Draw the major product of this reaction. Ignore inorganic byproducts and the phthalic acid side-product. O= N: K 1. PhCH2CH2CI 2. NaOH (excess) Q Atoms, Bonds and Rings Charges Draw or tap a new bond to see suggestions. Undo Reset Remove Done + Drag To Pan
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Draw the major product of this reaction. Ignore inorganic byproducts and the phthalic acid side-product.
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- Which of the following reactions will not produce a carboxylic acid? Benzaldehyde + hot, conc'd KMnO4 1) Ph-N2+ + KI 2) BrMgCH=CH2 in ether, followed by H3O+ sec-butylBenzene + hot, conc'd KMnO4 1) Benzene, HNO3 + H2SO4; 2) Fe, HCl; 3) NaNO2; 4) CuCN; 5) dilute acid and heat 1) Benzene and acetyl chloride & AlCl3; 2) HCl & Ni Thank you!select the most appropriate reagent(s) to effect the change. K2Cr2O7, H+ H2, Pd 1. Disiamylborane, 2. HO–, H2O, H2O2 NaOCl H2SO4, HgSO4From the reaction below what product would be formed and why?
- Organometallic reagents and alkoxides are strong bases as well as being good nucleophiles. Given the pKa values below, which reagent is the stronger base, nBuLi (CH3CH2CH2CH2LI) or the tert-butoxide anion ((CH3)3CO‒)? Explain your choice clearly in terms of structure andbonding. (Comparing the pKa values is not an explanation.) (CH3)3COH pka= 16.54CH3CH2CH2CH3 pka= 50Can you match best nucleophile / conditions that will give a successful hydrolysis reaction for each electrophile? 1. Acid chloride 2. Acetic anhydride 3. Ester 4. Amide a. LiAlH4; hydronium work-up b. PCC c. NaH d. CrO3 e. HOEt f. H3O+ or OH- g. H2O h. NaBH4;hydronium work-upI want the sequence of the reaction to get to the target for number 3 thank u
- Please give me a technique on how to do these. Should I just remove the halide (e.g. Cl) then attach the remaining figure to a benzene? Will that always work? Or do you have any techniques you can suggest? An easy one without much solving?Why we need step 3 before step 4? a. Because the nitro group increases the electrophilicity at the ortho positions which is where the bromine is added. b. Because the amine group is a strong ortho, para director which is what controls the regiochemical outcome of this bromination. c. Step 4 is unessesary. The symmetry of compound 3 allows for the bromination to be regioselective and give compound 5. 5. There will be a mixture of products because there is no selectivity for a major product.Can somone help me with these problems and let me know if I have did them right.