The main objective of this experiment is to determine the amount of nicotine in commercial brand cigarettes by means of a nonaqueous acid-base titration.A simple glass device simulating a smoker is proposed, which allows the determination of the volatilized, filter retained, and inhaled portions.Students will readily see that the amount of nicotine/cigarette stated on the label (∼0.5–1.0 mg) refers indeed to the inhaled portion only, rather than to the total amount/cigarette (usually more than 10 mg).Even so, values for inhaled nicotine may be significantly higher than those reported for several brands.Students will also be able to make a critical evaluation of the true content of nicotine in the inhaled portion and confront it with the reported value for a given brand.In addition, the theoretical approach, supported by HPLC data, provides an excellent experience on nonaqueous acid-base volumetric analysis.Tobacco is obtained from two vegetal species, namely Nicotiana tabacum and Nicotiana rustica, both of them native from Peruvian and Ecuadorian Andes.

It has been used in many ways: smoked in pipes, inhaled, chewed, eaten, ingested as tea, used for intestinal lavage (clyster), scraped on the skin to fight louses, instilled as eye drops, and used in ointments, analgesics, and antiseptics [1].The production of cigarettes started by the end of the 19th century, when the British government decided to cultivate the plant within several of its colonies.e cigarette bagneuxNowadays, there are more than a billion smokers all over the world [2].The cigarette fumes are known to contain more than 4,000 chemical compounds, among them nicotine, carbon monoxide, benzo [α] pyrene (a pre-carcinogen of the cigarette tar), respiratory irritants (as acrolein, formaldehyde, phenols, etc) and others.e cigarette before blood testA xenobiotic compound to be especially considered is benzo [α] pyrene, a common environmental contaminant produced from the combustion of plant materials in tobacco, which is metabolized in animals, giving rise to benzo [α] pyrene-7, 8-dihydrodiol-9,10-epoxide, a potent carcinogen.e cigarette bien vapoter

However, most of the clinical studies show that, nicotine is the main agent responsible for the development of dependence on tobacco [3, 4].Nicotine, muscarine, atropine, quinine, morphine, strychnine, and others are typical representative alkaloids (alkali-like), many of which exhibit characteristic physiological actions.The pleasant effect induced by nicotine results from neurochemical and molecular adaptations of the dopaminergic system [5].e cigarette cilex enoHowever, a decrease of 50% in the consumption of nicotine is able to evoke signs and symptoms, which cause discomfort (as bradycardia, gastroenterological discomfort, increased appetite, weight gain, difficulty to ponder, anxiety, dysphoria, depression, and insomnia), thus characterizing the physical dependence [6].e cigarette coca colaThe dramatic consequence of prolonged smoking induced by nicotine dependence is often carcinogenesis.e cigarette drummondville

Cigarette smoking is responsible for 30% of U.S.It is estimated that, there are 1,100,000 world-wide lung cancer deaths per year, 85% of which caused by tobacco [7].The main acute effects of nicotine on the cardiovascular system are as follows: peripheric vasoconstriction, increased systemic arterial pressure, and increased cardiac frequency.In the nerve endings, it stimulates release of the neurotransmitters acetylcholine, dopamine (DA), glutamate, serotonine, and gamma-aminobutyric acid (GABA) [8].Nicotine acts on the so called nicotinic receptors.Cholinergic synapses may contain nicotine or muscarine receptors.Those containing nicotinic receptors occur at all excitatory neuromuscular junctions in vertebrates and at numerous sites in the nervous system [9].Nicotine is the major alkaloid of tobacco.The fatal dose for humans is about 100 mg.It may be oxidized to nicotinic acid (niacin) by the action of chromic acid.1) is a tertiary amine displaying a pyridine and a pyrrolidine rings, namely, 3-(N-Methyl-2-pyrrolidyl) pyridine [10].It is a liquid, colorless, volatile basic alkaloid with pK1 = 6.16 (pyridine ring) and pK2 = 10.96 (pyrrolidine ring) at 15°C, in aqueous medium.Figure 1.

