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Chemistry Test - 8
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  • Question 1
    1 / -0

    For standardizing NaOH solution, which of the following is used as a primary standard?

    Solution

    Oxalic acid is used as a primary standard for NaOH standardizing.

    Main Concept :

    Acid-Base TitrationAn acid–base titration is the determination of the concentration of an acid or base by exactly neutralizing the acid or base with an acid or base of known concentration. This allows for quantitative analysis of the concentration of an unknown acid or base solution. It makes use of the neutralization reaction that occurs between acids and bases.

    The initial pH is approximated for a weak acid solution in water using the equation.

    Where Ka is the dissociation constant and C is the concentration of the acid.

    The pH before the equivalence point depends on the amount of weak acid remaining and the amount of conjugate base formed. The pH can be calculated by the following formula (which is a variation of the Henderson-Hasselbalch equation):

    Where:

    ● pKa is the negative log of the acid dissociation constant of the weak acid.

     is the number of moles of added strong base in the solution.

    ● nHA initial is the number of moles the weak acid initially present.

    When the numerator of the log term equals the denominator  then the ratio goes to 1 and the log term goes to zero. Thus the pH will equal the pKa which occurs half-way to the equivalence point.

    At the equivalence point, the weak acid is consumed and converted to its conjugate base. The pH will be greater than 7 and can be calculated from an equation derived from the following relationships:

    The previous 3 relationships are used to generate the equivalence point pH formula below:

    ● Ca = concentration of acid and Cb = concentration of base

    ● Kw = dissociation constant for water and Ka = for the acid

    Note that when an acid neutralizes a base, the pH may or may not be neutral (pH = 7). The pH depends on the strengths of the acid and base.

    After the equivalence point, the solution will contain two bases: the conjugate base of the acid and the strong base of the titrant. However, the base of the titrant is stronger than the conjugate base of the acid. Therefore, the pH in this region is controlled by the strong base. As such the pH can be found using the following:

  • Question 2
    1 / -0

    2-methyl pent-2-ene on ozonlysis will give

    Solution

    Main Concept :

    Ozonolysis of alkenes (Oxidation and Reduction)

    The reaction of alkenes with ozone:
    Ozonolysis of alkenes with reduction workup

    Note that the carbon-carbon double bond is broken and we are forming a carbon-oxygen double bond on each of the two carbons that originally composed the alkene. The second step in ozonolysis is called the "workup". There are two different types of "workup", and the most common is referred to as "reductive workup". In this step, we add a reducing agent (commonly zinc metal or dimethyl sulfide) that decomposes the intermediate formed at the end of the ozonolysis reaction (called an "ozonide" by the way). If you're wondering where the third oxygen of ozone went - it's now attached to what used to be our reducing agent (making either zinc oxide (ZnO) or dimethyl sulfoxide (DMSO). [For more details / mechanism everything is written out in this post.]

    Using "reductive workup" preserves all other aspects of the molecule save the double bond. So if we start with, say, a trisubstituted alkene, as in the example below, we will end up with a ketone and an aldehyde. [What happens if the alkene carbon is attached to two hydrogens? It becomes formaldehyde, which is then further converted to carbon dioxide]

    Note that although I've written (CH3)2S as the reductant here, it's essentially interchangeable with Zn for our purposes.

    An interesting consequence of ozonolysis is that if the alkene is within a ring, you end up with a chain containing two carbonyls:

    If your molecule has multiple alkenes, then you will end up with more than two fragments. For many years ozonolysis was used as a method for the structure determination of unknown molecules. By analyzing the fragments it is then possible to deduce what the original structure was, through "stitching" together the fragments. [This was particularly important in the case of unsaturated molecules known as terpenes]. Here's one example:

    Molecules with multiple alkenes are cleaved into fragments

    This isn't the end of the story with ozonolysis. There's a second type of workup that can be used, referred to as oxidative workup. Instead of using Zn or S(CH3)2, if we use the oxidant hydrogen peroxide [H2O2], any aldehydes that form will be oxidized to give carboxylic acids. Like in the example below - notice that the green C - H bond is oxidized to C - OH [but all the other hydrogens remain intact].

