IGNOU ASSIGNMENT | CHEMICAL ENERGETICS, EQUILIBRIA AND FUNCTIONAL | ORGANIC CHEMISTRY I

          

                                                      ASSIGNMENT

 

ORGANIC CHEMISTCHEMICAL ENERGETICS, EQUILIBRIA AND FUNCTIONAL RY I

 

Core Course in Chemistry

 

Course Code: BCHCT-133 Assignment Code: BCHCT-133/TMA/2020

Maximum Marks: 100

 

Note: Attempt all questions. The marks for each question are indicated against it.

 

PART A: CHEMICAL ENERGETICS AND EQUILIBRIA

 

1.            a)      Define and explain a thermodynamically reversible process.                                        (5)

b)            0.25 mol of an ideal monoatomic gas undergoes isothermal expansion from a volume of 2.0 dm³ to 10 dm³ at 27 ^oC. Calculate the maximum work that can be

obtained from this process.                                                                                           (5)

 

2.            a)      Define standard enthalpy of formation and describe a method for its direct

determination with the help of an example.                                                                   (5)

b)            Differentiate between enthalpy driven and entropy driven reactions with the help of

suitable examples.                                                                                                         (5)

 

3.            a)      Give the statements of Zeroth, First, Second and the Third laws of thermodynamics

and outline their significance.                                                                                        (5)

b)            What is reaction quotient and how is it helpful in determining the direction of a

given reaction?                                                                                                              (5)

4.            a)       In the following equilibrium, predict the direction of shift of equilibrium for each condition listed below:

N2(g)    +     3H2(g)                     2NH 3(g)   + 92 kJ

i)                    Addition of H₂

ii)                  increased pressure

iii)                lowering of temperature.

b)      Define degree of ionisation of a weak electrolyte and discuss the factors affecting

it.                                                                                                                                   (5)

5.            a)      Explain the effect of common ions on the ionisation equilibria of weak acids with

the help of a suitable example.                                                                                      (5)

b)      Define solubility product constant and derive the relationships between solubility

and solubility product constants for salts of AB₂, A₂B types.                                        (5)

 

PART B: FUNCTIONAL GROUP ORGANIC CHEMISTRY-I

 

 

6        a)       List various methods of preparation of benzene and also write one chemical reaction for each case.

(5)

b)      Write the mechanism of Friedel-Craft’s alkylation. What are its limitations?               (5)

 

7.              a)   Explain why Nitro group is meta-directing deactivator?                                                   (5)

 

b)      Which one of the following would undergo faster S_N2 reaction? Explain

CH₃ – CH = CH – Cl or CH₃CH₂CH₂ – Cl

 

8.               a)       In normal reaction conditions, chlorobenzene does not react with NaOH, but 1- chloro-4-nitrobenzene reacts with NaOH in these conditions. Explain.

(5)

 

 

(5)

 

b)      Write the chemical equation and mechanism of pinacol-pinacolone rearrangement.

(5)

 

9.              a)   How will you perform following conversions?                                                               (5)

 

i)          Phenol to p-bromophenol

 

ii)        1,3-dihydroxyphenol to 1-(2,4-dihydroxyphenyl)ethanal

 

b)      How will you prepare tert-butylmethyl ether?                                                              (5)

 

10.           a)       Arrange the following carbonyl compounds in the order of their favourability for formation of nitriles:

CH₃CH₂CHO, CH₃COCH₃, HCOH and PhCOCH₃

Justify your answer.

(5)

 

b)      Taking suitable example write the mechanism of Mannich reaction.                            (5)

ASSIGNMENT SOLUTION GUIDE (2019-2020)

BCHCT-133

CHEMICAL ENERGETICS, EQUILIBRIA AND FUNCTIONAL ORGANIC CHEMISTRY I

Disclaimer/Special  Note:  These  are  just  the  sample  of  the  Answers/Solutions  to  some  of  the Questions  given  in  the  Assignments.  These  Sample  Answers/Solutions  are  prepared  by  Private Teacher/Tutors/Authors for the help and guidance of the student to get an idea of how he/she can answer  the  Questions  given  the  Assignments.  We  do  not  claim  100%  accuracy  of  these  sample answers  as  these  are  based  on  the  knowledge  and  capability  of  Private  Teacher/Tutor.  Sample answers may be seen as the Guide/Help for the reference to prepare the answers of the Questions given in the assignment. As these solutions and answers are prepared by the private teacher/tutor so  the  chances  of  error  or  mistake  cannot  be  denied.  Any  Omission  or  Error  is  highly  regretted though every care has been taken while preparing these Sample Answers/Solutions. Please consult your  own  Teacher/Tutor  before  you  prepare  a  Particular  Answer  and  for  up-to-date  and  exact information,  data  and  solution.  Student  should  must  read  and  refer  the  official  study  material provided by the university.

