Animated Solution for Chemistry - Organic Chemistry: The major product formed in the following reaction is $C{H}_3-\underset{\stackrel{|}{C{H}_3}}{\stackrel{\underset{|}{C{H}_3}}{C}}-\underset{\stackrel{|}{\mathrm{OH}}}{\mathrm{CH}}-C{H}_3\underset{\text{a few drops}}{\overset{\text{Conc. }{H}_{2}S{O}_4}{\to }}\text{Major product}$

We are given an alcohol, specifically 3,3-dimethylbutan-2-ol, and we are treating it with concentrated sulfuric acid ($H_{2}S{O}_4$).

Whenever you see an alcohol paired with a strong, non-nucleophilic acid and heat, your mind should immediately jump to an acid-catalyzed dehydration reaction. This reaction proceeds via an $E1$ mechanism, which means carbocations will be involved. And where there are carbocations, there is always the potential for rearrangement!

The hydroxyl group ($-OH$) is a notoriously poor leaving group because the hydroxide ion is a strong base.

To make it leave, we need to turn it into a weaker base. This is where the acid comes in. The lone pair of electrons on the oxygen atom attacks a proton ($H^{+}$) from the sulfuric acid.

This protonation converts the $-OH$ group into $-O{H}_2^{+}$. Now, the leaving group is a neutral water molecule ($H_{2}O$), which is an excellent, highly stable leaving group.

With a great leaving group in place, the carbon-oxygen bond breaks heterolytically. The water molecule departs, taking the bonding electrons with it.

This leaves behind a positive charge on carbon-2, creating a secondary ($2^{\circ }$) carbocation.

However, this is not the end of the story. Right next to our secondary carbocation is a quaternary carbon (carbon-3) bonded to three methyl groups. Nature always seeks the lowest energy state. By undergoing a 1,2-methyl shift, one of the methyl groups migrates with its electrons to the positively charged carbon.

This transforms our secondary carbocation into a tertiary ($3^{\circ }$) carbocation, which is significantly more stable due to increased hyperconjugation and the $+I$ (inductive) effect of the surrounding alkyl groups.

Now that we have our most stable carbocation, the final step is elimination.

A weak base from the reaction mixture (such as the bisulfate ion, $HS{O}_4^{-}$) will abstract a proton from an adjacent carbon. The electrons from the broken $C-H$ bond will collapse to form a $\pi$ bond.

According to Zaitsev's Rule, the major product will be the most highly substituted alkene because it is thermodynamically the most stable. Abstraction of a proton from the adjacent $CH$ group yields a tetrasubstituted double bond.

The final major product is 2,3-dimethylbut-2-ene, which perfectly matches option (b).