Open FigureDownload Powerpoint slide3-[(2S)-1-Methyl-2-pyrrolidinyl] pyridine (nicotine) C10H14N2:Mr = 162.23:pK1 = 6.16:pK2 = 10.96:15°C.The theoretical approach usually applied in acid-base titrations may be extended to nonaqueous systems as well.Nicotine is an organic base, unable to allow an accurate titration in water.In nonaqueous solvents however, as anhydrous acetic acid, nicotine may be quantified by acid-base titration [11].A rather complex, somewhat unsuitable for routine practical purposes, spectroscopic method for quantification of nicotine in cigarette smoke was described, where substances interfering in the UV analysis are previously removed by steam distillation from an acid solution, whereas the remaining solution is then made alkaline and nicotine distilled to be later determined by UV spectroscopy at λ = 236, 259, and 282 nm [12].This project aimed at the quantification of nicotine, using a simple, practical and accurate acid-base titration, supported by High Performance Liquid Chromatography (HPLC), for the quantification of nicotine in: aCommercial brand cigarettes.bThe portion inhaled by the smoker.cThe portion retained in the filter.dThe portion lost by vaporization in the environment.PRE-LABORATORY PREPARATION AND GENERAL NOTESThe technique here described for evaluation of nicotine in cigarettes has important implications regarding smoking and health.Our experience of over 50 years teaching Biochemistry basically for Medicine, but eventually for Pharmacy, Nutrition, and Biological Sciences students has often induced us to introduce new practical classes able to allow students to get a broader understanding of analytical biochemical methods broaching fundamental concepts, such as “equivalent” in acid-base and oxidation-reduction reactions, mainly directed to the biological field.

The experiments here reported represent a worthy subject, given its interdisciplinary aspect linked to a world-wide health problem concerned with smoking and offers a rewarding learning assessment as was the case with our graduation students.Every titration should be based on at least two parallel determinations.The results should agree within ∼0.1–0.3% if the titrant volume is around 20 mL, but up to ∼1.2% for titrant volumes of ∼5 mL.In addition, the “Study questions” (section 7) should be used to complement students' learning.For practical purposes, we recommend two classes of 2 h each (20 to 30 students) to broach the subject, namely: 1st: Estimation of total nicotine.2nd: Distribution of nicotine in the volatilized, filter retained and inhaled portions.EXPERIMENTAL PROCEDUREAll reagents used were of analytical grade and included: oxalic acid dihydrate [HO2CCO2H.2H2O, FW 126.07] 99% purity; sodium hydroxide pellets [NaOH, FW 40.00] ≥ 98%; perchloric acid [HClO4, FW 100.46] 69–72% (m/m) solution, ACS reagent; glacial acetic acid [CH3COOH, FW 60.05] > 99%; barium hydroxide octahydrate [Ba(OH)2.8H2O, FW 315,46] ≥ 98%; potassium biphtalate [HOOCC6H4COOK, FW 204.22] > 99%; toluene, anhydrous 99.8%; acetic anhydride [(CH3CO)2, FW 102.09] > 98%; phenolphthalein, ACS reagent; crystal violet ≥ 90%; cyclohexanecarboxylic acid 98% (Sigma, Aldrich, St. Louis, MO, USA); nicotine 98%, d = 1.01 mg/mL (Merck, Darmstadt, Germany).The whole procedure, up to the quantification of nicotine, may be depicted as follows:The original 69–72% solution of HClO4 is previously dehydrated with a calculated amount of acetic anhydride (solution C).

To make sure that solution C is absolute anhydrous, ∼1 mL of it is dropped on ∼1 g of anhydrous CuSO4, which will remain white if solution C is anhydrous but will turn to light- blue when a drop of water is added, because of the formation of the hydrated cupric ion, [Cu(H2O)4]++, that occurs in aqueous or hydrated solutions of cupric salts and in some of the hydrated crystals, as CuSO4.5H2O.Anhydrous solution C is then titrated with potassium biphtalate (see details in the Results and Discussion section).Nicotine is finally titrated with solution C, using crystal violet, (hexamethylpararosaniline), an indicator for nonaqueous titrations, which turns from violet to greenish yellow when protonated.Calculation of the contents in nicotine was supported by the results from HPLC.As the mass of each cigarette in the same pack is not a constant, titration should be carried out after collecting the total tobacco of 20 cigarettes (a whole pack), homogenizing the pool, and weighing a sample of 1.0 g. This sample is then transferred to a 125 mL Erlenmeyer flask, and 8 mL of a saturated solution of Ba(OH)2 + 270 mg of solid Ba(OH)2 + 15.0 mL toluene is added.

