QAIPM-2015-Ch-1188: NEET 2015 Chemistry - Class 11 AIPM-2015-Ch-1188 Solution: can be prepared by heating which of the following compounds…

Step-by-Step Chemistry Explanation When alkenes are heated with a strong acid, they undergo protonation followed by carbocation rearrangement to yield the most stable, highly substituted alkene as the major product (Saytzeff's Rule). Let's trace out the mechanism for $\mathrm{(CH}_{\mathrm{3}}\mathrm{)}_{\mathrm{3}}\mathrm{CCH=CH}_{\mathrm{2}}$ to see how it yields 2,3-dimethyl-2-butene: 1. Protonation: The double bond of 3,3-dimethyl-1-butene attacks the $\mathrm{H}^{\mathrm{+}}$ from the strong acid to form a secondary ($2^{\circ }$) carbocation on Carbon-2: $\mathrm{(CH}_{\mathrm{3}}\mathrm{)}_{\mathrm{3}}\mathrm{C-}\overset{+}{C}\mathrm{H-CH}_{\mathrm{3}}$ 2. 1,2-Methyl Shift (Rearrangement): The secondary carbocation is adjacent to a quaternary carbon holding three methyl groups. To achieve greater stability, one of the methyl groups shifts over with its pair of bonding electrons (1,2-methyl shift). This converts the $2^{\circ }$ carbocation into a highly stable tertiary ($3^{\circ }$) carbocation: $\mathrm{(CH}_{\mathrm{3}}\mathrm{)}_{\mathrm{2}}\overset{+}{C}\mathrm{-CH(CH}_{\mathrm{3}}\mathrm{)}_{\mathrm{2}}$ 3. Deprotonation (Elimination): A weak base eliminates a proton ($\mathrm{H}^{\mathrm{+}}$) from the adjacent carbon to form the most substituted, most stable alkene possible (having 12 hyperconjugative $\alpha$-hydrogens): $\mathrm{(CH}_{\mathrm{3}}\mathrm{)}_{\mathrm{2}}\mathrm{C=C(CH}_{\mathrm{3}}\mathrm{)}_{\mathrm{2}}\quad \text{[2,3-dimethyl-2-butene]}$