1. Nature of reaction

Under $Br_2/h\nu$, bromination occurs by a free-radical substitution at allylic positions (not electrophilic addition across the double bond).

The substrate is 1-methylcyclohex-1-ene.

Let us number the double bond carbons as:

• $C_1$: one alkene carbon bearing the methyl group
• $C_2$: the other alkene carbon

So the structure is:

• double bond: $C_1=C_2$
• methyl attached to $C_1$
• ring continues through $C_3, C_4, C_5, C_6$

2. Find all distinct allylic positions

Allylic positions are carbons adjacent to the double bond.

These are:

• the methyl carbon attached to $C_1$
• $C_3$ (adjacent to $C_2$)
• $C_6$ (adjacent to $C_1$)

Because the double bond is unsymmetrical due to the methyl substituent, $C_3$ and $C_6$ are not equivalent.

Thus, bromination can occur at three different allylic sites.

3. Case 1: Bromination at the methyl group

Replacing one H of the methyl group gives an exocyclic substituent $-CH_2Br$ at $C_1$.

Product: $1$-(bromomethyl)cyclohex-1-ene type.

• No stereocenter is created.
• No geometrical isomerism arises because the double bond is inside a six-membered ring (trans-cyclohexene not feasible here, and no $E/Z$ alternative).

So this gives: $1 \text{ isomer}$

4. Case 2: Bromination at } C_3$

Substitution at $C_3$ gives 3-bromo-1-methylcyclohex-1-ene.

Now check stereochemistry:

• $C_3$ becomes a stereogenic center because it is attached to:
  1. $Br$
  2. $H$
  3. path toward $C_2$ (toward the double bond side)
  4. path toward $C_4$ (other side of ring)

These are all different, so $C_3$ is chiral.

Hence this product exists as: $2 \text{ enantiomers}$

5. Case 3: Bromination at } C_6$

Substitution at $C_6$ gives 6-bromo-1-methylcyclohex-1-ene.

Again, $C_6$ is attached to:

1. $Br$
2. $H$
3. path toward $C_1$ (methyl-substituted alkene carbon)
4. path toward $C_5$

These are different, so $C_6$ is also a stereogenic center.

Thus this gives: $2 \text{ enantiomers}$

6. Important extra possibility: allylic rearrangement

In allylic bromination, the allylic radical formed is resonance-stabilized, so bromination can occur at either end of the allylic radical system, giving products with shifted double bond as well.

We must therefore consider resonance-related products from each allylic radical.

(a) Radical from methyl group

Abstraction at methyl gives allylic radical: $\ce{CH2^. - C1 = C2 <-> CH2 = C1 - C2^.}$

This leads to two products:

• bromine at methyl carbon, double bond unchanged
• bromine at $C_2$, with double bond shifted to exocyclic position $\ce{CH2=C1}$

So from this radical we get:

• A: bromomethyl product = $1$ isomer
• B: $2$-bromo-1-methylenecyclohexane type product = at $C_2$, no stereocenter

Thus: $2 \text{ isomers}$

(b) Radical from } C_3$

Abstraction at $C_3$ gives allylic radical: $C_3^. - C_2 = C_1 <-> C_3 = C_2 - C_1^.$

This gives two constitutional products:

• bromine at $C_3$, double bond unchanged: 3-bromo-1-methylcyclohex-1-ene
• bromine at $C_1$, double bond shifted to $C_2=C_3$: 1-bromo-1-methylcyclohex-2-ene

Count stereoisomers:

For 3-bromo-1-methylcyclohex-1-ene:

• one stereocenter at $C_3$
• hence $2$ enantiomers

For 1-bromo-1-methylcyclohex-2-ene:

• At $C_1$, there are substituents $Br$, $CH_3$, and two ring paths.
• The two ring paths are different because one side goes toward alkene carbon $C_2$ and the other toward saturated carbon $C_6$.
• So $C_1$ is a stereogenic center.
• Hence again $2$ enantiomers.

Thus from this radical: $2+2=4 \text{ isomers}$

(c) Radical from } C_6$

Abstraction at $C_6$ gives allylic radical: $C_6^. - C_1 = C_2 <-> C_6 = C_1 - C_2^.$

This gives two products:

• bromine at $C_6$, double bond unchanged: 6-bromo-1-methylcyclohex-1-ene
• bromine at $C_2$, double bond shifted to $C_1=C_6$: an exocyclic/shifted alkene product

Count stereoisomers:

For 6-bromo-1-methylcyclohex-1-ene:

• one stereocenter at $C_6$
• so $2$ enantiomers

For shifted-double-bond product with Br at $C_2$:

• $C_2$ becomes $sp^3$
• attached to $Br$, $H$, path toward $C_1$ (alkenic/exocyclic side), and path toward $C_3$
• all different, so $C_2$ is stereogenic
• therefore $2$ enantiomers

Thus from this radical: $2+2=4 \text{ isomers}$

7. Total count

Adding all distinct products including stereoisomers: $2+4+4=10$

So the maximum number of mono-bromination products is: $\boxed{10}$

8. Comparison with stored answer

Stored correct answer = $13$

My derived answer is $10$, not $13$.

The likely reason for $13$ would be overcounting geometrical isomers or counting conformers separately, which should not be done. For cyclic alkenes of this type, only configurational stereoisomers are counted, not conformations.