    Typical oxidant used for "oxidative workup" is H2O2, this oxidizes any aldehydes to carboxylic acids.
    The same process can be performed by replacing Owith hot, acidic KMnO4KMnO4

    An alternative to using ozone for oxidative workup is to use the reagent MnO4, especially in the presence of hot acid; this will lead to the same result.

  • Question 3
    1 / -0

    A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity smell was formed. The liquid was:

    Solution

    Fruity smell is due to ester formation which is formed by reaction between ethanol and acetic acid.


    The Lewis or Brønstedt acid-catalyzed esterification of carboxylic acids with alcohols to give esters is a typical reaction in which the products and reactants are in equilibrium.

    The equilibrium may be influenced by either removing one product from the reaction mixture (for example, removal of the water by azeotropic distillation or absorption by molecular sieves) or by employing an excess of one reactant.

    Mechanism of the Fischer Esterification

    Addition of a proton  or a Lewis acid leads to a more reactive electrophile. Nucleophilic attack of the alcohol gives a tetrahedral intermediate in which there are two equivalent hydroxyl groups. One of these hydroxyl groups is eliminated after a proton shift (tautomerism) to give water and the ester.

  • Question 4
    1 / -0

    Solution


    It can be inferred that number of gram equivalents of a substance = n × number of moles and Normality= ‘n’ × Molarity.

    It is extremely convenient to use the law of gram equivalents to solve problems based on chemical reactions. According to this law the ‘number of gram equivalents of all reactants are equal to each other in a reaction assuming none of them are in excess and is also equal to number of gram equivalents of all products’ assuming that all the reactants are undergoing reaction in the reaction.

    We can use this law conveniently to solve problems without requiring to know much about the reactions. For this we need to have good understanding of the ‘n’ factor of a substance. ‘n’ factor is the valency factor or conversion factor.

    * n – factor of substance in redox reaction is equal to number of moles of lost or gain electron per molecule.

    * n – factor of substance in non – redox reaction is equal to the product of displaced mole and its charge.

    (1) Acids: The number of moles of replaceable H+ ions per mole of the acid e.g. for 

  • Question 5
    1 / -0

    Solution

    Main Concept :
    Test of Protein (Ninhydrin Test)

    Ninhydrin (1,2,3-Indantrione monohydrate, or triketohydrindene hydrate) is often used to detect α-amino acids and also free amino and carboxylic acid groups on proteins and peptides. When about 0.5 mL of a 0.1% solution of ninhydrin is boiled for one or two minutes with a few mL of dilute amino acid or protein solution, a blue colour develops.

    Ninhydrin degrades amino acids into aldehydes, ammonia, and CO2 through a series of reactions; the net result is ninhydrin in a partially reduced form hydrindantin:

    Ninhydrin then condenses with ammonia and hydrindantin to produce an intensely blue or purple pigment, sometimes called Ruhemann's purple:

  • Question 6
    1 / -0

    Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated?

    Solution

    In terms of VSEPR theory, the number of Hybridised orbitals around the central atom of BrF5 is

    Main Concept :
    Hybridisation of Atomic Orbitals sp, sp2, sp3, sp3d, sp3d2, dsp2 and sp3d3.

    The concept of hybridization was introduced by Pauling and Slater.

    Hybridization: It is defined as the intermixing of dissimilar orbitals of the same atom but having slightly different energies to form same number of new orbitals of equal energies and identical shapes. The new orbitals so formed are known as hybrid orbitals.

    How to determine type of hybridization: The structure of any molecule can be predicted on the basis of hybridization which in turn can be known by the following general formulation,

    Where; H = Number of orbitals involved in hybridization viz. 2, 3, 4, 5, 6 and 7, hence nature of hybridization will be sp, sp2, sp3, sp3d, sp3d2, sp3d3 respectively.
    V = Number of electrons in valence shell of the central atom,
    M = Number of monovalent atom
    C = Charge on cation,
    A = Charge on anion

    Characteristics of hybridization:
    (1) Only orbitals of almost similar energies and belonging to the same atom or ion undergoes hybridization.
    (2) Hybridization takes place only in orbitals, electrons are not involved in it.
    (3) The number of hybrid orbitals produced is equal to the number of pure orbitals, mixed during hybridization.
    (4) In the excited state, the number of unpaired electrons must correspond to the oxidation state of the central atom in the molecule.
    (5) Both half filled orbitals or fully filled orbitals of equivalent energy can involve in hybridization.
    (6) Hybrid orbitals form only sigma bonds.
    (7) Orbitals involved in pi bond formation do not participate in hybridization.
    (8) Hybridization never takes place in an isolated atom but it occurs only at the time of bond formation.

