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Perfect Chemistry #Connoisseur Bonjour, Let's Enjoy Chemistry!

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12/09/2020

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30/09/2018

CARBOCATIONS-1 (PROJECT IIT/NEET)-01

Carbocations (01102018)- SPSPIIT01

1, On the basis of hyperconjugation, among alkyl carbocations, tertiary butyl carbocation has been shown to be most stable, because in the following reaction, (CD3)3C⊕ + (CH3) 3CH = (CH3)3C⊕+ (CD3)3CH; the equilibrium constant, K = 1.97 shows that the reaction proceeds more towards right confirming that tertiary butyl carbocation is more stable than (CD3)3C⊕. There appears to be less hyperconjugation in deuterium containing carbocation (CD3)3C⊕ than (CH3)3C⊕. It has been shown that of all the alkyl carbocations, the ter-butyl carbocation is most stable. Even the relatively stable t-pentyl and t-hexyl cations break up at higher temperatures to produce the t-butyl cation. Even methane, ethane and propane in presence of superacids produce t-butyl cation as the main product. Even paraffin wax and polyethylene give t-butyl cation.

2. There appears to be less hyperconjugation in deuterium containing carbocation (CD3)3C⊕ than (CH3)3C⊕.

3. Of all the alkyl carbocations, the ter-butyl carbocation is most stable.

4. (CD3)3C⊕ + (CH3) 3CH = (CH3)3C⊕+ (CD3)3CH has value of the equilibrium constant, K >1.

5. Acetyl cation CH3CO⊕ is as stable as ter-butyl cation.

6. Tropylium cation is (1011 times) more stable than the Ph3C+.

7. MeOCH2Cl is solvolysed (atleast 1014 times) faster than CH3Cl.

8. 1,2,3-tripropylcyclopropenyl cation (is 103 times) more stable than tropylium cation.

9. Reactivity towards nucleophilic displacement by EtO- : MeCH2Br > Me CH2CH2Br > Me2 CH CH2Br > Me3CCH2Br.

10. The reaction of 1-aminopropane with sodium nitrite and dilute HCl gives 7% 1-propanol, 28% propene and 32% propan-2-ol (chlorides, nitro, nitrites, ethers, higher amines, higher cations MeCH+CH2R and so on etc.)

11. The reaction of aliphatic primary amine with nitrous acid is not a good synthesis of alcohols as it gives a complex mixture of products.

12. C6H5CH2C(OH)Me2 with FSO3H/SbF5 generates C6H5CH+CHMe2.

13. The direct ionisation of alkyl halides gives carbocations is accelerated due to the presence of heavy metal ions like Ag⊕. Hence, SN1 is favored.

14. MeCOF with BF3 gives MeCO+BF4- whereas Me3CCOCl with AlCl3 gives Me3C+, AlCl4- and CO(g).

15. RF and SbF5 gives R+SbF6- whereas MeCH2CH2Me with SbF5/FSO3H gives Me3C+SbF5FSO3-.

16. Cyclopropylmethyl cation has been found to be more stable than benzyl cation and the stability increases with each addition of cyclopropyl groups. The vacant p orbital of cationic carbon lies parallel to C2–C3 bond of the cyclopropane ring and not perpendicular to it. Thus the geometry becomes similar to that of a cyclopropane ring conjugated with an olefinic bond.

17. The presence of a heteroatom adjacent to the cationic centre and bearing an unshared pair of electrons, like nitrogen, oxygen or halogen increases stability (stabilized by resonance).

21/08/2018
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14/03/2018

Hydrogen spectrum- wavelengths

12/03/2018
Sharad Pratap Singh

Sharad Pratap Singh

Problem- Fact- Cr2+ is a reducing agent while Mn3+ is an oxidizing agent in spite of both having d4 configuration. When Cr2+ gets oxidized to Cr3+ (d3), it attains a stable half-filled t2g orbitals but when Mn3+ gets reduced to Mn2+ (d5), it attains a stable half-filled d orbitals. Question- If this is the case, why can't Cr2+ get reduced to Cr+ to attain a stable half-filled d orbitals and Mn3+ get oxidized to Mn4+ to attain a stable half-filled t2g orbitals?? [ 71 more words ]

https://sharadpra.wordpress.com/2018/03/12/cr2-is-a-reducing-agent-while-mn3-is-an-oxidizing-agent-transition-elements-12-03-18/

