An alkene X upon reductive ozonolysis ( O_3 , followed by Zn/H_2O )...

Reductive ozonolysis of an alkene cleaves the carbon-carbon double bond to form carbonyl compounds.

Since the products Y and Z are both ketones, the original alkene X must be a tetrasubstituted alkene. For both ketones to respond positively to the iodoform test, they must be methyl ketones, meaning each must contain a $CH_3-CO-$ group.

Furthermore, Y and Z are isomeric, which means they must possess the same molecular formula.

Evaluating the ozonolysis products for each given option:

For 2,3-dimethylhex-2-ene, the structure is $CH_3-C(CH_3)=C(CH_3)-CH_2-CH_2-CH_3$. Its ozonolysis yields propan-2-one ($C_3H_6O$) and pentan-2-one ($C_5H_{10}O$). These products are not isomers.

For 2,3,4-trimethylhept-3-ene, the structure is $CH_3-CH(CH_3)-C(CH_3)=C(CH_3)-CH_2-CH_2-CH_3$. Its ozonolysis yields 3-methylbutan-2-one ($C_5H_{10}O$) and pentan-2-one ($C_5H_{10}O$). Both are methyl ketones and have the same molecular formula, making them isomers. This satisfies all given conditions.

For 2,4-dimethylhex-3-ene, the structure is $CH_3-CH(CH_3)-CH=C(CH_3)-CH_2-CH_3$. Its ozonolysis yields 2-methylpropanal (an aldehyde) and butan-2-one. This does not result in a mixture of two ketones.

For 2,3,5-trimethylhex-2-ene, the structure is $CH_3-C(CH_3)=C(CH_3)-CH_2-CH(CH_3)-CH_3$. Its ozonolysis yields propan-2-one ($C_3H_6O$) and 4-methylpentan-2-one ($C_6H_{12}O$). These products are not isomers.

Therefore, the correct IUPAC name of the alkene is 2,3,4-trimethylhept-3-ene.