Organic THREAD

Stereoselectivity
In chemistry, stereoselectivity is the property of a chemical reaction in which a single reactant forms an unequal mixture of stereoisomers during the non-stereospecific creation of a new stereocenter or during the non-stereospecific transformation of a pre-existing one. The selectivity arises from differences in steric effects and electronic effects in the mechanistic pathways leading to the different products. Stereoselectivity can vary in degree but it can never be total since the activation energy difference between the two pathways is finite. However, in favorable cases, the minor stereoisomer may not be detectable by the analytic methods used.
An enantioselective reaction is one in which one enantiomer is formed in preference to the other, in a reaction that creates an optically active product from an achiral starting material, using either a chiral catalyst, an enzyme or a chiral reagent. The degree of selectivity is measured by the enantiomeric excess. An important variant is kinetic resolution, in which a pre-existing chiral center undergoes reaction with a chiral catalyst, an enzyme or a chiral reagent such that one enantiomer reacts faster than the other and leaves behind the less reactive enantiomer, or in which a pre-existing chiral center influences the reactivity of a reaction center elsewhere in the same molecule.
A diastereoselective reaction is one in which one diastereomer is formed in preference to another (or in which a subset of all possible diastereomers dominates the product mixture), establishing a prefered relative stereochemistry. In this case, either two or more chiral centers are formed at once such that one relative stereochemistry is favored, or a pre-existing chiral center (which needs not be optically pure) biases the stereochemical outcome during the creation of another. The degree of relative selectivity is measured by the diastereomeric excess.
Stereoconvergence can be considered an opposite of stereoselectivity, when the reaction of two different stereoisomers yield a single product stereoisomer.
Stereospecific reaction
In chemistry, stereospecificity is the property of a reaction mechanism that leads to different stereoisomeric reaction products from different stereoisomeric reactants, or which operates on only one (or a subset) of the stereoisomers.
In contrast, stereoselectivity is the property of a reactant mixture where a non-stereospecific mechanism allows for the formation of multiple products, but where one (or a subset) of the products is favored by factors, such as steric access, that are independent of the mechanism.

A stereospecific mechanism specifies the stereochemical outcome of a given reactant, whereas a stereoselective reaction selects products from those made available by the same, non-specific mechanism acting on a given reactant. Given a single, stereoisomerically pure starting material, a stereospecific mechanism will give 100% of a particular stereoisomer (or no reaction), although loss of stereochemical integrity can easily occur through competing mechanisms with different stereochemical outcomes. A stereoselective process will normally give multiple products even if only one mechanism is operating on an isomerically pure starting material.
The term stereospecific reaction is ambiguous, since the term reaction itself can mean a single-mechanism transformation (such as the Diels-Alder reaction), which could be stereospecific, or the outcome of a reactant mixture that may proceed through multiple competing mechanisms, specific and non-specific. In the latter sense, the term stereospecific reaction is commonly misused to mean 'highly stereoselective reaction'.
Chiral synthesis is built on a combination of stereospecific transformations (for the interconversion of existing stereocenters) and stereoselective ones (for the creation of new stereocenters), where also the optical activity of a chemical compound is preserved.
Examples
• Nucleophilic substitution at sp3 centres can proceed by the stereospecific SN2 mechanism, causing only inversion, or by the non-specific SN1 mechanism, the outcome of which can show a modest selectivity for inversion, depending on the reactants and the reaction conditions to which the mechanism does not refer. The choice of mechanism adopted by a particular reactant combination depends on other factors (steric access to the reaction centre in the substrate, nucleophile, solvent, temperature). For example, tertiary centres react almost exclusively by the SN1 mechanism whereas primary centres (except neopentyl centres) react almost exclusively by the SN2 mechanism. When a nucleophilic substitution results in incomplete inversion, it is because of a competition between the two mechanisms, as often occurs at secondary centres, or because of double inversion (as when iodide is the nucleophile).
• The addition of carbenes to alkenes is stereospecific in that the geometry of the alkene is preserved in the product. For example, dibromocarbene and cis-2-butene yield cis-2,3-dimethyl-1,1-dibromocyclopropane, whereas the trans isomer exclusively yields the trans cyclopropane.