23/02/2018

Sharad Pratap Singh

Starting from the most acidic form as given, keep removing the most acidic hydrogen then less acidic until the most basic form is obtained. Now isoelectric pH is the average of pKa on (just) either side of zwitterion (neutral) form.
The First hydrogen of COOH having least pKa will be lost and we get structure having net charge +1. Then hydrogen of NH3+ having pKa 8.95 is lost and we obtain the neutral form. Then in last step hydrogen of distant NH3+ is lost and we obtain the structure with net charge -1. So find the average of pKa on either side of neutral form that is second and third pKa here.
The scheme will be same for acidic amino acids having a greater number of carboxyl groups however here, isoelectric pH will be average of pKa of two COOH groups. Whereas in case of basic amino acids as in given question, the pI will be average of pKa of two NH3+ groups.
A zwitterion is a dipolar ion. So it must be neutral as the net charge in that form.

22/02/2018

Sharad Pratap Singh

Why do transition metals in zero or lower oxidation states form complexes with weak acidic ligands such as CO or NO?

22/02/2018

Sharad Pratap Singh

20/02/2018
Sharad Pratap Singh

Sharad Pratap Singh

Isomers Of tri-substituted cyclopropane e.g., tri-methyl cyclopropane or tri-chloro cyclopropane etc.---- (a)1,1,2- isomer contains a chiral carbon and hence exist as a pair of enantiomers. (b) 1,2,3-isomers -- There are three chiral centres, so the maximum number of stereoisomers is 2^3=8 (however due to identical chiral carbons, less number of stereoisomers). Here are the four mirror image pairs. Let's say that up is U and down is D and name the members of each pair as [ 162 more words ]

https://sharadpra.wordpress.com/2018/02/20/isomers-of-tri-substituted-cyclopropane/

20/01/2018

DO SOLVE IT and analyse

JEE ADVANCED 2018 CHEMISTRY MOCK TEST

22/12/2017

BITSAT-2018 online exam announcement.

The link to apply for writing BITSAT-2018 online exam for admission to Pilani, Goa and Hyderabad campuses for 1st semester 2018-19 is now open. The last date for applying online is 5 PM on 13th March 2018. CLICK HERE to apply online.

BITSAT-2018 online exam announcement.

The link to apply for writing BITSAT-2018 online exam for admission to Pilani, Goa and Hyderabad campuses for 1st semester 2018-19 is now open. The last date for applying online is 5 PM on 13th March 2018. CLICK HERE to apply online.

http://www.bitsadmission.com/bitsat/bitsatapply/OnlineReg.aspx

02/12/2017
GOC | Carbocation Cyclisation | IIT | JEE | AIIMS | NEET | Class-11 | with Sharad Pratap Singh

Generally, simple alcohol (-OH) is better protonation site than simple alkene.
However, alkene become better protonation site if it has greater electron density due to +R effect of oxygen or nitrogen.

Conc. H2SO4 cause dehydration. Dil. H2SO4 cause hydration.

Hydration of alkyne converts it into vinylic cation which is attacked by water to give enol and it further tautomerise into ketone.

Hydration of ethers -When positive charge is on oxygen, One alkyl group is lost and attacked by H2O in SN1 orSN2 fashion to give mixture of ROH.

Generally, simple alcohol (-OH) is better protonation site than simple alkene. However, alkene become better protonation site if it has greater electron dens...

23/11/2017
Isomers of di-halocyclobutane | Stereoisomerism for | IIT | JEE | AIIMS | with Sharad Pratap Singh

The correct statement regarding dihalogenation of cyclobutane is/are-
(A) It gives two pairs of geometrical isomers
(B) One of the dihaloderivative does not show geometrical isomerism
(C) It gives three meso compounds
(D) It produces only one pair of enantiomers..
(E) Total six products are formed.
(F) Four products are “achiral” and hence optically inactive

Isomers of di-substituted cyclobutane or di-halo cyclobutane or di-methyl cyclobutane. C4H6Y2--- [The cis 1,2- or cis-1,3- dichloro cyclobutane has two of su...

06/09/2017
Does cyanide show the ortho effect in benzoic acid? If not, then what are the exceptions of the ortho effect in benzoic acid?

Does cyanide show the ortho effect in benzoic acid? If not, then what are the exceptions of the ortho effect in benzoic acid?