 

Q1. a) Define and explain a thermodynamically reversible process.

Ans. In thermodynamics, a reversible process is a process whose direction can be returned to its original position by inducing infinitesimal changes to some property of the system via its surroundings. Throughout the entire reversible process, the system is in thermodynamic equilibrium with its surroundings.

A Reversible process (thermodynamics) is one you can make go in reverse by an infinitesimal change to some parameter. It's of interest because it's a proxy for a process that does not generate entropy. And by analyzing reversible processes you can nail down the properties of entropy, which is initially pretty mysterious. It turns out that what generates entropy is energy falling down finite (non-zero) gradients of temperature, pressure, chemical potential etc.

For example, the prototypical entropy-generating situation is heat flowing from hot to cold.

Per the dS=δQ/T formula for entropy, heat flowing out of a hot cup of coffee at say 90°C =

363.15 K takes away δQ/363.15 of entropy. The very same heat entering a cool room at 20°C

= 293.15 K adds δQ/293.15 of entropy, which is greater. Entropy had been created. Equally, high-quality energy that could have been used for work has been degraded to thermal energy.

And it was unnecessary, because at least in principle you could have converted some of that energy to work with a heat engine: compress the working fluid in the heat engine till it matches the temperature of the coffee, let the heat from the coffee flow into the working fluid across a negligible temperature gradient, and then let the working fluid expand. Because the working fluid is hotter, it has more pressure and does more work expanding than it took to compress it.

b) 0.25 mol of an ideal monoatomic gas undergoes isothermal expansion from a volume of 2.0 dm3 to 10 dm3 at 27 oC. Calculate the maximum work that can be obtained from this process.

Ans. W_max = nRTln(V2/V1) = 0.25 * 8.314 * (27+273) * ln(10/2) = 1004 J

Q2. a) Define standard enthalpy of formation and describe a method for its direct determination with the help of an example.

Ans. Standard Enthalpies of Formation: The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane.

The standard conditions for which most thermochemical data are tabulated are a pressure of 1 atmosphere (atm) for all gases and a concentration of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state, which is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called standard enthalpies of formation ( ΔHof ) The enthalpy change for the formation of 1 mol of a compound from its component elements when the component elements are each in their standard states. The standard enthalpy of formation of any element in its most stable form is zero by definition.

The standard enthalpy of formation of any element in its standard state is zero by definition.

For example, although oxygen can exist as ozone (O₃), atomic oxygen (O), and molecular oxygen (O2), O2 is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H₂(g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (Figure). Therefore, O2 (g), H₂ (g), and graphite have ΔHfo values of zero.

Figure : Elemental Carbon. Although graphite and diamond are both forms of elemental carbon, graphite is slightly more stable at 1 atm pressure and 25°C than diamond is. Given enough time, diamond will revert to graphite under these conditions. Hence graphite is the standard state of carbon.

The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy

change for the following reaction:

It is not possible to measure the value of ΔHof for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, O2, and H₂ and measuring the heat evolved as glucose is formed; the reaction shown in Equation does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of are obtained using Hess’s law and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of ΔHof for an extensive list of compounds are given in Table. Note that ΔHof values are always reported in kilojoules per mole of the substance of interest. Also notice in Table that the standard enthalpy of formation of O2(g) is zero because it is the most stable form of oxygen in its standard state.

b) Differentiate between enthalpy driven and entropy driven reactions with the help of suitable examples.

Ans. Entropy.

For a spontaneous process (chemical reaction), the change in entropy is always greater than

0. For example:

Cu + 2AgNO3 = 2Ag + Cu (NO₃)2

f the entropy change is less than 0, then the reaction goes in the opposite direction (or you

need to apply additional energy for this process to proceed).

