The structure of product C, formed by the following sequence of reactions is:
CH$$_3$$COOH + SOCl$$_2$$ $$\rightarrow$$ A $$\xrightarrow[AlCl_{3}]{Benzene}$$ B $$\xrightarrow[OH]{KCN}$$ C

Step 1 : Formation of the acyl chloride (compound A)
Acetic acid reacts with thionyl chloride as
$$CH_3COOH + SOCl_2 \rightarrow CH_3COCl + SO_2 + HCl$$
Thus $$A = CH_3COCl$$ (ethanoyl chloride or acetyl chloride).

Step 2 : Friedel-Crafts acylation on benzene (compound B)
In the presence of $$AlCl_3$$, acetyl chloride furnishes the acylium ion $$CH_3CO^+$$, which electrophilically substitutes a hydrogen atom of benzene:
$$C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl$$
Hence $$B = C_6H_5COCH_3$$, i.e. acetophenone (phenyl-methyl ketone).

Step 3 : Cyanohydrin formation with $$KCN/OH^-$$ (compound C)
Under basic conditions, $$CN^-$$ acts as a nucleophile. It adds to the carbonyl carbon of acetophenone and the resulting alkoxide ion is protonated to give a cyanohydrin:
$$C_6H_5COCH_3 + CN^- + H_2O \rightarrow C_6H_5C(OH)(CN)CH_3$$

Therefore compound C possesses a hydroxyl group and a cyano group on the same carbon, directly attached to the benzene-carbonyl carbon. Its condensed structure is
$$C_6H_5C(OH)(CN)CH_3$$.

Among the given drawings, this corresponds to Option C.

Final answer: Option C which is: $$C_6H_5C(OH)(CN)CH_3$$ (cyanohydrin of acetophenone).