Question

Introduction When an acid neutralizes a base, a salt and water are formed. Aqueous hydrogen ions (H+(aq)) from the acid react with the hydroxide ions (OH-(aq)) from the base, forming water, e.g. HCl + NaOH → NaCl + H2O The Na+ and Cl- ions remain unchanged; the only reaction is between the H+ and OH- ions: H+(aq) + OH-(aq) → H2O (l) The neutralization reaction will occur when H+ and OH- combine and energy is released until either H+ or OH- is entirely consumed. The reactant which is consumed first is called the limiting reactant. The molar enthalpy of neutralization is defined as ΔHneut = qneut / n, where qneut is the heat of neutralization, and n is the moles of the limiting reactant. qneut = c * m * ΔT, where m is the mass of the solution, C is the specific heat capacity of the solution (4.2 Jg-1K-1), and ΔT is the temperature difference. In this practical, we shall be measuring the enthalpy changes (ΔH) for different neutralization reactions. Because the number of moles of water formed varies according to the acids and bases used, it is the convention to measure enthalpies of neutralization (ΔH) in kilojoules per mole of water formed. Procedure Experiment 1: The reaction between Hydrochloric acid and sodium hydroxide Measure 50cm3 of 2.0M hydrochloric acid into a polystyrene cup and record its temperature. Pour 50cm3 of 2.0M sodium hydroxide into a measuring cylinder and record the temperature. Pour the sodium hydroxide into the acid, stir, and take the final temperature. (Note: if the acid and base were not at the same temperature initially, use the average of their initial temperature as the starting temperature) Experiment 2 Repeat experiment 1 using 2.0M potassium hydroxide solution instead of sodium hydroxide solution. Experiment 3 Repeat experiment 1 using 2.0M nitric acid (caution – corrosive) instead of hydrochloric acid. Experiment 4 Repeat experiment 1 using 2.0M sulfuric acid instead of hydrochloric acid. Experiment 5 Repeat experiment 1 using 2.0M sulfuric acid and 4.0M sodium hydroxide. Experiment 6 Repeat experiment 1 using 4.0M sodium hydroxide solution. Table of Results Acid Base Initial temp. °C (Average) Final temp. °C (Average) ΔT °C Expt. 1 2M HCl 2M NaOH 20 26 6 Expt. 2 2M HCl 2M KOH 19 32 13 Expt. 3 2M HNO3 2M NaOH 19 26 7 Expt. 4 2M H2SO4 2M NaOH 19 26 7 Expt. 5 2M H2SO4 4M NaOH 19 51 30 Expt. 6 2M HCl 4M NaOH 20 33 13 Treatment of results i) Write a balanced equation for each experiment, including the states. ii) How much energy (qneut) was given to the 100cm3 of final solution during each reaction? (Assume that the specific heat capacity of the plastic cup is negligible and the specific heat capacity of the solution is 4.2 Jg-1K-1) iii) How many moles of water were formed in each reaction (show your calculation). iv) Calculate the enthalpy change of neutralization for each reaction (show your calculation). Questions Accepted values for the enthalpy changes of neutralization of some acids and bases are shown below. Reaction ΔH kJmol-1 HCl + NaOH → NaCl + H2O -57.9 HNO3 + NaOH → NaNO3 + H2O -57.6 HBr + NaOH → NaBr + H2O -57.6 CH3COOH + NaOH → CH3COONa + H2O -56.1 1. Compare the experimental values to the values in the table for the first two reactions. Account for any errors in your results. 2. Why is the enthalpy change for ethanoic acid slightly lower than the values for the first three reactions? (Hydrochloric, nitric, and hydrobromic acids are all strong acids. Ka (ethanoic acid) = 1.7x10-5. 3. Compare the temperature rise when 4.0M sodium hydroxide is used instead of 2.0M. Explain the relative values.

Introduction
When an acid neutralizes a base, a salt and water are formed. Aqueous hydrogen ions (H+(aq)) from the acid react with the hydroxide ions (OH-(aq)) from the base, forming water, e.g.

HCl + NaOH → NaCl + H2O

The Na+ and Cl- ions remain unchanged; the only reaction is between the H+ and OH- ions:

H+(aq) + OH-(aq) → H2O (l)

The neutralization reaction will occur when H+ and OH- combine and energy is released until either H+ or OH- is entirely consumed. The reactant which is consumed first is called the limiting reactant.

