Question

part 5
1
2
3
4
oh
nc
5
1 = sp³; 2 = sp³; 3 = sp²; 4 = sp³; 5 = sp³
1 = sp³; 2 = sp²; 3 = sp; 4 = sp³; 5 = sp
1 = sp³; 2 = sp³; 3 = sp²; 4 = sp; 5 = sp
1 = sp³; 2 = sp²; 3 = sp²; 4 = sp³; 5 = sp
1 = sp³; 2 = sp³; 3 = sp²; 4 = sp³; 5 = sp

Explanation:

Step1: Determine hybridization for carbon in single - bonded chain (1 and 2)

Carbon atoms in single - bonded chains (like in the alkyl part for 1 and 2) have \(sp^{3}\) hybridization as they form four sigma bonds.

Step2: Determine hybridization for carbonyl carbon (3)

The carbonyl carbon (3) has \(sp^{2}\) hybridization. It forms three sigma bonds (two with carbon atoms in the ring and one with oxygen) and one pi bond (with oxygen).

Step3: Determine hybridization for hydroxyl - attached carbon (4)

The carbon attached to \(OH\) (4) has \(sp^{3}\) hybridization as it forms four sigma bonds (three with carbon atoms in the ring and one with oxygen of \(OH\)).

Step4: Determine hybridization for \(CN\) - attached carbon (5)

The carbon attached to \(CN\) (5) has \(sp\) hybridization. The carbon in \(CN\) forms one sigma bond (with the ring carbon) and two pi bonds (in \(C\equiv N\)).

Answer:

\(1 = sp^{3};2 = sp^{3};3 = sp^{2};4 = sp^{3};5 = sp\) (the last option in the given choices)