The number of -CH_2- (methylene) groups in th | vedclass

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(B) The starting material is $oct-4-ene$ $(CH_3CH_2CH_2CH=CHCH_2CH_2CH_3)$.Step $1$: Ozonolysis $(O_3, Zn/H_2O)$ cleaves the double bond to form two m

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(B) The starting material is $oct-4-ene$ $(CH_3CH_2CH_2CH=CHCH_2CH_2CH_3)$.Step $1$: Ozonolysis $(O_3, Zn/H_2O)$ cleaves the double bond to form two moles of butanal $(CH_3CH_2CH_2CHO)$.Step $2$: Oxidation with $KMnO_4$ converts butanal into butanoic acid $(CH_3CH_2CH_2COOH)$.Step $3$: Kolbe's electrolysis of the sodium salt of butanoic acid $(CH_3CH_2CH_2COONa)$ leads to the coupling of the propyl radical $(CH_3CH_2CH_2\cdot)$ to form $n-hexane$ $(CH_3CH_2CH_2CH_2CH_2CH_3)$.Step $4$: Aromatization of $n-hexane$ using $Cr_2O_3$ at $770 \ K$ and $20 \ atm$ yields benzene $(C_6H_6)$.In the final product,benzene,there are no $-CH_2-$ (methylene) groups as all carbon atoms are $sp^2$ hybridized and part of the aromatic ring.Therefore,the number of $-CH_2-$ groups is $0$.

The number of $-CH_2-$ (methylene) groups in the product formed from the following reaction sequence is. . . . . . .

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