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Born Haber Cycle

Mr Khemistry

Born Haber Cycle

Born Haber Cycle Diagram

Born Haber Cycle

Using the acronym F.A.I.L. where F stands for enthalpy change of formation of the ionic compound, A stands for atomisation of the constituent elements and I stands for ionisation of cation and anion and L stands for Lattice Energy.

We have four major components and this energy cycle is easy to draw with no worries about which species having a higher energy level. There is no need to include electrons as they are immediately used to form the anion.

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JC Chemistry Tuition in Bishan

Mr Khemistry

JC Chemistry Tuition in Bishan

We will be starting classes at Bishan this June, on Monday and Friday evenings. For a start, there will be two JC2 classes and one JC1 class. Went down to the area to take a professional photoshoot and it was quite good. 

JC Chemistry Tuition in Bishan

The location of the unit is very convenient, walking distance from the interchange. Do check out our class schedule if you are interested to sign up for H2 Chemistry Tuition.

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Organic Novel Questions Book

Mr Khemistry

Organic Novel Questions Book

Novel Organic Questions Book

By the time this post is published, hopefully we would have finished the proofreading and sent for printing. Distribution to the bookstores might take a month or so but i hope to make it in time for the 2019 A levels Examination. This is my second book and i hope it will cater to a different group of students. If you recall, my first book was to meet the demand from private candidates who needed practical worksheets for practice. This Organic Novel Questions Book hopes to address the needs of students who are competent doing the standard questions and want to stretch themselves. 

This doesn’t mean that average students shouldn’t buy the book. It only means they shouldn’t buy it to use as a guide book. This is like the apex of organic chemistry, students need to be familiar with the basics first. 

I already have parents and students asking when will the book be out, i can only hope soon. Hopefully students will find it a useful tool to sharpen their critical thinking. Instead of giving thousands of questions, i have chosen to select 30 of the most challenging ones. It is my belief that quality thinking is far more important than rote learning. Of course practising basic standard questions is a given but once passed that stage, students should not be looking for repetition. 

Side note on Organic Novel Questions Book

Just a note though, do not be disappointed even if you can’t do any of these questions (some of these reactions are only taught in University Level 1 modules). It is very likely none of these questions will come out in A levels! What i hope for is that it shows you how we can think about such novel questions. Obviously this thinking process is different for everyone, i’m just showing you a few ways to approach such questions. 

One last word, if you found the book useful, please recommend it to your friends! I would be most grateful 🙂

All the best for your A levels!

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FRS Chemical Environment

Mr Khemistry

FRS Chemical Environment

FRS Chemical Environment

U and V are structural isomers with molecular formula C7H16.

Both U and V produce the same number of possible structural isomers upon reaction with chlorine to form monochlorinated compounds with the formula C7H15Cl.

Which pairs are possible structures of U and V?

Or

Have a look at the different colour coded hydrogen atoms. Those with the same colour is the same type of hydrogen, i.e. they will form the same product if they are substituted by chlorine. 

Do look out for plane of symmetry as it will greatly accelerate the speed at which you identify FRS Chemical environment.

So answer is 1 and 3 only!

If you’ve enjoyed this organic MCQ, do share it with your friends and see if they are able to do it quickly.

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Free Radical Substitution MCQ Analysis

Mr Khemistry

Free Radical Substitution MCQ

This MCQ is about free radical substitution, be sure to read up before looking at the solution!

Which of the following is a possible product in the reaction between cyclohexane and bromine with UV light?

Free Radical Substitution

Analysis:  The common answer dicyclohexane isn’t one of the given options. So we will have to choose from the above options.

dicyclohexane

 

 

 

 

 

Solution:

A: Note that this option has a total of 14 carbon atoms, not 18. The cyclohexane radicals all has 6 carbon atoms so the products should be in multiples of 6.

B: Similarly, this option has a total of 13 carbon atoms, not 18.

C: This option has 16 carbon atoms.

D: This option has 18 carbon atoms and is the correct answer! Do remember that cyclohexane can undergo multiple abstraction of H atom by the bromine radical.

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Organic Chemistry MCQ Analysis

Mr Khemistry

Organic Chemistry MCQ

Organic Chemistry MCQ: Which of the following compounds exhibit cis–trans isomerism?

    1. organic molecule
    2. organic molecule 2
    3. organic molecule 3

Analysis: The question is straightforward. Which compounds exhibit cis-trans isomerism? In the Chemistry H2 syllabus, we learnt about the two criteria for cis-trans isomerism.

  • Presence of C=C bond
  • Different atoms/group of atoms bonded to each of the carbon atoms

Although in the H2 Chemistry syllabus, it is required that C=C bond cannot be in a cyclic structure.

Looking at the three options, looks like none of them can exhibit cis-trans isomerism! But this cannot be the case, let us examine the options carefully.

The three options examined

Option 1: Although there’s no C=C, there is a four-carbon ring structure which restricts rotation. Moreover, the two central carbon atoms each have different atoms bonded to it. So it is able to display cis-trans forms!

