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.

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.

particle   p is an electron and particle q is a neutron