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Question

(A) The increase in molecular weight upon oxidation is $60 - 44 = 16$. This indicates the addition of one oxygen atom $(16 	ext{ amu})$ to the origin

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An aldehyde

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(A) The increase in molecular weight upon oxidation is $60 - 44 = 16$. This indicates the addition of one oxygen atom $(16 	ext{ amu})$ to the original molecule. This transformation is characteristic of the oxidation of an aldehyde to a carboxylic acid $(RCHO \xrightarrow{[O]} RCOOH)$. For an aldehyde with molecular weight $44$,the formula is $CH_3CHO$ $(12+3+12+1+16+1 = 44)$. Upon oxidation,it forms acetic acid $(CH_3COOH)$,which has a molecular weight of $60$ $(12+3+12+16+16+1 = 60)$. Therefore,the original compound is an aldehyde.

$A$ compound,containing only carbon,hydrogen,and oxygen,has a molecular weight of $44$. On complete oxidation,it is converted into a compound of molecular weight $60$. The original compound is:

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