Animated Solution for Chemistry - Hydrocarbons: The maximum number of possible isomers (including stereoisomers) which may be formed on mono-bromination of 1-methylcyclohex-1-ene using Br${}_2$ and UV light is ______.

When you see a free radical halogenation reaction using $B{r}_2$ and UV light on an alkene, your brain immediately screams, "Allylic bromination!" And you would be absolutely right. The allylic radicals are highly stabilized by resonance, making them the major intermediates.

However, the examiners at JEE Advanced are masters of psychological warfare. They slipped in a crucial word: maximum. They didn't ask for the major products; they asked for the maximum number of possible isomers. This means we must cast our net wide and consider every single distinct mono-brominated product that could theoretically form, including the less favorable non-allylic ones. Let's break this down systematically.

Let's start with the most reactive sites. In 1-methylcyclohexene, we have three distinct allylic positions: the methyl group itself, the $C_3$ position, and the $C_6$ position.

1. The Methyl Group Radical: If the UV light knocks off a hydrogen from the methyl group, we get a primary allylic radical. Direct bromination here yields 1-(bromomethyl)cyclohexene. Because the carbon bearing the bromine is attached to two identical hydrogens, it is achiral. That's 1 isomer.

But wait, radicals love to dance! This radical can undergo resonance, shifting the double bond outside the ring (exocyclic) and moving the radical to $C_2$. Bromination of this resonance structure gives 3-bromo-2-methylidenecyclohexane. The carbon where the bromine attaches is now a chiral center, generating a pair of enantiomers (R and S). That's 2 isomers. Total from the methyl group: $1+2=3$ isomers.

2. The $C_3$ Radical: Abstracting a hydrogen from $C_3$ gives a secondary allylic radical. Direct bromination yields 3-bromo-1-methylcyclohexene. The $C_3$ carbon becomes a chiral center, giving us 2 isomers (R and S).

Resonance of this radical shifts the double bond to $C_2=C_3$ and places the radical at $C_1$. Bromination here yields 1-bromo-1-methylcyclohex-2-ene. The $C_1$ carbon is a chiral center, giving us another 2 isomers (R and S). Total from $C_3$: $2+2=4$ isomers.

3. The $C_6$ Radical: Abstracting a hydrogen from $C_6$ gives another secondary allylic radical. Direct bromination yields 6-bromo-1-methylcyclohexene. The $C_6$ carbon is a chiral center, giving us 2 isomers (R and S).

Now, let's look at its resonance structure. The double bond shifts to $C_1=C_6$, and the radical moves to $C_2$. If you brominate this and name the resulting molecule according to IUPAC rules, you will find it is exactly 6-bromo-1-methylcyclohexene! It is the exact same constitutional isomer, just drawn from a different perspective. Therefore, it does not contribute any new isomers to our count. Total from $C_6$: 2 isomers.

If we stopped here, we would have $3+4+2=9$ isomers. This is a very tempting trap, and many students fall for it.

To find the maximum possible isomers, we must remember that UV light is highly energetic. While allylic hydrogens are the easiest to remove, the non-allylic hydrogens at $C_4$ and $C_5$ can also be abstracted to form trace amounts of product.

4. The $C_4$ and $C_5$ Radicals: Bromination at $C_4$ yields 4-bromo-1-methylcyclohexene. The $C_4$ carbon is a chiral center, giving 2 isomers (R and S). Bromination at $C_5$ yields 5-bromo-1-methylcyclohexene. The $C_5$ carbon is also a chiral center, giving another 2 isomers (R and S). Total from non-allylic positions: $2+2=4$ isomers.

Now, we simply sum up all the unique stereoisomers we've discovered: - From the methyl group: 3 - From $C_3$: 4 - From $C_6$: 2 - From non-allylic positions: 4

Grand Total = $3+4+2+4=13$ isomers.

This problem is a beautiful test of systematic thinking. It rewards those who read the word "maximum" carefully and punishes those who blindly apply the "allylic only" rule without considering the full scope of free radical chemistry.