Q.Identify the products A and B formed in the following reaction: $CH_3\text{--}CH_2\text{--}CH=CH\text{--}CH_3 + HCl \rightarrow A + B$

Markovnikov addition of HI to 1-methylcyclohexene puts I on the more substituted carbon (C-1, which already bears –CH$_3$), via the more stable tertiary carbocation, giving 1-iodo-1-methylcyclohexane.

In 1-methylcyclohexene the double bond is between ring carbons C-1 and C-2, with a –CH$_3$ group on C-1. Electrophilic addition of HI proceeds in two steps:

Step 1 (protonation): H$^+$ adds to the alkene carbon that gives the more stable carbocation. Adding H$^+$ to C-2 places the positive charge on C-1, which is a tertiary carbocation (bonded to the ring's C-2 and C-6, plus the –CH$_3$ group). Adding H$^+$ instead to C-1 would place the charge on C-2, only a secondary carbocation. Since 3° carbocations are more stable (greater hyperconjugation/inductive stabilisation from three alkyl groups) than 2°, the reaction proceeds through the C-1 cation — this is Markovnikov's rule (the H goes to the carbon already bearing more hydrogens, i.e. C-2).

$\text{1-methylcyclohexene} + H^+ \rightarrow \text{1-methylcyclohexan-1-yl cation (3°, at C-1)}$

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Markovnikov's rule: with unsymmetrical alkenes, $H^+$ from $HX$ adds to the carbon that generates the more stable (here, tertiary) carbocation, and the halide ion then bonds to that carbon.

1-Methylcyclohexene has its ring double bond between $C_1$ (bearing the $CH_3$ substituent) and $C_2$ (bearing only $H$). Protonation of the alkene can occur in two ways:

• $H^+$ adds to $C_1$ $\Rightarrow$ carbocation forms at $C_2$, a secondary carbocation (flanked by $C_1$ and $C_3$, both ring carbons).
• $H^+$ adds to $C_2$ $\Rightarrow$ carbocation forms at $C_1$, a tertiary carbocation (bonded to $CH_3$, $C_2$ and $C_6$ — three carbon substituents). …

Ordinary Markovnikov addition of HBr to 2-methylpropene would put Br on the more-substituted carbon; getting the anti-Markovnikov product (Br on the terminal carbon) needs the peroxide-initiated free-radical mechanism (Kharasch/peroxide effect).

Target: 2-methylpropene, $(CH_3)_2C=CH_2$ (isobutylene), converted into 1-bromo-2-methylpropane, $(CH_3)_2CH\text{–}CH_2Br$ — i.e. Br ends up on the terminal (less substituted) carbon, which is the opposite regiochemistry to normal Markovnikov addition (which would instead place Br on the more substituted carbon, giving 2-bromo-2-methylpropane).

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$CH_3\text{–}CH_2\text{–}CH=CH\text{–}CH_3$ is pent-2-ene; electrophilic addition of $HCl$ proceeds via protonation to give whichever secondary carbocation is more stabilised, so the products are a mixture of 2-chloropentane (major) and 3-chloropentane (minor).

The alkene $CH_3\text{–}CH_2\text{–}CH=CH\text{–}CH_3$ is pent-2-ene, numbered $C_1H_3\text{–}C_2H_2\text{–}C_3H=C_4H\text{–}C_5H_3$, with the double bond between $C_3$ and $C_4$ (equivalently $C_2$–$C_3$ counting from the other end). Since this alkene is unsymmetrically substituted but both carbons of the double bond are internal, protonation of either carbon gives a secondary carbocation, so a mixture of two constitutional isomers is obtained:

• $H^+$ adds to $C_3$: leaves the cation at $C_4$, giving $CH_3\text{–}CH_2\text{–}CH_2\text{–}CH^+\text{–}CH_3$ — this secondary cation is flanked by a $-CH_2CH_2CH_3$ (propyl) group on one side and a $-CH_3$ on the other, i.e. more hyperconjugating $\alpha$-hydrogens overall, making it the somewhat more stabilised carbocation. $Cl^-$ then attacks here, giving $CH_3\text{–}CH_2\text{–}CH_2\text{–}CHCl\text{–}CH_3$, i.e. 2-chloropentane (major). …