Insecticides are substances preventing us from insects by destroying, repelling or mitigating them. — Organic Chemistry Chemistry Question
Chrysanthemic acid
Insecticides are substances preventing us from insects by destroying, repelling or mitigating them. The use of insecticides is one of the major factors behind the increase in agricultural productivity in the 20th century. Insecticides are also used in medicine, industry and housekeeping. Natural insecticides, such as nicotine and esters of chrysanthemic acid, are produced in plants. On the contrary to nicotine, esters of chrysanthemic acid are non-toxic to man and other mammals.
Many methods for chrysanthemic acid synthesis have been described to date. Two of these are presented in the hereunder scheme (the first step of both methods is the reaction discovered in 1905 by the Russian chemist A. Favorskii).
Write down the structural formulae of all compounds given in this scheme. Note that A is a gaseous hydrocarbon with the density lower than that of air, G is a natural alcohol, F’ is a mixture of isomers, whereas F’’ is formed only in trans-form.
[VISUAL]
Methods given in the scheme provide chrysanthemic acid as a mixture of stereoisomers, while natural chrysanthemic acid has (1R,3R)-configuration.
Model Answer
The compounds in the scheme are:
- A: acetylene (ethyne), C2H2
- B: acetone (propan-2-one), CH3COCH3
- C: 2,5-dimethylhex-3-yne-2,5-diol
- D: 2,5-dimethylhexane-2,5-diol
- E: 2,5-dimethylhex-2,4-diene, (CH3)2C=CH-CH=C(CH3)2
- F': ethyl chrysanthemate (mixture of isomers)
- G: prenol (3-methylbut-2-en-1-ol), (CH3)2C=CH-CH2OH
- H: prenyl bromide (1-bromo-3-methylbut-2-ene), (CH3)2C=CH-CH2Br
- I: mesityl oxide (4-methylpent-3-en-2-one), (CH3)2C=CH-CO-CH3
- J: senecioic acid (3-methylbut-2-enoic acid), (CH3)2C=CH-COOH (or sodium 3-methylbut-2-enoate)
- K: ethyl 3-methylbut-2-enoate, (CH3)2C=CH-COOCH2CH3
- L: 2-methylbut-3-yn-2-ol
- M: 2-methylbut-3-en-2-ol
- N: prenyl 4-toluenesulfone (prenyl p-tolyl sulfone)
- F'': ethyl trans-chrysanthemate
Chemical transformations described:
1. Acetylene (A) reacts with acetone (B) in a 1:2 ratio to form C, which is hydrogenated to D, followed by acid-catalyzed dehydration with heating to give E.
2. Diene E reacts with ethyl diazoacetate to give ethyl chrysanthemate F' as a mixture of isomers, which upon hydrolysis yields chrysanthemic acid.
3. Alternatively, E can be formed through the Grignard reaction of prenyl bromide (H) with acetone followed by elimination of water.
4. Acetylene (A) and acetone (B) react in a 1:1 ratio to form L, which is hydrogenated to M. M is reacted with HBr to give H (with allylic rearrangement), which is then converted into sulfone N with sodium 4-toluenesulfinate.
5. Self-condensation of acetone yields mesityl oxide I. Haloform (iodoform) reaction of I produces senecioic acid J, which is esterified with ethanol to K. The reaction of K with deprotonated sulfone N results in ethyl trans-chrysanthemate F''.
Write down the structural formulae of natural chrysanthemic acid.
Model Answer
Natural chrysanthemic acid has the (1R,3R)-configuration. The structure is (1R,3R-2,2-dimethyl-3-(2-methylprop-1-en-1-yl)cyclopropanecarboxylic acid, which features a trans-relationship between the carboxylic acid group and the isobutenyl substituent on the cyclopropane ring.
Tetramethrin is a key substance of many household insecticides. This compound belonging to pyrethroids of the 1st generation can be obtained by esterification of chrysanthemic acid with alcohol X. Synthesis of the latter is given below.