Enthalpy.

If the change in the enthalpy of the reaction is less than 0, then this reaction proceeds with

the release of heat. For example:

S + O₂ = SO2

If the change in the enthalpy of a reaction is greater than 0, then this reaction proceeds with

the absorption of heat. For example:

CaCO3 = CaO + CO₂

a)    Give the statements of Zeroth, First, Second and the Third laws of thermodynamics and outline their significance.

Ans. In order to avoid confusion, scientists discuss thermodynamic values in reference to a system and its surroundings. Everything that is not a part of the system constitutes its surroundings. The system and surroundings are separated by a boundary. For example, if the system is one mole of a gas in a container, then the boundary is simply the inner wall of the container itself. Everything outside of the boundary is considered the surroundings, which would include the container itself.

The boundary must be clearly defined, so one can clearly say whether a given part of the world is in the system or in the surroundings. If matter is not able to pass across the boundary, then the system is said to be closed; otherwise, it is open. A closed system may still exchange energy with the surroundings unless the system is an isolated one, in which case neither matter nor energy can pass across the boundary.

A Thermodynamic SystemA diagram of a thermodynamic system

The First Law of Thermodynamics

The first law of thermodynamics, also known as Law of Conservation of Energy, states that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another. For example, turning on a light would seem to produce energy; however, it is electrical energy that is converted.

A way of expressing the first law of thermodynamics is that any change in the internal energy (∆E) of a system is given by the sum of the heat (q) that flows across its boundaries and the work (w) done on the system by the surroundings:

[latex]\Delta E = q + w [/latex]

This law says that there are two kinds of processes, heat and work, that can lead to a change in the internal energy of a system. Since both heat and work can be measured and quantified, this is the same as saying that any change in the energy of a system must result in a corresponding change in the energy of the surroundings outside the system. In other

words, energy cannot be created or destroyed. If heat flows into a system or the surroundings do work on it, the internal energy increases and the sign of q and w are positive. Conversely, heat flow out of the system or work done by the system (on the surroundings) will be at the expense of the internal energy, and q and w will therefore be negative.

The Second Law of Thermodynamics: The second law of thermodynamics says that the entropy of any isolated system always increases. Isolated systems spontaneously evolve towards thermal equilibrium—the state of maximum entropy of the system. More simply put: the entropy of the universe (the ultimate isolated system) only increases and never decreases.

A simple way to think of the second law of thermodynamics is that a room, if not cleaned and tidied, will invariably become more messy and disorderly with time – regardless of how careful one is to keep it clean. When the room is cleaned, its entropy decreases, but the effort to clean it has resulted in an increase in entropy outside the room that exceeds the entropy lost.

The Third Law of Thermodynamics

The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero. The entropy of a system at absolute zero is typically zero, and in all cases is determined only by the number of different ground states it has. Specifically, the entropy of a pure crystalline substance (perfect order) at absolute zero temperature is zero. This statement holds true if the perfect crystal has only one state with minimum energy.

b)   What is reaction quotient and how is it helpful in determining the direction of a given reaction?

Ans. The expression for the reaction quotient, Q, looks like that used to calculate an equilibrium constant but Q can be calculated for any set of conditions, not just for equilibrium.

Q can be used to determine which direction a reaction will shift to reach equilibrium. If K > Q, a reaction will proceed forward, converting reactants into products. If K < Q, the reaction will proceed in the reverse direction, converting products into reactants. If Q = K then the system is already at equilibrium.

In order to determine Q we need to know: the equation for the reaction, including the physical states, the quantities of each species (molarities and/or pressures), all measured at the same moment in time.

To calculate Q: Write the expression for the reaction quotient.

Find the molar concentrations or partial pressures of each species involved. Substitute values into the expression and solve.

Example: 0.035 moles of SO₂, 0.500 moles of SO₂Cl₂, and 0.080 moles of Cl₂ are combined in an evacuated 5.00 L flask and heated to 100oC.  What is Q before the reaction begins?  Which direction will the reaction proceed in order to establish equilibrium?

 

SO2Cl₂(g)       SO2(g) + Cl₂(g)       K_c = 0.078 at 100oC

 

·         Write the expression to find the reaction quotient, Q.

