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.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.