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Question: $CH_2=CHCH_2CH=CH_2 \xrightarrow{NBS} X \text{ (Major), (X) is:}$ ...

CH2=CHCH2CH=CH2→NBSX (Major), (X) is:CH_2=CHCH_2CH=CH_2 \xrightarrow{NBS} X \text{ (Major), (X) is:}

A

CH2=CH−CH∣BrCH=CH2CH_2=CH-\underset{Br}{\underset{|}{CH}}CH = CH_2

B

CH2=CH−CH=CH−CH2−BrCH_2=CH-CH=CH-CH_2-Br

C

CH2=CHCH2CH=CHBrCH_2=CHCH_2CH = CHBr

D

CH2=CHCH2C∣Br=CH2CH_2=CHCH_2\underset{Br}{\underset{|}{C}} = CH_2

Answer

CH2=CH−CH=CH−CH2−BrCH_2=CH-CH=CH-CH_2-Br

Explanation

Solution

The reaction involves the allylic bromination of 1,4-pentadiene (CH2=CH−CH2−CH=CH2CH_2=CH-CH_2-CH=CH_2) using N-bromosuccinimide (NBS). NBS selectively brominates at allylic positions via a free radical mechanism.

1. Formation of Allylic Radical:

The allylic hydrogens are located on the carbon atom adjacent to a double bond. In 1,4-pentadiene, the central methylene group (CH2CH_2 at C3) has allylic hydrogens. Abstraction of a hydrogen atom from C3 by a bromine radical (Br⋅Br\cdot) leads to the formation of a resonance-stabilized allylic radical (pentadienyl radical):

CH2=CH−CH2−CH=CH2→−H⋅CH2=CH−C˙H−CH=CH2CH_2=CH-CH_2-CH=CH_2 \xrightarrow{-H\cdot} CH_2=CH-\dot{C}H-CH=CH_2

This radical can exist in multiple resonance forms:

(I) CH2=CH−C˙H−CH=CH2CH_2=CH-\dot{C}H-CH=CH_2 (Radical at C3)

(II) C˙H2−CH=CH−CH=CH2\dot{C}H_2-CH=CH-CH=CH_2 (Radical at C1, by shifting the π\pi bond from C1-C2 to C2-C3)

(III) CH2=CH−CH=CH−C˙H2CH_2=CH-CH=CH-\dot{C}H_2 (Radical at C5, by shifting the π\pi bond from C4-C5 to C3-C4)

Note that (II) and (III) are equivalent by symmetry.

2. Bromination of the Allylic Radical:

The bromine atom can add to any carbon atom bearing significant radical character (C1, C3, or C5).

  • Addition at C3:

    From resonance form (I), addition of Br⋅Br\cdot at C3 gives:

    CH2=CH−CH∣Br−CH=CH2CH_2=CH-\underset{Br}{\underset{|}{CH}}-CH=CH_2

    This product is 3-bromo-1,4-pentadiene. It contains two isolated double bonds.

  • Addition at C1 (or C5):

    From resonance form (II) or (III), addition of Br⋅Br\cdot at C1 (or C5) gives:

    Br−CH2−CH=CH−CH=CH2Br-CH_2-CH=CH-CH=CH_2 (or CH2=CH−CH=CH−CH2−BrCH_2=CH-CH=CH-CH_2-Br)

    This product is 1-bromo-1,3-pentadiene. It contains two conjugated double bonds.

3. Determining the Major Product:

In radical reactions involving resonance-stabilized intermediates, the major product is often the one that leads to the most stable final product.

Comparing the two possible products:

  • 3-bromo-1,4-pentadiene has isolated double bonds.
  • 1-bromo-1,3-pentadiene has conjugated double bonds.

Conjugated dienes are significantly more stable than isolated dienes due to the delocalization of π\pi electrons. Therefore, 1-bromo-1,3-pentadiene is the thermodynamically more stable product and will be the major product.