An organic compound A (C5H11Cl) is Optically Active and on treatment with ethanolic KOH solution yieldsB (C5H10) as Major product which does not show stereo-isomerism. Also A on treatment with (CH3)2 Cu yields C (C6H14) which is optically inactive. Deduce structures of A to C.
Molecular Formula of A = C5H11Cl
We need to find that whether compond A is saturated of unsaturated by using the formula to calculate double bond equivalent (D.B.E)
D.B.E = {(Ʃ n(v-2) / 2 } + 1
where, n = number of atoms of an element in the compound
v = valency of that element
D.B.E = [{ 5(4-2) + 11(1-2) + 1(1-2)} / 2] + 1
D.B.E = {(10-11-1) / 2} + 1
D.B.E = (-2/2) + 1
D.B.E = -1 + 1 = 0
Hence, compound A is a saturated aliphatic compound and since compound A is optically active it must have atleast one chiral carbon atom. There can be two possible structures for A :
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Therefore,
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