    Geometric Arrangements Charactristic of Hybrid Orbital Sets


    Other Concepts :

    Concept 1 :
    Valence Shell Electron Pair Repulsion theory [VSEPR] for Molecules

    The basic concept of the theory was suggested by Sidgwick and Powell (1940). It provides a useful idea for predicting shapes and geometries of molecules. The concept tells that, the arrangement of bonds around the central atom depends upon the repulsion operating between electron pairs(bonded or non bonded) around the central atom. Gillespie and Nyholm developed this concept as valence shell electron pair repulsions (VSEPR) theory. The main postulates of the VSEPR theory are :

    (1) For polyatomic molecules containing 3 or more atoms, one of the atoms is called the central atom to which other atoms are linked.

    (2) The geometry of a molecule depends upon the total number of valence shell electron pairs (bonded or not bonded) present around the central atom and their repulsion due to relative sizes and shapes.

    (3) If the central atom is surrounded by bond pairs only, then it gives the symmetrical shape to the molecule.

    (4) If the central atom is surrounded by lone pairs (lp) as well as bond pairs (bp) of electrons then the molecule has a distorted geometry (unsymmetrical shape).

    (5) The relative order of repulsion between electron pairs is as follows : 

    A lone pair is concentrated around the central atom while a bond pair is pulled out between two bonded atoms. As such repulsion becomes greater when a lone pair is involved.

    Lone pair-bond pair and their respective shapes table

  • Question 7
    1 / -0

    A certain zero order reaction has k = 0.025 Ms-1 for the disappearance of A. What will be the concentration of A after 15 seconds if the initial concentration is 0.5 M?

    Solution

    Main Concept :

    Zeroth order reactionIt is the duration of time after which the concentration of a reactant is reduced to half (50%) of its initial value. We shall determine the half-life for a zero order and first order reaction.

    (i) For zero order reaction,

    Thus, we can say that t1/2 for a zero order reaction is directly proportional to initial concentration of the reactant and inversely proportional to rate constant (k). ​
    Other Concepts :

    Concept 1 :
    Rate constant (k)The rate at which a substance reacts is directly proportional to its active mass and the rate at which a reaction proceeds is proportional to the product of the active masses of the reacting substances.

    Rate constant k is also called Specific reaction rate.

    The value of rate constant depends on, nature of reactant, temperature and catalyst. It is independent of concentration of the reactants.

  • Question 8
    1 / -0

    The IUPAC name of this compound is:

    Solution

    1.Find the Functional Group, here aldehyde is fuctional group.
    2.Name prefix according to the priority of other groups.

    Main Concept :
    IUPAC names

    1 Meth
    2 Eth
    3 Prop
    4 But
    5 Pent
    6 Hex
    7 Hept
    8 Oct
    9 None
    10 Dec

    In chemistry, a number of prefixes, suffixes and infixes are used to describe the type and position of functional groups in the compound.

    The steps to naming an organic compound are:

    I. Identification of the parent hydrocarbon chain. This chain must obey the following rules, in order of precedence:

    i. It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.

    ii. It should have the maximum number of multiple bonds.

    iii. It should have the maximum number of single bonds.

    iv. It should have the maximum length.

    II. Identification of the parent functional group, if any, with the highest order of precedence.

    III. Identification of the side-chains. Side chains are the carbon chains that are not in the parent chain, but are branched off from it.

    IV. Identification of the remaining functional groups, if any, and naming them by the their ion names (such as hydroxy for -OH, oxy for =O, oxyalkane for O-R, etc.).

    V. Different side-chains and functional groups will be grouped together in alphabetical order. (The prefixes di-, tri-, etc. are not taken into consideration for grouping alphabetically. For example, ethyl comes before dihydroxy or dimethyl, as the "e" in "ethyl" precedes the "h" in "dihydroxy" and the "m" in "dimethyl" alphabetically. The "di" is not considered in either case). When both side chains and secondary functional groups are present, they should be written mixed together in one group rather than in two separate groups.

