\(\Rightarrow \) \(\begin{matrix}
   O  \\
   ||  \\
   -C-  \\
\end{matrix}\) group

Negative iodoform test \(\Rightarrow \) \(\begin{matrix}
   \,\,\,\,\,\,\,\,\,O  \\
   \,\,\,\,\,\,\,\,\,||  \\
   C{{H}_{3}}-C-  \\
\end{matrix}\) group is absent

Negative Tollens' test \(\Rightarrow\) ketone

Hence, the compound is \(3 -\) pentanone.

\(\underset{\begin{smallmatrix} 
 3\,-\,Pen\tan one \\ 
 ({{C}_{5}}{{H}_{10}}O) 
\end{smallmatrix}}{\mathop{\begin{matrix}
   O  \\
   ||  \\
   C{{H}_{3}}C{{H}_{2}}-C-C{{H}_{2}}C{{H}_{3}}  \\
\end{matrix}}}\) \(\xrightarrow{\operatorname{Re}duction}\)  \(\underset{n\,-\,Pen\tan e}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\)