carbocation stability

Carbocation are highly reactive intermediate species, they have one an inherent positive charge and thus always in the look out for electrons, which tells us about their high reactivity. 
So the stabilization effects in a Carbocation are as follows: 
(1) Carbocations are stabilized by neighboring carbon atoms
Carbons (alkyl groups) are “electron-releasing” groups through inductive effects, thus stabilizing the system. The second is hyperconjugation, which stabilizes through donation of the electrons in C-H sigma bonds to the empty p orbital of the carbocation.

2) Carbocations are stabilized by neighboring carbon-carbon multiple bonds.
Any Double bond is a rich source of electrons, π clouds above and below having loosely attached electron cloud, this electron cloud, with the resonance effect can donate the electrons to the Carbocationic Carbon.This effect, called “Delocalization" where the charge “moves” from atom to atom. Delocalization of π electrons.

3) Carbocations are stabilized by adjacent lone pairs
The key stabilizing influence is a neighboring atom that donates a pair of electrons to carbocation. Note here that this invariably results in forming a double bond (π bond) and the charge will move to the atom donating the electron pair. Hence this often goes by the name of “π donation”.

4) Also Stabilized in presence of Polar Sovents, by solvating the carbocation species. This aids in increasing the reaction rate of the S1N reactions.

There are many factors,coming directly to the answer

1. Stability is directly proportional to the +I effect,that is the reason tertiary carbocation is more stable than secondary( explained by hyperconjugation or dispersal of charge)
2. Carbocation stability increases with increase in percentage of p character in hybridization
3. Carbocation stability increases with increasing alpha hydrogen atoms due hyperconjugation
4. Carbocation are stabilized by adjacent lone pairs of electron due to resonance
5. Carbocations are stabilized by neighbouring carbon carbon multiple bond due to resonance

There are 2 types of groups EDG (electron donating group) and EWG (electron withdrawing group).

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A carbonation is a centre of positive charge which will be stabilised when an EDG is bonded with it. This is known as +I effect of the EDG. Further it will be more stabilised when more number of EDG groups are attached. 

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Therefore the order of stability of carbonation is 3°>2°>1°.

How do I compare the stability of carbocation depending upon -

1. Hyperconjugation more the hyperconjugation more the stability of carbocation. This seen in 3° carbocation.
2. +I effect more the electron releasing group attached to carbocation more the density of electron increases on the carbocation so stability increases. So 3° carbocation is more stable.
3. +R effect group donate the lone paire to the near carbocation and stable the carbocation. So tripheny carbocation is more stable.
4. So stability order is 3°>tripheny carbocation> Allylic carbocation >2>1.
5. But most stable is tricyclo propane carbocation due to bend p- orbital of cyclo propane.

The general stability order of simple alkyl carbocations is: (most stable) 3o> 2o > 1o> methyl (least stable)

This is because alkyl groups are weakly electron donating due to hyperconjugation and inductive effects. 
Resonance effects can further stabilise carbocations when present (therefore allyl or benzyl carbocations are more stable than simple primary carbocations).

the stability of carbocation is decided by following factors = dancing resonance are most stable > part of aromatic system > +M stabilised > +H stabilised > +I stabilised > -I unstabilised > -H unstabilised > -M unstabilised > part of antiaromatic system

Note -i assume you know all these effect

ortho meta effect

Hypo is used when there is a single oxygen atom at the end of an ion. e.g FO-

hypo- + -ite

FO-, hypofluorite ClO-,

hypochlorite BrO-, hypobromite

IO-, hypoiodite

Pyro is a prefix for designated compounds formed by heating a compound, usually with the removal of water, carbon dioxide, or other simple molecule/substance. E.g. pyroglutamic acid from glutamic acid.

Meta indicates the positions of substituents in aromatic cyclic compounds. The substituents have the 1,3-positions, for example in resorcinol. (organic chemistry)

Ortho describes a molecule with substituents at the 1 and 2 positions on an aromatic compound. In other words, the substituent is adjacent or next to the primary carbon on the ring. (organic chemsitry)

The prefix thio-, when applied to a chemical, such as an ion, means that an oxygen atom in the compound has been replaced by a sulfur atom. e.g. Sodium Thiosulfate (organic chemistry)

This is a classic example of stabilization by hydrogen bonding.

In the series of hydroxy-benzoic acids the order is:

• The OH in salicylic acid (the ortho-derivative) will stabilize the anion due to hydrogen bonding.

