Difficulty: Intermediate | Prerequisites: Organic Chemistry I basics, familiarity with nucleophilic substitution and elimination concepts
This experiment sits at the heart of second-semester organic chemistry: understanding how and why alcohols react with hydrogen halides to form alkyl halides. You are looking at competing reaction pathways (SN1 substitution vs E1 elimination) and learning to predict which one dominates based on substrate structure and reaction conditions. If you missed the lectures on carbocation stability or nucleophilicity, go back to those first. The lab also introduces core purification techniques (liquid-liquid extraction, drying, simple distillation) that recur throughout the course.
t-Butyl alcohol reacts with concentrated HCl via an SN1 mechanism to form t-butyl chloride (2-chloro-2-methylpropane) as the major product, with isobutylene (2-methylpropene) as a minor E1 by-product. The reaction proceeds through a tertiary carbocation intermediate, and the SN1 pathway is favoured because chloride is a good nucleophile and the conditions do not promote elimination.
SN1 (Unimolecular Nucleophilic Substitution)
A two-step substitution mechanism in which the leaving group departs first to form a carbocation intermediate, followed by nucleophilic attack. The rate depends only on the concentration of the substrate. In simple terms, the molecule falls apart on its own, then the nucleophile swoops in.
E1 (Unimolecular Elimination)
A two-step elimination mechanism sharing the same carbocation intermediate as SN1, but a base removes a proton from the carbocation to form an alkene instead of a substitution product. Think of it as the carbocation losing a proton rather than gaining a nucleophile.
t-Butyl Alcohol (2-Methyl-2-propanol, tert-butanol)
A tertiary alcohol with the formula (CH₃)₃COH. The tertiary carbon is key: it stabilises the carbocation intermediate, making SN1 and E1 pathways accessible.
t-Butyl Chloride (2-Chloro-2-methylpropane)
The SN1 product of this reaction. Formula: (CH₃)₃CCl. Boiling point approximately 51 °C.
Isobutylene (2-Methylpropene)
The E1 elimination product. Formula: (CH₃)₂C=CH₂. A gas at room temperature (boiling point approximately −7 °C), which is one reason it may not be observed in the product collection.
Carbocation
A positively charged carbon intermediate. Tertiary carbocations (three alkyl groups attached) are the most stable due to hyperconjugation and inductive effects. In simple terms, more alkyl groups means more electron density shared with the positive carbon, which keeps it from being so desperate for electrons.
Nucleophile
A species that donates an electron pair to an electrophilic centre. In this reaction, Cl⁻ acts as the nucleophile, attacking the carbocation.
Leaving Group
The group that departs with the bonding electrons. Here, water (formed after protonation of the hydroxyl group) is the leaving group.
Theoretical Yield
The maximum mass of product calculated from stoichiometry, assuming complete conversion of the limiting reagent and no losses. In simple terms, the best you could possibly get on paper.
Limiting Reagent
The reactant that is completely consumed first, capping the maximum amount of product. Identified by comparing moles of each reactant relative to the stoichiometric ratio.
Percent Yield
The ratio of actual yield to theoretical yield, multiplied by 100. Tells you how efficient your reaction and purification were in practice.
Reactants: t-butyl alcohol + concentrated HCl (37% w/w aqueous)
SN1 product (major): t-butyl chloride, (CH₃)₃CCl
E1 product (minor): isobutylene, (CH₃)₂C=CH₂, plus water and HCl as by-products
The reaction is run at or near room temperature with an excess of HCl
Step 1, Protonation: HCl protonates the hydroxyl group of t-butyl alcohol, converting -OH into -OH₂⁺ (a good leaving group, water)
Step 2, Ionisation: The C-O bond breaks heterolytically. Water departs, generating a tertiary carbocation, (CH₃)₃C⁺. This is the rate-determining step.