This suspension is then magnetically stirred for 20 minutes to extract nicotine.The organic phase is filtered through a Whatman filter paper and two aliquots of 5.0 mL each are pipetted for titration (relative error between both values ≤ 0.3%), using the arithmetic mean as a final value for calculation.The amount of nicotine, chemically transformed during smoking, is unforeseeable and depends on the burning temperature.In view of this fact, a glass device simulating an artificial smoker was constructed, to avoid this destruction by reducing the oxygen supply.In this manner, the whole nicotine content in the tobacco could be preserved and its distribution into volatilized, filter retained, and inhaled portion could be quantitatively determined on basis of the total amount of nicotine/cigarette.All tobacco from 20 cigarettes was carefully removed, homogenized as before, and part of it was then redistributed among some of the original wrappings in equal amounts.To determine the distribution of nicotine among the portion lost in the environment through volatilization that retained on the filter and, finally, that which is inhaled by the smoker, the simple experimental glass device shown in Fig.

2 was used, consisting of a 150 mL Erlenmeyer, simulating the environment, with a 5 mm diameter hole, through which atmospheric air could freely flow, and a ground-glass mouth, where a glass coil, simulating the smoker, was tightely adapted.A cigarette was then fitted at the coil's end inside the Erlenmeyer, whereas a smooth vacuum device was adapted at the opposite terminal.For practical classes, water aspirate pumps, whose maxim vacuum is 100 mmHg, are recommended.When the cigarette burns, part of its nicotine volatilizes and condenses on the Erlenmeyer wall, a second one is retained on the cigarette filter and, finally, the remaining volatile nicotine condenses along the coil wall and corresponds to the smoker's inhalated fraction.These separate portions were then quantitatively removed by successive washings with small amounts of toluene, which were brought together for titration of each individual portion, after adding the proper amounts of Ba(OH)2 solution and solid Ba(OH)2 as described.Figure 2.

Open FigureDownload Powerpoint slideSmoker simulator with cigarette connected.As nicotine is reported to come along with small amounts of other similar alkaloids from tobacco, parallel analyzes were run by HPLC for quantification of nicotine in some cigarette brands, to compare the results, as HPLC is able to discriminate the peak of nicotine from those of other eventual contaminant tobacco alkaloids.Routine, daily calibration curves for HPLC assays were prepared by addition of 50 μL of a standard solution of nicotine at concentrations of 0.28, 0.57, 0.94, 1.13, 2.26, and 4.53 mg/mL in methanol and 25 μL of cyclohexanecarboxylic acid (Internal Standard, IS), 2.4 mg/mL in methanol.Gas chromatography (GC) analyzes were carried out with a Shimadzu – 17A series instrument (Kyoto, Japan) equipped with a flame-ionization detector (FID).A BP-20 fused silica capillary column (30 m × 0.25 mm i.d.× 0.25 μm f.t.)(Merck, São Paulo, Brasil) was used.The carrier gas was nitrogen at a constant flow of 1.0 mL/min and the column temperature was 170°C.

The injector and detector temperatures were set at 230°C, the injection volume was 2.0 μL (by manual injection), and the split was 1:50.After extraction with toluene, tobacco samples (50 μL) were mixed with 25 μL of IS (2.4 mg/mL in methanol) and after shaking for 10 s in a vortex-type shaker, 2 μL were injected into the GC-FID system.STATISTICAL ANALYSISResults are presented as the mean ± SD, using the Microcal Original Microcal Software, USA.STUDENT PITFALLSThis is an interesting subject for student understanding and learning, as smoking is indeed a world-wide problem and nicotine is the main tobacco component responsible for the induced dependence.Titration, as a rule, is a relatively simple technique that does not require expensive apparatuses and represents an appropriate laboratory exercise.However, students should be aware that, at the end point the addition of titrant must be stopped promptly and its volume read correctly to be used in the calculation.A careful examination of the calculation presented here will help in the understanding of fundamental concepts of stoichiometry involving molarity and normality, but students usually need help with these calculations.RESULTS AND DISCUSSION1Preliminary titration of a freshly prepared ∼0.1 N solution of NaOH with a standard 0.1000 N (0.0500 M) solution of oxalic acid, using phenolphthalein as indicator, gave 0.1010 for the normality of the NaOH solution (A).2An aliquot of 0.9 mL of the original solution of HClO4 was diluted to 100.0 mL (solution B) with water and titrated with solution A, giving 0.1040 for the normality (or molarity) of solution B. Therefore, the original HClO4 solution is 0.1040 × 100.0/0.9 = 11.56 N (or 11.56 M), which corresponds to, 11.56 × 100.46 (Mr of HClO4) = 1631.32 g of HClO4/L of the original solution or1631.32/1.764 (density of pure HClO4) = 658.3 mL/L.

