How to Synthesize o-ethoxybenzaldehyde from phenol, methanol, and hydrochloric acid

Take the methanol and add 10% hydrochloric. This will form methylchloride. Then, react the methyl chloride with phenol in aluminum chloride (AlCl3). This will form o-methylphenol. React with sodium hydride, which will remove the proton from the hydroxy group, and react with methylchloride to perform an SN2 reaction. React the compound with KMnO4, forming o-ethoxybenzoic acid. Then, react with LiAlH4 in water, turning the carboxylic acid to an alcohol. React again with PCC, and you have your compound- o-ethoxybenzaldehyde, a structural analogue to vanillin.

2-Heptanone

Hi,
In ochem lab, we had a bunch of unknowns to identify and I identified 2 of them (out of 3 thus far), but then the third one threw me for a loop cause the H NMR sucked. The integration units were totally off and I didn't know what to do. So I talked to the teacher- she's awesome by the way- I love her and I wish I had her for ochem I- and she told me that my H NMR was junk and that I probably had a methyl ketone. So, I looked and looked and looked, and compared the C NMR, which said that I had 7 different carbons, and the only one that sorta made sense was 2-heptanone, but what irks me is that it only had 5 peaks, when there was 6 types of protons. Anyways, I confirmed it with the teacher again and she said it was probably correct. I think that the signals from two adjacent types of protons were combining to give one weird signal. Interesting. Very interesting.