With every breath, I breathe in so much of inspiration. I feel if there is one thing as free and as — Organic Chemistry Chemistry Question
Oxidation and inspiration
With every breath, I breathe in so much of inspiration.
I feel if there is one thing as free and as important as oxygen, it's inspiration.
(Sharad Sagar)
Oxygen is not only chemical element but also the essential element of life. Its most stable form, dioxygen, O2, constitutes 20.8% of the volume of the Earth’s atmosphere. All forms of oxygen are able to oxidize various compounds, especially organic ones. The type of oxygen-containing functionalities in organic compounds could be efficiently determined by IR spectroscopy. Thus, C=O groups are characterized by intense absorption at 1750 - 1660 cm–1 and O–H group is characterized by absorption in the region of 3600-3000 cm–1.
The most typical oxidation of saturated hydrocarbons with O2 is the burning process. However, some reaction have been developed producing various oxygenated products. For example, Gif reaction (which received this name in honor of Gif-sur-Yvette where Prof. Derek Barton was working when he disclosed this transformation) represents the oxidation of saturated hydrocarbons with air oxygen at room temperature. Thus, the oxidation of adamantane leads to three products:
[VISUAL]
Products X and Y are isomers. The compound Y can be easily transformed to Z. IR spectra of the compounds X and Y contain a band around 3300 cm–1, while IR spectrum of compound Z contains band 1720 cm–1.
Write down the structural formulae of compounds X–Z.
Model Answer
From the IR spectroscopy data it is possible to conclude that adamantane is oxidized to two alcohols (X, Y) and one ketone (Z). The adamantane molecule has two non-equivalent carbon atoms – secondary one and tertiary one. Thus, we can conclude that one of the products is adamantan-1-ol, the second one is adamantan-2-ol. Only the last compound can be oxidized to ketone, adamantan-2-one (Z).
Without catalysts O2 has, fortunately, very low reactivity. Otherwise, all living organisms should be oxidized by air. A wide variety of oxidizing reagents have been developed for selective or undiscriminating oxidation of various functional groups in high yield. Below 7 organic molecules are given. They were synthesized by oxidation with 7 indicated reagents.
[VISUAL]
Products
Oxidants
a Me2SO + (COCl)2, then Et3N (Swern oxidation)
b CrO3 + H2SO4; acetone (Jones oxidation)
c SeO2
d KMnO4, 20 C, pH 7-7.5
e KMnO4, H2SO4,
f meta-chloroperbenzoic acid (mCPBA)
g Ag(NH3)2OH, then H+
These oxidation reactions proceed without C–C cleavage.
Put the correspondence between the products and the oxidants and determine the starting organic compounds.
Model Answer
The oxidation with KMnO4 at room temperature and pH 7-7.5 (system d) is the well-known process of the alkene dihydroxylation. The only compound containing 1,2-diol moiety is the compound G. It is the first “reagent-product” pair. System a (Swern oxidation) is used for the oxidation of alcohols to aldehydes or ketones. The compound F is the only product containing one of two of these functionalities. It is a second pair. The epoxide A can be formed by epoxidation of the corresponding alkenes with mCPBA only (reagent f). Other reagents are inappropriate for the preparation of this compound. Compound D contains hydroxyl group. It allows for excluding all oxidants except c and g. However, SeO2 is used for allylic oxidation of alkenes and related oxidations of ketones. It is not this case. Therefore, we can conclude that D was synthesized by oxidation of hydroxyaldehyde by Ag(NH3)2OH (system g). In turn, SeO2 (reagent c) was used for synthesis of allyl alcohol C. Two remaining products are B and E. Two reagents are KMnO4/H2SO4 under heating (system e) and CrO3/H2SO4 in acetone (system b). Even if we do not know about the Jones oxidation, we know that the system e should oxidize methyl groups in toluene derivatives. Therefore, phthalic acid (E) was formed by oxidation with system e and 2-methylbenzoic acid (B) was obtained from 2-methylbenzaldehyde or 2-methylbenzyl alcohol by Jones oxidation.
Therefore, 7 pairs are: A – f; B – b; C – c; D – g; E – e; F – a; G – d.
Examples of the chemoselective oxidation of the same substrate with different oxidants affording different reaction products are given below.
[VISUAL]
Compounds I, J, and K react with Ag(NH3)2OH solution producing the metallic silver precipitation. Compound H could be formed by oxidation of J. The treatment of 1.44 g of L with metallic sodium produces 0.224 L (p = 1 atm, T = 273 K) of hydrogen gas.
Write down the structural formulae of compounds H–L.
Model Answer
The molecular formula of the initial compound is C14H24O5. It contains the protected aldehyde group and three different alcohol groups: primary, secondary and tertiary ones; two of them are located in vicinal positions, i.e., form a 1,2-diol system. This system is known to be oxidized with NaIO4 producing compound J (C14H22O5) containing two carbonyl functions. Compound H has two hydrogen atoms less but two oxygen atoms more than compound J and can be obtained by oxidation of this compound. It allows for concluding that H is the corresponding diacid. Compounds I, K, L have both hydroxyl and carbonyl groups. The presence of two bands at 1730 and 1715 cm–1 indicates that compound I has two different carbonyl groups. Molecular formula of I differs from that of the initial compound by 4 hydrogen atoms. These data allow to conclude that I is the product of oxidation of primary and secondary alcohols to aldehyde and ketone, respectively. Compound K contains two hydrogen atoms more, than compound I. Therefore, only one alcohol group was oxidized to the carbonyl moiety. The selection of group is unambiguously determined by the fact that K is oxidized by Ag(NH3)2OH, i.e., it contains the aldehyde group. Therefore, only primary alcohol is oxidized.
Let us analyze the last product L (C14H24O6). Metallic sodium could react with alcohols and carboxylic groups. This –COOH group could be formed from primary alcohol or by C–C cleavage of the 1,2-diol. In both cases there is a loss of hydrogen atoms, however final product have the same number of hydrogen atoms as the initial substrate. It allows for concluding that there are no carboxylic groups in the molecule. Thus, compound L contains 4 –OH groups according to the quantity of H2 gas. The 4-th hydroxy group could be only formed by the opening of the ketal ring. Accounting for molecular formula, we can write structure of this compound.