(hydrogen bonding is also possible in the protonated form (the acid) but is stronger in the carboxylate anion.

• For the para isomer, the mesomeric effect makes the compound less acidic in comparison with benzoic acid.
• For the meta isomer a minor inductive effect operates, but no resonance effect, which makes it slightly more acidic.

Remark: in solution the carboxylic acid part of the molecule is far more acidic in comparison with the phenol part, but in the gas-phase there are exceptions.

Lets start with some information we know so far : OH- has +R and weak -I effect and we know that -I is distant dependent and +R don't Show any effect at meta position and to your note ortho substituted benzoic acids are stronger acids than benzoic acids regardless of the nature of the substituent ( be it electron donating or electron withdrawing ) due to stearic and electronic reasons.

Let's start solving :

So blindly ortho would be most acidic ( due to ortho effect as explained earlier ) now although you didn't asked but i found it quite important to share so what happens at meta is bit deep than comparing at other positions , like I told you that Oh has +R and -I and at meta +R won't show its effect so only -I left which again helps in dispersing the negative charge in other words it is more acidic than benzoic acid but less acidic than ortho one., and finally at para position -I due to its distance dependence is very weak and +R is quite strong so para is least acidic as +R intensifies negative charge.

Result : ortho hydroxy B.acid > meta hydroxy B.acid > benzoic acid > para hydroxy B. acids.

So Answer to your question should be ,

Para is least acidic.

Ortho hydroxy benzoic acid is more acidic than it's para isomer because the carboxylate ion in ortho hydroxy benzoic acid get stable with intermolecule H bonding. ….But in case of it's para isomer OH group shows +R effect ,which increase the elctron density and that's why the stability of coo - decrease …For this reason ortho hydroxy benzoic acid is more acidic than ortho hydroxy benzoic acid. ..

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 the strongest [similar to ortho-effect].

Acidic Character order:

o-Cyanobenzoic acid [2.8] >

p-Cyanobenzoic acid [3.55] >

m-Cyanobenzoic acid [3.60] > Benzoic acid [4.2]

(pKa values are given in brackets; less pKa means more acidic)

Here, the inductive effect of the cyano group pulls electron density away from the carboxyl group, making the acid proton more positively charged, and therefore more acidic. Additionally, the -R effect of the cyano group delocalizes the negative charge from the carboxylate ion, leading to greater stability for 2-cyanobenzoate ion.

• Due to ortho effect, in most cases,-
• “Ortho-isomer of Benzoic acid is strongest Acid as compared to Simple benzoic acid or meta & para-isomers & Ortho- isomer of Aniline is Weakest Base as compared to Simple Aniline or meta & para-isomers.”

• In case of ortho-substituted benzoic acids, due to steric inhibition, the -COOH group goes out of plane and hence decrease in resonance stabilisation of acid as compared to anion make it better acid.
• Acidic Character order:
• o-Bromobenzoic acid [3.1] > m-BromoBenzoic acid [3.93]> p-BromoBenzoic acid [4.1] > Benzoic acid [4.17]
• o-ChloroBenzoic acid [2.89] > m-ChloroBenzoic acid [3.82]> p-ChloroBenzoic acid [3.98]> Benzoic acid [4.17]
• o-FluoroBenzoic acid [3.27] > m-FluoroBenzoic acid [3.87]> p-FluoroBenzoic acid [4.14]> Benzoic acid [4.17]
• O-Toluic acid [3.89] > Benzoic acid [4.17]> m-Toluic acid [4.28] > p-Toluic acid [4.35]

• 2,6-diHydroxybenzoic acid [2.3] > o-HydroxyBenzoic acid [2.98] > m-HydroxyBenzoic acid [4.08] > Benzoic acid > p-HydroxyBenzoic acid [4.58]

• o-NitroBenzoic acid [2.17] > p-NitroBenzoic acid [3.44] > m-NitroBenzoic acid [3.45] > Benzoic acid

(Note- pKa values are given in brackets; less pKa means more acidic)

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When aniline acting as a base becomes NH3 + (on top of a benzene ring), it is usually stabilised by solvation. but if there is a substituent on the ortho position, it inhibits solvation. Thus the tendency to act like a base is reduced. Also, it can be explained by SIP (steric inhibition to protonation) effect.

