Chemical Synthesis and Its Various Mechanisms - Essay Example

www.academia-research.com Sumanta Sanyal d: 16/09/07 Chemistry: Mechanism and Synthesis Introduction This paper contrives to answer four questions specific to chemical synthesis and its various mechanisms. This is assignment S205 06 covering book 10. The questions are included in this paper to conserve brevity and only the answers are posited. Answer 1: (Part a): The reagents that may be needed to initiate the transformation in step a of scheme 1 are listed as hereunder: RMgX(a) RLi RNa Note: For the first reagent, X denotes the halides Cl, Br or I. (Taylor, p. 104) The reagents that may be needed to initiate the transformation in step b of scheme 1 are listed as hereunder: RMgX RLi Note: 1) The X in the first reagent, as before, denotes the halides Cl, Br, or I. 2) The R here, in the case of this set of reagents, is either an alkyl or aryl group. (Taylor, p. 111) (Part b): A detailed mechanism for step b is as follows: Diagram 1: Note to the above diagram: Though, in scheme 1, the starting material, as per diagram 1, is shown as an ethyl aldehyde, replacement of the ethyl group with a nitrile group is more difficult than replacing a methyl group with it. Thus, a methyl aldehyde is taken as starting material and made to react with an ethyl group Grignard’s reagent such that a secondary alcohol is formed. Next, FGI agents are used to convert the methyl group to a nitrile one and the final product – scheme 1, product 2 – is formed. Question 2: (Part a): 1) can be synthesised from using Grignard’s Reagent (Taylor, p. 86, 2002). 2) can also be synthesised from using an organolithium compound. In the first step of the reaction, the organolithium acts as a base with the acid to deprotonate it and form a hydrocarbon and the carboxylate salt. In the second step, it acts as nucleophile to attack the carbon atom of the carbonyl group and form the lithium salt. This lithium salt is undergoes acid hydrolysis to form the pentane-2-diol, the hydrate of the ketone, and this, in the absence of the organolithium any excess of which is destroyed by the addition of water, readily decomposes to form the ketone (Taylor, p. 95, 2002). 3) can be synthesised from using an organocopper (lithium dialkyl copper specifically) reagent. The reagent acts as a source for , that acts as a nucleophile and replaces the leaving group in the halide. This forms the ketone. The organocopper is not strong enough to attack the ketone and the reaction stops here (Taylor, p. 103, 2002). (Part b): Diagram 2: (Source: Taylor, p. 86) Note to the above diagram: Grignard reagents usually react with carboxylic acid derivatives to form ketones as intermediate substances but ketones cannot be prepared in this manner because they react further with more Grignard reagents to form alcohols. Usually, to prepare ketones, a less reactive organocopper reagent that reacts with the carboxylic derivative but not with the ketone is used (Taylor, p. 84, 2002). In this case, is a nitrile with a functional group that has similar polarisation characteristics to the carbonyl group. Thus, it can undergo addition reaction with the Grignard reagent and form a magnesium salt of an imine. It is notable that the salt has no leaving group and is also negatively charged and does not react further with the Grignard reagent. Thus, it is treated with aqueous acid and the excess Grignard reagent is destroyed and the salt is now converted to the imine – pentane-2-imine. The imine is unstable in the aqueous acidic conditions and readily hydrolyses to the ketone (Taylor, p. 85-86, 2002). Question 3: (Part a): This is the least stable radical as the relevant carbocation is flanked on either sides by other carbocations while only one side is somewhat stabilised by the alkyl electron-releasing group (Taylor, p. 126, 2002). This radical is second in being least stable as the relevant group is flanked by a carbocation and only one stabilising alkyl electron-releasing group (Taylor, p. 126, 2002). This radical is second in stability as the relevant group is flanked on one side by an electron-releasing alkyl group and on the other by an electron-withdrawing group (Taylor, p. 126, 2002). This radical is the most stable as it is a primary alkyl group (Taylor, p. 126, 2002) (Part b): The technical name of the more commonly known pinacol reaction is the ‘radical-radical coupling reaction’. The product formed in this case is the commonly called ketyl radical anion (diagram 3 below) (Taylor, p. 134, 2002). The alkenealkyl ketone is converted by metal application to the ketyl radical and it is assumed that dimerisation of the radical is effected to produce the 1,2-diol by using a solvent (protic or aprotic) with the metal (Taylor, p. 133, 2002) Diagram 3: (Taylor, p. 133, 2002). (Part c): In this case in the ‘McMurry Reaction’ (a type of radical-radical coupling) titanium metal dissolved in an aprotic solvent is used to form the alkene. The active titanium is formed by reducing titanium trichloride with lithium aluminium hydride (Taylor, p. 136, 2002). (Part d): 1) -azo-bis-isobutyronitrile or AIBN is an azo compound with a double nitrogen bond. It readily undergoes thermal homolysis to produce two nitrile-stabilized radicals that make it easier to abstract a hydrogen atom from the tributyltin hydride (). Though the tin-hydrogen bond of the butyl compound is weak it is not weak enough to dissociate under normal thermal conditions. The AIBN thus assists in easily extracting the hydrogen atom that is needed to replace the bromine one in the halo-ketone. Thus, AIBN is used here as an ‘initiator’ and the low molar equivalent is because of it initiates the reaction with a few radicals after which the tributyltin radicals take over and carry on the reaction (Taylor, p. 151, 2002). 