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Name the following organic compounds;
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- Like alcohols, ethers undergo a cleavage by breaking a carbon–carbon bond between an alkyl group and the carbon bonded to the ether oxygen atom; that is, the red C–C bond in R–CH2OR' is broken. With this in mind, propose structures for the fragments formed by a cleavage of (CH3)2CHCH2OCH2CH3. Suggest a reason why an ether fragments by acleavage.Explain how the reaction of (CH3)2CHCH(Cl)CH3 with H2O yields two substitutionproducts, (CH3)2CHCH(OH)CH3 and (CH3)2C(OH)CH2CH3Explain the order of boiling points of the isomeric alcohols but an_1_ol boiling point of 118°C,but an_2_al boiling point of 99°C and 2_methylpropan_2_al boiling point 82°C Menthone and menthal are both isolated from mint ,explain why menthal is a solid at room temperature while menthone is a liquid at room temperature The boiling point of cis _but_2_ene is 3.7°C whereas that of trans_but_2_ene is 0.9°C explain
- Oximene and myrcene, two hydrocarbons isolated from alfalfa that have the molecular formula C10H16, both yield 2,6- dimethyloctane when treated with H2 and a Pd catalyst. Ozonolysis of oximene forms (CH3)2C = O, CH2 = O, CH2(CHO)2, and CH3COCHO. Ozonolysis of myrcene yields (CH3)2C = O, CH2 = O, (two equiv), and HCOCH2CH2COCHO. Identify the structures of oximene and myrcene.Draw the products formed when p-methylaniline (p-CH3C6H4NH2) is treated with each reagent.a. HClb. CH3COClc. (CH3CO)2Od. excess CH3Ie. (CH3)2C = Of. CH3COCl, AlCl3g. CH3CO2Hh. NaNO2, HCli. Part (b), then CH3COCl, AlCl3j. CH3CHO, NaBH3CNWrite the structure of the compound E,E-2,4-Hexadien-1-ol and label each non-equivalent carbon with a letter, A,B,C..
- I cyclohexanone II cyclohexanol III methylcyclohexanol. (b) Arrange the above compounds in the order of increasing boiling point. Explain. thank you for helping me.Pick the reactant or solvent in each part that gives the faster SN2 reaction.a. reaction of -OH with CH3CH2Br or CH3CH2Clb. reaction of CH3CH2CH2Cl with NaOH or NaOCOCH3c. reaction of CH3CH2CH2I with -OCH3 in CH3OH or DMSOAn attempted dehydration of propan-1-ol (ch3ch2ch2oh) with H2SO4 at 135 degrees celcius gave a by product C6H14O that when tested gave the following results: Baeyer test: Negative Br2/CCl4 Test: Negative Chromic Acid Test: Negative Propse a strucure for the following odd product and explain its formation
- What organic product is formed when 1‑methylcyclopentene is treated with NMMO in the presence of H2O and a catalytic amound of OsO4? Clearly show stereochemistry by drawing a wedge and dashed bond for each chiral carbon. Only draw one stereoisomer if more than one can be formed.2-Methylpropane have more than one kind of hydrogen and will generate 2 different products with bromination in the presence of uv light. t or fGive reasons for the following :(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide.(ii) (±) 2-Butanol is optically inactive.(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH3—X.