2019 Paper 1 Suggested Solution

2019 A Level H2 Chemistry Paper 1 Question 22 - Oxidation of Alcohols for Compound Q

Here's 2019 A Level H2 Chemistry Paper 1 Question 22.

2019 p1 q22 question

We need to deduce the product formed when compound Q is heated under reflux with an excess of acidified potassium dichromate (VI).

Notice there are only alcohol functional groups in compound Q.

Different types of alcohols are oxidised to different extent when reacting with acidified dichromate, heated under reflux.

Primary alcohols are oxidised to carboxylic acids.

Note that primary alcohols can be oxidised to aldehydes but only with acidified dichromate, heating under reflux with immediate distillation.

Secondary alcohols are oxidised to ketones.

Tertiary alcohols cannot be oxidised.

2019 p1 q22 oxidation of alcohols

Check out my previous video lesson for the detailed discussion of oxidation of different types of alcohols.

So we can now run through the options and figure out our answer!

Option A is wrong as the primary alcohol is not oxidised to aldehyde.

2019 p1 q22 deduce A

Option B is wrong as the secondary alcohol should be oxidised.

2019 p1 q22 deduce B

Option C is the correct answer.

2019 p1 q22 deduce C

Option D is wrong as the tertiary alcohol should not be oxidised.

2019 p1 q22 deduce D

Hence the answer to this question is option C.

Topic: Alcohols, Organic Chemistry, A Level Chemistry, Singapore

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2019 A Level H2 Chemistry Paper 1 Question 21 - Determine Functional Groups that React with NaOH

Here's 2019 A Level H2 Chemistry Paper 1 Question 21.

2019 p1 q21 question

The functional groups present in thyroxine (and mentioned in the options) are:

- carboxylic acid
- phenol
- iodobenzene

The reactions to consider with NaOH are as follows:

1. Acid-base or neutralisation of acidic functional groups

We should be familiar with the acidity of carboxylic acids, phenols and alcohols.

For students who are interested, check out a previous video comparing acidity of organic compounds:

The reactions of acids, phenols and alcohols with bases are given below:

2019 p1 q21 reaction of acidic functional groups with bases

Acids and phenols are acidic enough to react with NaOH.

2. Nucleophilic Substitution of halogenoalkanes

Since the iodine atoms are attached to benzene, they are resonance stabilised hence will not undergo nucleophilic substitution.

3. Hydrolysis of nitriles, esters and amides

There are no functional groups in thyroxine that can be hydrolysed.

Hence we can conclude only -COOH and -OH(phenol) groups will react with NaOH and the answer to this question is option D.

Topic: Carboxylic Acids and Derivatives, Organic Chemistry, A Level Chemistry, Singapore

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2019 A Level H2 Chemistry Paper 1 Question 19 - Friedel Crafts Acylation of Methylbenzene

Let's take a look at 2019 A Level H2 Chemistry Paper 1 Question 19.

2019 p1 q19 question

We are required to predict the intermediates for the Friedel Crafts Acylation reaction.

This reaction is not in syllabus but we can apply what we know from the mechanism of electrophilic substitution for benzene to figure out the intermediates.

If you are not familiar with electrophilic substitution mechanism, do check out this video lesson that I've done previously.

Let's draw out the mechanism part by part.

1. Generate Electrophile

2019 p1 q19 generate electrophile

AlCl3 acts as a Lewis Acid and accepts electron pair from Cl- (of acid chloride) to form carbocation intermediate [CH3CO]+ and AlCl4-

The carbocation acts as the electrophile which will attract benzene which is electron rich.

Hence looking at the options we can eliminate B and D where the [CH3CO]- intermediates are negatively charged.

2. Electrophilic Attack

2019 p1 q19 electrophilic attack on methylbenzene

Benzene will use its delocalised pi system to attack the [CH3CO]+ intermediate and form a highly unstable intermediate.

Notice the [CH3CO]+ electrophile is added to position 4 with respect to methyl group since alkyl groups are 2,4-directing.

Also, the benzene carbon that is attached to the -COCH3 group is sp3 hybridised or saturated.

Hence the region around that sp3 carbon has no delocalisation of pi electrons, and we have to draw the "ring opening" around that carbon.

So we will eliminate option D where the "ring opening" is around methyl group.