2) The mechanism of the reaction in scheme 2 is given below. As stated earlier the reaction starts off with the homolysis of the initiator AIBN into two nitrile stabilized radicals. This is as per diagram 4 below. Nitrogen gas is let off (Taylor, p. 150, 2002). Diagram 4: (Taylor, p. 150, 2002) In the next stage of the reaction the 2-cyanopropyl radical extracts a hydrogen atom from the tributyltin hydride reagent to form a tin-centred radical as per diagram 5 below. Diagram 5: (Taylor, p. 150, 2002) It is notable now that the bromine-carbon bond enthalpy is much less than the carbon-hydrogen one. Diagram 6: Thus, as per diagram 6 above, the tin-centred radical will take the bromine atom from the alkene ketone compound and form a bond by giving one electron itself and taking another from the broken carbon-bromine bond. The broken carbon-bromine bond generates a carbon-centred radical that reacts with another molecule of tributyltin hydride and extracts a hydrogen atom from it to form a carbon-hydrogen bond. Thus, in this process, another tin-centred radical is formed and the reaction goes on as before without any further aid from the initiator. This is the main reason why AIBN is used in such low molar equivalent quantities (Taylor, p. 152, 2002). Thus, ultimately, the bromine atom from the alkene ketone is exchanged for a hydrogen atom. 3) The tin-mediated reaction here in scheme 2 has a high production yield because of the following five factors. 1. The AIBN initiator is needed in very small quantities. 2. It is highly efficient even at low thermal energy (>) conditions. 3. The reaction, once initiated by AIBN, goes on indefinitely till all the halogen-hydrogen conversion has been completed. This is because the AIBN only initiates the reaction and thereafter the tin-centred radical takes over the momentum of the reaction till the end. Thus, the reaction is a chain one driven by the tin-centred radicals. 4. Though a number of radicals are present within the reaction at any one time, there is no alternate products that are formed at any time. This ensures that all energy put into the reaction is used efficiently to produce the desired product. 5. The halogens to be substituted in this chain reaction are the less electrophilic bromine and iodine while chlorine and fluorine are avoided. The carbon-bromine or –iodine bonds are weak and can be easily broken for halogen-hydrogen substitution. (Taylor, p. 150-155, 2002) Question 4: (Part a): The synthesis is absolutely linear. The products are being formed in direct sequence from the ones before (Taylor, p. 249, 2002). The overall yield of synthesis is 0.103275. This is derived from calculating percentages of the percentages of the previous reaction products (Taylor, p. 249, 2002). The very low yield is conformant with the view that linear synthesis tends to diminish the yield dramatically which is primarily why synthesising efforts tend to be in the direction of convergent synthesis (Taylor, p. 249, 2002). (Part b): In scheme 3 the product 17 is an alkynide that tends to decompose into the following synthons as per diagram 7 below. Diagram 7: The synthon with the metallic component and an overall negative charge is a carbon nucleophil. Since the ion is a very strong base the synthon it is attached to, the nucleophil, is also a very strong base. Thus, a strongly basic organo-halide is necessary to replace the ion to form the product 18 in scheme 3 (Taylor, p. 28, 2002). This is done in such a manner that a nucleophilic halide part attaches to the nucleophilic synthon while a strongly acidic electrophilic part takes the ion away (Taylor, p. 28, 2002). (Part c): When there is possibility of producing stereo-isomers in a reaction, it sometimes necessary to create conditions where one isomer is produced in preference to another, if not in full proportion then at least in the highest possible proportions. This is known as stereoselectivity (Taylor, p. 241, 2002). In scheme 3, the step 6 is an example of stereoselectivity where the keto alcohol group is replaced by a hydroxyl one to produce the correct stereometric configuration so that in step 7 the ketone carbonyl group has the correct final stereometric configuration and produces the final product 23. (Part d): Steps 4 & 7 demonstrate examples of functional group conversion (FGI). In the first instance the cyanide group is converted to a keto alcohol while in the second one the hydroxyl group has been converted to a ketone carbonyl group. For the first FGI conversion either or can be used as conversion reagent as per Item 35, Table A.2, Taylor, p. 258). For the second FGI conversion together with conc. may be used as reagent. Other reagents that may used for the same conversion can found as per Item 5, Table A.2, Taylor, p. 255. Reference: Taylor, Peter (Ed.), The Molecular World, Mechanism and Synthesis, 2002, The Open University, Walton Hall, Milton Keynes, MK7 6AA. Read More