• Basic Character order:
• o-Nitroaniline < p-Nitroaniline < m-Nitroaniline < Aniline
• o-Toluidine [4.38] < Aniline < m-Toluidine [4.67] < p-Toluidine [5.07]
• o-Fluoroaniline [pKa = 2.96] < m-Fluoroaniline [pKa = 3.38] < p-Fluoroaniline [pKa = 4.52] < Aniline (pKa = 4.6)
• o-Chloroaniline [pKa = 2.62] < m-Chloroaniline [pKa = 3.32] < p-Chloroaniline [pKa = 3.8] < Aniline (pKa = 4.6)
• o-Bromoaniline [pKa = 2.6, pKb = 11.4] < m-Bromoaniline [pKa = 3.5, pKb = 10.5] < p-Bromoaniline [pKa = 3.9, pKb = 10.1] < Aniline (pKa = 4.6)

(Note- pKa values are given in brackets; more pKa mean more basic; pKa + pKb = 14)

An acid is called a comparatively stronger acid if its anion is stable after release of the proton.

In case of p-hydroxybenzoic acid, the anion is destabilized by the +M effect of the hydroxy group lying on the p group.

Whereas, in m-hydoxy benzoic acid, the hydroxy group does not provide +M effect to the C to which the acidic group is attached.

Hence, m-hydoxy benzoic acid is stronger than its para derivative.

You can draw the resonating structures to see where the negative charges are formed. If negative charge is formed the C where the acidic group is attached, then it will destabilize the anion formed after releasing the proton.

In most cases ortho substituted benzoic acids are most acidic than their para and meta counterparts due to ortho effect. Now the question lies whether it is stronger than benzoic acid or not?. First let us compare the acidity between meta-hydroxy benzoic acid and benzoic acid. OH group being both -I and +R group will act as - I group in meta position and will withdraw electron thus decreasing the electron density on COO- and thus its acidity will be more than benzoic acid which has no electron withdrawing effect. So meta-hydroxy benzoic acid is more acidic than benzoic acid. Compared to meta-hydroxy benzoic acid it's ortho counterpart is much more acidic due to ortho effect. Hence ortho-hydroxy benzoic acid is more acidic than benzoic acid.

periodic poem

The periodic table

The periodic table
The chemical elements table

First there is hydrogen,
Which is used for rocket propulsion

Then there is helium
Which floats balloons

Then comes lithium
Which is present in battery cells

Then there is beryllium
Which is toxic and forms sweet salts

Then comes boron
Which is found in every household

Now comes carbon
Which forms millions of useful compounds

This the periodic table
The metals and non metals table

There comes nitrogen
Which is present in laughing gas and explosives

Then there is oxygen
Which sustains life

Then comes fluorine
Which prevents dental cavities

Now comes neon
Which makes bulbs glow red

This is the periodic table
With periods and groups

Now comes sodium
Which is present in common salt

Then there is magnesium
Which is a important nutrient

Here comes aluminium
Which can be made into fools and cables

Next comes silicon
Which is present in computer chips

There comes phosphorus
Which is used in fireworks

Then comes sulphur
Which is part of proteins and antibiotics

This is the periodic table
With halogens and noble gases

Now there is chlorine
Which kills germs

Next comes argon
Which makes filaments last longer

Then there is potassium
Which is used in the fertilizer industry

And then comes calcium
Which keeps our bones and teeth healthy

STREOISOMERS

Introduction

The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and, as such, is also often called the Cahn-Ingold-Prelog rules. In addition to the Cahn-Ingold system, there are two ways of experimentally determining the absolute configuration of an enantiomer:

1. X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer.
2. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction.

However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of optical rotation, although different for the two enantiomers of a chiral molecule,at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes.

Stereocenters are labeled R or S

The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S.

Consider the first picture: a curved arrow is drawn from the highest priority (1) substituent to the lowest priority (4) substituent. If the arrow points in a counterclockwise direction (left when leaving the 12 o' clock position), the configuration at stereocenter is considered S ("Sinister" → Latin= "left"). If, however, the arrow points clockwise,(Right when leaving the 12 o' clock position) then the stereocenter is labeled R ("Rectus" → Latin= "right"). The R or S is then added as a prefix, in parenthesis, to the name of the enantiomer of interest.