Step 3, Nucleophilic attack: Chloride ion (Cl⁻) attacks the carbocation, forming t-butyl chloride
Curved arrows: each step should show electron-pair movement from the electron-rich site to the electron-poor site. In Step 2, the arrow goes from the C-O bond to oxygen. In Step 3, from a lone pair on Cl⁻ to the positive carbon.
Two reasons commonly cited in this course:
Chloride is a good nucleophile, not a strong base. Cl⁻ is large, polarisable, and nucleophilic, so it attacks the carbocation carbon rather than abstracting a proton. Stronger bases (e.g. alkoxide) would favour elimination.
Low temperature. The reaction is carried out near room temperature. Elimination reactions generally have higher activation energies and are favoured at elevated temperatures. Keeping the temperature low tilts the competition toward substitution.
Key values you need for the reagent/product table:
t-Butyl alcohol: MW = 74.12 g/mol, density = 0.775 g/mL, bp = 82 °C
HCl (gas): MW = 36.46 g/mol (the aqueous solution is 37% w/w with density approximately 1.19 g/mL)
t-Butyl chloride (SN1 product): MW = 92.57 g/mol, bp approximately 51 °C
Isobutylene (E1 product): MW = 56.11 g/mol, bp approximately −7 °C
Finding moles: Convert grams (or mL via density) of each reactant to moles. For the HCl solution, use: grams of solution × 0.37 = grams of HCl, then divide by 36.46 g/mol.
Limiting reagent: Compare moles of t-butyl alcohol to moles of HCl in a 1:1 ratio. The one with fewer moles limits the reaction.
Theoretical yield: Moles of limiting reagent × MW of product (92.57 g/mol for t-butyl chloride)
Percent yield: (actual yield in grams / theoretical yield in grams) × 100
The observed boiling point range of the distilled product is compared to literature values. If it falls near 51 °C, the product is t-butyl chloride. Isobutylene (bp −7 °C) is a gas at room temperature, so it would have escaped during the reaction or workup, which is precisely why it may form but not be collected.
Isobutylene is a gas at room temperature. Any that forms simply bubbles out of the reaction mixture and is lost. To detect it, you would need to redesign the experiment with a gas collection apparatus (e.g. an inverted graduated cylinder over water, or a cold trap) to capture and quantify the gaseous alkene.
\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\text{Theoretical yield (g)} = \text{moles of limiting reagent} \times \text{MW of product}\% \text{ Yield} = \frac{\text{actual yield (g)}}{\text{theoretical yield (g)}} \times 100\text{Grams HCl from solution} = \text{mass of solution (g)} \times 0.37\text{Mass of solution (g)} = \text{volume (mL)} \times \text{density (g/mL)}Quick Reference Table:
Compound | MW (g/mol) | Density (g/mL) | bp (°C) |
|---|---|---|---|
t-Butyl alcohol | 74.12 | 0.775 | 82 |
HCl (gas) | 36.46 | n/a | n/a |
37% HCl (aq) | n/a | 1.19 | n/a |
t-Butyl chloride | 92.57 | 0.842 | 51 |
Isobutylene | 56.11 | gas at RT | −7 |
The SN1 reaction of tertiary alcohols with HX is one of the simplest routes to tertiary alkyl halides, which are useful synthetic intermediates in pharmaceutical and agrochemical manufacturing. Understanding competing substitution and elimination pathways is also central to designing selective syntheses in industry, where unwanted elimination products represent wasted material and lower yields.
"SN1 and E1 have different intermediates." They do not. Both pathways share the same carbocation intermediate. The difference is what happens next: nucleophilic attack (SN1) or proton loss (E1).
"A stronger nucleophile always means more substitution." For SN1 reactions this is less relevant because the rate-determining step is carbocation formation, not nucleophilic attack. Nucleophile strength matters more in SN2.
"Tertiary substrates always give SN1." Tertiary substrates favour SN1 over SN2, but they can still give E1 or E2 products depending on the base strength and temperature. The competition between SN1 and E1 is controlled by conditions.
"Percent yield above 100% means you made extra product." A yield above 100% indicates impurities or residual solvent in your product, not a miraculous surplus.