The molar enthalpy of neutralization is defined as ΔHneut = qneut / n, where qneut is the heat of neutralization, and n is the moles of the limiting reactant.

qneut = c * m * ΔT, where m is the mass of the solution, C is the specific heat capacity of the solution (4.2 Jg-1K-1), and ΔT is the temperature difference.

In this practical, we shall be measuring the enthalpy changes (ΔH) for different neutralization reactions. Because the number of moles of water formed varies according to the acids and bases used, it is the convention to measure enthalpies of neutralization (ΔH) in kilojoules per mole of water formed.

Procedure
Experiment 1: The reaction between Hydrochloric acid and sodium hydroxide
Measure 50cm3 of 2.0M hydrochloric acid into a polystyrene cup and record its temperature. Pour 50cm3 of 2.0M sodium hydroxide into a measuring cylinder and record the temperature. Pour the sodium hydroxide into the acid, stir, and take the final temperature. (Note: if the acid and base were not at the same temperature initially, use the average of their initial temperature as the starting temperature)

Experiment 2
Repeat experiment 1 using 2.0M potassium hydroxide solution instead of sodium hydroxide solution.

Experiment 3
Repeat experiment 1 using 2.0M nitric acid (caution – corrosive) instead of hydrochloric acid.

Experiment 4
Repeat experiment 1 using 2.0M sulfuric acid instead of hydrochloric acid.

Experiment 5
Repeat experiment 1 using 2.0M sulfuric acid and 4.0M sodium hydroxide.

Experiment 6
Repeat experiment 1 using 4.0M sodium hydroxide solution.

Table of Results
Acid Base Initial temp. °C (Average) Final temp. °C (Average) ΔT °C
Expt. 1 2M HCl 2M NaOH 20 26 6
Expt. 2 2M HCl 2M KOH 19 32 13
Expt. 3 2M HNO3 2M NaOH 19 26 7
Expt. 4 2M H2SO4 2M NaOH 19 26 7
Expt. 5 2M H2SO4 4M NaOH 19 51 30
Expt. 6 2M HCl 4M NaOH 20 33 13

Treatment of results
i) Write a balanced equation for each experiment, including the states.
ii) How much energy (qneut) was given to the 100cm3 of final solution during each reaction? (Assume that the specific heat capacity of the plastic cup is negligible and the specific heat capacity of the solution is 4.2 Jg-1K-1)
iii) How many moles of water were formed in each reaction (show your calculation).
iv) Calculate the enthalpy change of neutralization for each reaction (show your calculation).

Questions
Accepted values for the enthalpy changes of neutralization of some acids and bases are shown below.

Reaction ΔH kJmol-1
HCl + NaOH → NaCl + H2O -57.9
HNO3 + NaOH → NaNO3 + H2O -57.6
HBr + NaOH → NaBr + H2O -57.6
CH3COOH + NaOH → CH3COONa + H2O -56.1

1. Compare the experimental values to the values in the table for the first two reactions. Account for any errors in your results.
2. Why is the enthalpy change for ethanoic acid slightly lower than the values for the first three reactions? (Hydrochloric, nitric, and hydrobromic acids are all strong acids. Ka (ethanoic acid) = 1.7x10-5.
3. Compare the temperature rise when 4.0M sodium hydroxide is used instead of 2.0M. Explain the relative values.

Added by Laura G.