Sharad Pratap Singh's answer: Cyano-group is a linear group and hence is not supposed to cause any steric hinderance to the group present at ortho-position on the aromatic ring. However, all three mono-substituted cyano-benzoic acids are more acidic than simple benzoic acid, ortho-isomer being t...

22/08/2017
For a 109% labelled oleum, if the number of moles of H2SO4 and free SO3 are x and y respectively, then what will be the approximate value...

For a 109% labelled oleum, if the number of moles of H2SO4 and free SO3 are x and y respectively, then what will be the approximate value...

Sharad Pratap Singh's answer: Oleum is mixture of H2SO4 and free SO3. Labeling of oleum is based on the total amount of H2SO4 formed after reacting 100 g oleum with just enough H2O. The additional mass comes form amount of H2O required to convert all SO3 present in 100 g oleum into complete H2SO...

27/05/2017
What do " cross-over products " mean in organic chemistry?

What do " cross-over products " mean in organic chemistry?

Sharad Pratap Singh's answer: These are most often used to distinguish between intramolecular and intermolecular reactions In a crossover experiment, two similar but distinguishable reactants simultaneously undergo a reaction as part of the same reaction mixture. The products formed will either ...

27/05/2017
How can one explain the ring expansion effect from pyrrole into 3 chloro pyridine?

How can one explain the ring expansion effect from pyrrole into 3 chloro pyridine?

Sharad Pratap Singh's answer: This is ‘abnormal” Reimer Tiemann reaction product”. (1) The alkaline chloroform(CHCl3 + alc. KOH or t-BuO-/t-BuOH) generates intermediate dichlorocarbene, :CCl2. (2) Carbenes give variety of reactions, like insertion reactions etc. (these behave as both an electro...

21/04/2017
What is the degree of unsaturation of cubane? How?

What is the degree of unsaturation of cubane? How?

Sharad Pratap Singh's answer: “We do the difficult immediately. The impossible takes a little time.” Cubane(C8H8) appears to possess six rings, corresponding to the six faces of a cube. However, in the case of hydrocarbons, you can use a simple trick- count the deficiency of hydrogens as compare...

27/03/2017
How do I study Ionic Equilibrium for JEE Advanced?

How do I study Ionic Equilibrium for JEE Advanced?

Sharad Pratap Singh's answer: > * “The present is the equilibrium of the past and the future. (Le présent, c'est l'équilibre - Du passé et du futur)” ― Charles de Leusse * Remember, * * CH3COOH has Ka = 1.8*10^(–5) and pKa = 4.74; * For NH3, pKb = 4.77 * Sq.rt. of 1.8 and 18 are 1.34 an...

27/03/2017
How can we identify/differentiate nucleophile from bases?

How can we identify/differentiate nucleophile from bases?

Sharad Pratap Singh's answer: A base is the reagent which abstracts a proton (acidic hydrogen) from the reactant. A nucleophile is a reagent which attacks on any other electrophilic atom in the reactant. [code]The Nucleophilic character is a kinetic term. Nucleophilic character is proportional ...

19/03/2017
Sharad Pratap Singh

Sharad Pratap Singh

Unless rings are large enough (Bredt’s Rule), At small bicyclic bridgehead positions, planarity is difficult to attain, in other words, bridgehead carbon cannot have a pure “p” orbital. Hence, Bridgehead free radicals are pyramidal. Bridgehead carbocations are not possible. At bridgehead carbons, the double bond is not possible. Unlike I, II (being a bridgehead olefin) does not exist. II has excessive steric strain as the rigid cage structure prevents the bridgehead carbon from being planar (flatten out). [ 146 more words ]

https://sharadpra.wordpress.com/2017/03/18/bredts-rule-and-decarboxylation

09/03/2017
Sharad Pratap Singh

Sharad Pratap Singh

I just love to break norms prudently and so in love with judicious exceptions. Anyway it's awesome to know secrets. Isn't it? Inorganic chemistry is complex, I agree. It gets difficult to decide which factor predominates in a particular question. Infact it is a highly practical science and that makes it interesting for some persons like me. Three sources are basic necessity to go through once- ncert books, exemplar problems and all last year questions. [ 471 more words ]

https://sharadpra.wordpress.com/2017/03/08/how-to-read-inorganic-chemistry

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