Consider the organic molecule provided in the figure. It is...

The given starting material is pentanamide, which has a 5-carbon chain: $CH_3CH_2CH_2CH_2CONH_2$.

Reaction with $LiAlH_4$ in ether followed by $H_2O$ reduces the amide to a primary amine. The carbonyl group ($-C=O$) is converted to a methylene group ($-CH_2-$).
$CH_3CH_2CH_2CH_2CONH_2 \xrightarrow{LiAlH_4} CH_3CH_2CH_2CH_2CH_2NH_2$

Reaction of the resulting primary aliphatic amine (pen\tan^{-1}-amine) with $NaNO_2$ and $HCl$ at $273-278$ K generates nitrous acid ($HNO_2$), which forms an unstable aliphatic diazonium salt.
$CH_3(CH_2)_4NH_2 \xrightarrow{NaNO_2, HCl} [CH_3(CH_2)_4N_2^+Cl^-]$

Warming this diazonium salt with water causes it to decompose, releasing nitrogen gas and forming the corresponding primary alcohol.
$[CH_3(CH_2)_4N_2^+Cl^-] \xrightarrow{H_2O, \Delta} CH_3(CH_2)_4OH + N_2 + HCl$

The final major organic product is pen\tan^{-1}-ol.

Answer: Pen\tan^{-1}-ol