FRIEDEL-CRAFTS REACTIONS Friedel-Crafts reactions provide a method for the preparation of alkylbenzenes (ArR) and acylbenzenes (ArCOR). These reactions are called Fricdel-Crafts alkylation and Friedel-Crafts acylation. A Friedel-Crafts Alkylation The following is a general equation for a Friedel-Crafts alkylation reaction: (FIGURE CANNOT COPY) The mechanism for the reaction starts with the formation of a carbocation. The carbocation then acts as an electrophile and is attacked by the benzene ring to form an arenium ion. The arenium ion then loses a proton. This mechanism is illustrated below using 2 -chloropropane and benzenc. (FIGURE CANNOT COPY) This is a Lewis acid-base reaction (see Section $15.3)$ The complex dissociates to form a carbocation and $\mathrm{AlCl}_{4}^{-}.$ (FIGURE CANNOT COPY) The carbocation, acting as an electrophile, reacts with benzene to produce an arenium ion. A proton is removed from the arenium ion to form isopropylbenzene. This step also regenerates the AlCl_ and liberates HCI. When $R-X$ is a primary halide, a simple carbocation probably does not form. Instead, the aluminum chloride forms a complex with the alkyl halide, and this complex acts as the electrophile. The complex is one in which the carbon-halogen bond is nearly broken-and one in which the carbon atom has a considerable positive charge: (equation can't copy) Even though this complex is not a simple carbocation, it acts as if it were and it transfers a positive alkyl group to the aromatic ring. These complexes react so much like carbocations that they also undergo typical carbocation rearrangements (Sction $15.6 \mathrm{C}).$ Friedel-Crafts alkylations are not restricted to the use of alkyl halides and aluminum chloride. Other pairs of reagents that form carbocations (or species like carbocations) may be used in Friedel-Crafts alkylations as well. These possibilities include the use of a mixture of an alkene and an acid: (FIGURE CANNOT COPY) Propene Isopropylbenzene (cumene) $(84 \%)$ (FIGURE CANNOT COPY) Cyclohexene Cyclohexylbenzene $(62 \%)$ A mixture of an alcohol and an acid may also be used: (FIGURE CANNOT COPY) Cyclohexanol Cyclohexylbenzene $(56 \%)$ There are several important limitations of the Friedel-Crafts reaction. These are discussed in Section $15.6 \mathrm{C}.$ Outline all steps in a reasonable mechanism for the formation of isopropylbenzene from propene and benzene in liquid HF (just shown). Your mechanism must account for the product being isopropylbenzene, not propylbenzene. THE CHEMISTRY OF.... industrial Styrene Synthesis Styrene is one of the most important industrial chemicalsmore than 11 billion pounds is produced each year. The starting material for a major commercial synthesis of styrene is ethylbenzene, produced by Friedel-Crafts alkylation of benzene: (FIGURE CANNOT COPY) Styrene $(90-92 \% \text { yield })$ Most styrene is polymerized (Special Topic C) to the familiar plastic, polystyrene: (FIGURE CANNOT COPY) Polystyrene B Friedel-Crafts Acylation The $R$ group is called an acyl group, and a reaction whereby an acy group is introduced into a compound is called an acylation reaction. Two common acyl groups are the acetyl group and the benzoyl group. (The benzoyl group should not be confused with the benzyl group, $\left.-\mathrm{CH}_{2} \mathrm{C}_{6} \mathrm{H}_{5} ; \text { see Section } 14.2 .\right)$ (FIGURE CANNOT COPY) Acetyl group (ethanoyl group) Benzoyl group The Friedel-Crafts acylation reaction is often carried out by treating the aromatic compound with an acyl halide (often an acyl chloride). Unless the aromatic compound is one that is highly reactive, the reaction requires the addition of at least one equivalent of a Lewis acid (such as $\mathrm{AlCl}_{3}$ ) as well. The product of the reaction is an aryl ketone: (FIGURE CANNOT COPY) Acetyl chloride Acetophenone (methyl phenyl ketone) $(97 \%)$ Acyl chlorides, also called acid chlorides, are easily prepared (Section 18.5 ) by treating carboxylic acids with thionyl chloride (SOCI_) or phosphorus pentachloride (PCI_): (FIGURE CANNOT COPY) Acetic acid Thionyl chloride Acetyl chloride $(80-90 \%)$ (FIGURE CANNOT COPY) Benzoic acid Phosphorus pentachloride Benzoyl chloride $(90 \%)$ Friedel-Crafts acylations can also be carried out using carboxylic acid anhydrides. For example, (FIGURE CANNOT COPY) Acetic anhydride (a carboxylic acid anhydride) Acetophenone $(82-85 \%)$ In most Friedel-Crafts acylations the electrophile appears to be an acylium ion formed from an acyl halide in the following way: (FIGURE CANNOT COPY) (FIGURE CANNOT COPY) An acylium ion (a resonance hybrid) Show how an acylium ion could be formed from acetic anhydride in the presence of $\mathrm{AlCl}_{3}.$ STRATEGY AND ANSWER: We recognize that AICl_, is a Lewis acid and that an acid anhydride, because it has multiple unshared electron pairs, is a Lewis base. A reasonable mechanism starts with a Lewis acid-base reaction and proceeds to form an acylium ion in the following way. (FIGURE CANNOT COPY) Acylium ion The remaining steps in the Friedel-Crafts acylation of benzene are the following: (FIGURE CANNOT COPY) other resonance structures (draw them for practice) The acylium ion, acting as an electrophile, reacts with benzene to form the arenium ion. (FIGURE CANNOT COPY) A proton is removed from the arenium ion, forming the aryl ketone. (FIGURE CANNOT COPY) The ketone, acting as a Lewis base, reacts with aluminum chloride (a Lewis acid) to form a complex. (FIGURE CANNOT COPY) Treating the complex with water liberates the ketone and hydrolyzes the Lewis acid. Limitations of Friedel-Crafts Reactions Several restrictions limit the usefulness of Fricdel-Crafts reactions: 1. When the carbocation formed from an alkyl halide, alkene, or alcohol can rearrange to one or more carbocations that are more stable, it usually does so, and the major products obtained from the reaction are usually those from the more stable carbocations. When benzene is alkylated with butyl bromide, for example, some of the developing butyl cations rearrange by a hydride shift. Some of the developing $1^{\circ}$ carbocations (see following reactions) become more stable $2^{\circ}$ carbocations. Then benzene reacts with both kinds of carbocations to form both butylbenzene and sec-butylbenzene: (FIGURE CANNOT COPY) Butylbenzene $(32-36 \% \text { of mixture })$ sec-Butylbenzene $(64-68 \% \text { of mixture })$ 2. Friedel-Crafts alkylation and acylation reactions usually give poor yields when powerful electron-withdrawing groups (Section $15.8 \mathrm{B}$ and Table 15.1 ) are present on the aromatic ring. (FIGURE CANNOT COPY) These usually give poor yields in Friedel-Crafts reactions because the ring is electron deficient. Poor yiclds are also the case when the ring bears an $-\mathrm{NH}_{2},-\mathrm{NHR},$ or $-\mathrm{NR}_{2}$ group because they become electron-withdrawing when they react with the Lewis acid in the reaction mixture. (FIGURE CANNOT COPY) Does not undergo a Friedel-Crafts reaction These groups are changed into powerful electron-withdrawing groups by the Lewis acids used to catalyze Friedel-Crafts reactions. 3. Aryl and vinylic halides cannot be used as the halide component because they do not form carbocations readily (see Section $6.14 \mathrm{A}$ ): (FIGURE CANNOT COPY) No Friedel-Crafts reaction because the halide is at an $s p^{2}$ carbon. 4. Polyalkylations often occur. Alkyl groups are inductive electron-donating groups (Sections $15.8 \text { and } 15.9),$ and once one is introduced into the benzene ring, it activates the ring toward further substitution: (FIGURE CANNOT COPY) Isopropylbenzene $(24 \%)$ $p$ -Diisopropylbenzene $(14 \%)$ Polyacylations are not a problem in Friedel-Crafts acylations. The acyl group (-COR) by itself is an electron-withdrawing group, and when it forms a complex with AlCl in the last step of the reaction (Section $15.6 \mathrm{B}),$ it is made even more electron withdrawing. This strongly inhibits further substitution and makes monoacylation easy. When benzene reacts with 1-chloro-2,2-dimethylpropane (neopentyl chloride) in the presence of aluminum chloride, the major product is 2 -methyl-2-phenylbutanc, not 2,2 -dimethyl-1-phenylpropane (neopentylbenzenc). Explain this result. STRATEGY AND ANSWER: The carbocation formed by direct reaction of AICl_, with 1-chloro-2,2-dimethylpropane would be a primary carbocation; however, it rearranges to the more stable tertiary carbocation before it reacts with the benzene ring. (FIGURE CANNOT COPY) (FIGURE CANNOT COPY) Provide a mechanism that accounts for the following result. (FIGURE CANNOT COPY)

Question

Please give Ace some feedback