Acetone = A, n – butane = B, methoxy ethane = C, Protan – 1 – ol = D ; Propanal = E. Decreasing order of boiling point is
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a
B < C < E < A < D
b
D > A > E > C > B
c
D > E > A > B > C
d
B < C < A < E < D
answer is B.
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Detailed Solution
Intermolecular attactions in alcohol is H-boning.Intermolecular attactions in carbonyl compounds are dipole-dipole attractions.Ketones are more polar than aldehydes. Inter molecular attractions in ethers are weak dipole -dipole attraction.intermolecular attractions in hydrocarbons are very weak London dispersion forces.∴ Boiling point of (nearly same molecular weight) alcohols > ketones > aldehydes > ethers > alkane. But as molecular weight ethers increases they become more unsymmetrical. In such cases the dipole -dipole attractions are very weak. They are weaker than dispersion forces in alkanes.CH3CH2OH>CH3CHO>CH3-O-CH3>CH3CH2CH3 CH3CH2CH2OH>CH3CHOH>CH3>CH3COCH3>CH3CH2CHO>CH3-O-C2H5>CH3CH2CH2CH3but CH3CH2CH2CH2OH>CH3COCH2CH3>CH3CH2CH2CHO>CH3CH2CH2CH2CH3>C2H5-O-C2H5It is intereseting to note that boiling point of ethoxyethane (307K) < n-pentane(309K)