 

·         Since K_c is given, the amounts must be expressed as moles per liter (molarity). The amounts are in moles so a conversion is required.

 

0.500 mole SO2Cl2/5.00 L = 0.100 M SO2Cl2

0.035 mole SO2/5.00 L = 0.070 M SO2

0.080 mole Cl2/5.00 L = 0.016 M Cl2

 

·         Substitute the values in to the expression and solve for Q.

·         Compare the answer to the value for the equilibrium constant and predict the shift. 0.078 (K) > 0.011 (Q)

Since K >Q, the reaction will proceed in the forward direction in order to increase the

concentrations of both SO2 and Cl₂ and decrease that of SO2Cl₂ until Q = K.

 

 

Q4. a) In the following equilibrium, predict the direction of shift of equilibrium for each condition listed below:

N₂ (g) + 3H2 (g)«2NH₃ (g) + 92 kJ

i)   Addition of H₂

ii)   Increased pressure

iii)   Lowering of temperature. Ans.

N2 (g) + 3H₂ (g) « 2NH3 (g)         ∆H = ‐92 kJ (exothermic)

This is an exothermic reaction (negative value) add it to the products side N2 (g) + 3H₂ (g) « 2NH3 (g)+92 kJ (exothermic)

Disturb equilibrium by increasing Temp – Shift Left, away from addition (just like adding

ammonia)

- Decrease Temp – Shift Right, toward removal

b) Define degree of ionisation of a weak electrolyte and discuss the factors affecting it. Ans.

 

5. a) Explain the effect of common ions on the ionisation equilibria of weak acids with the help of a suitable example.

Ans. The common ion effect refers to adding to a solution at equilibrium, a salt which contains an ion in common with one of the products of that equilibrium. The effect is to shift the equilibrium toward the reactant side of the equation.

Explanation:

The common ion effect is used to reduce the concentration of one of the products in an aqueous equilibrium. This may mean reducing the concentration of a toxic metal ion, or controlling the pH of a solution. The latter case is known as buffering.

Here are two examples:

Barium sulfate is given to a patient prior to abdominal x-rays, as it blocks the rays, enabling the image of the gut to be seen clearly.

 

b) Define solubility product constant and derive the relationships between solubility and solubility product constants for salts of AB₂, A₂B types.

Ans. The Solubility product constant, Ksp , is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in

solution. The more soluble a substance is, the higher the Ksp

value it has.

Consider the general dissolution reaction below (in aqueous solution):

aA(a) ƒ cC(cq ) + dD(aq )

............................(i)

Q6 a) List various methods of preparation of benzene and also write one chemical reaction for each case.

Ans.  Benzene  History, Chemical Properties                                                                                     Benze

arene and cyclic hydrocarbon. In 1825 benzene was first isolated by Faraday. Benzene is a natural component of crude oil, and is one of the most elementary petrochemicals. In 1845,

Hoffmann  isolated  benzene  from  coal-tar.                                                                                      Benzen

benzene structure is having alternative double bonds with hexagon shape. Hydrogens present in the benzene can be replaced by some of the other functional groups. As a result number benzene derivatives will be generated. Derivatives of benzene are ethylbenzene, cumene.

Benzene structure, Benzene ring structure was deduced by Friedrich August Kekule. The carbons present in benzene ring are arranged in a hexagon, and he suggested alternating double and single bonds between them. Each carbon atom has two hydrogens instead of four hydrogens. To satisfy the tetravalency of carbon, the benzene ring consisted of alternate single and double bonds.

August Kekule in dream saw a snake coil up, and grabs its own tail then he proposed benzene might be a ring structure.

