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Question
2.2 g of an alcohol (A) when treated with CH3-MgI liberates 560 mL of CH4 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 .
Structure of (A) is ?
Structure of (A) is ?
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Solution
The correct option is B CH3−C|CH3H−CH−C|OHH−CH3
Given that:A(ROH)2.2g+CH3MgI→CH4560mL at STPA(ROH)(i)H+(ii)Ozonolysis→B(ketone)+CB+NH2OH→O(oxime)19.17NA(ROH)[O]→D(ketone560 mL of CH4 at STP:number of moles of CH4, n=560/22400=0.025 mol1 mol of CH4 will be obtained by reaction of 1 mol of CH3MgI with 1 mol of Athus 0.025 mol of CH4 is obtained from 0.025 mol of Aif 0.025 mol of A=2.2gmass of 1 mol of A=2.2/0.025=88g/molMolecular formula of an alcohol=CnH2n+1OHmolar mass of alcohol=12n+2n+1+16+1For A, molar mass=88,thus, 12n+2n+1+16+1=88and n=5Thus A=C5H11OHSince A on dehydration will form an alkene which on ozonolysis gives a ketone. Therefore the alkene formed should have two alkyl groups attached on one of the doubly bonded carbon so as to get a ketone. Thus possible structure of alkenes are:CH2=C|CH3−CH2−CH3
CH3−C|CH3=CH−CH3
Thus, possible structure of A are:OH|CH2−C|CH3−CH2−CH3 OR CH3−OH|C|CH3−CH2−CH3CH3−C|CH3H−OH|CH−CH3Thus possible B are:O=C|CH3−CH2−CH3
CH3−C|CH3=O
and corresponding oxime are:HON=C|CH3−CH2−CH3=X
CH3−C|CH3=NOH=Ypercent N in X is:(mass of N/molar mass of X)×100=14(12×4+9+16+14)×100=16.1%percent N in Y is(:mass of N/molar mass of Y)×100=14(12×3+7+16+14)×100=19.18%
Therefore according to given data oxime obtained is CH3−C|CH3=NOHfrom ketone =B= CH3−C|CH3=Oand possible alcohol are:CH3−OH|C|CH3−CH2−CH3
CH3−C|CH3H−OH|CH−CH3Given that A gives ketone D on oxidation and with same number of carbons, thus it can only be a secondary alcohol as tertiary alcohol is resistant to oxidation. Thus A can only be:CH3−C|CH3H−OH|CH−CH3
overall reaction can be represented as:
CH3−C|CH3H−OH|CH−CH32.2g+CH3MgI→CH4560mL at STPCH3−C|CH3H−OH|CH−CH3(i)H+(ii)Ozonolysis→CH3−C|CH3=O+O=CHCH3CH3−C|CH3=O+NH2OH→CH3−C|CH3=NOH19.17NCH3−C|CH3H−OH|CH−CH3[O]→CH3−C|CH3H−O||C−CH3
Given that:
A(ROH)2.2g+CH3MgI→CH4560mL at STP
A(ROH)(i)H+(ii)Ozonolysis→B(ketone)+C
B+NH2OH→O(oxime)19.17N
A(ROH)[O]→D(ketone
560 mL of CH4 at STP:
number of moles of CH4, n=560/22400=0.025 mol
1 mol of CH4 will be obtained by reaction of 1 mol of CH3MgI with 1 mol of A
thus 0.025 mol of CH4 is obtained from 0.025 mol of A
if 0.025 mol of A=2.2g
mass of 1 mol of A=2.2/0.025=88g/mol
Molecular formula of an alcohol=CnH2n+1OH
molar mass of alcohol=12n+2n+1+16+1
For A, molar mass=88,
thus, 12n+2n+1+16+1=88
and n=5
Thus A=C5H11OH
Since A on dehydration will form an alkene which on ozonolysis gives a ketone. Therefore the alkene formed should have two alkyl groups attached on one of the doubly bonded carbon so as to get a ketone. Thus possible structure of alkenes are:
CH2=C|CH3−CH2−CH3
CH3−C|CH3=CH−CH3
Thus, possible structure of A are:
OH|CH2−C|CH3−CH2−CH3 OR CH3−OH|C|CH3−CH2−CH3
CH3−C|CH3H−OH|CH−CH3
Thus possible B are:
O=C|CH3−CH2−CH3
CH3−C|CH3=O
and corresponding oxime are:
HON=C|CH3−CH2−CH3=X
CH3−C|CH3=NOH=Y
percent N in X is:(mass of N/molar mass of X)×100=14(12×4+9+16+14)×100=16.1%
percent N in Y is(:mass of N/molar mass of Y)×100=14(12×3+7+16+14)×100=19.18%
Therefore according to given data oxime obtained is CH3−C|CH3=NOH
from ketone =B= CH3−C|CH3=O
and possible alcohol are:
CH3−OH|C|CH3−CH2−CH3
CH3−C|CH3H−OH|CH−CH3
Given that A gives ketone D on oxidation and with same number of carbons, thus it can only be a secondary alcohol as tertiary alcohol is resistant to oxidation. Thus A can only be:
CH3−C|CH3H−OH|CH−CH3
overall reaction can be represented as:
CH3−C|CH3H−OH|CH−CH32.2g+CH3MgI→CH4560mL at STP
CH3−C|CH3H−OH|CH−CH3(i)H+(ii)Ozonolysis→CH3−C|CH3=O+O=CHCH3
CH3−C|CH3=O+NH2OH→CH3−C|CH3=NOH19.17N
CH3−C|CH3H−OH|CH−CH3[O]→CH3−C|CH3H−O||C−CH3
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