This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Okay, let's pick up where we left off. 5 (a) iii) The reaction of compound E (CH_3CH=CH_2, propene) with Br_2/CCl_4 is an electrophilic addition reaction. Step 1: The electrons of the alkene attack one bromine atom, forming a cyclic bromonium ion and releasing a bromide ion. CH_3CH=CH_2 + Br_2 CH_3CH \\ CH_2 ^+ Br + Br^- Step 2: The bromide ion (Br^-) acts as a nucleophile and attacks one of the carbons of the cyclic bromonium ion from the opposite side, opening the ring. CH_3CH \\ CH_2 ^+ Br + Br^- CH_3CH(Br)CH_2Br The product is 1,2-dibromopropane. 6 (a) i) Step 1: Calculate the moles of each element in 100 g of the compound. Given percentages: C = 35.0%, H = 6.6%, Br = 58.4%. R.A.M: C = 12, H = 1, Br = 79.9. Moles of C = 35.0 g12 g/mol = 2.9167 mol Moles of H = 6.6 g1 g/mol = 6.6 mol Moles of Br = 58.4 g79.9 g/mol = 0.7309 mol Step 2: Divide by the smallest number of moles to find the simplest mole ratio. Smallest number of moles is 0.7309 (for Br). Ratio of C = (2.9167)/(0.7309) ≈ 3.99 ≈ 4 Ratio of H = (6.6)/(0.7309) ≈ 9.03 ≈ 9 Ratio of Br = (0.7309)/(0.7309) = 1 Step 3: Write the empirical formula. The empirical formula is C_4H_9Br. ii) Step 1: Calculate the empirical formula mass. Empirical formula mass of C_4H_9Br = (4 × 12) + (9 × 1) + (1 × 79.9) = 48 + 9 + 79.9 = 136.9 g/mol. Step 2: Compare the empirical formula mass with the given molecular weight. Given molecular weight = 136.9. Since the empirical formula mass (136.9) is equal to the molecular weight (136.9), the molecular formula is the same as the empirical formula. The molecular formula is C_4H_9Br. 6 (b) i) To distinguish between CH_3CH_2CHO (propanal, an aldehyde) and CH_3COCH_3 (propanone, a ketone): Use Tollens' reagent. • Propanal (aldehyde) will react with Tollens' reagent (ammoniacal silver nitrate) upon warming to form a silver mirror on the inside of the test tube. • Propanone (ketone) will show no reaction. ii) To distinguish between CH_3CH_2C (but-1-yne, a terminal alkyne) and CH_3C_3 (but-2-yne, an internal alkyne): Use ammoniacal silver nitrate solution. • But-1-yne (terminal alkyne) will react to form a white precipitate of silver acetylide. • But-2-yne (internal alkyne) will show no reaction. 6 (c) Organic chemistry is the branch of chemistry that studies the structure, properties, composition, reactions, and preparation of carbon-containing compounds. Two unique properties of carbon: 1. Catenation: Carbon atoms can form strong covalent bonds with other carbon atoms, leading to the formation of long chains, branched chains, and rings. 2. Tetravalency: Carbon atoms can form four covalent bonds, allowing for a vast diversity of stable molecular structures. 6 (d) i) A functional group is a specific group of atoms within a molecule that is responsible for the characteristic chemical reactions of that molecule. ii) The systematic name of the compound CH_3CH(CH_3)CONH_2 is 2-Methylpropanamide. iii) The compound CH_3CH(CH_3)CONH_2 exhibits optical isomerism because it contains a chiral carbon atom (the carbon bonded to the methyl group, hydrogen, and the amide group). 6 (e) i) The asymmetric carbon atom in CH_3CH(NH_2)COOH is the central