⚠️ You will almost certainly be asked to draw the full SN1 mechanism with curved arrows. Practise until you can do it from memory: protonation, ionisation (rate-determining step), nucleophilic attack.
⚠️ Be prepared to explain why SN1 is favoured over E1 for this specific reaction. The two standard reasons are: Cl⁻ is a good nucleophile (not a strong base), and low reaction temperature favours substitution.
⚠️ Know how to calculate theoretical yield and percent yield from scratch, including handling the 37% w/w HCl solution (converting solution mass to HCl mass).
⚠️ Expect a question on why the E1 product might not appear in your collected product. The answer: isobutylene is a gas at room temperature and escapes during the reaction.
True or False: The SN1 mechanism is a one-step process. (False, it is two steps: ionisation then nucleophilic attack, with protonation as a preliminary step.)
Fill in the blank: The rate-determining step in SN1 is the formation of the ________. (carbocation)
True or False: Isobutylene is a liquid at room temperature. (False, it is a gas with a boiling point of approximately −7 °C.)
Fill in the blank: To find the grams of HCl in a 37% w/w solution, multiply the mass of the solution by ________. (0.37)
True or False: Higher temperatures would increase the proportion of E1 product. (True.)
Q: Draw the complete SN1 mechanism for the reaction of t-butyl alcohol with HCl, including all intermediates and curved arrows.
A: Step 1, protonation of the -OH by HCl to form (CH₃)₃COH₂⁺. Step 2, departure of water to form the tertiary carbocation (CH₃)₃C⁺ (rate-determining step). Step 3, attack of Cl⁻ on the carbocation to give (CH₃)₃CCl.
Q: Give two reasons why SN1 is favoured over E1 in this experiment.
A: (1) Chloride ion is a good nucleophile but a weak base, so it preferentially attacks the carbocation rather than abstracting a proton. (2) The reaction is run at low (near room) temperature, which kinetically favours substitution over elimination.
Q: A student uses 2.0 mL of t-butyl alcohol (density 0.775 g/mL) and excess HCl. Calculate the theoretical yield of t-butyl chloride.
A: Mass of t-butyl alcohol = 2.0 mL × 0.775 g/mL = 1.55 g. Moles = 1.55 / 74.12 = 0.0209 mol. Theoretical yield = 0.0209 × 92.57 = 1.94 g.
Q: If the student collects 1.40 g of product, what is the percent yield?
A: (1.40 / 1.94) × 100 = 72.2%.
Q: Why might E1 product form during the reaction but not appear in the distilled product?
A: The E1 product (isobutylene) has a boiling point of −7 °C and is a gas at room temperature. It escapes from the reaction vessel and is not captured by the simple distillation setup.
Q: How could you modify the experiment to detect and isolate the E1 product?
A: Use a gas collection apparatus such as a cold trap or an inverted graduated cylinder over water to capture gaseous isobutylene as it forms.
This material connects directly to SN2 and E2 mechanisms covered earlier in the course. The key insight is that substrate structure (primary, secondary, tertiary) determines which mechanism dominates. Tertiary substrates favour SN1/E1; primary substrates favour SN2/E2. This experiment also ties into the broader theme of selectivity in organic synthesis: choosing reagents and conditions to steer a reaction toward the desired product.
The purification techniques used here (extraction, drying, distillation) reappear in nearly every subsequent lab, so mastering them now pays off throughout the course.
Related Terms / Search Tags
t-butyl alcohol, tert-butanol, 2-methyl-2-propanol, t-butyl chloride, 2-chloro-2-methylpropane, isobutylene, 2-methylpropene, SN1 mechanism, E1 elimination, unimolecular nucleophilic substitution, carbocation intermediate, tertiary carbocation, nucleophilic attack, leaving group, theoretical yield, percent yield, limiting reagent, HCl reaction with alcohol, alkyl halide synthesis, CHEM 2311, organic chemistry lab