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In benzoic acid, when it has a deactivating group in position "para" (relative to carboxylate position) , it attracts electrons towards it, weakening the acid and an activator pushes the electrons making the acid stronger.
What happens to this benzoic acid if it has a substituent at the "meta" or "ortho" position as an activator or a deactivator?
I need an explanation of what happens in each case please
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- Alcohols react with sulfonyl chlorides to form sulfonate esters. Only the O-H bond of the alcohol is broken in the reaction, and so no inversion of configuration occurs. The resulting sulfonate esters are reactive in SN1 and SN2 reactions since the sulfonate group is a very weak base and is therefore a good leaving group.Draw curved arrows to show the movement of electrons in this step of the mechanism.NO2 being a deactivating group, increase the acidic nature of alcohol, by increasing the stability of the corresponding conjugate base. True FalseThe product in this reaction is basic enough to be protonated by a dilute HCl solution. Draw the protonated species, clearly showing where protonation occurs. Draw all possible resonance structures of the conjugate acid of the product, and use these to explain why the product is so much more basic than a typical ester, like ethyl acetate.
- Using your knowledge of general chemistry and organic chemistry, answer the following questions: Predict the relative acidities within each of the following groups, then identify the conjugate bases, and finally predict their relative leaving group abilities HClO3 and HClO2 PH3 and H2S [NH4]+ and [H3O]+ Explain why the trifluoromethane sulfonate (TfO-) ion is a better leaving group than the methanesulfonate (MsO-) ion.Show the interconversion of the functional group and where the compound disconnects (show where the disconnection occurs). And show the donor synthone and acceptor synthone and the appropriate reagents for the compoundThe trimethylsilyl protecting group is easily removed in aqueous solution containing a trace of acid. Propose a mechanism for this reaction. (Note that a TBDMS protecting group is stable under these conditions because of the greater steric crowding around silicon created by the t-butyl group.)
- Enolates are formed by deprotonation of an α-carbon hydrogen. Answer the following questions about enolate formation. In the molecule shown, select the α-carbon hydrogen that would be removed to form an enolate when NaOEt is used as a base. Draw the thermodynamic enolate that results for the molecule in Part 1. Draw only the enolate resonance form that includes a formal charge on the α carbon. Be sure to indicate that formal charge as well as any lone pair of electrons in your answer.Which of the following is a correct statement regarding electrophilic aromatic substitution? A. Many suitable electrophiles are unreactive and can be stored for long periods of time prior to use. B. The carbocation intermediate will lose a proton to regain aromaticity, usually from a position other than the site of the electrophilic attack. C. Formation of the carbocation intermediate has a high activation barrier due to loss of aromaticity. D. Reformation of the aromatic ring has a low activation barrier and therefore occurs slowly. E. The carbocation intermediate has several resonance structures and is negatively charged.Enolates are formed by deprotonation of an α-carbon hydrogen. Answer the following questions about enolate formation. In the molecule shown, select the α-carbon hydrogen that would be removed to form an enolate when sodium hydroxide is used as a base. (use photo to answer this question). Then, Draw the thermodynamic enolate that results for the molecule in Part 1. Draw only the enolate resonance form that includes a formal charge on the α carbon. Be sure to indicate that formal charge as well as any lone pair of electrons in your answer.
