22) Select the appropriate synthetic sequence to accomplish the transformation shown. Br
The Correct Answer and Explanation is:1
Correct Answer: D) 1. KOtBu; 2. HBr, ROOR; 3. CH₃CH₂CCNa
Explanation
The transformation shown is the conversion of a tert-butyl bromide on a cyclopentane ring into a cyclopentyl group with a side chain containing a terminal alkyne (triple bond). Let’s analyze this transformation step by step:
Step 1: Elimination Reaction
Reagent: KOtBu (potassium tert-butoxide)
- This strong, bulky base promotes E2 elimination, particularly favoring Hofmann product (less substituted alkene).
- From the tert-butyl bromide, KOtBu will abstract a β-proton and eliminate HBr, forming 1-cyclopentylethene (CH₂=CH–cyclopentyl).
Step 2: Anti-Markovnikov Hydrohalogenation
Reagents: HBr + ROOR (peroxides)
- The alkene undergoes radical addition of HBr in the presence of ROOR, giving anti-Markovnikov addition.
- This results in the bromine adding to the terminal carbon of the double bond (CH₃–CHBr–cyclopentyl), forming a 1-bromoethyl-cyclopentane.
Step 3: Nucleophilic Substitution with Sodium Acetylide
Reagent: CH₃CH₂C≡CNa (sodium propyne or sodium alkyne anion)
- This is a strong nucleophile and will undergo SN2 substitution with the 1° alkyl bromide made in step 2.
- The result is the substitution of the Br with CH₃CH₂C≡C– group (a terminal alkyne side chain).
Summary of the Transformation:
- KOtBu → E2 elimination: forms alkene.
- HBr/ROOR → Anti-Markovnikov radical addition: adds Br to terminal carbon.
- CH₃CH₂C≡CNa → SN2 attack: installs alkyne.
Why the Other Choices Are Incorrect:
- A involves Br₂/hν (radical bromination) which doesn’t fit the need.
- B & C introduce a Grignard or perform addition without radical conditions, not suitable here.
- E ends with H₃O⁺ (acidic workup), which doesn’t install an alkyne.
Thus, choice D correctly sequences elimination → anti-Markovnikov addition → SN2 substitution to give the desired alkyne product.
