Given below are two statements :
Statement I : Ozonolysis followed by treatment with Zn, H$$_2$$O of cis-2-butene gives ethanal.
Statement II : The production obtained by ozonolysis followed by treatment with Zn, H$$_2$$O of 3, 6-dimethyloct-4-ene has no chiral carbon atom.
In the light of the above statements, choose the correct answer from the options given below

Reductive ozonolysis means adding $$O_3$$ to an alkene, cleaving the $$C=C$$ bond, and then reducing the unstable ozonide with $$Zn / H_2O$$.
Each carbon of the original double bond becomes the carbon of a carbonyl compound (aldehyde or ketone depending on whether the carbon originally carried an $$H$$ atom or not).

Case 1:

cis-2-butene is $$CH_3-CH=CH-CH_3$$.
Breaking the $$C_2=C_3$$ bond places one carbonyl on $$C_2$$ and another on $$C_3$$, giving two identical molecules of ethanal:

$$CH_3-CH \;=\; CH-CH_3 \;\xrightarrow[\;Zn/H_2O\;]{\;O_3\;} 2\,CH_3-CHO$$

Thus ozonolysis of cis-2-butene indeed yields ethanal.
Statement I is therefore true.

Case 2:

3,6-dimethyloct-4-ene possesses the skeleton

$$CH_3-CH_2-CH(CH_3)-CH_2-CH=CH-CH(CH_3)-CH_3$$

The double bond lies between $$C_4$$ and $$C_5$$ (counting from the left).
Ozonolysis cleaves this bond and converts $$C_4$$ and $$C_5$$ into carbonyl carbons:

Left fragment (carbons $$1\rightarrow4$$): $$CH_3-CH_2-CH(CH_3)-CHO$$  = 3-methylbutanal

Right fragment (carbons $$5\rightarrow8$$): $$CH_3-CH(CH_3)-CH_2-CHO$$  = 2-methylbutanal

Examine chirality:
• In 3-methylbutanal the carbon bearing the methyl group is attached to $$H,\;CH_3,\;CH_2CH_3,$$ and $$CHO$$ - four different groups ⇒ one chiral centre.
• In 2-methylbutanal the carbon bearing the methyl group is attached to $$H,\;CH_3,\;CHO,$$ and $$CH_2CH_3$$ - again four different groups ⇒ one chiral centre.

Hence the product set contains chiral carbon atoms.
Statement II is therefore false.

Conclusion: Statement I is true but Statement II is false ⇒ Option C.