Option 2: This is outright wrong as it violates the rule that C=C bond cannot be in a cyclic structure. (But we shall see that the cyclic structure is only limited to 7 carbon ring and below)

Option 3: This is not in the H2 syllabus as cis-trans isomerism can be exhibited by C=C bond in a cyclic structure having more than 7 carbon atoms. Click here to find out more

This Organic Chemistry MCQ isn’t the run-of-the-mill type. Two of the options involves knowledge outside of the H2 syllabus. Most likely, this question is meant to challenge the stronger ability students to read beyond their syllabus requirements. Also, it expose students to a novel context that is slightly outside of their comfort zone.

In our weekly classes, we let students have a chance to try out these challenging questions to stretch them. If you liked doing this question, come join us and broaden your horizon!

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Solving Organic Deductive Questions

Mr Khemistry

Solving Organic Deductive Questions

Compound J, C8H6O4, forms a molecule K, C8H10O3, when reacted with LiAlH4 in dry ether. However, when J was reacted with NaBH4 instead, a product of molecular formula C8H8O4 was formed.

J gives an orange precipitate with 2,4-dinitrophenylhydrazine but does not give any precipitate when warmed with Fehling’s reagent. K reacts with bromine water to form L, C8H7O3Br3.

Suggest the structures of J, K and L. 

Solution:

First information is the molecular formula, it has almost the same C to H ratio and has more than six carbons => likely to have benzene ring C6H6

(Note that there is still two carbon atoms not accounted for)

When it undergoes reduction with LiAlH4, 4 hydrogen atoms were added but 1 oxygen atom was removed. => likely to have one COOH group, accounts for 2 oxygen atoms

But when it undergoes reduction with NaBH4, only 2 hydrogen atoms were added. 
=> confirmed one of the reduction by LiAlH4 was on COOH

J undergoes condensation with 2,4-DNPH but does not undergo oxidation with Fehling’s reagent. => carbonyl but not aliphatic aldehyde. Possible to be aromatic aldehyde or ketone.

K  undergoes electrophilic substitution (tri-substitution) with bromine water => phenol present and also 2,4,6 positions does not have any substituent groups. This means that the two remaining carbon atoms must be at 3 and 5 positions. Additionally, this means that ketone is no longer possible as we would need two carbon on the same substituent group.

Summary: From the above deductions, J has phenol, aromatic aldehyde and carboxylic acid on positions 1,3,5.

K has phenol and two primary alcohols.

L has tri-substituted Br on positions 2,4,6, phenol, aromatic aldehyde and carboxylic acid on position 1,3,5.

deductive question

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Distinguishing Organic Compounds

Mr Khemistry

Distinguishing Organic Compounds

Distinguishing alkenes

 

 

 

 

Suggest one simple chemical test to distinguish them from each other. State the reagents and conditions needed and give the expected observations for each compound.

 

Analysis:

This pair of compounds are both dienes, meaning they both contain two C=C bonds. 

We will not be able to distinguish them by using chemical tests for alkenes. This is true for any pair of compounds with the same functional group(s). For example, esters, secondary alcohols, amides etc.

The tip is to distinguish them base on their structures. Use structure-specific tests such as iodofoam test and KMnO4.

For the pair of compounds above, note that the double bonds are closer for cyclohexa-1,3-diene. Using hot acidified KMnO4 will yield carboxylic acid for both compounds but one of the product for cyclohexa-1,3-diene is is ethandioic acid which will further oxidize to carbon dioxide and water.

Answer:

Add KMnO4(aq), H2SO4(aq) and heat to both compounds.

Observation: Both solutions will change from purple to colourless but the solution of cyclohexa-1,3-diene will produce effervescence of CO2 that form white ppt with limewater

Extra practise:

distinguishing esters

 

 

Distinguishing organic compounds is one of the easier types of questions in H2 Chemistry, be sure to score here!

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How to solve Organic Synthesis Question

Mr Khemistry

Organic Synthesis Question

Organic synthesis question

In any synthesis questions, we first need to note the number of carbon atoms, is there an increase or decrease in the carbon chain?

In the above question, yes the carbon chain increased from 8 to 9 carbons. This informs us that somewhere in the intermediate steps, a step-up reaction involving nitrile group will be needed. 

Then we take a look at the changes in the functional groups. We are starting with alkene and ending with alkene and carboxylic acid group.

To insert a nitrile group, we need to first add halogenoalkane functional group as a precursor. This is a common way to introduce the CN group via nucleophilic substitution in the H2 Chemistry Syllabus.

Next we need to consider how to keep the C=C group in the target molecule. Inevitably, when we add the halogen to the C=C, the alkene functional group will disappear. So how do we reform the C=C after adding the halogen? 

There are 3 options,

  1. HCl
  2. Cl-Cl in tetrachloromethane
  3. Cl-Cl in aq

First option is out as it will convert the functional group to halogenoalkane, which leaves us no way to reform the alkene through elimination.

Second option is also out as both Cl will be substituted by CN group.

Option 3 is our answer as Cl will be added to the the terminal carbon atom and OH will be added to the more substituted carbon atom (Markovnikov Rule) Choosing the correct reagents and condition for the first step is crucial to solving the question as it will set the course of synthesis.

Sequencing the Synthesis Route

After which we can just substitute the terminal Cl with CN and hydrolyse it to COOH. The alkene group can also be reformed via elimination of water.

Organic Synthesis Question

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