Question

The 1H NMR spectrum of compound A (C15H14O) shows only two signals: a multiplet at 7.15 ppm and a singlet at 3.55 ppm in a 5:2 ratio. The IR spectrum has no absorption in the 3200-4000 cm^-1 region, but strong peaks can be found near 1700 cm^-1. Compound A reacts with NaBH4 followed by acidification to give compound B of the molecular formula C15H16O. The reaction of A with CH3MgBr, and then with H3O^+, gives C, with a molecular formula of C16H18O. Suggest structures for B and C.

Answer 1

Answer:

Both B and C are alcohols and their structures are in the attached figure.

Explanation:

Compound A: C15H14O

NMR: 2 signals → 2 types of hydrogen surroundings; symetric molecule

7.15 ppm  → indicates the presence of benzene

3.55 ppm  → indicates the presence of a methyl (-CH3) substituted benzene

IR: no absorption in the 3200-4000 cm⁻¹ region → no hydroxyl groups (-OH)

strong peaks can be found near 1700 cm⁻¹ region → possible carbonyl group (-C=O) and alkenes (C=C).

C₁₅H₁₄O + NaBH₄ → C₁₅H₁₆O

NaBH₄: reduces aldehydes and ketones to alcohols.

C₁₅H₁₄O + CH₃MgBr → C₁₆H₁₈O

CH₃MgBr: reacts with ketones and aldehydes to produce terciary or secundary alcohols.

Considering all the information above, the structures attached were discovered.