Designing Exercises to Determine the Structural Formula of Organic Compounds Based on the Experimental Data

Abstract This paper introduces the way to design exercises to determine the structural formula of organic compounds based on the experimental data. They are structure, physical properties, typical chemical properties, synthesis methods and application. From that the authors have built the scheme to design exercises. On that basis, the authors compiled some exercises which are used for teaching and learning organic chemistry in university and high school.


Introduction
Designing exercises is essential in teaching chemistry, an indispensable activity of chemistry teacher at the universities and high schools.Organic chemistry exercises have more interesting contents, special exercises for identified structural formula of organic compounds will require students to have a basic knowledge of organic chemistry such as stereochemistry, reaction mechanisms or organic synthesis, etc [1].
Exploiting the experimental data to determine the structure of organic compounds are important skills to help students practice on scientific research in the university.Solving this exercises will help students to form experimental thinking in teaching of chemistry and research of organic chemistry.

Method -Desining Exercises
In order to determine the structural formula of organic compounds must first identify the components of qualitative and quantitative elements present in organic compounds, mean that molecular formula.Since it is based on experimental data on the molecular structure, chemical and physical properties of organic compounds and synthetic methods, we can argue to determine correctly structural formula of the organic compounds [2].
General scheme to design exercises as shown below.
-Homology of benzene.Summarizing some typical chemical properties in Table 1.

Sample Problems Example 1
From pepper isolated alkaloid piperine A (C 17 H 19 NO 3 ) is a neutral substance [3] For M Y -M X = 288 -282 = 6 → The molecular formula of Y compared to X to increase 6H after the Meerwein-Ponndorf-Verley (MPV) reduction [5], proving that X has 3 carbonyl groups (C = O).
We have transformed the following: b) Mechanism of forming X 1 from X according to Michael addition reaction through 2 steps: Step (1): NaNH 2 attacks on H α to form carbanion.
Step (2): Carbanion attack on carbon to produce ring with 6 sides.

Example 3
Following is the 1 H-NMR spectrum [6] for compound X of molecular formula C 7 H 14 O.
Propose a structural formula for compound X.

Solution:
From the 1 H-NMR spectrum to the structural formula for compound X.

-Examine the molecular formula, calculate the degree of unsaturation (DU), and deduce what information you can
about the presence or absence of rings or π -ponds.The molecular formula C 7 H 14 O that has DU = 1 → the compound contains 1 ring or 1 π -pond.

-Count the number of signals to arrive at a minimum number of different sets of hydrogens in the compound.
There are three signals, therefore, three sets of equivalent hydrogens.
-Use the line of integration and the molecular formula to determine the number of hydrogens in each set.Reading from right to left, signals are in the ratio 9H : 3H : 2H.-Examine the NMR spectrum for signals characteristic of the following types of hydrogens.Keep in mind that these are broad ranges and that hydrogens of each type may be shifted either farther upfield or farther downfield depending on details of molecular structure.
The singlet at δ 1.01 is characteristic of a methyl group adjacent to an sp 3 hybridized carbon.The singlets at δ 2.11 and 2.32 are characteristic of alkyl groups adjacent to a carbonyl group.
-Examine splitting patterns for information about the number of nonequivalent neighboring hydrogens.All signals are singlets (s).
-Write a structural formula consistent with the previous information.The compound X is 4,4-dimethylpentan-2-one.DU (degree of unsaturation) = 6, four of which are due to a benzene ring.
Since (A) does not dissolve in NaOH or give a color with FeCl 3 , it is not a phenol.
Formation of H 2 C=O on ozonolysis means that A has a chain with a terminal =CH 2 grouping, and since B is an aldehyde (positive Tollens' test), the grouping is -CH=CH 2 .The double bond accounts for the fifth DU.
E is a monocarboxylic acid formed by complete oxidation of the alkenyl side chain, and its molecular weight is 166.
No other C can be directly attached to the ring because it would have oxidized, giving a dicarboxylic acid.
The two remaining Os must be present as ether linkages probably present as a ring, the sixth DU, fused to the benzene ring.This is confirmed by isolating H 2 C=O and 3,4-dihydroxybenzoic acid on cleavage with HI.This fused ring is actually a stable acetal.
The molecular formula of E is C 8 H 6 O 4 .The acetal ring and the benzene ring account for seven Cs, leaving three Cs for the alkenyl side chain.The structures for A, B and E are: Example 5 A pleasantly smelling, optically active compound F (MW = 186) [9].It does not react with Br 2 in CCl 4 .Hydrolysis of F gives two optically active compounds, G, which is soluble in NaOH, and H. Compound H gives a positive iodoform test, and on warming with conc.H 2 SO 4 gives I with no diastereomers.When the Ag + salt of G is reacted with Br 2 , racemic J is formed.Optically active J is formed when H is treated with tosyl chloride (TsCl) and then NaBr.The IR spectrum of F shows a single C=O stretching peak.Give structures of F through J and explain your choices.
Solution: F is a saturated monoester with MW = 186 (no Br 2 reaction).We can logically determine the number of Cs and the molecular formula by subtracting the mass of the two Os and dividing the remainder by 14, the mass of CH 2 : (186 -32)/14 = 11.To complete the mass there must be 22 Hs.The molecular formula is C 11 H 22 O 2 .The acid G has one more C than the alcohol H because it is degraded by one C in the Hunsdiecker reaction [7,8] (RCOO -Ag + + Br 2 ) to J which is also made from H with no change in C content.H is a methyl carbinol, CH 3 CH(OH)R, because it gives a positive iodoform reaction and, in order to be chiral, must have at least four Cs.However, H has five Cs because the alkene, I, obtained on dehydration (warm conc.