This addition remains stereospecific even if the starting alkene is not isomerically pure, as the products' stereochemistry will match the reactants'.
• The disrotatory ring closing reaction of conjugated trienes is stereospecific in that isomeric reactants will give isomeric products. For example, trans,cis,trans-2,4,6-octatriene gives cis-dimethylcyclohexadiene, whereas the trans,cis,cis reactant isomer gives the trans product and the trans,trans,trans reactant isomer does not react in this manner.

Regioselective reaction

In chemistry, regioselectivity is the preference of one direction of chemical bond making or breaking over all other possible directions . It can often apply to which of many possible positions a reagent will affect, such as which proton a strong base will abstract from an organic molecule, or where on a substituted benzene ring a further substituent will add.
A specific example is a halohydrin formation reaction with 2-propenylbenzene :

The reaction product is a mixture of two isomers and the regioselectivity is said to be poor.
Regioselectivity in ring-closure reactions is subject to Baldwin's rules.

A synthon is a concept in retrosynthetic analysis. It is defined as a structural unit within a molecule which is related to a possible synthetic operation. The term was coined by E.J. Corey. It is noted that the phrase does not feature very prominently in Corey's book, The Logic of Chemical Synthesis, as it is not included in the index.

In planning the synthesis of phenylacetic acid, two synthons are identified: a nucleophilic "-COOH" group, and an electrophilic "PhCH2+" group. Of course, both synthons do not exist per se; synthetic equivalents corresponding to the synthons are reacted to produce the desired product. In this case, the cyanide anion is the synthetic equivalent for the -COOH synthon, while benzyl bromide is the synthetic equivalent for the benzyl synthon.
The synthesis of phenylacetic acid determined by retrosynthetic analysis is thus:
1. PhCH2Br + NaCN → PhCH2CN + NaBr
2. PhCH2CN + 2 H2O → PhCH2COOH + NH3

victor meyer test

BT se nikala hai na??

ya FROM BT ........WILL ADD MORE ..ATLEAST KUCH TO HELP HOGA

ya manmay >>>>>>>>>

tab to bohat sare jo iodoform nahin denge unhe haam negative test ke andar la sakte hain na [7]

No. of possible stereoisomerism in compounds having n-chiral carbon atoms
Case 1:-
wen der is no symmetry
number of d and l forms, a(say) = 2^n
no. of meso forms, m(say) = 0
total no of stereoisomerism = a+m = 2^n

Case 2
wen the molecule hs symmetry and n is even
den d no. of d & l forms , a = 2^(n-1)
the total no. of meso forms , m =2^(n/2 - 1)
total no. of meso forms = a+m

Case 3
wen d molecule can b divided into two parts & n is odd then
no. of d & l forms, a = 2^(n-1) -2^(n-1)/2
no.of meso forms,m = 2^(n-1)/2
total no. of stereoisomerisms = a+m

THE DOUBLE BOND EQUIVALENCE CONCEPT:-

Consider a general organic compound having molecular formula as CaHbNcOdXe ,

where,

C-Carbon,

H-Hydrogen,

N-Nitrogen,

O-Oxygen,

X-Halogen.

Using the Double Bond Equivalence we can find the number of the pi bonds present in the compound and in most of the cases also predict the exact structural formula of that compound.

D.B.E = (a+1)-{(b+e-c)/2} [D.B.E :- Double Bond Equivalence]

For Ex.

(1) D.B.E for CH4 is 0 hence no pi bond is present.

(2) D.B.E for ethane is 0 hence no pi bond is present.

(3) D.B.E for C2H5O is 0 hence no pi bond is present.

If while calculating for D.B.E. the second term in the equation i.e. {(b+e-c)/2} comes out to be fractional then round it up. For ex. 3.5 to 4, 5.5 to 6. In most of such cases the compound will be a intermediate as described below.

While predicting the structures using the above concept sometimes you will be getting the structures in which the carbon valencies is not satisfied. In such cases the formula may represent intermediate species such as carbocation , carbanion or carbon free radicals. For Ex. C7H7 .It represents a Benzyl anion.

Manmay you didn't mention this in your elimination reactions -:

E₁CB Reaction

This stands for unimolecular elimination through conjugate base. When the base is strong enough and the leaving group is very poor(such as fluorine or hydroxyl groups), E₁CB is preferred.

For example, let the substrate be pentan-2-ol.

The important requirement here is that the beta-hydrogen must be acidic. It is indeed so for the -OH group is present.

