Question 4.7: Identify degrees of unsaturation (index of hydrogen deficien......
Identify degrees of unsaturation (index of hydrogen deficiency) in the following.
(a) C _{17} H _{18} F _3 NO when treated with H _2 / Ni gives a structure with molecular formula C _{17} H _{30} F _3 NO . It contains no triple bonds. How many rings are in C _{17} H _{18} F _3 NO?
(b) C _{17} H _{18} FN _3 O _3 contains four rings and no triple bonds.How many double bonds does it contain?
(c) C _{28} H _{35} NO _2 contains five double bonds and one triple bond. How many rings are there in it?
(d) C _{17} H _{14} O _4 S contains three rings and no triple bonds. Howmany double bonds are there? (For each sulphur atom,four hydrogen atoms are subtracted to arrive at the equivalent hydrocarbon formula.)
(e) C _{19} H _{17} C\text{l}FN _3 O _5 S \text {, } contains eight double bonds. How many rings are present? (In this case, treat sulphur as you would treat oxygen.)
(f) C _{19} H _{20} FNO _3 upon exhaustive hydrogenation \left( H _2 / Pt \right) a compound C _{19} H _{32} FNO _3 is formed. How many double bonds and how many rings are in C _{19} H _{20} FNO _3 \text { ? }
(a) Molecular formula: C17H18F3NO Molecular formula after treatment: C17H30F3NO Number of rings: ? Number of triple bonds: 0
(b) Molecular formula: C17H18FN3O3 Number of rings: 4 Number of double bonds: ?
(c) Molecular formula: C28H35NO2 Number of rings: ? Number of double bonds: 5 Number of triple bonds: 1
(d) Molecular formula: C17H14O4S Number of rings: 3 Number of double bonds: ?
(e) Molecular formula: C19H17ClFN3O5S Number of rings: ? Number of double bonds: 8
(f) Molecular formula: C19H20FNO3 Molecular formula after hydrogenation: C19H32FNO3 Number of double bonds: ? Number of rings: ?
Final Answer
The degree of unsaturation (or index of hydrogen deficiency) is defined as the number of pairs of hydrogen atoms that must be subtracted from molecular formula of the corresponding alkane to get the compound. On hydrogenation, each double bond consumes one molar equivalent of hydrogen and each triple bond consumes two. Rings do not undergo hydrogenation at room temperature but each ring can be equated to one degree of unsaturation. For compounds containing halogen atoms, the halogen atoms are counted as though they were hydrogen atoms. For compounds containing oxygen, the oxygen atom is ignored and degree of hydrogen is calculated from the remaining formula. For compounds
containing nitrogen atoms, one hydrogen is subtracted for each nitrogen and the nitrogen atoms are then ignored.
(a) The degree of unsaturation:
C _{17} H _{36}- C _{17} H _{20(18+3-1)}= H _{16}= H _{16} / 2=8^{\circ} .Hydrogenation: C _{17} H _{30} F _3 NO – C _{17} H _{18} F _3 NO = H _{12}= H _{12} / 2=6 \text { double bond. }
Number of rings = 8 − 6 = 2
(b) Number of degree of unsaturation:
C _{17} H _{36}- C _{17} H _{16(18+1-3)}= H _{20}= H _{20} / 2=10^{\circ}Number of double bond = 10 − 4 rings = 6
(c) Number of degree of unsaturation:
C _{28} H _{58}- C _{28} H _{34(35-1)}= H _{24}= H _{24} / 2=12^{\circ} .Number of rings = 12 − 5 double bond−1 triple bond (= 2 double bond) = 5
(d) Number of degree of unsaturation:
C _{17} H _{36}- C _{17} H _{10(14-0-4)}= H _{26}= H _{26} / 2=13^{\circ} \text {. }Number of double bond = 13 − (3 rings) = 10
(e) Number of degree of unsaturation:
C _{19} H _{40}- C _{19} H _{16(17+2-3)}= H _{24}= H _{24} / 2=12^{\circ} \text {. }Number of rings = (12−8) double bond = 4
(f) Number of degree of unsaturation:
C _{19} H _{40}- C _{19} H _{20(20+1-1)}= H _{20}= H _{20} / 2=10^{\circ} \text {. }
\text { Hydrogenation: } C _{19} H _{32} FNO _3- C _{19} H _{20} FNO _3= H _{12}= H _{12} / 2=6 \text { double bond }Number of rings = (10−6) double bond = 4