Conclusions
Above are some examples of building exercises to determine the structure of organic compounds based on empirical data.Because empirical data on the structure is very rich so there are many forms of this type of exercise.In resolving the types of exercises will improve the students' thinking, especially the analysis of the reaction mechanism and the structure of organic compounds.

-
. Ozonolysis of compound A formed substances: ethanedial (ethane-1,2-dione), B, D. Hydrolysis of B formed OHC-COOH and six-member heterocyclic compounds piperidine (C 5 H 11 N).Given D reacts with concentrated solution of HI to form 4,5-dihydroxybenzandehyde. Identify structural formulae of A, B, D. How many stereoisomers of A are there?Solution: -Hydrolysis of B formed OHC-COOH and piperidine, inferred B to contain the O=C-N-bond (nitrogen atom belongs to six-member ring).Structural formula of B is Ozonolysis of compound A formed ethanedial, inferred A contains the =CH-CH= group.O O CH = CH -CH = CH N (A)

→
The oxidation ozonolysis of X to form the only compound Z which has M Z -M X = 32 → Proving that oxidation ozonolysis of X which had gained more 2 oxygen atoms to form Z. Therefore, X has a double bond (C = C) in the ring and in this double bond is no longer hydrogen atom.The structural formula of X, Y, Z, X 1 and X 2 are determined by the following diagram: H 2 SO 4 ), must have two Mes on one of the doubly bonded Cs to avoid cis-trans isomerism.I is CH 3 CH=C(CH 3 ) 2 with five Cs and G had six Cs.H is CH 3 CHOHCH(CH 3 ) 2 and is converted to J, CH 3 CHBrCH(CH 3 ) 2 , through the tosylate, CH 3 CH(OTs)CH(CH 3 ) 2 , with no change in configuration, by an S N 2 reaction with Br -.Consequently, H and J have inverted configurations.The skeleton of the alkyl group of G is the same as H. Replacing Br of J by COOH gives the structure of G, (CH 3 ) 2 CHCH(CH 3 )COOH.F is one of the four possible enantiomers of

'
17(A) + soln. of HNO 3 /H 2 SO 4 (conc.)→ yellow liquid that does Given an organic compound X is created by three chemical elements, has a molar mass of 282 g/mol.Mass analysis of X shows 17.02% oxygen.Complete combustion X formed moles of CO 2 twice as much as moles of H 2 O. Let X reacts with excess (i-C 3 H 7 O) 3 Al / i-C 3 H 7 OH to form compound Y which has a molar mass of 288 g/mol.When reduction or oxidation ozonolysis to form the only compound Z which has a molar mass of 314 g/mol.Reduction Z by NaBH 4 , then the resultant product is oxidized with NaIO 4 to form mixture of o-C 6 H 4 (CHO) 2 and OHC(CH 2 ) 5 CH(OH)(CH 2 ) 2 CHO.On the other hand, treatment X with NaNH 2 / DMF to form X 1 (C 18 H 18 O 3 ).Let X 1 reacts with H 2 / Pd, then heated resultant product with conc.H 2 SO 4 to give X 2 (C 18 H 20 O).
-Compound A has 2 double bonds C=C, so there are 4 geometric isomers: ZZ, EE, ZE, EZ.Example 2 a) Draw the structural formula of X, Y, Z, X 1 and X 2 .b)Explain the formation mechanism of X 1 from X.
C 10 H 10 O 2 , a sweet-smelling liquid isolated from oil sassafrass, given the following properties: -It does not dissolve in NaOH or give a color with FeCl 3 .-It adds 1 eq of H 2 on catalytic hydrogenation.Reductive ozonolysis affords H 2 C=O and B (C 9 H 8 O 3 ) that gives a positive Tollens' test.-Oxidation of A with KMnO 4 gives an acid E (MW = 166) which gives no color with FeCl 3 .-When E is refluxed with conc.HI, H 2 C=O and 3,4dihydroxybenzoic acid are isolated and identified.Deduce the structures of A, B and E.