    VI. Identification of double/triple bonds.

    VII. Numbering of the chain. This is done by first numbering the chain in both directions (left to right and right to left), and then choosing the numbering which follows these rules, in order of precedence:

    i. Has the lowest-numbered locant (or locants) for the suffix functional group. Locants are the numbers on the carbons to which the substituent is directly attached.

    ii. Has the lowest-numbered locants for multiple bonds (The locant of a multiple bond is the number of the adjacent carbon with a lower number).

    iii. Has the lowest-numbered locants for double bonds.iv. Has the lowest-numbered locants for prefixes.

    VIII. Numbering of the various substituents and bonds with their locants. If there is more than one of the same type of substituent/double bond, a prefix is added showing how many there are.The numbers for that type of side chain will be grouped in ascending order and written before the name of the side-chain. If there are two side-chains with the same alpha carbon, the number will be written twice. Example: 2,2,3-trimethyl- . If there are both double bonds and triple bonds, "en" (double bond) is written before "yne" (triple bond). When the main functional group is a terminal functional group (A group which can only exist at the end of a chain, like formyl and carboxyl groups), there is no need to number it.

    I. Arrangement in this form: Group of side chains and secondary functional groups with numbers made in step 3 + prefix of parent hydrocarbon chain (eth, meth) + double/triple bonds with numbers (or "ane") + primary functional group suffix with numbers.

    II. Wherever it says "with numbers", it is understood that between the word and the numbers, the prefix(di-, tri-) is used.

    III. Adding of punctuation:

    i. Commas are put between numbers (2 5 5 becomes 2,5,5)

    ii. Hyphens are put between a number and a letter (2 5 5 trimethylheptane becomes 2,5,5-trimethylheptane)

    iii. Successive words are merged into one word (trimethyl heptane becomes trimethylheptane)

    iv. Note: IUPAC uses one-word names throughout. This is why all parts are connected.


    The finalized name should look like this:#,#-di-#--#--#,#,#-tri-#,#-di-#--#-Note: # is used for a number. The group secondary functional groups and side chains may not look the same as shown here, as the side chains and secondary functional groups are arranged alphabetically. The di- and tri- have been used just to show their usage. (di- after #,#, tri- after #,#,#, etc.)

    Example: Here is a sample molecule with the parent carbons numbered:

    For simplicity, here is an image of the same molecule, where the hydrogens in the parent chain are removed and the carbons are shown by their numbers:

    Now, following the above steps:I. The parent hydrocarbon chain has 23 carbons. It is called tricosa-.II. The functional groups with the highest precedence are the two ketone groups.i. The groups are on carbon atoms 3 and 9. As there are two, we write 3,9-dione.ii. The numbering of the molecule is based on the ketone groups. When numbering from left to right, the ketone groups are numbered 3 and 9. When numbering from right to left, the ketone groups are numbered 15 and 21. 3 is less than 15, therefore the ketones are numbered 3 and 9. The smaller number is always used, not the sum of the constituents numbers.III. The side chains are: an ethyl- at carbon 4, an ethyl- at carbon 8, and a butyl- at carbon 12. IV. Note: The -O-CH3 at carbon atom 15 is not a side chain, but it is a methoxy functional groupi. There are two ethyl- groups. They are combined to create, 4,8-diethyl.ii. The side chains are grouped like this: 12-butyl-4,8-diethyl. (But this is not the final grouping, as functional groups may be added in between.)V. The secondary functional groups are: a hydroxy- at carbon 5, a chloro- at carbon 11, a methoxy- at carbon 15, and a bromo- at carbon 18. Grouped with the side chains, this gives 18-bromo-12-butyl-11-chloro-4,8-diethyl-5-hydroxy-15-methoxyVI. There are two double bonds: one between carbons 6 and 7, and one between carbons 13 and 14, They would be called "6,13-diene", but the presence of alkynes switches it to 6,13-dien. There is one triple bond between carbon atoms 19 and 20. It will be called 19-yneVII. The arrangement (with punctuation) is: 18-bromo-12-butyl-11-chloro-4,8-diethyl-5-hydroxy-15-methoxytricosa-6,13-dien-19-yne-3,9-dioneVIII. Finally, due to Cis-trans isomerism, we have to specify the relative orientation of functional groups around each double bond. For this example, we have (6E,13E) The final name is (6E,13E)-18-bromo-12-butyl-11-chloro-4,8-diethyl-5-hydroxy-15-methoxytricosa-6,13-dien-19-yne-3,9-dione.