Benzene preparation methods. Sodium benzoate is heated with soda-lime (NaOH) and when it gets decarboxylated benzene is obtained. Phenol vapors are passed over heated zinc dust, benzene is formed. When ethyne is passed through a red hot copper tube, it polymerizes to benzene Benzene is formed on the reduction of benzene diazonium chloride with sodium stannite or hypophosphorus acid. Benzene sulphonic acid on hydrolysis with superheated steam gives benzene. C6H5.SO₃H + H₂O —> C6H6 + H₂SO4

5 major processes for production of benzene , Following are the other major processes for production of benzene , Catalytic reforming , Toluene hydrodealkylation , Toluene disproportionation , Pyrolysis gasoline, Production from coal tar

Pyrolysis gasoline. Pyrolysis gasoline is the by-product of steam cracking of petroleum by products like paraffin gases, naphthas, gas oils. Pyrolysis gasoline contains 5 per cent diolefins. In addition it also contains 60 per cent aromatic compounds, 50 per cent of benzene. Different techniques are applied on diolefins to produce benzene, these are. Distillation of diolefins to olefins. Saturation of olefins to remove sulfur content. Execution of solvent extraction and distillation process to obtain benzene

Production from coal tar. This process improved methods of recovery and purification that coke-oven benzene has been able to withstand the competition of petroleum-derived benzene as well as it has. Production of benzene from coal tar involves recovering benzene

from coal tar.                                                                                                                                             Extrac

the   removal   of tar  acids                                                                                                                    Cru d e

hydrodealkylation

Toluene Hydrodealkylation - Toluene hydrodealkylation reaction takes place is as follows    Mixing of toluene with aromatics or paraffins , At specific pressures these  mixtures  are heated in the presence of hydrogen gas The steam which is formed in previous step then moved to reactor containing dealkylation catalyst In this reactor toluene reacts  with  hydrogen as a result, benzene and methane products will form At high pressures benzene

is separated from methane Then methane also removed from reactor Here in this step

benzene can be recovered from fractionalization column and then be stored.

Catalytic Reforming - Catalytic reforming involves the dehydrogenation of naphthenes to aromatics, or the isomerizatoin of alkylnaphthenes and it follows dehydrogenation process. The feed for this process is naptha. First the naptha is hydrotreated to remove sulfur contaminant. Recycled hydrogen is then added, mixed and heated. Conversion of paraffins to aromatic compounds in catalytic reactors and in this reactors platinum or rhenium chloride is acts as catalyst. In further step a stream is formed which is rich in aromatic compounds. Then stream is sent to separation section to separate hydrogen and this hydrogen recycled as basic feedstock. Liquid portion of stream fed to a stabilizer which separates hydrocarbons from liquids. The liquid is then sent to a debutanizer Benzene, toluene and xylenes are then extracted using glycol and sulfonate solvents

b) Write the mechanism of Friedel-Craft’s alkylation. What are its limitations? Ans.

This Lewis acid-catalyzed electrophilic aromatic substitution allows the synthesis of alkylated products via the reaction of arenes with alkyl halides or alkenes. Since alkyl substituents activate the arene substrate, polyalkylation may occur. A valuable, two-step alternative is Friedel-Crafts Acylation followed by a carbonyl reduction.

 

Several restrictions limit the usefulness of Friedel - Crafts reactions.

1. Rearrangement of carbocation-carbocation formed from an alkyl halide, alkene, or alcohol can rearrange to a more stable carbocation, it usually does so and the major product obtained from the reaction is usually the one from the more stable carbocation. When benzene is alkylated with butyl bromide.

Propyl caution rearrange by a hydride shift some developing primary carbocation's become more stable sec. carbocation's. Then benzene reacts with both kinds of carbocation's to form both propyl-benzene and isopropyl benzene:

This rearrangement can be minimized if we do the reaction at very low temperature

2-Polyalkylations-Polyalkylations often occur because Alkyl groups are electron - releasing groups, and once one is introduced into the benzene ring it activates the ring toward further substitution.

Polyacylations are not a problem in Friedel-Crafts acylations, because the acyl group (RCO–

) by itself is an electron - withdrawing group, and when it forms a complex with AlCl3 in the last step of the reaction, it is made even more electron withdrawing. This strongly inhibits further substitution and makes monoacylation easy.

3.   Friedel: Crafts reactions do not occur when powerful electron-withdrawing groups are present on the aromatic ring or when the ring bears an –NH₂, –NHR, or –NR2 group. This applies to alkylations and acylations.

