Double bond equivalents examples

From Cg/ZgNO (problem 4), for example, the empirical formula C Hjo is derived and compared with the alkane formula Cc>H2o, a hydrogen deficit of ten and thus of five double-bond equivalents is deduced. If the NMR spectra have too few signals in the shift range appropriate for multiple bonds, then the double-bond equivalents indicate rings (see, for example, a-pinene. Fig. 2.4). 

The last example was fully saturated but it is usually a help in deducing the structure of an unknown compound if, once you know the atomic composition, you immediately work out how much unsat-uration there is. This is usually expressed as double bond equivalents . It may seem obvious to you that, if C4H11NO has no double bonds, then C4H9NO (losing two hydrogen atoms) must have one double bond, C4H7NO two double bonds, and so on. Well, it s not quite as simple as that. Some possible structures for these formulae are shown below. 

Checking for octets, we find that both oxygen atoms have filled octets, but the sulfur has only six electrons. It needs two more, which can be obtained by forming a double bond. But on which side should we place this double bond Unlike Example Problem 7.6, here we are confronted with two equivalent choices for the position of the double bond. We could draw either of the two structures. 

For example, for C7H7ON, the formula gives 7 + 0.5 - 3.5 + 1 = 5 double bond equivalents. This corresponds to benzamide, C5H5CONH2, the aromatic ring being counted as three double bonds plus one for the ring. 

The hydrazone structure 40 can be eliminated at once many examples of this class of compounds are known and their properties are completely different from the diaziridines. For example, 3,3-dimethyldiaziridine has a heat of combustion of about 35 kcal higher than the isomeric acetone hydrazone. Further pairs of isomers of diaziridines and hydrazones are known. The spectrum eliminates both the hydrazone structure and the betaine structure 41. The diaziridines do not absorb in the UV range. In the infrared spectrum, absorption is completely absent in the double-bond region. - The NMR spectrum of 3,3-dimethyldiaziridine is in agreement with a formulation that has two equivalent iV-protons. ... 

Rule 2 If a decision can t be reached by ranking the first atoms in the substituent, look at the second, third, or fourth atoms away from the double-bond carbons until the first difference is found. A — CH2CH3 substituent and a -CH3 substituent are equivalent by rule 1 because both have carbon as the first atom. By rule 2, however, ethyl receives higher priority than methyl because ethyl has a carbon as its highest second atom, while methyl has only hydrogen as its second atom. Look at the following examples to see how the Tule works ... 

When an alkene reacts with BH3 in THF solution, rapid addition to the double bond occurs three times and a tricilkylborcme, R3B, is formed. For example, 1 molar equivalent of BH3 adds to 3 molar equivalents of cyclohexene to yield tricyclohexylborane. When tricvclohexylborane is then treated with aqueous hydrogen peroxide (H2C>2) in basic solution, an oxidation takes place. The three C-B bonds are broken, -OH groups bond to the three carbons, and 3 equivalents of cyclohexanol are produced. The net effect of the... 

In fl-trimethylsilylcarboxylic acids the non-Kolbe electrolysis is favored as the carbocation is stabilized by the p-effect of the silyl group. Attack of methanol at the silyl group subsequently leads in a regioselective elimination to the double bond (Eq. 29) [307, 308]. This reaction has been used for the construction of 1,4-cyclohexa-dienes. At first Diels-Alder adducts are prepared from dienes and P-trimethylsilyl-acrylic acid as acetylene-equivalent, this is then followed by decarboxylation-desilyl-ation (Eq. 30) [308]. Some examples are summarized in Table 11, Nos. 12-13. 

Scheme 6.15 gives some representative examples of the orthoester Claisen rearrangement. Entry 1 is an example of the standard conditions for the orthoester Claisen rearrangement using triethyl orthoacetate as the reactant. The allylic alcohol is heated in an excess of the orthoester (5.75 equivalents) with 5 mol % of propanoic acid. Ethanol is distilled from the reaction mixture. The E-double bond arises from the chair TS. 

Based on the few reported examples, the pattern of ring cleavage that accompanies the ionic hydrogenation of alkylidenencyclopropanes seems to be related to the pattern and degree of substitution on both the ring and the double bond.233 Thus, treatment of l,l-dimethyl-2-methylenecyclopropane with two equivalents of triethylsilane and four equivalents of trifluoroacetic acid for 90 hours at room temperature yields 65% of 2,3-dimethylbutane (Eq. 114).229 Exposure of 1,1-dimethyl-2-isopropylidenecyclopropane to the same ratio of reactants at 50° for 16 hours produces a complex mixture containing 63% of 2,5-dimethylhexane, 18.5% of 2,5-dimethyl-3-hexene, 1.6% of 2,5-dimethyl-2-hexene, and 7% of 2,5-dimethyl-2-hexyl trifluoroacetate (Eq. 115).229... 

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