The setup
Part 2: Distillation
- Organic products from a few groups were poured into a measuring cylinder and the volume measured was recorded.
- After that, a distillation setup was made and the products were poured into the pear-shaped flask.
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The pear-shaped flask was started to be heated and product from the fraction in the range of 47oC and 53oC was collected.
- The volume of collected 2-chloro-2methylpropane was measured again using a measuring cylinder.
The setup:
Observations:
- Concentrated hydrochloric acid, 2-methylpropan-2-ol and 2-chloro-2methylpropane were all colourless.
- The concentrated hydrochloric acid had an irritating smell.
- There were some dense white fumes above the surface of the concentrated HCl.
- There were 2 immiscible colourless layers in the separation funnel.
- The anhydrous sodium sulphate was in white granule form.
- There were colourless gaseous bubbles formed when sodium hydrogencarbonate was added into the separation funnel.
- The separation funnel turned warm when concentrated hydrochloric acid was added into the separation funnel with the alcohol.
- There was gas pressure built up in the separation funnel when concentrated HCl was added.
- After the pouring the organic product into anhydrous sodium sulphate, the boiling tube turned cool.
Data:
Volume of 2-methylpropan-2-ol used = 9.0 cm3
Initial weight of the boiling tube with stopper = 74.4g
Final weight of the boiling tube with stopper and organic product = 80.5g
Volume of product collected for the final distillation = 24cm3
Volume of cyclohexene obtained = 23cm3
Calculations:
Initial weight of the boiling tube with stopper = 74.4g
Final weight of the boiling tube with stopper and organic product = 80.5g
So, the weight of product obtained = 80.5 – 74.4 = 6.1g
Average portion of 2-chloro-2-methylpropane in the product mixture =
Molar mass of 2-chloro-2-methylpropane = (35.5 + 12.0×4 + 9.0) = 92.5 gmol-1
Number of moles of 2-chloro-2-methylpropane obtained = × = 0.063 mol
Since the equation of the reaction is
So, the mole ratio of 2-methylpropan-2-ol and 2-chloro-2-methylpropane is 1:1.
Volume of 2-methylpropan-2-ol used = 9.0 cm3
Density of 2-methylpropan-2-ol = 0.7809 gcm-3
Weight of 2-methylpropan-2-ol used = 9.0 × 0.7809 = 7.0g
Molar mass of 2-methylpropan-2-ol = (12.0×4 +16.0 +10×1.0) = 74.0 gmol-1
The theoretical number of moles of 2-chloro-2-methylpropane obtained = Number of moles of 2-methylpropan-2-ol = = 0.095 mol
Percentage yield of the 2-chloro-2-methylpropane = × 100% = 66.3%
Discussion:
Mechanism of the nucleophilic substitution reaction
Why is the percentage yield not 100%?
- There was a loss of product during the extractions.
First, when we distilled out the 2-chloro-2-methylpropane from the reaction mixture, some 2-chloro-2-methylpropane might be vapoured and condense and stuck to the wall of the pear-shaped flask, the thermometer and other apparatus it passed through. So, the product obtained was slightly reduced.
Second, the boiling point of cyclohexene is only 51oC, so it is quite volatile. During the transfer of the product between boiling tubes and the separation funnel, some of the cyclohexene might be vapourized and lost. The vapourization could also occur in the separation funnel where the cyclohexene turned to vapour inside the separation funnel.
Third, we added anhydrous sodium sulphate to the product after using the separation funnel. There might be some product that stayed with the sodium sulphate granules after pouring the product out which led to a loss of product.
- There was a side reaction.
2-methylpropan-2-ol may undergo elimination which forms methylpropene. The equation is as below:
How to suppress the elimination reaction?
Nucleophilic substitution and elimination are actually 2 competing reactions undergoing at the same time and there are some factors affecting which reaction dominates the other.
- Concentration of nuclephile
Substitution is preferred when a concentrated nucleophile is used. In this experiment, concentrated acids were used, so the concentration of the nucleophiles is very high which favours elimination.
- Temperature
High temperature favours elimination reaction, that is one of the reasons why we add concentrated HCl little by little instead of adding 20cm3 altogether, as the reaction is exothermic which release larger amount of heat, then the rate of reaction of elimination may increase (although it is still very slow).
The other reason for adding concentrated HCl little by little will be discussed later.
- Distillation of cyclohexene from the reaction mixture when it was formed. This is important because the cyclohexene formed may react with water to form back cyclohexanol. Using high temperature can ensure that cyclohexene reaches its boiling point immediately after formation and being distilled out.
- Polarity of the solvent.
Low polarity solvent favours elimination which high polarity solvent favours substitution.
In this experiment, we used water as a solvent which is of a high polarity, so this increases the rate of substitution.
- Structure of the reactant
3o carbon increases the stability of the carbonium ion formed in the intermediate.
With the SN1 pathway, a carbonium ion is formed as an intermediate. The carbonium ion has higher stability with a tertiary carbon because of inductive effect. R groups (CH3 in this experiment) are electron donating groups which disperse the charge of the carbonium ion, hence increasing the stability. Therefore, reaction through this passway is more likely to happen.
- Strength of nucleophile
Strong nucleophile such as KOH and RO- favours elimination while weak nucleophile such as ROH favours nucleophilic substitution.