 

We know that groups present on an aromatic ring can have large effect on the reactivity of the ring towards electrophilic aromatic substitution, Electron - withdrawing groups make the ring less reactive by making it electron deficient. Any substituent more electron withdrawing (or deactivating) than a halogen, that is, any meta-directing group, makes an aromatic ring too electron deficient to undergo a Friedel - Crafts reaction. The amino groups, –NH2, –NHR, and –NR2, are changed into powerful electron - withdrawing groups by the Lewis acids used to catalyze Friedel - Crafts reactions. For example:

4.Aryl and vinylic halides cannot be used as the halide component because they do not

form carbocations readily.

 

Q7. a) Explain why Nitro group is meta-directing deactivator?

Ans. Deactivating groups (that is, deactivating compared to benzene) deactivate because they withdraw electron density from the ring and make it a poorer nucleophile. This can be

either an inductive or a resonance effect, and shows up in the rate of the reaction (slower than for plain benzene).

Ortho, para-directing groups donate electrons by resonance, or sometimes (alkyl and silyl groups) by hyperconjugation or induction. This stabilizes resonance structures for o,p- substitution, as Harry Peters showed in his answer.

Most o,p-directing groups are also activating, but halogens are deactivating even though they are o,p-directing. This is because, although they can donate electrons by resonance, the inductive electron-withdrawing effect dominates for halogens.

Deactivating groups that either do not donate electrons by resonance (sulfonic acid groups and ammonium ion groups) or actually withdraw electrons by resonance (carbonyl, nitrile and nitro groups) are meta-directing.

The reason that they are meta-directing is not that they stabilize the intermediate for meta- substitution: far from it. Instead, they destabilize the intermediates for o,p-substitution relative to the meta intermediate.

This is true in general:

o,p-directing groups stabilize the intermediates for o,p substitution relative to the meta intermediate.

m-directing groups destabilize the intermediates for o,p substitution relative to the meta intermediate.

b) Which one of the following would undergo faster SN₂ reaction? Explain CH₃ – CH = CH – Cl or CH₃CH₂CH₂ – Cl

Ans. In vinyl chloride (CH3-CH=CH-Cl) lone pair of chlorine is in resonance so chlorine will not get removed easily while in case of alkyl chloride resonance is not present.There's no such resonance in propylchloride. Hence, chlorine reacts readily with the nucleophile. CH3-CH₂-CH₂-Cl is more reactive in sn₂ reactions.

Q8. a) In normal reaction conditions, chlorobenzene does not react with NaOH, but 1- chloro-4-nitrobenzene reacts with NaOH in these conditions. Explain.

Ans. When the reaction with chloro-4-nitrobenzene occurs, SNAr mechanism takes place, because Meisenheimer's intermediate is stabilized by resonance, which is impossible with regular chlorobenzene.

 

b) Write the chemical equation and mechanism of pinacol-pinacolone rearrangement. Ans.

Q9. a) How will you perform following conversions?

i)   Phenol to p-bromophenol

ans. You have to brominate Phenol in carbon disulfide to obtain 4-Bromophenol. At -30° C, 4-bromophenol is yielded by 97% and at +30° C, the yield is 82%. You will need a 5 litre flask fitted with a mechanical stirrer, a reflux condensor and a separatory funnel. To the top of the tunnel, a calcium chloride tube is attached and from this, a glass tube leading into a beaker holding about 1200ml of cracked ice and water for the absorption of hydrobromic acide evolves. In a separatory funnel approximately 546 ml of bromine which is dissolved in equal volume of carbon disulfide is placed. The flask is cooled in salt and ice mixture while the stirring starts and bormine solution starts to run in. This takes approximately 2 hours. Post 2 hours, the flask is detached and a condensor is attached in downward distillation. The flask is then heated and the carbon disulfide is distilled of. The carbon disulfide collects as much as 1200ml. The residual liquid is then vaccum distilled yiekding 1660gs of p-bromophenol boiling 145-150/25-30mm and 136 gms of lower boiling material consisting of a mixture of o- and p-bromophenol. The reaction mixture should be cooled with salt and ice, when making 4-bromophenol. 25gms of 1,4 dioxane dibromide is gradually added to 9.4 gms of phenol with water cooling. Then the mixture is poured with water and extracted with ether. The ether is dried with sodium sulfate, filtered and evaporated yielding 88% of 4-bromophenol.