[VISUAL]
Write down the structural formulae of O-R, and X. Note that the transformation of O into P is an isomerization with retention of the carbocyclic skeleton leading to the most stable isomer.
Model Answer
The compounds involved are:
- O: cyclohex-4-ene-1,2-dicarboxylic anhydride (1,2,3,6-tetrahydrophthalic anhydride), formed by Diels-Alder reaction of butadiene and maleic anhydride.
- P: cyclohex-1-ene-1,2-dicarboxylic anhydride (3,4,5,6-tetrahydrophthalic anhydride), which is the most stable isomer of O with a tetrasubstituted double bond.
- R: cyclohex-1-ene-1,2-dicarboximide (3,4,5,6-tetrahydrophthalimide), obtained by heating P with ammonia.
- X: N-(hydroxymethyl)cyclohex-1-ene-1,2-dicarboximide [N-(hydroxymethyl-3,4,5,6-tetrahydrophthalimide], formed by reaction of R with formaldehyde (CH2O).
Which of the following acid derivatives could easily form esters in reaction with alcohols?
a) anhydride; b) methyl ester; c) amide; d) hydrazide
Model Answer
The correct answers are a) anhydride and b) methyl ester.
Explanation: Amides (c) and hydrazides (d) are highly stable and do not readily react with alcohols to form esters. On the other hand, acid anhydrides (a) are highly reactive acylating agents and easily form esters. Methyl esters (b) can easily undergo transesterification (re-esterification) with high-boiling alcohols when the equilibrium is shifted by distilling off the volatile methanol.
The 1stgeneration pyrethroids are photochemically unstable, which stimulated development of new types of pyrethroids (of the 2nd and 3rd generations). In particular, substitution of the CH=C(CH3)2 fragment in chrysanthemic acid by the CH=CHal2 moiety increases photostability of pyrethroids. Thus, three compounds (cis-permethrin, Y, cypermethrin, Z, and deltamethrin, W) were prepared from cis-2-(2,2-dihalovinyl-3,3-dimethylcyclopropane-1-carboxylic acid and 3-phenoxybenzaldehyde according to the scheme below.
[VISUAL]
Write down the structural formulae of S, T, W, Y, Z. Note that the halide content in W, Y, Z is 31.6, 18.1, and 17.0 %, respectively.
Model Answer
The compounds are:
- S: (3-phenoxyphenyl)methanol (3-phenoxybenzyl alcohol), formed by the reduction of 3-phenoxybenzaldehyde.
- T: 2-hydroxy-2-(3-phenoxyphenyl)acetonitrile (3-phenoxybenzaldehyde cyanohydrin), formed by reaction of 3-phenoxybenzaldehyde with NaCN.
- Y: cis-permethrin (C21H20Cl2O3), formed by esterification of S with cis-2-(2,2-dichlorovinyl-3,3-dimethylcyclopropanecarboxylic acid. Calculated chlorine content: 18.12%.
- Z: cypermethrin (C22H19Cl2NO3), formed by esterification of T with cis-2-(2,2-dichlorovinyl-3,3-dimethylcyclopropanecarboxylic acid. Calculated chlorine content: 17.03%.
- W: deltamethrin (C22H19Br2NO3), formed by esterification of T with cis-2-(2,2-dibromovinyl-3,3-dimethylcyclopropanecarboxylic acid. Calculated bromine content: 31.63%.
Halide Content Calculation:
Esters of S have the general formula C21H20Hal2O3, wt% Hal = 2*M_Hal / (2*M_Hal + 320.3).
Esters of T have the general formula C22H19Hal2NO3, wt% Hal = 2*M_Hal / (2*M_Hal + 345.3).
Using M_Cl = 35.45 g/mol:
- S ester: 70.9 / (70.9 + 320.3) = 18.12% → matches Y (18.1%)
- T ester: 70.9 / (70.9 + 345.3) = 17.03% → matches Z (17.0%)
Using M_Br = 79.90 g/mol:
- T ester: 159.8 / (159.8 + 345.3) = 31.63% → matches W (31.6%)