Question

How many NMR signals does $C{H_3}C{H_2}C{H_2}OH$ show ?

Answer 1

$^1H$ is in dissimilar environments that give different NMR signals.
Protons in dissimilar environments give different NMR signals. Equivalent protons give the similar NMR signal. The number of NMR signals equals the number of dissimilar types of protons in a compound.

Complete step by step answer:
The H NMR spectrum of $propan - 1 - ol$ expresses four signals.
The molecule has not one symmetry element. So the three carbon atoms and the oxygen atom signify four different hydrogen environments.
We should get four signals with area ratios $3{\text{ }}:{\text{ }}2{\text{ }}:{\text{ }}2{\text{ }}:{\text{ }}1$.
The OH signal contrasts from $\delta {\text{ }}1$ to $\delta {\text{ }}6$, depending on the solvent and dilution, so we should assume to see a $^1H$ signal anywhere in this region.
The electronegative O atom pulls complete signals downfield. The effect decreases speedily with distance.
The group at $C - 1$ is pulled about $2$ ppm downfield after its normal value of $\delta {\text{ }}1.4$. We should get a ${\;^2}H$ signal at about $\delta {\text{ }}3.4$.
The group at $C - 2$ is pulled near $0.1$ ppm downfield. We should get a ${\;^2}H$ signal at about $\delta {\text{ }}2.5$.
The group should be at its usual location. We should get a${\;^3}H$ signal at $\delta {\text{ }}0.9$.
All signals but the O-H signal should show spin-spin splitting.
Here is a definite spectrum of $propan - 1 - ol$.

Note:
$^1H$ NMR (proton NMR) is used to define the number and form of hydrogen atoms in a molecule; and $^{13}C$ NMR (carbon NMR) is used to define the type of carbon atoms in a molecule.
Simply nuclei that hold odd mass numbers (such as $^1H{,^{13}}C{,^{19}}F,{\text{ }}an{d^{31}}P$) or odd atomic numbers (such as $^2H{\text{ }}an{d^{14}}N$) give increase to NMR signals. Because both $^1H$ and $^{13}C$, the less plentiful isotopes of carbon, are NMR active, NMR agrees with us to map the carbon and hydrogen basis of an organic molecule.