Counting total covalent bonds in each compound

Steps:

• CH3 (methyl radical): 3 C-H single bonds, total 3 (odd).
• CH3OH (methanol): 3 C-H, 1 C-O, 1 O-H single bonds, total 5 (odd).
• CH3CHO (acetaldehyde): 3 C-H (methyl), 1 C-C, 1 C-H (aldehyde), 1 C=O sigma + 1 C=O pi, total 8 (even).
• CH2OH (hydroxymethyl radical): 2 C-H, 1 C-O, 1 O-H single bonds, total 4 (even).

Why B is correct:

• Exactly two compounds (CH3 and CH3OH) have an odd total number of covalent bonds, per Lewis structure bond counting where multiple bonds contribute fully.

Why the others are wrong:

• A is wrong: Two compounds have odd bonds, not one.
• C is wrong: Only two have odd bonds, not three.
• D is wrong: Two have even bonds, so not all four.

Final answer: B