$\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.

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