Pre-requisite Reading- Type of reactions
Identify from the images, which of the following is an alkyl or an aryl halide?
An equimolar (1:1) mixture of a Lewis acid anhydrous ZnCl2 and concentrated HCl- Lucas Reagent, is used to identify and classify unknown alcohol (R-OH) as primary (1o), secondary (2o) or tertiary (3o). The Lucas test is based on the speed at which corresponding insoluble alkyl chlorides (R-Cl) are formed post reaction.
A sp3 hybridized carbon is a tetravalent carbon that forms single covalent bonds (sigma bonds) with atoms of other p-block elements- Hydrogen, Oxygen, Carbon, Nitrogen, Halogens, etc. The bonds formed are of equal strength and at an angle of 109.5o due to which the central carbon atom is tetrahedral in shape. Example, carbon of an alkane or an alkyl group.
Pre-requisite Reading- Valency of Elements, Modern Electronic configuration, Atomic orbitals, Concept of Hybridization
sp3 hybridization of Carbon
Ethanal is a two-carbon aldehyde, and propanone is a three carbon ketone. The first step is to make a Carbon-Carbon bond, and the Grignard reagent would be the best choice. The aldehyde ethanal is less of a methyl carbon so to introduce methyl substituent, methyl magnesium bromide is preferred. The carbon of the Grignard reagent is a nucleophile (CH3-) and will effortlessly attack the electrophilic carbonyl carbon (-CHO) of the aldehyde.
A Haloalkane is the halogen derivative of an alkane obtained by the replacement of one or more hydrogen atom. For example, when one hydrogen of an alkane, methane is replaced with a halogen, say chlorine, the haloalkane obtained is called as the chloromethane.
The Benzene is a planar, six-membered cyclic ring structure with alternating double bonds that are responsible for its aromaticity. The Benzene is a simplest aromatic hydrocarbon made up of only carbon and hydrogen without any other substituents. The molecular formula of Benzene is C6H6.
A primary carbon/nitrogen is denoted by numeral 1, secondary with numeral 2, tertiary with numeral 3 and quaternary with numeral 4 with a degree (o) sign in the superscript. Example, primary (1o), secondary (2o), tertiary (3o), quaternary (4o).
The Lewis dot structures are used to show the shared electron pairs between the bonded atoms in the molecule and the lone pairs of electrons if any. The representative structures follow the octet rule wherein the atoms combine by either transfer of electrons (loss or gain) or by sharing of valence electrons in a way that the valence shell of the atoms attains the octet configuration.
There are different types of arrow notations that are frequently encountered in Chemistry, mainly Organic Chemistry. Each one has a specific purpose and cannot be used interchangeably.
Few of the most common ones are:
A targeted approach to answering your 12th CBSE Board Question Paper can ensure good marks. Given below are a few tips, tricks, and strategies for scoring well in your Chemistry Board Exam.
1) For comparison of the relative oxidizing and reducing powers of various metals: A Positive number indicates stronger reduction potential of the electrode, and it functions as a stronger oxidizing agent. A negative number means weaker reduction potential of the electrode, and hence it is a more powerful reducing agent.
The electrochemical series or the activity series is the arrangement of various electrodes in the increasing order of their standard reduction potential. The reduction potential values are seen to increase from negative to zero and positive real numbers. When arranged linearly, it would look like a number line with the positive numbers on the right side of zero and the negative numbers on the left side of zero.
An electrode when in contact with an electrolyte solution of the similar ionic nature (Example, Cu electrode in CuSO4 solution, Zn electrode in ZnSO4 solution) tends to either undergo Oxidation (loss of electrons) or reduction (gain of electrons).
Due to this oxidation or reduction, there develops a charge separation between the metal electrode and its ions in the solution creating a potential difference.
If all the elements were to create a gated community for themselves; they would build four apartment blocks namely, the s-block, p-block, d-block and the f-block. The elements having similar property would be grouped together occupying one block and this behavior is based on the orbital the last electron chooses to enter.
Markovnikov’s rule was proposed by Russian chemist Vladimir Markovnikov in the year 1869. It is the condition that applies to the unsymmetrical alkene or alkyne predicting the regiochemistry of the addition of hydrogen halide to give alkyl halide. As per the Markovnikov rule,
‘For an unsymmetrical alkene or alkyne, when treated with hydrogen halide HX, the negative part of the reagent attaches to carbon having less number of hydrogen across the double bond giving alkyl halide.’
In short, the negative ion attacks the most substituted carbon with less hydrogens across the double bond. A mnemonic or shortcut to remember this is-
‘Markovnikov says - NO Member Shall Cheat.'
NO stands for Negative iOn
Member Shall Cheat for Most Substituted Carbon.
The negative ion comes from the hydrogen halide (HX) that is polar in nature. The electronegative halogen carries a partial negative charge and the hydrogen a partial positive charge. Amongst the hydrogen halides, the reactivity order is-
The acidity of HI is highest in the group as it is easy to break the HI bond. The conjugate base I- is more stable and nucleophilic than HBr, HCl and least is for HF.
