These are the elements which invove the filling of d-subshell. And these elements lie in between the s-block and p-block elements so these are known as transition elements.
The rate constant of a first order reaction is 3.0 x 10 -8 s-1 . what is the rate of the reaction when the concentration of reactant is 1.5 mol L-1 ?
Question : For the reaction : N2 + 3H2 → 2NH3 , the rate of reaction measured as ∆[NH3]/∆t was found to be 2.4 x 10-4 mol L-1 s -1 . Calculate the rate of reaction expressed in terms of (i) N2 and (ii) H2 .
Calculation of the rate of reaction.
Here we have some examples
- Calculate the rate of reaction expressed in terms of N2O5 for the following reaction
here is the solution
Students find it difficult to learn the various topics related to periodic classification of elements. We have worked on it and made it easy. just hit the play button and understand how , what and why .
ਆਵਰਤੀ ਸਾਰਣੀ ਨੂੰ ਆਸਾਨੀ ਨਾਲ ਯਾਦ ਕਰਨ ਲਈ ਆਪ ਸਭ ਲਈ ਇਹ ਵੀਡੀਓ ਬਣਾਈ ਗਈ ਹੈ। ਇਸ ਨੂੰ ਧਿਆਨ ਨਾਲ ਦੇਖਣਾ ਅਤੇ ਕਾਪੀ ਤੇ ਨੋਟ ਵੀ ਕਰ ਲੈਣਾ।
ਇਸ ਵਿੱਚ ਪੂਰੇ ਪਾਠ ਬਾਰੇ ਜਾਣਕਾਰੀ ਦਿੱਤੀ ਗਈ ਹੈ।
Most of the students find it difficult to learn and apply the knowledge related to the organic nomenclature.
Here in this post you simply need to memorise the following table and after that there are a few steps related to that.
It can be done by the following reaction.
The direct hydration of alkene leads to the formation of alcohols.
The reaction takes place according to the markovnikov’s rules for addition reaction.
we can make 2-methyl propan-2-ol (tertiary – butyl alcohol) from 2- methyl propene (Isobutylene) as shown in the above reaction.
The ethylene chlorohydrin can be prepared by the reaction of ethylene with Hypochlorus acid (Hypohalus acid)
It is a kind of addition reaction.
It is a type of addition reaction in which the hydroxyl and halogen groups are introduced in alkene molecule .
The alkenes react with freshly prepared solution of hypohalus acid. The acid is formed by the reaction of dihalogen with water.
The hypohalus acid gives Halohydrin on reacting with alkene.
The mechanism for this kind of reaction involves the formation of free radicals, and the relative stability of different molecules will decide the formation of major and minor products.
Step 1 , Chain initiation :- First of all , the homolytic cleavage takes place in peroxide molecule as shown below :-
Step2 Chain propagation :-
In this step , the bromine free radicals formed in first step 1 attack on the alkene
Step 3 Termination step:-
In this step all the free radicals are consumed and so we get different products.
It is the phenomenon which is seen in case of addition of hydrogen halide in the presence of a peroxide like benzoyl peroxide.
The major product in this kind of reaction is the alkane in which the halide is attached to terminal or the less substituted carbon atom as shown below.
This is due to the following reason :-
This is in accordance with sytzeff rule that whenever two alkenes are theoretically possible during a dehydrohalogenation reaction , it is always the more highly substituted alkene which predominates .
It can also be best understood by the following illustrations showing the stabilizes of primary(1°) , secondary (2°) and tertiary (3°) carbocations.
The reaction involves the formation of carbocation as shown above. The stabilities of the relative carbocations are shown below.
It is clear from the above structures that the uppermost carbocation is most stable and it leads to the product formation.
Thus from complete discussion we come to the conclusion that it gives following reaction .
The haloalkanes give alkenes on undergoing beta elimination. But in case if we there is unsymmetrical haloalkane , the product formation follows Sytzeff rule as shown below.
In case of unsymmetrical alkenes the addition of hydrogen halide takes place according to the markovnikov’s rule .
During this reaction under markovnikov’s rule the positive part of the attacking reagent joins with the carbon with more number of Hydrogen and the negative part of attacking reagent joins with the more substituted carbon.
We can make haloalkanes from the alkenes by treating them with rhe hydrogen halides HX .
Here the double bond is converted into single bond , hydrogen and halogen gets attached to the carbons involved in double bond.
We can make the dibromoethane from ethene by the addition of halogen in ethene , in the presence of carbon tetrachloride.
In the presence of ether as a solvent the above reaction takes place. The dihalo products are formed in which the halogen groups are present on adjacent carbon atoms.
By catalytic hydrogenation we can get the alkanes from the alkenes.
The reaction takes place by reacting ethene with hydrogen in the presence of platinum and pladium and Rany nickel (active form of Nickel)
Here is the reaction :-
The mechanism involves the following steps :-
Step 1 :- In first step the polarity is developed on the bromine molecule , and the partial positive side of the bromine molecule attacks on the carbon atom of the double bond. This leads to the formation of carbocation.
Step 2 :- In second step the negative part of the attacking reagent attacks on the carbon with positive charge.
Alkenes give electrophillic addition reactions because of the following reasons:-
1. The pie electrons of the double bond attract electrophiles .
2. The electron density is more on the C-C double bond , so the electrophiles preferably attack on it as compared to the electrophiles which are repelled.
In this reaction the decarboxylation of dicarboxylic acids take place .
The potassium salts of dicarboxylic acid give alkene as the COO- group is removed as CO2 , this leads to the formation of double bond between two adjacent carbon atoms.