Question

Draw the product formed when the compound shown below undergoes a reaction with hbr in ch2cl2. 2-methylbut-2-ene

Answer 1

Methylbut-2-ene undergoes asymmetric electrophilic addition with hydrogen bromide to produce two products:

• $\text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}$, 2-bromo-2-methylbutane;
• $\text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}$, 2-bromo-1-methylbutane.

It is expected that $\text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}$ would end up being the dominant product.

Explanation

Molecules of methylbut-2-ene contains regions of high electron density at the pi-bonds. Those bonds would attract hydrogen atoms with a partial positive charge in polar hydrogen bromide molecules and could occasionally induce heterolytic fission of the hydrogen-bromide bond to produce positively-charged hydrogen ions $\text{H}^{+}$ and negatively-charged bromide ions $\text{Br}^{-}$.

$\text{H-Br} \to \text{H}^{+} + \text{Br}^{-}$

The positively-charged hydrogen ion would then attack the methylbut-2-ene to attach itself to one of the two double-bond-forming carbon atoms. It would break the pi bond (but not the sigma bond) to produce a carbocation with the positive charge centered on the carbon atom on the other end of the used-to-be double bond. The presence of the methyl group introduces asymmetry to the molecule, such that the two possible carbocation configurations are structurally distinct:

• $\text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3}$;
• $\text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3}$.

The carbocations are of different stabilities. Electrons in carbon-carbon bonds connected to the positively-charged carbon atom shift toward the electron-deficient atom and help increase the structural stability of the molecule. The electron-deficient carbon atom in the first carbocation intermediate shown in the list has three carbon-carbon single bonds after the addition of the proton $\text{H}^{+}$ as opposed to two as in the second carbocation. The first carbocation- a "tertiary" carbocation- would thus be more stable, takes less energy to produce, and has a higher chance of appearance than its secondary counterpart. The polar solvent dichloromethane would further contribute to the stability of the carbocations through dipole-dipole interactions.  

Both carbocations would then combine with bromide ions to produce a neutral halocarbon.  

• $\text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}$
• $\text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}$

The position of bromine ions in the resultant halocarbon would be dependent on the center of the positive charge in the carbocation. One would thus expect 2-bromo-2-methylbutane, stemming from the first carbocation which has the greatest abundance in the solution among the two, to be the dominant product of the overall reaction.