All groups will be using the data from all the labs for your post lab. You will need to use EXCEL or another type of database software to complete this lab.

Assigned readings:

Both eds. - Chapter 9, sections 9.3 and 9.4 and Chapter 5, sections 5.7 and 5.10.

Lecture Materials:

For this lab you will need a spreadsheet program, EXCEL and Google sheets are two examples but you are not limited to these. There are other good free online programs available.

For a tutorial on EXCEL - https://www.excel-easy.com/

For a tutorial on Google sheets - https://zapier.com/learn/google-sheets/google-sheets-tutorial

The calculations needed and how to set up the spread sheet are included in the link below.

Calculation work sheet for EXCEL or other spreadsheet software and Expt-2.docx

There is a bit of confusion on the calculation for the line C. It should be -

ln ( [HCl]_inf- [HCl]_t ) be careful with the parentheses!

Additional Materials:

Polar Protic? Polar Aprotic? Nonpolar? All About Solvents

http://www.masterorganicchemistry.com/

Solvents can cause considerable confusion in reactions, because they’re listed along with the reagents of a reaction but often don’t actually participate in the reaction itself.

A solvent is a liquid that serves as the medium for a reaction. It can serve two major purposes:

(Non-participatory) to dissolve the reactants. Remember “like dissolves like” ? Polar solvents are best for dissolving polar reactants (such as ions); nonpolar solvents are best for dissolving nonpolar reactants (such as hydrocarbons).
Participatory: as a source of acid (proton), base (removing protons), or as a nucleophile(donating a lone pair of electrons). The only class of solvents for which this is something you generally need to worry about are polar protic solvents (see below).

OK. So what does “polar” and “non-polar” mean?

Polar solvents have large dipole moments (aka “partial charges”); they contain bonds between atoms with very different electronegativities, such as oxygen and hydrogen.
Non polar solvents contain bonds between atoms with similar electronegativities, such as carbon and hydrogen (think hydrocarbons, such as gasoline). Bonds between atoms with similar electronegativities will lack partial charges; it’s this absence of charge which makes these molecules “non-polar”.

There are two common ways of measuring this polarity. One is through measuring a constant called “dielectric constant”. The greater the dielectric constant, the greater the polarity (water = high, gasoline = low). A second comes from directly measuring the dipole moment.

There’s a final distinction to be made and this causes confusion. Some solvents are called “protic” and some are called “aprotic”. What makes a solvent a “protic” solvent, anyway?

Protic solvents have O-H or N-H bonds. Why is this important? Because protic solvents can participate in hydrogen bonding, which is a powerful intermolecular force. Additionally, these O-H or N-H bonds can serve as a source of protons (H+).
Aprotic solvents may have hydrogens on them somewhere, but they lack O-H or N-H bonds, and therefore cannot hydrogen bond with themselves.

For the average first semester student, these distinctions come up the most in substitution reactions, where hydrogen bonding solvents tend to decrease the reactivity of nucleophiles; polar aprotic solvents, on the other hand, do not.

Sn1 overview:

The general form of the S_N1 mechanism is as follows:

Because the mechanism goes through a carbocation, the leaving group must be attached to either a tertiary or secondary carbon to stabilize the intermediate. A methyl or primary leaving group will not form a carbocation. Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction. The polarity and the ability of the solvent to stabilize the intermediate carbocation is very important. So for our purposes, the more polar the solvent and in the case of SN1 reactions, the faster the rate.

Understanding the energy diagram for this reaction will help to understand the lab. Click on the link below for a video lecture on this reaction.

http://www.masterorganicchemistry.com/videos/sn1-reaction-energy-diagram/

Elements of Titration

The standard solution is the solution of known concentration. An accurately measured amount of standard solution is added during titration to the solution of unknown concentration until the equivalence or endpoint is reached. The equivalence point is when the reactants are done reacting.
The solution of unknown concentration is otherwise known as the analyte. During titration the titrant is added to the analyte in order to achieve the equivalence point and determine the concentration of the analyte.
The equivalence point is the ideal point for the completion of titration. In order to obtain accurate results the equivalence point must be attained precisely and accurately. The solution of known concentration, or titrant, must be added to the solution of unknown concentration, or analyte, very slowly in order to obtain a good result. At the equivalence point the correct amount of standard solution must be added to fully react with the unknown concentration.
The end point of a titration indicates once the equivalence point has been reached. It is indicated by some form of indicator which varies depending on what type of titration being done. For example, if a color indicator is used, the solution will change color when the titration is at its end point. Care must be taken not to over titrate. Pay attention to the desired color and proceed slowly as you near the end point.

