over substitution reactions (SN1 and SN2).

Practice Problems

Now, let’s tackle some practice problems to reinforce your understanding of these reactions. Each problem will be followed by a detailed explanation.

Problem 1

Predict the major product(s) of the following reaction:

(CH3)3C-Br + CH3OH → ?

Explanation:

• Substrate: The substrate is a tertiary alkyl halide, (CH3)3C-Br.

• Reagent: The reagent is methanol, CH3OH, which is a weak nucleophile and a polar protic solvent.

• Reaction Type: Since the substrate is tertiary and the solvent is polar protic, the reaction is likely to proceed via an SN1 or E1 mechanism.

• Major Product(s):

  • SN1 Product: (CH3)3C-OCH3 (tert-butyl methyl ether)
  • E1 Product: (CH3)2C=CH2 (2-methylpropene)

  Both SN1 and E1 products can form, but E1 is favored due to the stability of the alkene and the higher temperature generally favoring elimination.

Problem 2

Predict the major product(s) of the following reaction:

CH3CH2CH2CH2-Cl + NaCN → ?

Explanation:

• Substrate: The substrate is a primary alkyl halide, CH3CH2CH2CH2-Cl.
• Reagent: The reagent is sodium cyanide, NaCN, which provides a strong nucleophile (CN-).
• Reaction Type: Primary alkyl halides favor SN2 reactions, especially with strong nucleophiles. E2 is possible but less favored.
• Major Product: CH3CH2CH2CH2-CN (pentanenitrile)

Problem 3

Predict the major product(s) of the following reaction:

(CH3)2CHCHBrCH3 + KOH (alcoholic) → ?

Explanation:

• Substrate: The substrate is a secondary alkyl halide, (CH3)2CHCHBrCH3.

• Reagent: The reagent is potassium hydroxide in alcohol, KOH (alcoholic), which is a strong base and favors elimination reactions.

• Reaction Type: This favors E2 reactions due to the strong base.

• Major Product(s): There are two possible alkenes:

  • (CH3)2C=CHCH3 (2-methyl-2-butene) - more substituted (Zaitsev's rule)
  • (CH3)2CHCH=CH2 (3-methyl-1-butene) - less substituted

  According to Zaitsev's rule, the more substituted alkene is the major product: (CH3)2C=CHCH3.

Problem 4

Predict the major product(s) of the following reaction:

(CH3)3C-Cl + H2O → ?

Explanation:

• Substrate: The substrate is a tertiary alkyl halide, (CH3)3C-Cl.

• Reagent: The reagent is water, H2O, which is a weak nucleophile and a polar protic solvent.

• Reaction Type: This favors SN1 and E1 reactions due to the tertiary substrate and polar protic solvent.

• Major Product(s):

  • SN1 Product: (CH3)3C-OH (tert-butanol)
  • E1 Product: (CH3)2C=CH2 (2-methylpropene)

  Both products are possible, but the relative amounts depend on the temperature. Higher temperatures favor elimination (E1).

Problem 5

Predict the major product(s) of the following reaction:

CH3CH2CHBrCH3 + NaOCH2CH3 → ?

Explanation:

• Substrate: The substrate is a secondary alkyl halide, CH3CH2CHBrCH3.

• Reagent: The reagent is sodium ethoxide, NaOCH2CH3, which is a strong base and a good nucleophile.

• Reaction Type: This can lead to both SN2 and E2 reactions. However, since ethoxide is a strong, bulky base, E2 is more favored.

• Major Product(s):

  • E2 Product (Major): CH3CH=CHCH3 (2-butene) - more substituted, and CH2=CHCH2CH3 (1-butene) - less substituted. 2-butene is usually the major product due to Zaitsev's rule.
  • SN2 Product (Minor): CH3CH2CH(OCH2CH3)CH3 (2-ethoxybutane)

Problem 6

Which reaction will occur faster:

a) (CH3)2CHCH2Br + NaOH in H2O

b) (CH3)3CBr + NaOH in H2O

Explanation:

• Reaction a: The substrate is a primary alkyl halide, (CH3)2CHCH2Br. Primary alkyl halides favor SN2 reactions.
• Reaction b: The substrate is a tertiary alkyl halide, (CH3)3CBr. Tertiary alkyl halides favor SN1 or E1 reactions.

In the presence of NaOH in H2O, reaction (a) will occur faster via an SN2 mechanism. Tertiary alkyl halides are sterically hindered, making SN2 reactions slow or impossible.

Problem 7

Draw the mechanism for the following reaction and predict the product:

CH3CH2CH2OH + H2SO4 (conc.) → ?

Explanation:

• Reagent: Concentrated sulfuric acid (H2SO4) acts as a strong acid and dehydrating agent.
• Substrate: The substrate is a primary alcohol, CH3CH2CH2OH.
• Reaction Type: This reaction is an E1 reaction, which leads to the formation of an alkene.

Mechanism:

1. Protonation: The alcohol is protonated by sulfuric acid to form an oxonium ion.

   CH3CH2CH2OH + H2SO4 ⇌ CH3CH2CH2OH2+ + HSO4-
   

2. Loss of Water: The oxonium ion loses water, forming a primary carbocation.

   CH3CH2CH2OH2+ → CH3CH2CH2+ + H2O
   

3. 1,2-Hydride Shift (Optional): A 1,2-hydride shift can occur to form a more stable secondary carbocation. However, in this case, it is not significant.
4. Deprotonation: A proton is removed from a carbon adjacent to the carbocation, forming an alkene.

   CH3CH2CH2+ → CH3CH=CH2 + H+
   

Major Product: The major product is propene, CH3CH=CH2.

Problem 8

Predict the product(s) of the following reaction and provide the mechanism:

cyclohexyl chloride + NaOH (aq) → ?

Explanation:

• Substrate: Cyclohexyl chloride is a secondary alkyl halide.
• Reagent: Aqueous sodium hydroxide is a strong nucleophile and base.
• Reaction Type: This reaction can proceed via both SN2 and E2 mechanisms.

Mechanisms and Products:

• SN2 Mechanism:

  • The hydroxide ion (OH-) acts as a nucleophile and attacks the carbon bearing the chlorine atom, displacing the chloride ion in a single step.
  • Product: Cyclohexanol.

  cyclohexyl-Cl + OH- → cyclohexyl-OH + Cl-
  

• E2 Mechanism:

  • The hydroxide ion acts as a base and removes a proton from a carbon adjacent to the carbon bearing the chlorine atom, leading to the formation of a double bond and the elimination of HCl.
  • Product: Cyclohexene.

  cyclohexyl-Cl + OH- → cyclohexene + H2O + Cl-
  

Major Product: Whether the major product is cyclohexanol (SN2) or cyclohexene (E2) depends on the conditions. At lower temperatures and with a less bulky base, the SN2 reaction is favored, yielding cyclohexanol. At higher temperatures and with a more bulky base, the E2 reaction is favored, yielding cyclohexene.

Problem 9

Identify which alkyl halide in each pair will react faster in an SN2 reaction:

a) CH3Br or CH3I

b) (CH3)2CHBr or (CH3)3CBr

Explanation:

a) CH3I will react faster in an SN2 reaction because iodide (I-) is a better leaving group than bromide (Br-). The larger size and weaker bond strength of C-I compared to C-Br make iodide a better leaving group.

b) (CH3)2CHBr will react faster in an SN2 reaction. SN2 reactions are favored by less sterically hindered substrates. (CH3)2CHBr (secondary) is less hindered than (CH3)3CBr (tertiary).

Problem 10

Predict the major product of the following reaction:

CH3CH2CH2CH2Br + (CH3)3COK → ?

Explanation:

• Substrate: The substrate is a primary alkyl halide, CH3CH2CH2CH2Br.
• Reagent: The reagent is potassium tert-butoxide, (CH3)3COK, which is a strong, bulky base.
• Reaction Type: Bulky bases favor E2 reactions because they are sterically hindered from acting as nucleophiles in SN2 reactions.
• Major Product: CH3CH2CH=CH2 (1-butene).