Example 1

(R)-2-Bromobutane

(S)-2,3- Dihydroxypropanal

Sequence rules to assign priorities to substituents

Before applying the R and S nomenclature to a stereocenter, the substituents must be prioritized according to the following rules:

Rule 1

First, examine at the atoms directly attached to the stereocenter of the compound. A substituent with a higher atomic number takes precedence over a substituent with a lower atomic number. Hydrogen is the lowest possible priority substituent, because it has the lowest atomic number.

1. When dealing with isotopes, the atom with the higher atomic mass receives higher priority.
2. When visualizing the molecule, the lowest priority substituent should always point away from the viewer (a dashed line indicates this). To understand how this works or looks, imagine that a clock and a pole. Attach the pole to the back of the clock, so that when when looking at the face of the clock the pole points away from the viewer in the same way the lowest priority substituent should point away.
3. Then, draw an arrow from the highest priority atom to the 2nd highest priority atom to the 3rd highest priority atom. Because the 4th highest priority atom is placed in the back, the arrow should appear like it is going across the face of a clock. If it is going clockwise, then it is an R-enantiomer; If it is going counterclockwise, it is an S-enantiomer.

When looking at a problem with wedges and dashes, if the lowest priority atom is not on the dashed line pointing away, the molecule must be rotated.
Remember that

• Wedges indicate coming towards the viewer.
• Dashes indicate pointing away from the viewer.

Rule 2

If there are two substituents with equal rank, proceed along the two substituent chains until there is a point of difference. First, determine which of the chains has the first connection to an atom with the highest priority (the highest atomic number). That chain has the higher priority.

If the chains are similar, proceed down the chain, until a point of difference.

For example: an ethyl substituent takes priority over a methyl substituent. At the connectivity of the stereocenter, both have a carbon atom, which are equal in rank. Going down the chains, a methyl has only has hydrogen atoms attached to it, whereas the ethyl has another carbon atom. The carbon atom on the ethyl is the first point of difference and has a higher atomic number than hydrogen; therefore the ethyl takes priority over the methyl.

Rule 3

If a chain is connected to the same kind of atom twice or three times, check to see if the atom it is connected to has a greater atomic number than any of the atoms that the competing chain is connected to.

• If none of the atoms connected to the competing chain(s) at the same point has a greater atomic number: the chain bonded to the same atom multiple times has the greater priority
• If however, one of the atoms connected to the competing chain has a higher atomic number: that chain has the higher priority.

Example 2

A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below:

However:

Remember that being double or triple bonded to an atom means that the atom is connected to the same atom twice. In such a case, follow the same method as above.

Caution!!
Keep in mind that priority is determined by the first point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant.

When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.

After all your substituents have been prioritized in the correct manner, you can now name/label the molecule R or S.

1. Put the lowest priority substituent in the back (dashed line).
2. Proceed from 1 to 2 to 3. (it is helpful to draw or imagine an arcing arrow that goes from 1--> 2-->3)
3. Determine if the direction from 1 to 2 to 3 clockwise or counterclockwise.

i) If it is clockwise it is R.
ii) if it is counterclockwise it is S.

USE YOUR MODELING KIT: Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S).

IF YOU DO NOT HAVE A MODELING KIT: remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models.

If you have a modeling kit use it to help you solve the following practice problems.

Problems

Are the following R or S?

Solutions

1. S: I > Br > F > H. The lowest priority substituent, H, is already going towards the back. It turns left going from I to Br to F, so it's a S.
2. R: Br > Cl > CH₃ > H. You have to switch the H and Br in order to place the H, the lowest priority, in the back. Then, going from Br to Cl, CH₃ is turning to the right, giving you a R.
3. Neither R or S: This molecule is achiral. Only chiral molecules can be named R or S.
4. R: OH > CN > CH₂NH₂ > H. The H, the lowest priority, has to be switched to the back. Then, going from OH to CN to CH₂NH₂, you are turning right, giving you a R.
5. (5) S: −COOH$-\text{COOH}$ > −CH2OH$-{\text{CH}}_2^{}\text{OH}$ > C≡CH$\text{C}\equiv \text{CH}$ > H$\text{H}$. Then, going from −COOH$-\text{COOH}$ to −CH2OH$-{\text{CH}}_2^{}\text{OH}$ to −C≡CH$-\text{C}\equiv \text{CH}$ you are turning left, giving you a S configuration.

1. • Chirality
    
   • Chirality and Stereoisomers

Recommended articles

1. 1.19: R-S Sequence Rules
2. 6.2: R-S Sequence Rules
3. Stereoisomers
4. Meso Compounds
5. Optical Activity