As 1.66 g/mL = density of the original solution of HClO4, 1631.32 g/L = 1631.32 g/1660 g or 69.96 g/100.0 g, therefore within the limits established by the furnisher, namely 69–71% (m/m).3Based on the results above, we may conclude that 1000 mL of the original solution displays theoretically 1000–658.3 = 341.7 mL of water or 34.17% (v/v).Therefore to prepare 1 L of ∼0.1 N HClO4 in acetic acid as solvent we will need: 11.56 V = 1000 × 0.1 and V = 8.65 mL of the original solution containing 34.17% (v/v) water or 2.96 mL of H2O, to be eliminated by acetic anhydride (d = 1.08 g/mL, Mr = 102.09), which corresponds to 16.8 mL.Concluding, a ∼0.1 N anhydrous solution of HClO4 in acetic acid may be prepared containing: 8.65 mL of the original solution + 16.8 mL of acetic anhydride + glacial acetic acid up to 1 L. This solution (C) will be used next to titrate nicotine.To be sure that solution C was absolutely anhydrous and at least 0.1 N, the above solute volumes of solution C were increased to 9.0 mL and 18.0 mL, respectively, thus slightly increasing their concentrations as well as the ratio acetic anhydride/HClO4.

Complete absence of water in solution C was proved with the anhydrous CuSO4 test as described.4Titration of solution C was carried out with potassium biphtalate (here abbreviated as KHP, Mr = 204.22), which was previously dried at 110°C for 2 h. For the titration, 0.1125 g of KHP = 0.5509 mMoles (or mEq.)in 25 mL of glacial acetic acid, using now crystal violet as indicator, consumed 5.4 mL of solution C, which corresponds to N = 0.1020.In a second titration, 0.1253 g of KHP = 0.6136 mEq.consumed 6.0 mL of solution C, thus giving N = 0.1023, and finally 0.1022 in a third similar titration.The normality of solution C, to be used now as titrant of nicotine, was therefore assumed as 0.1021, the arithmetic mean of the three values above.On basis of the total mass of tobacco from 20 cigarette and the corresponding average mass/cigarette, the amount of nicotine in 14 cigarette brands, expressed in mg/g tobacco and mg/cigarette, is shown in Table I, whereas distribution of nicotine in the inhaled, vaporized and filter retained portions, as determined by use of our glass device (Fig.

