Question

Identify A, B, and C in the following reactions:

• A.
• B.
• C.
• D.

Hint:

In nucleophilic substitution and elimination reactions, consider the mechanism (S\(_\text{N}\)1 or S\(_\text{N}\)2) when identifying the nature of the products and intermediates.

Answer 1

The Correct Option is A

Reaction Pathways for 2-Bromo-2-methylbutane

Starting Material:

2-Bromo-2-methylbutane (tertiary alkyl halide)

CH₃-CH₂-C(Br)(CH₃)-CH₃

1. SN1 Reaction (Forming B)

Conditions: Ethanol (C₂H₅OH)

Mechanism:

1. Slow dissociation of Br^- to form carbocation
2. Nucleophilic attack by ethanol
3. Proton transfer to solvent

Product B: 2-ethoxy-2-methylbutane

CH₃-CH₂-C(OCH₂CH₃)(CH₃)-CH₃

2. E1 Elimination (Forming C)

Conditions: Ethanol (C₂H₅OH)

Mechanism:

1. Same initial carbocation formation as SN1
2. Base (EtOH) abstracts proton from β-carbon
3. Forms most stable alkene (Zaitsev's rule)

Product C: 2-methyl-2-butene (major) and 2-methyl-1-butene (minor)

CH₃-CH=C(CH₃)-CH₃ (major)
CH₂=C(CH₃)-CH₂-CH₃ (minor)

3. E2 Elimination (Forming A)

Conditions: Strong base (e.g., KOH/ethanol)

Mechanism:

1. Concerted removal of β-hydrogen and Br^- departure
2. Forms same alkenes as E1 but with different selectivity

Product A: Mixture of 2-methyl-2-butene and 2-methyl-1-butene

Key Observations:

• Tertiary halides favor elimination over substitution
• SN2 is impossible with tertiary substrates
• Ethanol promotes both SN1 and E1 pathways
• Strong base promotes E2 pathway

Correct Identification:

• A: 2-methyl-2-butene (E2 product)
• B: 2-ethoxy-2-methylbutane (SN1 product)
• C: 2-methyl-2-butene (E1 product)

The correct option is therefore A (2-methyl-2-butene).

Answer 2

Detailed Reaction Analysis: CH₃C(CH₃)₂CH₂Br Transformations

Substrate Structure:

CH₃C(CH₃)₂CH₂Br - 1-bromo-2,2-dimethylpropane (neopentyl bromide)

• Primary alkyl halide (Br attached to primary carbon)
• Adjacent quaternary carbon creates significant steric hindrance

Pathway A: Substitution with C₂H₅OH

Expected SN2 Product: CH₃C(CH₃)₂CH₂OC₂H₅

Reality: The reaction is extremely slow/unlikely because:

• Severe steric hindrance from the adjacent (CH₃)₂C- group
• Neopentyl systems are notoriously resistant to SN2
• Ethanol is a weak nucleophile for SN2

Pathway B: Elimination with C₂H₅OH

Expected E2 Product: CH₃C(CH₃)=CH₂ (2-methylpropene)

Mechanism:

1. Base (EtO^-) abstracts β-hydrogen
2. Simultaneous Br^- departure
3. Forms most stable (tetrasubstituted) alkene possible

Pathway C: Elimination with Strong Base

Product: CH₃C(CH₃)=CH₂ (same as Pathway B)