Finally we can conclude the answer to this question is option A.

Topic: Arenes, Organic Chemistry, A Level Chemistry, Singapore

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2019 A Level H2 Chemistry Paper 1 Question 20 - Deduce Chloroalkane Forming Racemic Mixture via SN1 Mechanism

Here's 2019 A Level H2 Chemistry Paper 1 Question 20.

2019 p1 q20 question

Given chloroalkane X undergoes nucleophilic substitution to form a racemic mixture.

A racemic mixture is an equimolar mixture of optical isomers.

This means the alcohol product must contain a chiral carbon, and chloroalkane X must also have a chiral carbon.

Also, we know that there are 2 pathways that nucleophilic substitution can occur - SN1 or SN2 mechanism.

Only SN1 mechanism can form a racemic mixture, so the mechanism for this question has to be via SN1.

2019 p1 q20 deduction

For a detailed discussion of this mechanism, check out this previous video that I've done on nucleophilic substitution mechanism of halogenoalkanes.

Primary halides will favour SN2 mechanism, while tertiary halides will favour SN1 mechanism.

SN1 mechanism is a two-step mechanism where the C-X bond is broken to form carbocation and halide X- in the first step.

Tertiary halides form tertiary carbocations which are bonded to 3 electron-donating alkyl groups.

2019 p1 q20 stability of tertiary carbocation

This will stabilise the carbocation significantly, which favours its formation and in turn favours SN1 mechanism.

Since we know that the mechanism for this question is via SN1, we would expect the chloroalkane X to be a tertiary chloroalkane.

Finally we can run through the options keeping in mind X must have the following criteria:

- chiral carbon
- tertiary chloroalkane

Option A - CH3(CH2)6Cl

2019 p1 q20 option A

This compound has no chiral carbon so will not be the answer.

Option B - (CH3CH2)3CCl

2019 p1 q20 option B

This compound has no chiral carbon so will not be the answer.

Option C - (CH3)2CHCH2C(CH3)2Cl

2019 p1 q20 option C

This compound has no chiral carbon so will not be the answer.

Option D - (CH3)2CHC(CH3)(CH2CH3)Cl

2019 p1 q20 option D

Finally we found our chiral carbon which is also a tertiary chloroalkane.

So the answer to this question will have to be option D.

Topic: Halogenoalkane, Organic Chemistry, A Level Chemistry, Singapore

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2019 A Level H2 Chemistry Paper 1 Question 18 - Deduce Free Radical Substitution Products

Let's take a look at question 18 for 2019 A Level H2 Chemistry Paper 1.

2019 p1 q18 question

Propane undergoes free radical substitution with chlorine and we want to predict the possible products.

We need to be familiar with this mechanism to deduce the products quickly.

I have a previous video lesson on free radical substitution mechanism for those who are interested.

Let's go through each compounds and see if they can be formed.

1. CH3CH2CH2Cl, 1-chloropropane

2019 p1 q18 1 chloropropane

This is a monosubstituted product which can be formed in the propagation step of FRS.

2. CH2ClCH2CH2Cl, 1,3-dichloropropane

2019 p1 q18 13 dichloropropane

This is a disubstituted product which can also be formed in the propagation step.

Remember free radical substitution is a random process and multiple substitution at any carbon is possible.

3. CH3CH2CH2CH2CH2CH3, hexane

2019 p1 q18 hexane

Hexane can be formed when a propyl radical meets another propyl radical and forms a carbon-carbon bond in the termination step.

2019 p1 q18 form c c bond in termination step

Since the probability of termination step is very low, hexane is found in trace amounts.

4. CH3CH2CH2CH2CH3, pentane

2019 p1 q18 pentane

We can only form pentane from propane if a propyl (3C) radical meets an ethyl (2C) radical.

Since carbon-carbon bonds are not broken during free radical substitution, ethyl radicals cannot be formed.

This means it's not possible to form pentane.

In fact the alkane products that can be formed must be a multiple of the starting unit.

If propane is the starting alkane, the alkane products that can be formed will be a multiple of 3-carbon, ie 6 carbon, 9 carbon, 12 carbon and so on.

Finally we can conclude that only compounds 1, 2 and 3 are possible products.

Hence the answer to this question will be option A.

Topic: Alkanes, Organic Chemistry, A Level Chemistry, Singapore

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