Final answer:
The synthesis of 3,4-dimethyl-cyclopentene would typically involve an E2 mechanism using an appropriate dihalocyclopentane as the reactant and a strong base. Exact details would depend on the starting halide structure, which isn't provided, but would aim to minimize side products and isomer formation.
Step-by-step explanation:
To synthesize 3,4-dimethyl-cyclopentene from an appropriate halide in a one-step reaction, you would generally look to perform an elimination reaction. Specifically, this would often involve an E2 mechanism where the halide, in this case, a dihalocyclopentane where the two halogen atoms are vicinal (on adjacent carbon atoms), would undergo dehydrohalogenation in the presence of a strong base.
Unfortunately, without the proper reactant information such as the starting halide structure, it is not possible to provide the exact reaction. Typically one might consider using a reactant such as 3,4-dibromo-1-methylcyclopentane with a strong base such as sodium ethoxide or potassium tert-butoxide to form 3,4-dimethyl-cyclopentene via E2 elimination.
It is important to choose a reactant such that the resulting elimination forms the alkene as the major product with minimal formation of substitution side products or other isomers. In the case of forming 3,4-dimethyl-cyclopentene, regioselectivity and stereoselectivity are critical to consider to achieve the desired product as the only outcome.