Compound (A) C_(10)H_(22)0_(2) is insoluble in aq. NaOH bu not is NaHCO_(3). Treatment of (A) with DMSO (CH_(3)-overset(S)overset(||)(S)-CH_(3)) in alkali give (B) C_(11)H_(14)O_(2). Treatment of (A) with strong alkali alone give an isomeric compound (C). When (A) is reflux with HI, CH_(3)I is obtained, compound (B) is insoluble in alkali and decolurises Br_(2)//C Cl_(4). (B) on treating with strong base gives (D), an isomer of (B). Ozonolysis (C) of gives (E), C_(8)H_(8)O and isomer of vanilline. Ozolysis of (D) gives (F) C_(9)H_(10)O_(3), which is identical with product of methylation of vanilline (4-hydroxy-3-methoxy benzaldehyde). Compound (B) is:

Compound (A) C_(10)H_(22)0_(2) is insoluble in aq. NaOH bu not is NaHC

1.	Compound (A) C_(10)H_(22)0_(2) is insoluble in aq. NaOH bu not is NaHCO_(3). Treatment of (A) with DMSO (CH_(3)-overset(S)overset(\|\|)(S)-CH_(3)) in alkali give (B) C_(11)H_(14)O_(2). Treatment of (A) with strong alkali alone give an isomeric compound (C). When (A) is reflux with HI, CH_(3)I is obtained, compound (B) is insoluble in alkali and decolurises Br_(2)//C Cl_(4). (B) on treating with strong base gives (D), an isomer of (B). Ozonolysis (C) of gives (E), C_(8)H_(8)O and isomer of vanilline. Ozolysis of (D) gives (F) C_(9)H_(10)O_(3), which is identical with product of methylation of vanilline (4-hydroxy-3-methoxy benzaldehyde). Compound (B) is:
Answer» ANSWER :B