Propane: A Metabolite Precursor

$Product \space A:$ Step 1) Bromine atom is added to $2^{\circ}$ carbon $$ $Product \space B:$ Step 2) Removal of halogen forms a double bond $$ $Product \space C:$ Step 3) Osmium Tetraoxide acts on the alkene to form a diol $$ $Product \space D:$ Step 4) Dichromate acts as a powerful oxidant that oxidized the $2^{\circ}$ alcohol to a ketone and the $1^{\circ}$ alcohol to a carboxylic acid to form the metabolite pyruvate $$ $Product \space E:$ Step 5) The carboxylic acid becomes a methyl ester through the process of transesterification $$ The final product $E$ is methyl pyruvate ( $C_4O_3H_6$ ), a.k.a methyl-2-oxopropanoate.

$Answer: 4C+3O+6H=\boxed{13 \space atoms}$