Problem 11

What conditions favor SN1 over SN2 reactions?

Explanation:

Conditions favoring SN1 reactions over SN2 reactions include:

• Tertiary or secondary alkyl halides: These substrates form more stable carbocations.
• Weak nucleophiles: Weak nucleophiles do not readily participate in the SN2 reaction.
• Polar protic solvents: These solvents stabilize the carbocation intermediate.
• Good leaving groups: Good leaving groups facilitate the formation of the carbocation.

Problem 12

What conditions favor E2 over E1 reactions?

Explanation:

Conditions favoring E2 reactions over E1 reactions include:

• Strong, bulky bases: Strong bases promote the E2 mechanism.
• High concentrations of base: High base concentrations favor bimolecular reactions like E2.
• Higher temperatures: Elimination reactions are generally favored at higher temperatures.
• Substrates with anti-periplanar hydrogens: E2 reactions require the hydrogen and leaving group to be anti-periplanar.

Problem 13

Draw the structure of the major product formed when 2-bromobutane is treated with sodium ethoxide. Show stereochemistry where appropriate.

Explanation:

• Substrate: 2-bromobutane is a secondary alkyl halide.
• Reagent: Sodium ethoxide (NaOCH2CH3) is a strong base.
• Reaction Type: Since we have a secondary halide and a strong base, this reaction will proceed via the E2 mechanism.

E2 Mechanism: The ethoxide base will remove a proton from a carbon adjacent to the carbon with the bromine, leading to the formation of a double bond. There are two possible alkenes:

• 2-butene (major product, more substituted)
• 1-butene (minor product, less substituted)

The major product, 2-butene, exists as cis and trans isomers. In general, the trans isomer is more stable due to reduced steric hindrance, so it will be the major product.

Major Product: trans-2-butene.

Problem 14

Predict the major product of the reaction of 2-methyl-2-chloropentane with ethanol.

Explanation:

• Substrate: 2-methyl-2-chloropentane is a tertiary alkyl halide.
• Reagent: Ethanol is a weak nucleophile and a polar protic solvent.
• Reaction Type: Tertiary halides in polar protic solvents favor SN1 and E1 reactions.

Products:

• SN1 product: 2-ethoxy-2-methylpentane
• E1 product: Mixture of alkenes (Zaitsev's rule will apply)

Major Product: The E1 reaction will produce a mixture of alkenes, but the major product will be the more substituted alkene, according to Zaitsev's rule. In this case, that's 2-methyl-2-pentene.

Problem 15

Arrange the following alkyl halides in order of increasing reactivity in an SN1 reaction:

• 1-bromobutane
• 2-bromobutane
• 2-bromo-2-methylpropane

Explanation:

SN1 reactions are favored by the formation of stable carbocations. The stability of carbocations follows the order: tertiary > secondary > primary. Therefore, the order of increasing reactivity in an SN1 reaction is:

1. 1-bromobutane (primary)
2. 2-bromobutane (secondary)
3. 2-bromo-2-methylpropane (tertiary)

Problem 16

Predict the product of the following reaction, including stereochemistry:

(R)-2-bromobutane + NaOH (aq) → ?

Explanation:

• Substrate: (R)-2-bromobutane is a chiral secondary alkyl halide.
• Reagent: Aqueous sodium hydroxide (NaOH) provides a strong nucleophile (OH-).
• Reaction Type: This can proceed via SN2 or E2. The choice of reaction will dictate the stereochemical outcome.

Possible Reactions and Outcomes: Since NaOH is not a particularly bulky base in aqueous solution, SN2 is favored. The SN2 reaction will lead to an inversion of configuration.

• SN2: (S)-butan-2-ol

Problem 17

What products are formed when tert-butyl chloride reacts with ethanol? Write out the detailed mechanism.

Explanation:

• Substrate: tert-butyl chloride is a tertiary alkyl halide.
• Reagent: Ethanol is a weak nucleophile and a polar protic solvent.
• Reaction Type: Tertiary alkyl halides in polar protic solvents typically undergo SN1 and E1 reactions.

