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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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- Why would concentrated HBr be an inappropriate catalyst for the dehydration of alcohols? The conjugate base Br- is a good nucleophile and would attack the carbocation to form an alkyl bromide O HBr is strongly acidic, so the water molecule would not be a good leaving group after alcohol protonation. O HBr would be more likely to promote rearrangement of the carbocation intermediate HBr is too weakly acidic to protonate alcoholsThe 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.In the following reaction Series, write down the appropriate reagents that can be used where there are question marks
- Two methods convert an alkyl halide to a carboxylic acid having onemore carbon atom. Depending on the structure of the alkyl halide, one or both of thesemethods may be employed. For each alkyl halide, write out a stepwisesequence that converts it to a carboxylic acid with one more carbonatom. If both methods work, draw both routes. If one method cannot be used, state why it can't.Like acid chlorides, acid anhydrides react with alcohols in the presence of pyridine to form esters. For the following reaction : (a) Draw a circle around the nucleophile ; (b) Draw a square around the electrophile; (c) Put a star next to the base; (d) Draw the structure of the tetrahedral intermediate in the box below. Then draw a reasonable 3-step arrow pushing mechanism for the reaction.Two methods convert an alkyl halide to a carboxylic acid having one more carbon atom. Depending on the structure of the alkyl halide, one or both of these methods may be employed. For each alkyl halide, write out a stepwise sequence that converts it to a carboxylic acid with one more carbon atom. If both methods work, draw both routes. If one method cannot be used, state why it can't.
- 1-Arrange the compounds according to the decrease in acidity? (2,4,6-Trihydroxyphenol) or (2,4,6-Tribromophenol )or (3-Methylphenol) or (phenol) Arrange the compounds according to the increase in acidity? (4-Aminophenol) or (4-Chlorophenol) or (4-Methoxyphenol) or (phenol)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.Acyl transfer (nucleophilic substitution at carbonyl) reactions proceed in two stages via a "tetrahedral intermediate." Draw the tetrahedral intermediate as it is first formed in the following reaction. H3C NH₂ HCI/H₂O reflux • You do not have to consider stereochemistry. • Include all valence lone pairs in your answer. • Do not include counter-ions, e.g., Na+, I, in your answer. • In cases where there is more than one answer, just draw one.
- Choose the correct statement(s) regarding Aryl halides participation in NAS reactions. The rate of nucleophilic substitution increases with increasing numbers of electron withdrawing substituents A strong deactivating group in the para /ortho position is necessary A strong deactivating group in the meta position is necessary. The rate of nucleophilic substitution increases with increasing numbers of electron donating substituents. A strong activating group in the para and/or ortho is necessaryWhich is the stronger base: N-methylaniline (C6H5NHCH3) or benzylamine (C6H5CH2NH2). Explain your reasoning, supporting it with appropriate resonance contributors.Consider the SN2 reaction of butyl bromide with OH- ion. CH3CH2CH2CH2B + OH- → CH3CH2CH2CH2OH + Br Assuming no other changes, what effect on the rate would result from simultaneously doubling the concentrations of both butyl bromide and the OH- ion? Draw the curved arrow mechanism for the SN1 reaction of methanol with tert-butyl iodide. Next, draw an Energy vs. Reaction Coordinate diagram for this reaction, labeling all transition states as "TS" and intermediates as appropriate for those drawn in the mechanism above, showing the relative energies on the diagram.