PREPRATION OF THYMOLPHTHALEIN BY FREIDAL CRAFTS REACTION

dissepearing ink EXPERIMENT

1.    Add 1 gram of thymolphthalein into 100 mL of ethyl alcohol. The solution will require stirring to dissolve all of the powder.

2.   Add 900 mL of water to the solution and stir. Don't worry if the solution looks white - that's because the thymolphthalein indicator is not soluble in water. No worries, the next step will fix everything.

3.   Slowly add 10 mL of 3 molar sodium hydroxide to the solution to turn the liquid a dark blue.

Caution: Sodium hydroxide (commonly known as lye) is a caustic solution and must be handled by an adult. Once the sodium hydroxide is diluted with water, the solution is safe to use as disappearing ink, but care must be taken because the pH of the ink solution is about 10 (basic). Be sure to wash your hands well with water after handling the more concentrated solution.

Always test the disappearing ink on a small piece of white fabric to make sure that it actually disappears. In a few seconds, the ink stain will disappear. The color will vanish more quickly if you apply a cotton ball dampened with vinegar or if you blow on the spot (the carbon dioxide in your breath is the secret!). The pH of the ink solution is 10-11, but after exposure to air the pH will drop to 5-6. The damp spot will eventually dry. A white residue may be visible on some dark fabrics. Be sure to store the disappearing ink solution in a sealed container. All of the materials may be safely poured down the drain.

HOW DOES IT WORK?

The secret to making the ink disappear is carbon dioxide in the air which reacts with the water in the solution to form carbonic acid. The carbonic acid then reacts with the sodium hydroxide in a neutralization reaction to form sodium carbonate. This lowers the pH of the solution with the alcohol acting as an acid to turn the indicator colorless and the ink stain magically disappears. The "fading time" can be prolonged by adding a small amount (use drops to make these adjustments) of sodium hydroxide. But care should be taken not to add too much sodium hydroxide. Here's a fun fact... red disappearing ink can be made using phenolphthalein (a very common acid-base indicator) in place of thymolphthalein.

thymolphthalein

Type: HIn + H₂O  In^- + H₃O⁺
pK: 9.7
Approximate pH range for color change: 9.3-10.5
Color of acid form: colorless
Color of base form: blue

PREPRATION OF THYMOLPHTHALEIN

BY FREIDAL CRAFTS REACTION 

FREIDEL CRAFTS REACTION

electrophillic aromatic substitution

A cascade reaction or tandem reaction or domino reaction

A cascade reaction or tandem reaction or domino reaction is a consecutive series of intramolecular organic reactions which often proceed via highlyreactive intermediates. 

It allows the organic synthesis of complex multinuclear molecules from a single acyclic precursor. 

The substrate contains manyfunctional groups that take part in chemical transformations one at the time. 

Often a functional group is generated in situ from the previous chemical transformation. 

The definition includes the prerequisite intramolecular in order to distinguish this reaction type from a multi-component reaction. 

In this sense it differs from the definition of a biochemical cascade. 

The main advantages of a cascade reaction in organic synthesis are that the reaction is often fast due to its intramolecular nature, the reaction is also clean, displays high atom economy, does not involve workup and isolation of many intermediates, and adds much complexity in effectively one step.

oxidation reaction of alkene

Oxidation reactions of alkenes give cyclic ethers in which both carbons of a double bond become bonded to the same oxygen atom. 

These products are called epoxides or oxiranes.

 An important method for preparing epoxides is by reaction with peracids, RCO₃H.

The epoxidation reaction occurs in a single step with a transition state.

 Consequently, epoxidations by peracids always have syn-stereoselectivity

Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the syn-hydroxylation reaction described above.

Proton transfer from the acid catalyst generates the conjugate acid of 

the epoxide, which is attacked by nucleophiles such as water .

The result is anti-hydroxylation of the double bond .

