Answer to Question #165380 in Organic Chemistry for Desmond

Question

An organic compound A made up of 87.8% Carbon and 12.2% Hydrogen; reacts with cold dilute KMnO₄ to yield D (C₆H₁₂O₂). A takes up only one mole of hydrogen gas to form B when hydrogenated. A also decolourises aqueous Bromine to form C. (RAM of C = 12, H = 1.00, O = 16)

1. In one experiment, 4.1g of A gave 4.0g of D. Calculate the percentage yield.
2. Compound A is found to be optically inactive. Explain
3. Refluxing D with acidified K₂Cr₂O₇ give E. Give the structure and name of E

Answer 1

C = 87.8%/12 = 7.32/7.32 = 1 × 3 = 3

H = 12.2%/1 = 12.2/7.32 = 1.67 × 3 = 5

Therefore the empirical formula of the compound "A" is C₃H₅

$A+ KMnO_4\to C_6H_{12}O_2\\ A + H_2 \to B\\ A + Br_2 \to C$

from the 3rd reaction above, A is an alkene

from the 1st equation, the compound that reacts with potassium permanganate to produce C₆H₁₂O₆ is cyclohexene.