Drawing the Lewis structure is mostly just arithmetic before it gets visual
The way I always approach any small organic molecule is to count valence electrons first, place the skeleton, then distribute. For methanol you have carbon with 4, oxygen with 6, and four hydrogens each contributing 1. That gives you 14 valence electrons total. You put carbon in the center bonded to three hydrogens and to oxygen, then oxygen bonds to the remaining hydrogen. That leaves two lone pairs sitting on the oxygen. You can check the math quickly: four C-H bonds plus one C-O bond plus one O-H bond equals 6 bonds times 2 electrons, which is 12, plus the 4 nonbonding electrons on oxygen gets you back to 14. Everything checks out. I keep a spreadsheet open when I am teaching this to undergrads because the stepwise version prevents the usual mistakes people make. Here is the sequence I use without variation. Step one: Write the molecular formula, CH3OH, and confirm it is not an ion. If the charge were not zero, you would adjust the electron count immediately. Methanol is neutral, so the count stays at 14.
Step two: Choose the central atoms. Carbon is less electronegative than oxygen, so carbon takes priority as the primary backbone atom. Oxygen bridges between carbon and hydrogen in the final arrangement. Step three: Draw single bonds only at first. Connect the three hydrogens to carbon and the fourth hydrogen to oxygen. Then connect carbon to oxygen. That is five single bonds. Step four: Subtract bonding electrons from the total. Five bonds use 10 electrons, leaving 4 electrons to place as lone pairs on the most electronegative atom that still needs octet completion. That is oxygen.
Step five: Place two lone pairs on oxygen. Verify every atom. Carbon has 8 electrons around it. Oxygen has 8. Each hydrogen has 2. The structure is complete. One detail that trips people up regularly is the placement of the hydrogens. The formula CH3OH already tells you which hydrogen belongs to oxygen, but beginners sometimes draw a linear chain like H-C-O-H with an extra hydrogen floating somewhere, or they put all four hydrogens on carbon. The condensed formula is not optional notation. It encodes the connectivity. Treat it as part of the problem statement, not as shorthand you can ignore. I encountered a specific issue once when a student submitted a structure where the C-O-H angle was drawn as 180 degrees with no lone pairs shown on oxygen, and they claimed the octets were satisfied. The octet on oxygen was technically met with four bonds in their drawing, which meant they had placed 16 electrons instead of 14. The error came from treating the line drawing as a skeletal structure rather than a full Lewis diagram. I had them recount from the molecular formula and redraw with all lone pairs explicit. The fix took about three minutes once they understood that methanol requires both bonding pairs and lone pairs to be shown in a Lewis structure, unlike a skeletal formula where hydrogens on heteroatoms are often implied.
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The geometry around carbon is tetrahedral with bond angles close to 109.5 degrees. The geometry around oxygen is bent, approximately 104.5 degrees, because of the two lone pairs. This matters if you are moving on to VSEPR or trying to predict dipole moments. The C-O bond is polar, the O-H bond is polar, and the molecule has a net dipole pointing toward the oxygen. That polarity is why methanol mixes with water and why it is a useful solvent in extraction work. A counter-intuitive point that most introductory texts do not emphasize enough is that resonance is not relevant here. Some students try to draw a double bond between carbon and oxygen to satisfy octets more elegantly, but that creates a formal charge problem. If you make C=O, carbon would have five bonds, which violates the octet rule for a second-row element. You cannot expand the octet on carbon. The structure with all single bonds and lone pairs on oxygen is the only valid Lewis representation. Another thing worth noting is formal charge. In the correct structure, every atom has a formal charge of zero. Carbon has 4 valence electrons, owns 4 in bonds. Oxygen has 6 valence electrons, owns 6 when you count 4 nonbonding plus 2 from bonds. Hydrogen has 1 valence electron and owns 1 from its bond. Zero formal charges everywhere is a good sign you have the right structure. If you ever calculate nonzero formal charges on a neutral methanol diagram, go back and recount.
The Lewis structure itself does not tell you everything. It does not show bond angles precisely, it does not show molecular orbital character, and it does not indicate that the O-H bond is significantly more acidic than the C-H bonds. For reactivity predictions you need more than this diagram. It is useful for counting electrons, checking formal charges, and understanding basic connectivity, but do not treat it as a substitute for thinking about electronegativity differences or reaction mechanisms. If you want a downloadable version for study purposes, you can find clean SVG and PNG renders on chemical drawing sites like PubChem or ChemDraw templates, though I usually just redraw it by hand because doing it manually forces you to go through the counting steps each time and prevents you from treating the image as something you can just copy without understanding.
