Asked by | 29th Dec, 2008, 04:11: PM
1. The compound contains 66.97% C , 11.63% H and 21.4% O. The atomic mass of C is 12, O is 16 and H is1. The molecualr formul of teh compound is 86.Therefore, the formula of the compound is C5H10O.
1. The compound reacts with sodium bisulphite to give an addition product. Therefore, it is a carbonyl compound.
2. The compound does not reduce tollen's reagent. Therefore, it is a ketone.
3. To identify the position of carbonyl group it is given that the compound gives a positive iodoform test. So, it is a methyl ketone. Also, it is given that on vigorous oxidation , it gives ethanoic acid and propanoic acid.
Therefore, the formula of the compound is CH3CH2CH2COCH3
Teh reactions are:
CH3CH2CH2COCH3 + NaHSO3 CH3CH2CH2C(OH)(OSO2Na)CH3
CH3CH2CH2COCH3 CH3COOH + CH3CH2CH2COOH
Answered by | 2nd Jan, 2009, 11:48: AM
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