The addition of haloacid (HX) to the double bond is a two step process and goes via the electrophilic addition mechanism.
In the first step, the π bond of the nucleophilic alkene first picks up the positively charged proton to give a carbocation intermediate. As the 2o carbocations are stabilized by hyperconjugation and induction it is more stable than 1o hence when a double bond picks up a proton; it tends to form more stable 2o carbocation than 1o.
After the formation of the carbocation at most substituted carbon, the attack of the negatively charged halogen takes place giving the alkyl halide. The loss of one π bond and formation of two σ bonds makes the reaction exothermic and energetically favorable.
If two carbon atoms of the double bond are equally substituted by hydrogen, then two products would be formed in equal ratios.
When hydrogen halide adds to the alkyne, it gives vinyl halide. In the presence of an excess of Hydrogen halide, a second addition of HX results in geminal dihalides. The mechanism is similar to the alkene addition forming a stable carbocation followed by the attack of the bromide ion.
Let’s look at few examples of alkenes following Markovnikov's Rule for the addition of HBr wherein the bromide will add to the most substituted carbon or the carbon carrying less hydrogens across the double bond-
From 1-Propene after Markovnikov’s addition of HBr, we get 2-bromopropane.
2-methyl-prop-1-ene on the addition of HBr according to the rule gives 2-bromo-2-methyl-propane.
2- butene is a symmetrical alkene (equally substituted) and it need not follow a rule for the addition of HBr across the double bond. Addition from either side would give the same product.
The symmetrical nature of 2-butene is lost on substituting a methyl at the 3rd position to give 3-methyl-but-2-ene and it now it obeys Markovnikov’s rule to give 2-bromo-2-methylbutane.
Remember, the formation of most stable carbocation is preferred by Markovnikov’s rule.
The Markovnikov reaction rule for addition to the unsymmetrical alkene is not only followed by the hydrogen halide but also by other reagents such as water, halogen, and water, Water and H2SO4, Iodine monochloride, with the more electronegative atom carrying the negative charge.
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The crystalline solid is made up of the unit cells. Therefore, the properties of the unit cell turn out to be the properties of the crystalline solid. In a unit cell, the length of the edges and the angle between the edges are fixed. Therefore, the entire crystalline solid would exhibit fixed value for both these parameters.
Differentiation between Crystalline and Amorphous solids based on the Cleavage Property
If a crystalline solid is cut with a sharp object, it would always give parts with smooth edges whereas an amorphous solid would cut into surfaces with rough, uneven edges. Crystalline solids are therefore said to have cleavage property, and amorphous solids do not show cleavage property.
Imagine you are passing a raw potato through a slicer blade; what you get is thin slices of potato with smooth edges. But if you pass a boiled potato through a slicer, it would crumble and fall apart giving small uneven pieces with rough edges. The raw potato represents a crystalline solid and the boiled potato acts like an amorphous solid.
Why Crystalline Solids show Cleavage Property?
The cleavage property is shown by crystalline solids because they possess cleavage planes. In a crystalline solid, the cells are neatly stacked. The cleavage planes are areas where the crystal structure is the weakest. It is only along these planes that a crystalline solid can be cut. Therefore, a cut from a sharp object would give two smooth parts. Amorphous solids do not show any cleavage planes.
If you look at the crystal structure, you will notice a constant arrangement of the unit cells. In the two-dimensional diagram, it is seen as a lattice. Usually, the cuts made in the direction of the linear sequence of points shown in green are preferred over other cuts. The cuts along the direction of red are not allowed. These are the cuts that can shatter or disintegrate the crystal. In other words, the cuts that preserve the arrangement of the particles are preferred in a crystalline solid.
An example is of the diamond, a crystalline solid that when cut along the cleavage planes gives small diamonds with smooth edges having the same arrangement of the particles as that of the parent diamond.
Let me give you another example. A NaCl crystal structure has a cubic arrangement. Its cleavage plane is parallel to the cube faces. If cut along any other plane as shown in red, the crystal structure will shatter.
Every crystal has a unique cleavage plane depending on the arrangement of the particles. For example, crystals gypsum, feldspar and calcite have one, two and three cleavage planes.
The amorphous solid quartz does not have well-defined cleavage planes. It does not show cleavage property and breaks unevenly giving rough edges.
In summary, crystalline solids show cleavage property and is the reason why crystalline solids are smoother than the amorphous solids. Cut along the cleavage plane results in getting crystalline solid with smooth edges. Cleavage planes exist due to the ordered arrangement of the atoms thereby giving smaller crystalline solids of same geometric arrangement as the parent. On the other hand, the constituent particles of the amorphous solids are randomly arranged and do not show cleavage property. They break into uneven pieces with rough edges when cut.
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