Friday, 1 December 2017

Keto-Enol Tautomerism: Key Points


The reason for the equilibrium lying to the left is due to bond energies. The keto form has a C–H, C–C, and C=O bond whereas the enol has a C=C, C–O an O–H bond. The sum of the first three is about 359 kcal/mol (1500 kJ/mol) and the second three is 347 kcal/mol (1452 kJ/mol). The keto form is therefore more thermodynamically stable by 12 kcal/mol (48 kJ/mol).
Although the keto form is most stable for aldehydes and ketones in most situations, there are several factors that will shift the equilibrium toward the enol form.  The same factors that stabilize alkenes or alcohols will also stabilize the enol form. There are two strong factors and three subtle factors.
1. Aromaticity.  Phenols can theoretically exist in their keto forms, but the enol form is greatly favored due to aromatic stabilization.
2. Hydrogen Bonding. Nearby hydrogen bond acceptors stabilize the enol form. When a Lewis basic group is nearby, the enol form is stabilized by internal hydrogen bonding.

Here are three more subtle effects in keto-enol tautomerism:
3. Solvent. Solvent can also play an important role in the relative stability of the enol form. For example, in benzene, the enol form of 2,4-pentanedione predominates in a 94:6 ratio over the keto form, whereas the numbers almost reverse completely in water. What’s going on? In a polar protic solvent like water, the lone pairs will be involved in hydrogen bonding with the solvent, making them less available to hydrogen bond with the enol form.
 4. Conjugation . π systems are a little like Cheerios in milk: given the choice, they want to connect together than hang out in isolation.  So in the molecule depicted, the more favorable tautomer will be the one on the left, where the double bond is a connected by conjugation to the phenyl.
 5. Substitution. In the absence of steric factors, increasing substitution at carbon will stabilize the enol form. Enols are alkenes too – so any factors that stabilize alkenes, will stabilize enols as well. All else being equal, double bonds increase in thermodynamic stability as substitution is increased. So in the above example, the enol on the left should be the more stable one. As you might suspect, “all things being equal” sounds like a big caveat. It is – all else is rarely equal. But that’s a topic for another day – or, more likely, another course.
 
Sources: “March’s Advanced Organic Chemistry”, “Solvents and solvent effects in Organic Chemistry”, by Christian Riechart. EDIT: Commenter Natalia helpfully points out that Carey & Sundberg A is a great resource for this topic (section 7.3 in my 4th edition) and she is right.

 One really interesting observation is that when you dissolve acetone in D2O, you slowly get incorporation of deuterium at the alpha carbon. This is the enol tautomer at work – it reacts with a proton/deuteron source at the alpha carbon and regenerates the ketone.
Under normal conditions, this is a pretty slow reaction. But when you add a bit of acid, suddenly the rate of deuterium incorporation drastically increases. What’s going on here?
Let’s call the rate for conversion of ketone into enol K1 and the rate for conversion of enol into ketone K2.  The cool thing is, acid speeds up BOTH reactions for keto-enol tautomerism – both the forward and the reverse reaction. Here’s how it works.
1)  For K1 (keto to enol), acid protonates the carbonyl, making the carbonyl carbon more electrophilic. The more electrophilic the carbonyl  the stronger an acid it becomes at the alpha carbon (remember, trifluoroacetone is more acidic than acetone). So deprotonation at the alpha carbon becomes much easier in the presence of acid, which results in the enol form.
2) For K2, acid also speeds up the conversion of the enol to the keto form. The enol form is nucleophilic at the alpha carbon, and reacts with the strongest electrophile around. An acid – say, D2SO4  – is a much, much stronger electrophile than D2O, so this is going to increase the rate of the enol to ketone conversion.
ring chain tautomerism




tautomerism mcq



Wednesday, 5 April 2017

Electrophilic substitution reactons of haloarenes

Electrophilic substitution reactons of haloarenes


The halogen atom is ortho and para 

directing and therefore, substitution takes 

place at ortho and para positions giving 

electrophilic substitution reactions at the benzene ring.
In the resonance structures of chlorobenzene the ortho and para positions get negatively charged, i.e. the electron density is relatively more at ortho and para positions. The incoming electrophile is more likely to attack these positions. But, because of steric hindrance at the ortho position, the para-product usually predominates over the ortho product. Also, the halogen atom has -I effect because it is a electron withdrawing group. As a result, it tends to deactivate the benzene ring. Therefore, the electrophilic substitution reactions of chlorobenzene occur slowly and under drastic conditions compared to benzene. Thus,
  • Haloarenes undergo electrophilic substitution reactions slowly as compared to benzene.
  • Halogen group is ortho and para directing (para-product usually predominates over the ortho product).

Halogenation

Haloarenes react with halogens in the presence of ferric salt as catalyst to give ortho para isomers.
Haloarenes react with halogens  to give ortho para isomers

Nitration

Haloarenes with concentrated HNO3 in the presence of concentrated H2SO4.
nitration of haloarenes

Sulphonation

Sulphonation occurs when haloarene is treated with concentrated H2SO4.
sulphonation of haloarene

Alkylation and acylation

The alkylation and acylation reaction, known as 'Friedel-Craft reaction', is carried by treating haloarene with alkyl chloride or acyl chloride in the presence of a catalyst like anhydrous aluminium chloride. For example,
alkylation of chlorobenzene
acylation of chlorobenzene

Friday, 31 March 2017

Haloform reaction

Haloform reaction

NETT       REACTION :

Haloform Reaction is a type of organic reaction where haloform is produced by halogenation of methyl ketone in the presence of a base. It is a kind of nuclophilic substitution reaction. This reaction is used in quanitatve analysis to indicate the presence of methyl ketone.



If the hypohalite used in this reaction is sodium hypoiodite, the product is iodoform (CHI3). Such a haloform reaction is also known as iodoform test. As iodoform is a yellow crystalline solid with a characteristic odor and melting point it can be handled much more easily handled in the the laboratory.

                              Haloform reaction scheme


                          Haloform Reaction


  • When methyl ketones are treated with the halogen in basic solution, polyhalogenaton followed by cleavage of the methyl group occurs.

  • The products are the carboxylate and trihalomethane, otherwise known as haloform.

  • The reaction proceeds via successively faster halogenations at the α-position until the 3 H have been replaced.

  • The halogenations get faster since the halogen stablises the enolate negative charge and makes it easier to form.

  • Then a nucleophilic acyl substitution by hydroxide displaces the anion CX3 as a leaving group that rapidly protonates.

  • This reaction is often performed using iodine and as a chemical test for identifying methyl ketones. Iodoform is yellow and precipitates under the reaction conditions.

MECHANISM OF THE HALOFORM REACTION OF METHYL KETONES

Step 1:
First, an acid-base reaction. Hydroxide functions as a base and removes the acidic α-hydrogen giving the enolate.

Step 2:
The nucleophilic enolate reacts with the iodine giving the halogenated ketone and an iodide ion.

Step 3:
Steps 1 and 2 repeat twice more yielding the trihalogenated ketone.
Step 4:
The hydroxide now reacts as a nucleophile at the electrophilic carbonyl carbon, with the C=O becoming a C-O single bond and the oxygen is now anionic.
                         Haloform Schritt 1.svg
Step 5:
Reform the favourable C=O and displace a leaving group, the trihalomethyl system which is stabilised by the 3 halogens. This gives the carboxylic acid.
Step 6:
An acid-base reaction. The trihalomethyl anion is protonated by the carboxylic acid, giving the carboxylate and the haloform (trihalomethane).

            Haloform Schritt 2.svg