ii)   1, 3-dihydroxyphenol to 1-(2, 4-dihydroxyphenyl) ethanol

Ans. Hydroxytyrosol is a phenolic phytochemical naturally occurring in extra virgin olive oil, with potential antioxidant, anti-inflammatory and cancer preventive activities. Although the mechanisms of action through which hydroxytyrosol exerts its effects have yet to be fully determined, this agent affects the expression of various components of the inflammatory response, possibly through the modulation of the nuclear factor-kappa B (NF-kB) pathway. The effects include the modulation of pro-inflammatory cytokines, such as the inhibition of interleukin-1alpha (IL-1a), IL-1beta, IL-6, IL-12, and tumor necrosis factor-alpha (TNF-a); increased secretion of the anti-inflammatory cytokine IL-10; inhibition of the production of certain chemokines, such as C-X-C motif chemokine ligand 10 (CXCL10/IP-10), C-C motif chemokine ligand 2 (CCL2/MCP-1), and macrophage inflammatory protein-1beta (CCL4/MIP-1b); and inhibition of the expression of the enzymes inducible nitric oxide synthase (iNOS/NOS2) and prostaglandin E2 synthase (PGES), which prevent the production of nitric oxide (NO) and prostaglandin E (PGE2), respectively. In addition, hydroxytyrosol is able to regulate the expression of other genes involved in the regulation of tumor cell proliferation, such as extracellular signal-regulated and cyclin-dependent kinases. Also, hydroxytyrosol scavenges free radicals and prevents oxidative DNA damage. This induces apoptosis and inhibits proliferation in susceptible cancer cells.

Hydroxytyrosol is a member of the class of catechols that is benzene-1,2-diol substituted by a 2-hydroxyethyl group at position 4. Isolated from Olea europaea, it exhibits antioxidant and antineoplastic activities. It has a role as a metabolite, an antioxidant and an antineoplastic agent. It is a member of catechols and a primary alcohol. It derives from a 2- (4-hydroxyphenyl) ethanol.

b) How will you prepare tert-butylmethyl ether?

Ans. The invention provides a preparation method and an application of methyl sec-butyl ether. The methyl sec-butyl ether (MSBE) is synthesized through an intermolecular dehydration by using sec-butyl alcohol and methanol. The method comprises the step of performing the intermolecular dehydration of sec-butyl alcohol and methanol at the temperature of 20 DEG C-180 DEG C for 1-20 hours by using an N-methylimidazole p- toluene sulfonate ionic liquid as a catalyst to obtain the methyl sec-butyl ether (MSBE), wherein the molar ratio of methanol to sec-butyl alcohol is 1 : 1-12 : 1; and the volume ratio of sec-butyl alcohol to the catalyst is 1 : 3-12 : 1. The method has the advantages that the catalyst has high reaction activity, is easy to recycle, and can be used repeatedly; and that the reaction process is simple in operations, etc.

Table 1: under the differing temps, the equilibrium conversion of methyl tertiary butyl

ether (MTBE).

Temperature/℃ 50 60 70 80 90 Transformation efficiency/% 96.8 95.8 94.6 93.1 91.4

Rising temperature of reaction energy fast reaction speed, but unfavorable to transformation efficiency, in order to obtain higher product yield, requirement is carried out etherification reaction under suitable temperature condition, when the concentration of iso-butylene in the raw material was reduced to finite concentration, the n-butene that is adsorbed onto on the catalyst surface then generated methyl sec-butyl ether (MSBE) with the methyl alcohol reaction. Reduce temperature of reaction and be conducive to reduce the growing amount that reaction generates methyl sec-butyl ether (MSBE), but in order not reduce the etherification reaction speed of iso-butylene, need to select the high catalyzer of low temperature active.

Correspondingly, methyl sec-butyl ether (MSBE) can adopt n-butene and methyl alcohol to make through etherificate, but the per pass conversion of reaction process n-butene lower (generally being lower than 5%), need a large amount of n-butenes to be separated and circulate, energy consumption is larger, and production cost is high, and complex technical process.

At present, preparation methyl tertiary butyl ether (MTBE) adopts strongly acidic ion- exchange resin catalyst, molecular sieve catalyst, heteropolyacid catalyst and modified catalyst thereof etc. mostly, and there is no the bibliographical information of relevant preparation methyl sec-butyl ether (MSBE) aspect both at home and abroad.