Detailed Mechanism:

1. Ionization (Rate-Determining Step): The tert-butyl chloride spontaneously ionizes to form a carbocation intermediate and a chloride ion.

   (CH3)3C-Cl  ⇌  (CH3)3C+  +  Cl-
   

2. Nucleophilic Attack (SN1): The ethanol molecule (nucleophile) attacks the carbocation, forming a protonated ether.

   (CH3)3C+  +  CH3CH2OH  →  (CH3)3C-O+HCH2CH3
   

3. Deprotonation (SN1): Another ethanol molecule removes a proton from the protonated ether, yielding the final SN1 product, tert-butyl ethyl ether.

   (CH3)3C-O+HCH2CH3  +  CH3CH2OH  →  (CH3)3C-O-CH2CH3  +  CH3CH2O+H2
   

4. Elimination (E1): Alternatively, a molecule of ethanol can act as a base and remove a proton from a carbon adjacent to the carbocation, forming isobutene.

   (CH3)3C+  +  CH3CH2OH  →  (CH3)2C=CH2  +  CH3CH2O+H2
   

Products:

• SN1 product: tert-butyl ethyl ether
• E1 product: isobutene

Problem 18

How would you carry out the following transformation?

CH3CH2CH2CH2Br → CH3CH2CH=CH2

Explanation:

This transformation involves converting an alkyl halide to an alkene, which can be achieved through an elimination reaction.

Reagents and Conditions: To perform this transformation, we need a strong base to promote an E2 reaction. A suitable reagent is potassium tert-butoxide (t-BuOK) in tert-butanol. The bulky base will favor elimination over substitution.

Procedure: React 1-bromobutane with potassium tert-butoxide in tert-butanol under reflux.

Problem 19

Predict the major product of the following reaction:

CH3CH2CH(CH3)CH2Br + NaOH (ethanolic) -> ?

Explanation:

• Substrate: The substrate is a secondary alkyl halide, CH3CH2CH(CH3)CH2Br.
• Reagent: The reagent is sodium hydroxide in ethanol, NaOH (ethanolic), which is a strong base.
• Reaction Type: A strong base with a secondary halide favors E2 reactions.

Products and Major Product: E2 elimination will lead to two possible alkenes:

1. CH3CH=C(CH3)CH2 (major, more substituted, Zaitsev's rule)
2. CH3CH2CH=CHCH3 (minor, less substituted)

Thus, the major product is 2-methyl-2-butene.

Problem 20

For each of the following pairs of reactions, indicate which reaction would proceed faster and explain why.

a) (CH3)3CBr + H2O vs. CH3Br + H2O

b) CH3CH2Br + NaCN in DMSO vs. CH3CH2Br + NaCN in ethanol

Explanation:

a) (CH3)3CBr + H2O vs. CH3Br + H2O: (CH3)3CBr + H2O will react faster. This reaction proceeds via an SN1 mechanism because (CH3)3CBr is a tertiary halide, which forms a stable carbocation intermediate. CH3Br + H2O proceeds via an SN2 mechanism, but primary halides react much slower than tertiary halides in SN1 conditions.

b) CH3CH2Br + NaCN in DMSO vs. CH3CH2Br + NaCN in ethanol: CH3CH2Br + NaCN in DMSO will react faster. The reaction is an SN2 reaction. DMSO (dimethyl sulfoxide) is a polar aprotic solvent, which favors SN2 reactions by not solvating the nucleophile (CN-). Ethanol is a polar protic solvent, which solvates the nucleophile and slows down the reaction.

Conclusion

Mastering substitution and elimination reactions is crucial for any student of organic chemistry. By working through these practice problems, you should have gained a better understanding of the factors that influence these reactions and how to predict their outcomes. Remember to focus on the substrate structure, nucleophile/base strength, leaving group ability, and solvent effects to accurately determine the major products of these reactions. Keep practicing, and you'll become proficient in predicting the outcomes of organic reactions!