Titration is an analytical chemistry technique used to find an unknown concentration of an analyte (the titrand) by reacting it with a known volume and concentration of a standard solution (called the titrant). Titrations are typically used for acid-base and redox reactions. Here's an example problem determining the concentration of an analyte in an acid-base reaction:

Titration Overview (For a chem 1 review see Chapter 8 in Bettleheim)

A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. Note the volume of HCl. In Expt. 10 you will need to figure out the volume of the HCl solution. No worries it is in the reading!

NaOH + HCl --> Na⁺_(aq) + Cl^-_(aq) + H₂O_(l)

What was the concentration of the HCl in Molarity?

Step-by-Step Solution

Step 1 - Determine [OH^-]

Every mole of NaOH will have one mole of OH^-. Therefore [OH^-] = 0.5 M.

Step 2 - Determine the number of moles of OH^-

Molarity = # of moles/volume

# of moles = Molarity x Volume

# of moles OH^- = (0.5 M)(.025 L)
# of moles OH^- = 0.0125 mol

Step 3 - Determine the number of moles of H⁺

In the balanced equation, the stoiciometry of the NaOH to HCl is 1:1 so when the base neutralizes the acid, the number of moles of H⁺ = the number of moles of OH^-. Therefore the number of moles of H⁺ = 0.0125 moles.

Step 4 - Determine the concentration of HCl in Molarity

Every mole of HCl will produce one mole of H⁺, therefore the number of moles of HCl = number o fmoles of H⁺.

Molarity = # of moles/volume

Molarity of HCl = (0.0125 mol)/(0.050 L)
Molarity of HCl = 0.25 M

Answer

The concentration of the HCl is 0.25 M.

Experimental Notes and Changes:

Pages 172 and 173 show how the equations are derived. This is not part of the experimental procedure and is for information only.

We will prepare the solutions in the preceding lab sections and store them until needed. I will assign the solvent mixes to each group at that time. You will make 125 ml of your solvent mix, stopper the flask and label it with your names, the date and the contents. The flask will be stored until the following week.

You will add 1.25 g of the t-butylchloride to the reaction flask, not 1.0g.

Prelab questions:

In this lab you will calculate the final concentration of your product by indirect measurement. We will never actually measure the product formed. Using the two balanced equations for this lab, explain how you can calculate product concentration. (Hint: one equation is the reaction in the labbook, the other is the titration reaction)

Postlab Materials:

Data analysis for Expt 10.

You need to calculate Theoretical, actual and % yields as well as atom economy and costs for this lab.

Some notes to help you solve this:

1. [HCl] means the molarity of the HCl solution.

quick note - to calculate your molarity of HCl - remember that you used 10 ml of solution to titrate each time. so after you calculate the number of moles of HCl you need to divide by 0.01 L to get molarity!

2. Use the infinity average to calculate your actual yield in moles of t-butyl alcohol. Not molarity here just moles.

3. When calculating your actual yield remember you analyzed 10 ml of your reaction mixture. Do not forget to correct for the total volume of your reaction mixture - 125 ml.

4. You get the theoretical yield in the usual way, calculating from the amount of reactants in your first balanced equation.

You will need to create 1 graph

When you create the graphs - use the scatter plot function and draw the trendline through those points.

1. Plot all the data from your lab section on one graph according to the instructions in the lab.

You do not have enough information to answer postlab question 3.

You will need both of these graphs for the post lab questions. Remember slope = rate!

As always - come see me if you have questions. None of these reactions are difficult but putting them together to get your answers can get confusing.