Comparison of the mechanisms of the nucleophilic substitution and elimination through SN1 and E1 pathway.
SN1 against SN2
There are actually 2 pathways to undergo nucleophilic substitution which is SN1 and SN2. These two reactions actually take place at the same time. The mechanism of SN1 is discussed above which involves a carbonium ion. SN2 does not require the formation of carbonium ion but it requires the collision of the nucleophile and the OH group.
For SN2, the mechanism is as below,
This pathway also forms 2-chloro-2-methylpropane
In this experiment, the rate of SN1 reaction dominates. The reasons are:
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The carbonium ion formed from 2-methylpropan-2-ol is a tertiary carbonium ion which increases its stability, so SN1 pathway is favoured.
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As 2-methylpropan-2-ol is tertiary, there are 3 bulky CH3 groups blocking which caused steric hindrance. This is very difficult for the nucleophile to perform a back side attack which makes the rate of SN2 very slow.
Difference in the products in these 2 pathways
For SN1, after the carbonium ion is formed. The chance for the nucleophile to undergo a backside attack and a front-side attack is about 50-50. This is not exactly 50-50 because the leave OH- gives a little bit repulsion to the coming Cl-. So, the chance of a backside attack is slightly higher. However, this effect is negligible as the ratio is only around 53:47.
If the product has an optical center (which is not in this experiment), the final product can be considered as optical inactive and it forms a racemic mixture.
For SN2, the attack of the nucleophile must be a backside attack because a half-bond is formed in the intermediate where the carbon becomes sp2 hybridized and the shape becomes trigonal planar. This pathway does not form mirror image but only one kind of product.
If the product has an optical center, the final product is considered to be optical active as the product is not a racemic mixture.
Other ways of doing this experiment
PCl3, PCl5 and SOCl2 can be to to replace HCl with the formation of different side products.
Why do we need to add concentrated HCl little by little?
- As mentioned above, high temperature favours elimination reaction. If we add the HCl too quickly, the temperature of the mixture will become very high. This increases the rate of the unwanted side reaction which wastes the reagent.
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The reaction is exothermic, so after adding HCl, the temperature of the mixture will increase quite readily. However, the boiling point of 2-methylpropan-2-ol is 82oC while the boiling point of 2-chloro-2-methylpropane is only 51oC. Adding HCl too quickly will result in the temperature approaching their boiling points or even reach their boiling points. The organic liquid will then vapours very quickly and there will be a massive loss of product and reagants which will surely make percentage yield very low or even make the experiment fail.
Why are there some dense white fumes above the HCl liquid surface?
At room temperature, HCl is a colorless , which forms white fumes of upon contact with atmospheric . These fumes are highly acidic and should be avoided breathing in or contacting. This is why the transfer of concentrated HCl is suggested to be done in the fume cupboard.
After removing the aqueous layer, we pour the organic layer from the top. Why?
As discussed in the previous experiment, when we poured the final organic product out of the separation funnel, we poured them out from the top instead of opening. This is because there are some aqueous solution which may contain HCl and other impurities which may contaminate the organic product.
What is the use of wire gauze?
The wire gauze is made of metal which is a very good conductor of heat, so the whole wire gauze maintain a quite even temperature as heat is quickly transfer to the end of the wire gauze. This helps in the heating of the reaction mixture as this ensures that the reaction mixture is evenly heated.
This can prevent cracking of glass and uneven temperature in the reaction mixture which cause unwanted materials to be distilled out or not enough heat for other reagents to react.
Actually, the pear-shaped flask also helps with heating evenly, as the end of the flask is quite point-shaped which heats better than a flat-shaped base.
Other errors in this experiment
Besides loss of product and reagants, there were also some errors.
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The percentage of the final yield was just an average of some groups, which did not necessarily mean that the fraction of cyclohexene in our group’s product was . There might be some error in this part of the calculation.
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Moreover, although the distillation products were collected from 47oC to 53oC. This does not necessarily mean that the distillation product was purely 2-chloro-2methylpropane. There might be some impurities distilled out together with the product we want.
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We were only using a measuring cylinder to measure each reagent, which was not quite accurate. Therefore, there might be a maximum error of 0.25cm3 as the measuring cylinder only had markings every 0.5cm3.
When that measuring cylinder was used to measure 9.0cm3 of 2-methylpropan-2-ol, the maximum percentage error was × 100% = 2.78%
In addition, the measuring cylinder was used to measure volumes 3 times. This might contributed quite a lot on the error in this experiment.
Further experiments: Adding AgNO3 solution
With a little bit of left-over product (2-chloro-2-methylpropane) in a boiling tube, a few drops of AgNO3 were added into the boiling tube to test for whether the product is a halogen compound.
After adding the AgNO3 solution, white ppt. was formed which confirmed that the product was actually a chloro compound.
This can be confirmed as AgCl(s) is white in colour, AgBr(s) is pale yellow, while AgI(s) is yellow.
After the ppt. was formed, the boiling tube was brought to direct sunlight. This caused the white ppt. to turn grey and slightly purple. This was actually a thermal decomposition and the equation is:
2AgCl 2Ag + Cl2
White Grey
The product was slightly purple because the silver was not evenly deposited. There were many cavities which diffracts lights, so a slightly different colour may be observed.
However, this test is slightly not too accurate because carboxylic acids have been known to react in this test, giving false positives.
Reference: