1. Find the degree of unsaturation (DBE)

For a neutral acyclic compound with formula $C_4H_5N$,

So the molecule has 3 degrees of unsaturation.

Since the compound is acyclic, these 3 must come from only:

• double bonds, and/or
• triple bonds.

2. Use the valency of nitrogen

A neutral nitrogen in an acyclic compound typically has valency 3. With formula $C_4H_5N$ and DBE $=3$, a very natural arrangement is a nitrile group:

A nitrile contributes:

• one carbon,
• one nitrogen,
• and 2 degrees of unsaturation.

That leaves 1 more degree of unsaturation to be accommodated elsewhere in the 4-carbon chain, i.e. one more $\mathrm{C=C}$ bond.

Thus the acyclic neutral structure must be of the type:

or an isomer with the same bonding pattern of one $\mathrm{C=C}$ and one $\mathrm{C\equiv N}$.

3. Count $sp^3$ carbons

In the example

• the two alkene carbons are $sp^2$,
• the nitrile carbon is $sp$,
• the carbon between alkene and nitrile, $\mathrm{-CH_2-}$, is $sp^3$.

So the number of $sp^3$ hybridised carbons is

4. Check if any other acyclic neutral isomer can change this count

For $C_4H_5N$ with DBE $=3$ and no ring, the stable neutral arrangements are acyclic unsaturated nitriles such as:

• $\mathrm{CH_2=CH-CH_2-CN}$
• $\mathrm{CH_3-CH=C-CN}$ (with connectivity adjusted appropriately)

In each valid acyclic neutral isomer, exactly:

• one carbon is nitrile carbon ($sp$),
• two carbons are involved in one extra unsaturation ($sp^2$ or one more $sp$ pair),
• leaving one carbon as saturated $sp^3$.

Hence the required number is 1.

Final Answer