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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4.1.1 Cracking is the process of breaking down long-chain hydrocarbons into shorter, more useful hydrocarbons (alkanes and alkenes) using heat and/or a catalyst. 4.1.2 Step 1: Determine the molecular formula of compound P. Heptane (C_7H_16) is cracked into a FIVE CARBON alkane P and compound Q. The general formula for an alkane is C_nH_2n+2. For a five-carbon alkane, n=5. So, compound P is C_5H_2(5)+2 = C_5H_12. Step 2: Determine the molecular formula of compound Q. The total number of carbon atoms must be conserved: 7 = 5 + C_Q C_Q = 2. The total number of hydrogen atoms must be conserved: 16 = 12 + H_Q H_Q = 4. Therefore, the molecular formula for compound Q is C_2H_4. The molecular formula for compound Q is C_2H_4. 4.1.3 The most important use of alkanes is as fuels. 4.1.4 Step 1: Write the unbalanced combustion reaction for compound P (C_5H_12). Complete combustion of a hydrocarbon produces carbon dioxide (CO_2) and water (H_2O). C_5H_12(g) + O_2(g) CO_2(g) + H_2O(g) Step 2: Balance the carbon atoms. There are 5 carbon atoms on the left, so there must be 5 CO_2 molecules on the right. C_5H_12(g) + O_2(g) 5CO_2(g) + H_2O(g) Step 3: Balance the hydrogen atoms. There are 12 hydrogen atoms on the left, so there must be 6 H_2O molecules on the right (6 × 2 = 12). C_5H_12(g) + O_2(g) 5CO_2(g) + 6H_2O(g) Step 4: Balance the oxygen atoms. On the right side, there are (5 × 2) + (6 × 1) = 10 + 6 = 16 oxygen atoms. To balance this, there must be 8 O_2 molecules on the left (8 × 2 = 16). C_5H_12(g) + 8O_2(g) 5CO_2(g) + 6H_2O(g) The balanced equation for the complete combustion of compound P is: C_5H_12(g) + 8O_2(g) 5CO_2(g) + 6H_2O(g) 4.2.1 The flow diagram shows a straight chain secondary haloalkane reacting to form a 5-carbon alcohol (Alcohol P), which then undergoes elimination to form C_5H_10 (a 5-carbon alkene). This means the haloalkane must also have 5 carbon atoms. A straight chain secondary haloalkane with 5 carbons would have the halogen on the second or third carbon. Given the handwritten notes on the left, the halogen is on the second carbon. Since NaBr is a product of Reaction 1, the halogen is bromine. The IUPAC name is 2-bromopentane. 4.2.2 Reaction 1 is the conversion of a haloalkane to an alcohol. This is a hydrolysis reaction (or nucleophilic substitution). 4.2.3 Besides heat, another reaction condition for the hydrolysis of a haloalkane is an aqueous strong base (e.g., NaOH(aq) or KOH(aq)). 4.2.4 If the secondary haloalkane is 2-bromopentane, then Alcohol P, formed by replacing the bromine with a hydroxyl group, is pentan-2-ol. The structural formula of alcohol P is: CH_3CH(OH)CH_2CH_2CH_3 4.2.5 Reaction 2 is the conversion of an alcohol to an alkene (C_5H_10). This is an elimination reaction where water is removed from the alcohol. The name of this type of elimination reaction is dehydration. 4.2.6 Alcohol P is pentan-2-ol (CH_3CH(OH)CH_2CH_2CH_3). During dehydration, the -OH group and a hydrogen atom from an adjacent carbon are removed to form a double bond. According to Zaitsev's rule, the major product is the more substituted alkene. Removing H from C1 gives pent-1-ene (CH_2=CHCH_2CH_2CH_3). Removing H from C3 gives pent-2-ene (CH_3CH=CHCH_2CH_3). Pent-2-ene is the more substituted alkene and thus the major product. The structural formula for the product that was produced (pent-2-ene) is: CH_3CH=CHCH_2CH_3 Send me the next one 📸