Identify the reaction type

Addition of HBr to $1,3$-butadiene is an electrophilic addition to a conjugated diene. Such reactions can give:

• 1,2-addition product (kinetic product)
• 1,4-addition product (thermodynamic product)

Write the substrate

$\ce{CH2=CH-CH=CH2}$

This is $1,3$-butadiene.

First step: protonation

Proton adds to one end of the diene to form an allylic carbocation:

$\ce{CH2=CH-CH=CH2 + H+ -> CH3-CH^+-CH=CH2}$

This carbocation is resonance stabilized:

$\ce{CH3-CH^+-CH=CH2 <-> CH3-CH=CH-CH2^+}$

Possible attack of $\ce{Br^-}$

Because of resonance, bromide can attack at two positions:

• Attack at C-2 gives 1,2-addition product: $\ce{CH3-CHBr-CH=CH2}$ This is 3-bromobut-1-ene (often written as 3-bromobutene).

• Attack at C-4 gives 1,4-addition product: $\ce{CH3-CH=CH-CH2Br}$ This is 1-bromobut-2-ene.

Kinetic vs thermodynamic control

• At low temperature, the 1,2-product forms faster, so it is the kinetic product.
• At higher temperature (such as $40^\circ\mathrm{C}$), equilibrium control becomes important, and the more stable alkene predominates.

Determine the more stable product

Compare the alkenes:

• $\ce{CH3-CHBr-CH=CH2}$ has a terminal double bond
• $\ce{CH3-CH=CH-CH2Br}$ has an internal double bond

Internal alkenes are more stable than terminal alkenes. Therefore,

$\ce{CH3-CH=CH-CH2Br}$

i.e. 1-bromo-2-butene, is the thermodynamic product.

Match with options

• A: 3-bromobutene under kinetically controlled conditions → true statement generally, but not for $40^\circ\mathrm{C}$ predominant product
• B: 1-bromo-2-butene under thermodynamically controlled conditions → correct
• C: 3-bromobutene under thermodynamically controlled conditions → incorrect
• D: 1-bromo-2-butene under kinetically controlled conditions → incorrect

Final answer

At $40^\circ\mathrm{C}$, the reaction is predominantly under thermodynamic control, giving 1-bromo-2-butene.