Compound AA, C4H®O, gives a positive result (formation of a silver mirror) when reacted with Tollen's reagent. Reduction of compound AA with lithium aluminium hydride, LIAIH4 followed by acidified water, H3O* produces an alcohol, BB. Oxidation of BB with hot acidified potassium dichromate, K2Cr2O7 gives butanoic acid, C3H7COOH. Esterification between butanoic acid and methanol, CH3OH produces compound Cc. Dehydration of compound BB using concentrated sulphuric acid, H2SO4 at a temperature of 180°C produces compound DD.
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- An unknown organic compound P has a molecular formula of CaHi60. P reacts with 2,4- dinitrophenylhydrazine but not with Tollen's reagent. P also reacts with iodine in sodium hydroxide to give yellow precipitate. On reduction, compound P produces a new compound Q. On heating the compound Q with concentrated sulphuric acid produces R which decolourises bromine in tetrachloromethane. Identify P, Q and R. Explain your reasoning. 10.Compound AA has a molecular formula of C3H6O and gives a positiveresult using Tollen’s reagent. The reaction of compound AA with hotacidified potassium permanganate, KMnO4 gives compound BB. Thecatalytic hydrogenation of compound AA with nickel, Ni producedcompound CC. The reaction of compound BB with ethanamine,CH3CH2NH2 produces compound DD I) Draw the structural formula of compounds AA, BB, CC and DD. 2)Name the type of chemical reaction for the formation of compound CC.7. Compound AA, C4H8O, gives a positive result (formation of a silver mirror) when reacted with Tollen’s reagent. Reduction of compound AA with lithium aluminium hydride, LiAlH4 followed by acidified water, H3O+ produces an alcohol, BB. Oxidation of BB with hot acidified potassium dichromate, K2Cr2O7 gives butanoic acid, C3H7COOH. Esterification between butanoic acid and methanol, CH3OH produces compound CC. Dehydration of compound BB using concentrated sulphuric acid, H2SO4 at a temperature of 180°C produces compound DD. Draw the structural formula of compounds AA, BB, CC, and DD
- X Upon ozonolysis, Compound X produces two compounds: Compound Y and Compound Z. Compound Y can also be prepared from the following synthetic route: PCC 1. R₂BH, THF 1. Mg. Et₂O PCC Compound Y 2. CH₂Cl₂ 2. NaOH, HO CH₂Cl₂ 3. H₂O* From this information, draw the structures of Compounds X, Y, and Z. For Compounds X and Z, different substituents are possible. For grading purposes, just use hydrogens as the substituents. Br مرد →] ►An unknown hydrocarbon A with the formula C6H10 reacts with 1 molar equivalent of H2 over a palladium catalyst to give B C6H12 (Rxn 1). Hydrocarbon A also reacts with OsO4 to give the glycol C (Rxn 2). A gives 5-oxohexanal on ozonolysis (Rxn 3). Draw the structures of A, B, and C. Give the reactions.7. Compound EE, CSH100 gives a positive result (formation of silver mirror) when reacted with Tollen's reagent. Reduction of EE with sodium borohydride, NaBH4 followed by acidified water, H30 produces compound FF. Dehydration of compound FF uses concentrated sulphuric acid, H2SO4 at a temperature of 180°C produces compound GG. Compound FF also reacts with hot acidified potassium dichromate, K2Cr207 to produce 2-methylbutanoic acid, CaHsCOOH. Esterification between 2-methylbutanoic acid and methanol, CH3OH produces compound HH. Sebatian EE, CsH1,0 memberikan keputusan positif (pembentukan cermin perak) apabila bertindak balas dengan reagen Tollen. Penurunan EE dengan natrium borohidrida, NaBH4 diikuti dengan air berasid, H30* menghasilkan sebatian FF. Pendehidratan oleh sebatian FF dengan menggunakan asid sulfurik, H2SO. pekat pada suhu 180°C menghasilkan sebatian GG. Sebatian FF juga bertindak balas dengan larutan panas kalium dikromat, K2Cr2O7 berasid menghasilkan 2- metilbutanoik…
- Compound EE, C5H10O gives a positive result (formation of silver mirror) when reacted with Tollen’s reagent. Reduction of EE with sodiumborohydride, NaBH4 followed by acidified water, H3O+ produces compound FF. Dehydration of compound FF uses concentrated sulphuric acid, H2SO4 at a temperature of 180°C produces compound GG. Compound FF also reacts with hot acidified potassium dichromate, K2Cr2O7 to produce 2-methylbutanoic acid, C4H9COOH. Esterification between 2-methylbutanoic acid and methanol, CH3OH produces compound HH. Draw the structural formula of compounds EE, FF, GG, and HH.8. An optically active alcohol, A, C,H10O reacts with concentrated H2SO4 at 180 °C to form compound B which can exist as cis-trans isomers. Compond A reacts with hot acidified KMNO4 to produce compound C which can reacts dinitrophenylhydrazine but does not react with Tollen's reagent. The reaction between C and Grignard reagent D yields compound E upon hydrolysis. The with 2,4 structural formula for E is as follows: OH H3C-CH2-C-CH3 H3C-C-CH3 Determine the structure A, B, C and D with appropriate reason.Dehydration of 2,2,3,4,4,-pentamethyl-3-pentanol gives two alkenes A and B. Ozonolysis of A gives formaldehyde and 2,2,4,4-tetramethyl-3- pentanone. Ozonolysis of B gave formaldehyde and 3,3,4,4-tetramethyl-2-pentanone. Identify and draw the structures of A and B, and explain how B could be formed in the dehydration reaction.
- Using Friedel Crafts alkylation of benzene, a group of scientists discovered compound A, C3H3O, an optically inactive compound. They discovered that compound B, CsH;BrO, is produced during the halogenation process. When compound B is reacted with nitric acid, compound C, C3H,B1NO3, is formed. While Friedel Crafts alkylation of compound A produces compound D. This compound D was oxidised to produce compound E, C9H§O3 as a major product. Compound F, C10H1002, was produced by Friedel Crafts acylation of compound A. This compound F reacts positively with alkaline iodine and 2,4- dinitrophenylhydrazine. However, it returns negative results for Tollens' and Benedict's tests. Identify the possible structural formulae compounds A to F and explain your answers.Compound EE, C4H8O shows a positive result (formation of yellow precipitate) when reacted with lodoform reagent. Reduction of compound EE with sodium boron hydride, NABH4 followed by acidified water, H3O* produces an alcohol FF. Compound FF is treated with phosporus pentachloride, PCI5 to form compound GG. Prolong heating of compound EE with concentrated solution of acidified potassium permanganate, KMNO4 will form compound HH and II. Sebatian EE, C4H8O menunjukkan keputusan positif (pembentukkan mendakan kuning) apabila bertindak balas dengan reagen lodoform. Penurunan sebatian EE dengan natrium boron hidrida, NaBH4 diikuti dengan air berasid, H30* menghasilkan satu alkohol FF. Sebatian FF telah dirawat dengan fosforus pentaklorida, PCI5 untuk membentuk sebatian GG. Pemanasan yang berterusan bagi sebatian EE dengan larutan pekat kalium permanganat, KMNO4 berasid akan membentuk sebatian HH dan II. Draw the structural formula of EE, FF, GG, HH and II. Lukiskan formula struktur bagi EE,…Compound H with molecular formula C6H12O showed no reaction in chromic acid test. Compound H undergo dehydration reaction to produce compound I. Compound J is produced when compound I react with HBr. Reaction of compound J with methanol produced compound K and I. Draw the structure of H, I, J, and K.