 In the following equation this procedure is illustrated for a cis-

disubstituted epoxide, which, of course, could be prepared from the

 corresponding cis-alkene. This hydration of an epoxide does not change 

the oxidation state of any atoms or groups.

Solution to problem given in my last post .

The given compound after reacting with NBS undergoes radical reaction.

In this radical reaction .Free radical is formed at benzyllic or allylic positions prefrentially.

why?

Since these radicals are resonance stabilized.This resonance stabilization is shown in the figure.

After bromination ,next step is elimination.

Ths gives an alkene.

This alkene has also carbonyl functional group.

This carbonyl functional group resonates to give a charged species.

This charged species is more stable,since this is resonance stablised.

a challenging problem ---------- in the preparation of an alkene or what is the mechanism ?

What is the reaction of the following 

compound with NBromoSuccanamide NBS 

in first step and to the following compound

formed with alk KOH ?

Is the final product stable or its resonance structure stable ?

oxymercuration reaction ---- an animation video

oxymercuration reaction

you tube url : http://youtu.be/u4v9GPYhPYE

Oxymercuration is a special electrophilic addition

The oxymercuration reaction is an electrophilic 

addition organic reactionthat transforms an alkene into a 
neutral alcohol.

 In oxymercuration, the alkene reacts with mercuric acetate (AcO–Hg–OAc) in aqueous solution to yield the 
addition of an acetoxymercuri (HgOAc) group and a hydroxy

 (OH) group across the double bond.

 Carbocations are not formed in this process and thus rearrangements are not observed.

 The reaction follows Markovnikov's rule (the hydroxy group 
will always be added to the more substituted carbon) and it

 is an anti addition (the two groups will be trans to each 

other)

. It is anti-stereospecific and regioselective. Regioselectivity 

is a process in which the substituents choses one direction it

 prefers to be attached to over all the other possible 

directions. The good thing about this reaction is that there are no carbocation rearrangement.

http://youtu.be/u4v9GPYhPYE

HYDRBORATION OF 1-methylcyclopentene

MECHANISM FOR REACTION OF ALKENES WITH BH₃

Step 1: 

A concerted reaction. The p electrons act as the nucleophile with the electrophilic B and the H is transferred to the C with syn stereochemistry .

Step 2: 

First step repeats twice more so that all of the B-H bonds react with C=C

Step 3: 

Peroxide ion reacts as the nucleophile with the electrophilic B atom.

Step 4: 

Migration of C-B bond to form a C-O bond and displace hydroxide. Stereochemistry of C center is retained.

Step 5: 

Attack of hydroxide as a nucleophile with the electrophilic B displacing the alkoxide.

Step 6: 

An acid / base reaction to form the alcohol.

HYDROBORATION / OXIDATION

HYDROBORATION   /    OXIDATION

Hydroboration-Oxidation is a two step pathway used to produce alcohols.

The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen

 (from BH₃ or BHR₂) attaches to the more substituted carbon and the

 boron attaches to the least substituted carbon in the double bond.

  Borane acts as a lewis acid by accepting two electrons in its empty p orbital from an alkene that is electron rich.

This process allows boron to have an electron octet. 

                        

 The Hydroboration mechanism has the elements of both hydrogenation

 and electrophilic addition and it is a stereospecific (syn addition), meaning 

that the hydroboration takes place on the same face of the double bond,

 this leads cis  stereochemistry.

THE OVERALL REACTION OF HYDROBORATION AND OXIDATION IS SHOWN BELOW:

refer to my post on hydration reacction an electrophillic addition for further clarity url: http://chemistryjee.blogspot.in/2014/01/hydration-of-alkenes.html

The reagent is usually borane, BH₃, but an organoborane can also be used, as long as at least one B-H bond is present.

 In the laboratory, the borane-THF complex, dissoved in tetrahydrofuran (THF) is often used.

 All three hydrogens of borane are usable. 

For simplicity, we will usually designate the borane as R₂BH, where R can be alkyl or hydrogen.

