Upon heating, compound X is converted to Y, which has the same molecular formula as that of X. The ${ }^{1} \mathrm{H}$ NMR spectrum of Y shows two doublets centered at 3.0 ppm (separation of two lines ~ 20 Hz) and 4.0 ppm (separation of two ~ 15 Hz) respectively. Compound Y is:
(a)$\mathrm{(CH_3O)_2P(O)(OH)}$
(b)$\mathrm{(CH_3O)_3P(O)}$
(c)$\mathrm{(CH_3O)_2(CH_3)P(O)}$
(d)$\mathrm{(CH_3O)_2(CH_3)P(OH)}$
Answer
Answer: C ✓ checked by 4AB · confidence high
The source book printed B; on checking, C is correct — see the explanation.
Printed B; the isomer of P(OMe)₃ with a P–CH₃ doublet is the phosphonate.
Explanation
Heating trimethyl phosphite (X) causes the Arbuzov rearrangement to dimethyl methylphosphonate, (MeO)₂P(O)CH₃ (Y), an isomer. Its ¹H NMR shows two doublets: P–CH₃ (²J_PH ≈ 17–20 Hz) and OCH₃ (³J_PH ≈ 11–15 Hz).