That's not really broad enough for O-H, it looks more like C-H stretching i.e. not very helpful. Yeah, the C=O stretch is significant.
See what I mean that with IR? You can pretty much see whatever you want to see
NMR and MS are much more useful, I'd strongly suggest you use these two first then use IR to confirm you're interpretation.
It suggests some kind of C=O environment
It's not COOH...
No, 151 is the [M+1]+ peak, 150 is the M+ ion you're interested in. The m/z is a scale, not the tallest peak
But it is significant because 105 = 77(benzene) + 28 (16O + 12C) suggesting that you have a C6H5CO+ fragment which goes along with what we've deduced so far - a C=O group
You've miscounted the number of carbon environments - every peak is a different environment, unlike 1H-NMR. A small peak is distinctive of a carbon with no hydrogens directly attached.
In your 1H-NMR you have two peaks at around 7-8 ppm (you should have counted from 0 on the left. so your ppms are 1 off
) these are your aromatic peaks from the benzene ring. You've correctly identified the CH2CH3 pattern form the triplet-quartet pair