The initial product of addition of borane or an organoborane across a carbon-carbon pi bond is an organoborane, where a new B-C bond has been made, along with a new C-H bond.

 These two bonds are formed and the B-H and C-C pi bonds are broken, all in concert, i.e., in a single reaction step with no intermediates being involved.

These organoboranes are not stable in air, reacting rather rapidly with oxygen. 

Instead of isolating them, they are normally treated in situ (i.e., in place) with alkaline hydrogen peroxide, a treatment which converts the B-C bond to a C-O bond (and a B-O bond). 

But we will note that the overall result of these two steps (hydroboration plus oxidation) is to convert an alkene to an alcohol. 

This reaction was discovered by Professor H.C. Brown of Purdue University and is an important enough synthetic conversion that he was awarded the Nobel Prize for Chemistry primarily based upon this work.

We should note that the net addition of water which occurs during hydroboration/oxidation is in the anti-Markovnikov regiochemical sense, with propene giving 1-propanol, rather than the 2-propanol which is generated by the acid catalyzed, electrophilic hydration mechanism.

The second step (oxidation of the borane to the alcohol) takes place with retention  of configuration.

 Hydroperoxide ion adds to the borane, causing the alkyl group to

migrate from boron to oxygen.

The alkyl group migrates with retention of configuration  because it moves with its electron pair and does not alter the tetrahedral structure of the migrating carbon atom.

Hydrolysis of the borate ester gives the alcohol.

ELECTROPHILLIC ADDITION ------HALOGEN ADDITION

     

v Addition of Bromine and Chlorine  Cl₂ or Br₂ is used to convert an

alkene into a vicinal dihalide.

v The mechanism involves attack by the alkene pi  bond on one atom of X₂ to give a bridged halonium ion intermediate (a cation) that is, in turn, attacked by X₂ from the backside to give the vicinal dihalide. Rearrangements do not occur.

Not all electrophilic additions necessarily involve carbocations, although they  involve the development of positive charge on the alkene, because it is serving as a nucleophile.

The addition of bromine, chlorine , or iodine to an alkene pi bond proceeds via an intermediate which has the positive charge mainly on the halogen.

This is because the positive charge is more stable on the halogen than on carbon.

The primary reason is that , an extra bond (a carbon-halogen bond) is formed in this intermediate, which is called an EPIHALONIUM ION .

v The reaction displays anti addition stereoselectivity because of the halonium ion intermediate .

v The reaction is stereospecifi c because Z alkenes give different products  than E alkenes.

The addition of bromine,  to trans-2-butene yields only meso-2,3-dibromobutane and no traces of the enantiomeric pair.

In contrast, cis-2-butene yields only the enantiomeric pair as a racemate, and none of the meso isomer.

By comparing the structure of the alkene (which we have drawn as a Newman projection) with that of the product corresponding to it, we can see that the bromines had to add to opposite faces of the double bond. This is termed anti stereospecific addition.

The addition of bromine,  in the presence of solvents such as WATER

Or METHANOL will also occur via a cyclic bromonium intermediate.

But, instead of bromine nucleophile attacking ,due to high concentration of solvent ,the solvent used acts as anucleophile ,and in this case WATER or METHANOL .

Now that we have established that bromine addition to alkenes is anti ,let us explore bromine addition to cyclic compounds .

v Usually the halogen is dissolved in some inert solvent such as tri- or tetrachloromethane,and then this solution is added dropwise to the alkene.

v Reaction is nearly instantaneous, even at room temperature or below. No light or heat is required, as in the case of substitution reactions.

v The addition of bromine can be used as a chemical test for the presence of unsaturation in an organic compound. Bromine solutions in tetrachloromethane are dark reddish-brown, and both the unsaturated compound and its bromine adduct are usually colorless.

v As the bromine solution is added to the unsaturated compound,

the bromine color disappears.

v    If the compound being tested is saturated, it will not

react with bromine under these conditions, and the color will persist.