2), is shown in Table II.It is clear now that, the amount of nicotine that commercial brand labels give (usually less than to 1 mg/cigarette) refers to the inhaled portion only, rather than to the total mass of nicotine/cigarette (around 10 mg).Even so, our data show (Table II) that the inhaled portion is much higher than that announced, at least for the 5 brands assayed.This difference may however be attenuated, if we consider that, in a completely open system, where access of atmospheric oxygen is maximal, part of the cigarette nicotine is destroyed during combustion.In our glass device however, the cigarette is confined inside the Erlenmeyer flask and access of oxygen is reduced.As the sum C + E + F in Table II practically equals the value found for total nicotine in mg/cigarette, the proposed system model allows a rather accurate analysis of nicotine distribution.Even considering that about 35% of the cigarette nicotine might be destroyed by combustion, the values in the Coil (C) column would be 1.07, 1.14, 1.06, 1.16, and 1.48 respectively, therefore still significantly higher than those given by the respective labels.Contents of total nicotine in commercial brand cigarettes as determined by acid-base titration(1) 0.918.29 ± 0.0314.05 ± 0.03(2) 0.714.73 ± 0.0310.93 ± 0.03(3) 0.519.92 ± 0.0412.02 ± 0.03(4) 0.817.22 ± 0.0312.01 ± 0.03(5) 0.918.86 ± 0.0412.64 ± 0.03(6) 0.615.33 ± 0.0310.53 ± 0.03(7) 0.713.99 ± 0.0310.22 ± 0.01(8) 0.715.75 ± 0.0311.31 ± 0.01(9) 0.816.66 ± 0.0411.16 ± 0.03(10) 0.717.17 ± 0.0311.08 ± 0.04(11) 0.617.50 ± 0.0310.27 ± 0.04(12) 0.818.47 ± 0.0410.58 ± 0.03(13) 0.718.31 ± 0.0111.94 ± 0.03(14) 0.812.37 ± 0.038.58 ± 0.03Contents of nicotine corresponding to the inhaled (coil), vaporized (Erlenmeyer), and retained (filter) portions(9) 0.81.64 ± 0.012.59 ± 0.016.91 ± 0.0111.1 ± 0.0411.16 ± 0.032.05 ± 0.01(10) 0.71.76 ± 0.001.48 ± 0.007.79 ± 0.0311.03 ± 0.0311.08 ± 0.012.11 ± 0.00(11) 0.61.63 ± 0.012.00 ± 0.016.68 ± 0.0010.31 ± 0.0010.27 ± 0.043.33 ± 0.02(12) 0.81.79 ± 0.002.59 ± 0.016.21 ± 0.0110.59 ± 0.0310.58 ± 0.032.24 ± 0.00(13) 0.72.28 ± 0.013.17 ± 0.006.46 ± 0.0311.91 ± 0.0411.94 ± 0.043.26 ± 0.02Figure 3, 4A–4C show, respectively, the HPLC calibration curve and the solvent, standard nicotine and sample (9) chromatograms for nicotine estimation.

The calibration curve shows an excellent linearity between concentration and peak height ratio along the range 0.28–9.07 mg/mL.Very important to our titration data is the observation that, in 8 tobacco brand extracts, none of them revealed any trace of contaminant alkaloids, and therefore our results represent practically the true nicotine contents of each tobacco analyzed.In addition, a preliminary evaluation of nicotine in three cigarette brands from Table I, namely (3), (9), and (11), revealed 16.44, 19.93, and 17.48 mg/g tobacco, therefore practically in agreement with the values found by titration, namely 16.66, 19.92, and 17.50 mg/g tobacco, respectively, which means that both nitrogen atoms of nicotine are protonated during titration before the indicator turning point is reached, because our calculations were based on the assumption of two equivalents/mole of nicotine.Figure 3.Open FigureDownload Powerpoint slideCalibration curve for HPLC determination of nicotine at concentration from 0.28 to 9.07 mg/mL.Figure 4.

Open FigureDownload Powerpoint slideA) Chromatogram of solvent; B) Chromatogram of standart nicotine (0.94 mg/mL) and cyclohexanecarboxylic acid (IS); C) Chromatogram of cigarette tobacco sample, showing no contaminant alkaloid.Peak identification: 1, nicotine; 2, IS (internal standard).A definitive confirmation of the validity and accuracy of the method was achieved by titration of a freshly prepared aqueous solution of nicotine, which was prepared by direct dissolution of an analytical standard nicotine (Merck, 98% purity, d = 1.01 mg/mL).For that, 100 μL of the standard nicotine were dissolved in 10 mL toluene + indicator and titrated with C. The volume of C consumed was 11.95 mL, exactly that theoretically expected (11.94 mL), as shown: 100 uL standard nicotine (d = 1.01 g/mL) = 101 mg.Correcting for 98% purity: 101 × 0.98 = 99 mg = 99/162.23 or o.61 mMoles = 1.22 mEq., which corresponds to 11.94 mL of C.STUDY QUESTIONS1How do you define equivalent and normality in acid-base neutralization reactions?2Extrapolate this concept, using the Avogadro number, to an oxidation-reduction equation, balancing it by matching up the electron (s) transfer.3Convert the molarity (M) of a nicotine solution into its corresponding normality (N) in the above shown nonaqueous titration.