Tutor profile: Vishakha G.
Subject: Organic Chemistry
Ques1. Why hot concentrated KMnO4 is a better oxidizing agent than dilute basic KMnO4? Ques2. Identify and describe the differences between oxymercuration-demercuration and hydroboration oxidation reactions.
Ans1. Treatment of alkenes with cold, dilute basic KMnO4 leads to 1,2-diols (vicinal diols) Whereas treatment of alkenes with hot, concentrated KMnO4 leads to formation of either ketones or carboxylic acids or both. So, hot, concentrated KMnO4 results in a more oxidized product. So, it is a better oxidizing agent than cold, dil KMnO4. Ans2. oxymercuration-demercuration Reagents: Hg(OAc)2/H2O; NaBH4 E+, Nu-: H, OH Major product: Markonikov Stereochemical orientation: Anti, formation of a three membered cyclic intermediate and ring opening from opposite side. hydroboration oxidation Reagents: BH3/THF; Basic Peroxide E+, Nu-: H, OH Major product: Anti-Markonikov Stereochemical orientation: Syn addition, see mechanism *I would have been able to provide sample examples of these reactions if attachment option would have been there.
Ques: When 3.2g of NaHCO3 reacts with excess of HC2H3O2, what mass of CO2 is produced?
Ans. Let us first write a balanced chemical equation for this reaction: NaHCO3 (s) + HC2H3O2 (aq) ⟶ NaC2H3O2 (aq) + H2O (l) + CO2 (g) Let us find the moles of baking soda (NaHCO3), since it is the limiting reagent. Moles of NaHCO3 = mass/Molecular mass Moles of NaHCO3 = 3.2/84 = 0.038 Now, we should calculate the moles of CO2 first to calculate its mass From the balanced chemical equation, it is evident that 1 mole of NaHCO3 gives -----------> 1 mole of CO2 0.038 moles of NaHCO3 will give ------> 1x 0.038 moles of CO2 = 0.038 moles of CO2 Now, let us calculate mass of CO2 using the formula Moles = mass/molecular mass Mass = Moles x molecular mass Mass of CO2 = 0.038 x 44 = 1.67g
Subject: Basic Chemistry
Ques: Calculate the pH and pOH of each solution given the following [H3O+] and [OH-] values: a. [H3O+] = 1.0 x 10-4 M b. [H3O+] = 3.0 x 10-9 M c. [OH-] = 1.0 x 10-5 M
Ans. For finding answers to such questions, first of all you should know the formula for calculating pH, pOH and the relation between pH and pOH. So, pH = -log[H+], pOH = -log [OH-], and pH + pOH = 14 Using these three relations, let us see how to calculate the values: a. pH = -log[H+] = -log [1.0 x 10-4] = 4 pH + pOH = 14 => 4+ pOH = 14 pOH = 10 b. pH = -log [3.0 x 10-9] = -(log 3 + log 10-9) = -(0.477 – 9) = 8.52 pOH = 14-pH = 5.4 c. pOH = -log [OH-] = -log [10-5] = 5 pH= 14-pOH =14-5 = 9
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