91Ó°ÊÓ

The organic reactions involving the removal of the loosely held proton by a strong base are known as deprotonation reactions.

Sodium hydride (NaH) is a strong base that would abstract a proton from the nitrogen. In the presence of a strong base, the C-Br bond breaks, decreasing the electron density on the carbon atom.

Deprotonation by the base

The electron-rich nitrogen atom would attack the electrophilic center, forming a heterocyclic compound.

Ring-closure

The IR spectra of the product have a strong peak at ${\text{3000-2850 c³¾}}^{\text{-1}}$ corresponding to N-H vibrations, and medium peaks at ${\text{3500-3200 c³¾}}^{\text{-1}}$ that correspond to the vibration of the ${\text{C}}_{{\text{sp}}^{\text{3}}}\text{-H}$ bond.

From the mass spectra, the molecular formula of the compound can be deducted as follows:

Mass spectra of the molecular ion= 71

A value of the m/z ratio indicates the presence of a nitrogen atom in the compound.

The number of possible carbon atom$\begin{array}{l}\\ \text{=}\frac{71}{\text{12}}\\ \text{=5 r±ð³¾²¹¾±²Ô»å±ð°ù }9\end{array}$

a. From above, 5 carbons atoms with 9 hydrogen atoms would be present in the compound.

b. Replacing one carbon atom with a nitrogen atom gives the formula ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}\text{N}$