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SUBSTITUTION REACTIONS:

WEAKER A BASE, STRONGER WILL BE ITS LEAVING TENDENCY.

STRONG BASE, WILL REPLACE A WEAK BASE.

COMPARISON FOR SN1 and SN2:

SN1 SN2

1. One step Rxn Two step Rxn

2. Electronic factors r significant. Not Significant.

3. Carbocation Formation Transition State

4. Rearrangement Possible Not Possible

5. Elemental Effect shown HERE also Shown

6. Nature of nucleophile Dependent on nature of nucleophile

independent

7. Concentration of Nu- independent Direct Relation

8. Polar Protic Favours Polar Aprotic Favours.

SOLVATING ABILITY OF POLAR PROTIC SOLVENTS FOR IONIC COMPOUNDS:

H20:

Polar Protic solvent's NEGATIVE PART will solvate the Cation nicely.

IONIC COMPOUND ke (hindi) ANIONIC PART Ka(hindi) SOLVATION ESSENTIALLY HAPPENS THROUGH H-BONDING.

THUS POLAR PROTIC SOLVENTS EFFECTIVELY SOLVATE THE IONIC BOND.

Q.H3COH (SOLVATING ABILITY NOT AS GOOD AS H2O) WHY...????

A. 1. Size is much more big, O can't project itself that nicely...ie the charge density will become less, so it won't be able to show its -ve charge to solvate the cation nicely.

2. No. Of Hydrogen, which can form H-bonds are less. THERFORE ANIONIC PART WON'T BE STABILIZED THAT NICELY.

Hence they are favourable for SN1 reactions, BECAUSE THEY CAN SOLVATE THE CARBOCATION AND THE LEAVING GROUP EFFICIENTLY.

DO REMEMBER: SN1 REACTIONS ARE INDEPENDENT OF NATURE OF NUCLEOPHILE.

POLAR PROTIC DON'T FAVOUR SN2 BECAUSE, THEY'LL SOLVATE THE NUCLEOPHILE and HENCE ITS SIZE WILL INCREASE AND HENCE CHARGE DENSITY WILL DECREASE, WHICH'LL ULTIMATELY MAKE IT A WEAK NUCLEOPHILE.

POLAR APROTIC SOLVENTS (DMS:Dimethyl Sulphoxide):

Due to electronegativity of O, S will gain a partial +ve charge. But due to large size of S the charge density will be very very less. Therefore, This Compound can't solvate the anion effectively.

THEY FAVOUR SN2 Because:

THEY CAN'T EFFECTIVELY SOLVATE THE CARBOCATION, WHICH WILL BE FORMED, SO THE CARBOCATION WILL REMAIN HIGHLY UNSTABLE WITHOUT SOLVATION, SAME REASON FOR THE LEAVING GROUP, if IT'LL NOT BE SOLVATED, ANION WILL BE HIGHLY UNSTABLE AND "INTERNAL RETURN" WILL OCCUR.

ORDER OF BASICITY: F- > Cl- >Br-> I-

It is a THERMODYNMICALLY CONTROLLED PROPERTY.

NUCLEOPHILICITY (Kinetic rate ControL): It is the measurement of rate constant of a reaction of a nucleophile with some electron deficient centre.

HOW TO CHECK NUCLEOPHILICITY:

IT REMAINS SOLVENT DEPENDENT, Can't say anything about the order, unless the solvent is specified.

JUST AS THE SIZE OF A ANION INCREASES DUE TO SOLVATION, ITS charge density decreases and hence does nucleophilicty.

MORE SOLVATION------>>>> LESS THE NUCLEOPHILICITY.

THE GOLDEN RULE:

BASICITY AND NUCLEOPHILICTY ARE OPPOSITE TO EACH OTHER IN POLAR PROTIC SOLVENTS.

order of nucleophilicity: F- < Cl- <Br- < I- (polar protic solvent)

BASICITY AND NUCLEOPHILICTY ARE PARALLEL TO EACH OTHER IN POLAR APROTIC SOLVENTS.

order of nucleophilicity: F- > Cl- >Br- > I- (polar aprotic solvent) (exactly reversed)

1. If we go down the group the bond strength of HYDRIDES decreases and hence acidity of Hydrides increases , as we go down.

Ex: Acidic strenth of Hi > HBr >HCl > HF

2. COVALENT BOND STRENGTH is proportional to 1/STRENGTH

3. IF the HYDROGEN responsible for acidic character, is involved in H-Bonding, the acidic strenth decreases.

4. After the loss of H+ from an acid, the anion formed stabilizes itself by intramolecular H-bonding, the acidic strenth of the acid increases.

5. H-bond can be only formed when:

A HYDROGEN ATTACHED TO F,O, N is bonded to F,O,N.

6. Acidic strength of CH3OH > H2O exception.

7. FOR PI BOND FORMATION: ATLEAST ONE ELEMENT SHOULD BE OF 2nd PERIOD.

The order of Dipole Moment:



CH3Cl > CH3F > CH3Br > CH3I



There is no such conceptual reason for this exception...

because the explanation will contradict other explanations and concepts.



As per the boiling point is concerned:



CH3I > CH3Br > CH3Cl > CH3F



This is so, because boiling point is proportional to molecular mass...

and the increment is very high as we go on from CH3F to CH3I therefore, this is the correct order...



And as per reactivity goes:



CH3I > CH3Br > CH3Cl > CH3F



Because CH3F is the most stable of all...

due to high polarity b/w the bonds, the ionic character increases and hence the bond strength...

remember More polarity...more bond strength..

WELL HOPE ALL THIS HELPS YOU....DO COMMENT...