Sulphonated polystyrene resin take divinylbenzene as linking agent for the preparation of the strongly acidic ion-exchange resin catalyst of methyl tertiary butyl ether (MTBE), such as Amberlyst-15, D-72 etc.

Q10. a) Arrange the following carbonyl compounds in the order of their favourability for formation of nitriles:

CH₃CH₂CHO, CH₃COCH₃, HCOH and PhCOCH₃

Justify your answer.

Ans. The R−CHO group suggests an aldehyde, and since there are 3 carbon atoms in the structure, the IUPAC name is Propanal. Also known as propionaldehyde.

The given compound is an aldehyde. According to IUPAC system for aldehydes, the carbon of CHO is also considered in the longest chain. So there are three carbon in the longest chain hence the parent alkane is propane (since all carbonyl compounds are considered as derivatives of alkane). In the IPAC system, for naming aldehyde the ‘e' of alkane is replaced by ‘al’ Therefore the name of given compound is ‘Propanal'

 

Preferred IUPAC name

Propan-2-one Other names Acetone Dimethyl ketone

Dimethyl carbonyl β-Ketopropane Propanone

2-Propanone

Dimethyl formaldehyde Pyroacetic spirit (archaic)

Ketone propane

b) Taking suitable example write the mechanism of Mannich reaction.

Ans. The Mannich reaction is the organic reaction in which an acidic H+ ion (proton),  which is positioned next to a carbonyl group, undergoes an amino alkylation with the help of formaldehyde and ammonia (a primary or secondary amine can be used instead of NH3). The product of this reaction is a beta-amino carbonyl compound.

It is an organic chemical coupling reaction which is named after the German chemist Carl Mannich. The beta-amino-carbonyl compound, which is the final product, is also referred  to as a Mannich base. Mannich reactions also include the reactions between an aldimine (imines that are analogs of an aldehyde, general chemical formula R-CH=N-R’) and an alpha-methylene carbonyl.

An example of the Mannich reaction between an amine or ammonia with formaldehyde and an alpha acidic proton from a carbonyl compound to give a beta-amino carbonyl compound is illustrated below.

It can be noted that the Mannich reaction can also be considered as a condensation reaction.

The  reason  primary/secondary  amines  (or  NH3)  are  used  for  the  activation  of  the

formaldehyde is that tertiary amines would lack the N-H proton which is required for the

formation of the enamine intermediate. Mannich Reaction Mechanism

The Mannich reaction mechanism proceeds via two steps:

Step 1:

The reaction between formaldehyde and the amine leads to the formation of the iminium ion. This formation of an iminium ion is illustrated below.

Step 2:

The compound containing the carbonyl group (which is a ketone in this illustration for the mechanism of the Mannich reaction) undergoes tautomerization to give its enol form. This enol form of the compound with a carbonyl functional group now proceeds to execute an attack on the iminium ion. This attack finally yields the required beta-amino-carbonyl compound or Mannich base. The illustration for this step is given below.

Thus, the amino alkylation of an acidic proton which is placed alongside a carbonyl group by formaldehyde and a primary or secondary amine or ammonia is achieved, and the required beta-amino carbonyl compound (or Mannich base) is produced. From the mechanism illustrated above, it can be observed that the Mannich reaction is an example of the nucleophilic addition of an amine to a carbonyl group.

Applications: This organic reaction has numerous applications, primarily in the synthesis of specific organic compounds. Some of these compounds and their applications are listed below.

This reaction is used in the preparation of alkyl amines, which are employed in the production of pesticides.

Many antibiotics are Mannich bases. For example, the Mannich base of tetracycline is Rolitetracycline, a broad-spectrum antibiotic.

Many catalysts and polymers are produced with the help of this reaction.

The Mannich reaction is used in the synthesis of many pharmaceutical drugs. One such example is the use of this reaction in the production of fluoxetine, a powerful antidepressant.

The nonsteroidal anti-inflammatory drug tolmetin is synthesized with the help of this reaction. The reaction is also used in the production of some soaps and detergents.

The reaction details and the mechanism of the Mannich reaction are briefly described in  this article along with few applications. To learn more about this reaction and other important named reactions in organic chemistry, register with BYJU’S and download the mobile application on your smartphone.