Answer:
Ethoxybenzene
Step-by-step explanation:
1. Index of hydrogen deficiency
Formula = C₈H₁₀O
Saturated = C₈H₁₈
IHD = (18 - 10)/2 = 8/2 = 4
The compound contains four rings and/or double bonds.
Whenever IHD ≥ 4, I immediately think of a benzene ring. That immediately accounts for all four units of unsaturation (1 ring + 3 double bonds).
There are 5H in the aromatic region (δ = 7.0 - 7.4), so I suspect a phenyl group.
The rest of the molecule must be saturated.
2. Triplet-quartet
The 3H:2H triplet-quartet is the classic signature of an ethyl group.
3. Structure
The only atom left is the O atom, and it must go between the ethyl and the phenyl group.
The compound is ethoxybenzene.
4. Confirmatory evidence
The CH₂ group is quite far downfield. The normal range for RCH₂-OR is δ = 3.5 - 3.9. I would expect RCH₂-OPh to be shifted even further downfield by the phenyl group.
The diagram below shows my interpretation of the spectrum.