Statement I: Ozonolysis of cis-2-butene

The structure of cis-2-butene is $\mathrm{CH_3-CH=CH-CH_3}$

On ozonolysis followed by reductive workup with $\mathrm{Zn/H_2O}$, the double bond is cleaved and each alkene carbon becomes a carbonyl carbon.

Since both double-bond carbons in cis-2-butene have one hydrogen each, each side forms an aldehyde: $\mathrm{CH_3-CH=CH-CH_3 \xrightarrow[Zn/H_2O]{O_3} 2\ CH_3CHO}$

The product is ethanal.

Therefore, Statement I is true.

Statement II: Ozonolysis of 3,6-dimethyloct-4-ene

First write the structure: $\mathrm{CH_3-CH_2-CH(CH_3)-CH=CH-CH(CH_3)-CH_2-CH_3}$

The double bond is between C4 and C5. On ozonolysis with $\mathrm{Zn/H_2O}$, this bond cleaves to give two aldehydes because each double-bond carbon has one hydrogen.

Products:

• Left fragment: $\mathrm{CH_3-CH_2-CH(CH_3)-CHO}$
• Right fragment: $\mathrm{OHC-CH(CH_3)-CH_2-CH_3}$

These are actually the same compound written from opposite ends: $\mathrm{2 ext{-}methylbutanal}$

So the ozonolysis gives two molecules of 2-methylbutanal.

Check chirality of 2-methylbutanal

Structure: $\mathrm{CHO-CH(CH_3)-CH_2-CH_3}$

The carbon next to the aldehyde group is attached to four different groups:

• $\mathrm{H}$
• $\mathrm{CH_3}$
• $\mathrm{CH_2CH_3}$
• $\mathrm{CHO}$

Hence this carbon is chiral.

Therefore, the product does have a chiral carbon atom.

So Statement II is false.

Conclusion

• Statement I: True
• Statement II: False

Hence the correct option is: $\boxed{\text{C}}$