2.2 g of an alcohol (A) when treated with CH_3 -Mgl liberates 560 mL of CH_4 at STP. Alcohol (A) on dehydration followed by ozonolysis gives ketone (B) along with (C). Oxime of ketone (B) contains 19.17% N. (A) on oxidation gives ketone (D) having same number of carbon atom. Molecular mass of (A) is

2.2 g of an alcohol (A) when treated with CH_3 -Mgl liberates 560 mL o

1.	2.2 g of an alcohol (A) when treated with CH_3 -Mgl liberates 560 mL of CH_4 at STP. Alcohol (A) on dehydration followed by ozonolysis gives ketone (B) along with (C). Oxime of ketone (B) contains 19.17% N. (A) on oxidation gives ketone (D) having same number of carbon atom. Molecular mass of (A) is
Answer» 74 88 60 102 Solution :`ROH+CH_(3)-MGL rarr CH_(4)+ROMgl` Number of MOLES of `CH_(4)`= number of moles of alcohol `560/22400=2.2/M rArr`M=[where M =MOLECULAR weight ] Thus molecular formula of alcohol A is `C_(5)H_(11)OH`. Molecular mass of `= 14/19.17 xx 100 =73` `R_1+R_2=73-(12+14+16+1) =73-43=30` `R_1=R_2=15 CH_3` Thus ketone is `CH_3-underset(CH_3)underset(\|)C=O` and hence C is `CH_3-CH=O`. on this basic alkene is `CH_3-underset(CH_3)underset(\|)C=CH-CH_3`and alcohol A will be `CH_3underset(CH_3)underset(\|)"CH"-underset(OH)underset(\|)"CH"-CH_3` because it is `2^@` alcohole which on oxidationgives ketone (D) having same number of C -atom.