The fast step in the reaction is the abstraction of the beta hydrogen(note that in E1, the leaving group departs first, forming a cation and in E2, the leaving group and beta hydrogen simultaneously depart).

Now a carbanion intermediate forms. Whenever a leaving group is present beta to the carbanion(however poor it is, but not Ph), it is ejected and a pi bond is forcefully formed(double bond formation). This is the slow step and is the RDS of the reaction(the carbanion intermediate step applies to all situations in general).

Thus the carbanion intermediate is the "conjugate base" of the original substrate, through which elimination has occurred.
Note that the methyl group on the left side of -CH(OH)- is more acidic than the methylene group(-CH₂-) on the right. Hence the product would be pentene and not pent-2-ene.
Thus in this mode of elimination, the beta hydrogen's departure is the starting step

The most common example of the E₁CB mode of elimination in a name reaction is in the aldol dehydration. Note that it always produces an alpha,beta-unsaturated carbonyl(this is because the most acidic hydrogen is the hydrogen beta to (>C=O) group).

Another example which may not be in our course(but the reaction concepts involved are very much there) is this http://en.wikipedia.org/wiki/Boord_olefin_synthesis

There are also instances, Wikipedia has given, in which E₁CB eliminations are involved in light-catalyzed decarboxylation reactions. A carbanion intermediate is said to be involved, and unlike the usual reversibility of proton abstractions(unless very strong bases like LDA or NaH are used), light catalyzed carbanion formation is irreversible.

Source - 90% Myself and 10% Wiki.

And there's the Lossen reaction also rocky, for prep of amines...
BITS ke baad will upload everything :)

13) Dehydration incase of 1° , 2° , 3° alcohol
Because the rate-determining step in the dehydration of a secondary or a tertiary
alcohol is formation of a carbocation intermediate, the rate of dehydration parallels the
ease with which the carbocation is formed. Tertiary alcohols are the easiest to dehydrate
because tertiary carbocations are more stable and therefore are easier to form than
secondary and primary carbocations . In order to undergo dehydration,
tertiary alcohols must be heated to about 50 °C in 5%H₂SO₄ secondary alcohols must
be heated to about 100 °C in 75%H₂SO₄ and primary alcohols can be dehydrated only
under extreme conditions (170 °C in 95%H₂SO₄ ) and by a different mechanism
because primary carbocations are too unstable to be formed

The relatively harsh conditions (acid and heat) required for alcohol dehydration and
the structural changes resulting from carbocation rearrangements may result in low
yields of the desired alkene. Dehydration, however, can be carried out under milder
conditions by using phosphorus oxychloride( POCl₃ ) and pyridine.

Reaction with POCl₃ converts the OH group of the alcohol into a good
leaving group OPOCl₂. The basic reaction conditions favor an E2 reaction, so a carbocation is
not formed and carbocation rearrangements do not occur. Pyridine serves as a base to
remove the proton in the elimination reaction and to prevent the buildup of HC1,
which would add to the alkene.

4) Summary of Reactions
1. Alkanes undergo radical substitution reactions with Cl₂ or Br₂ in the presence of heat or light.
2. Alkyl-substituted benzenes undergo radical halogenation at the benzylic position.
3. Alkenes undergo radical halogenation at the allylic positions. NBS is used for radical bromination at the allylic position.

5) S_N2 inversion of configuration
The carbon at which substitution occurs has inverted its configuration during the
course of the reaction, just as an umbrella has a tendency to invert in a windstorm. This
inversion of configuration is called a Walden inversion, after Paul Walden, who first
discovered that the configuration of a compound was inverted in S_N2 an reaction.

Because an S_N2 reaction takes place with inversion of configuration, only one substitution
product is formed when an alkyl halide—whose halogen atom is bonded to an asymmetric
carbon—undergoes an S_N2 reaction. The configuration of that product is inverted
relative to the configuration of the alkyl halide. For example, the substitution product of the
reaction of hydroxide ion with (R)-2-bromopentane is (S)-2-pentanol. Therefore, the
proposed mechanism accounts for the observed configuration of the product.

6) Leaving grp in S_N2 reaction
The weaker the basicity of a group, the
better is its leaving ability. The reason leaving ability depends on basicity is because
weak bases are stable bases—they readily bear the electrons they formerly shared
with a proton . Because weak bases don’t share their electrons well, a
weak base is not bonded as strongly to the carbon as a strong base would be and
a weaker bond is more easily broken.