Monobromination Chiral Products Count for Enantiomerically Pure Alkane

To determine the total number of chiral products formed during the monobromination of enantiomerically pure 2-bromopentane, we analyze the reaction at each position along the pentane carbon chain.

Let the starting material be $(2R)$-2-bromopentane, where the carbon atom C2 has a fixed configuration.

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1. Monobromination at C1 ($\text{CH}_3$ group)

• Product: 1,2-dibromopentane ($\text{BrCH}_2-\text{CH(Br)}-\text{CH}_2-\text{CH}_2-\text{CH}_3$)
• Chirality Analysis:
  • C2 retains its stereochemistry ($2R$).
  • C1 has two hydrogen atoms and is achiral.
• Chiral Products Count: 1 (a single enantiomer)

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2. Monobromination at C2 ($\text{CH(Br)}$ group)

• Product: 2,2-dibromopentane ($\text{CH}_3-\text{C(Br)}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$)
• Chirality Analysis:
  • C2 now has two identical bromine atoms, destroying the chiral center.
• Chiral Products Count: 0 (achiral product)

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3. Monobromination at C3 ($\text{CH}_2$ group)

• Product: 2,3-dibromopentane ($\text{CH}_3-\text{CH(Br)}-\text{CH(Br)}-\text{CH}_2-\text{CH}_3$)
• Chirality Analysis:
  • C2 retains its configuration ($2R$).
  • Reaction at C3 proceeds via a planar $sp^2$-hybridized free radical intermediate, allowing attack from both faces to give both $(3R)$ and $(3S)$ configurations.
  • This generates two diastereomers: $(2R, 3R)$ and $(2R, 3S)$.
  • Since the two terminal groups attached to the C2 and C3 chiral centers are different ($-\text{CH}_3$ vs. $-\text{CH}_2\text{CH}_3$), the molecule lacks symmetry, meaning no meso form exists.
• Chiral Products Count: 2

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4. Monobromination at C4 ($\text{CH}_2$ group)

• Product: 2,4-dibromopentane ($\text{CH}_3-\text{CH(Br)}-\text{CH}_2-\text{CH(Br)}-\text{CH}_3$)
• Chirality Analysis:
  • C2 retains its configuration ($2R$).
  • Monobromination at C4 via a planar free radical creates a new chiral center at C4, forming two diastereomers: $(2R, 4R)$ and $(2R, 4S)$.
  • 2,4-dibromopentane is a symmetrical molecule. The $(2R, 4S)$ isomer contains an internal plane of symmetry passing through C3 and is an achiral (meso) compound.
  • The $(2R, 4R)$ isomer has no plane or center of symmetry and is chiral.
• Chiral Products Count: 1

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5. Monobromination at C5 ($\text{CH}_3$ group)

• Product: 1,4-dibromopentane ($\text{CH}_3-\text{CH(Br)}-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{Br}$)
• Chirality Analysis:
  • C2 retains its configuration ($2R$).
  • C5 has two hydrogen atoms and is achiral.
• Chiral Products Count: 1

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Total Calculation:

$\text{Total Chiral Products} = 1 (\text{at C1}) + 0 (\text{at C2}) + 2 (\text{at C3}) + 1 (\text{at C4}) + 1 (\text{at C5}) = 5$

The total number of possible chiral products is 5.