    Other Concepts :

    Concept 1 :
    IUPAC Rules for polyfunctional compoundsRules for IUPAC names of poly functional organic compounds

    Organic elements which have two or more functional groups are called poly functional compounds. Their IUPAC names are obtained as follows,

    (i) Principal functional group: If the organic compound contains two or more functional groups, one of the functional sets is selected as the principal functional group while all the remaining functional groups (also called the secondary functional groups) are treated as substituents. The following order of preference is used while selecting the principal functional group.

    Sulphonic acids > carboxylic acids > anhydrides > esters > acid chlorides > acid amides > nitriles > aldehydes > ketones > thiols > alcohols >alkenes > alkynes.

    All the remaining functional sets such as halo (fluoro, chloro, bromo, iodo), nitroso (- NO) -nitro (-NO2), amino (- NH2) and alkoxy (-OR) are treated as substituents.

    (ii) Selecting the principal chain: Select the longest continuous chain of carbon atoms containing the principal functional group and maximum number of secondary functional groups and multiple bonds, if any.

    (iii) Numbering the principal chain: Number the principal chain in such a way that the principal functional group gets the lowest possible number followed by double bond and triple bond and the substituents, i.e.

    Principal functional set > double bond > triple bond > substituents

    (iv) Alphabetical order: Identify the prefixes and the positional numbers (also called locants) for the secondary functional groups and other substituents and place them in alphabetical order before the word root.

    IV. Polyfunctional compounds containing more than two like functional groups : According to latest convention (1993 recommendations for IUPAC nomenclature), if an unbranched carbon reaction chain is directly linked to more than two like functional sets, the organic element is called as a derivative of the parent alkane which does not include carbon atoms of the functional sets. As given,

  • Question 9
    1 / -0

    When HCI reacts with finely powdered iron, the chloride formed is:

    Solution

    The reaction of HCl with finally powdered iron produces hydrogen gas:

    Liberation of hydrogen prevents the formation of ferric chloride FeCl3.

    Main Concept :
    Examples on Properties of Compounds of Iron

    Properties

    I. Hydrated ferrous sulphate (FeSO4.7H2O) is a green crystalline compound whereas anhydrous FeSOis colourless or white. The crystals acquire brownish yellow colour due to formation of basic ferric sulphate on exposure to air.

    II. It is isomorphous with Epsom salt MgSO4.7H2O and white vitriol ZnSO4.7H2O. If effloresce on exposure to air.

    III. Action of heat: Hydrated salt loses water of crystallization at 300°C and on strong heating breaks up to form ferric oxide with the evolution of SO2 and SO3.

  • Question 10
    1 / -0

    Which of the following carbocation would have the greatest stability?

    Solution

    Nitrogen lone pair stabilizes the positive charge to the most through resonance.
    Main Concept :
    Application of inductive, Electromeric, Mesomeric, Resonance, Steric Effects, Hyperconjugation and H-bonding In Acidic and Basic Characters of Organica. EWGs (-I, -R, -M, -E) increase the acidic character and decrease the basic character.

    b. EDGs (+I, +R, +M, +E) decrease the acidic character and increase the basic character.

    c. Inductive effect (-I or +I) operates at o-, m-, and p- positions, and the effect at o- > m- > p- positions.

    d. Resonance effect (+R or -R) operates t o- and p- positions and not at  m- positions and the effect  is equal at o- and p- positions.

    e. Hyperconjugation effect operates at o- and p- positions and not at m- position, and the effect is equal to o- and p- positions.

    f. Ortho effect operates in o- substituted aromatic acid and o- substituted aniline and not in o- substituted phenol.

    g. Halo- substituted aromatic acid, phenols, and aromatic amines.
    Halogens are  withdrawing groups (-I) and while considering the acidic or basic character of halo-substituted aromatic acids, phenol, and aromatic amines, only -I effect is considered and not +R effect, except in case of fluoro substituted compounds in which both -I and +R effects are considered.

    h. Bulky substituted in the ortho- position of aromatic acid effects the acidic character by steric inhibition of resonance, but this effect is not observed in bulky subsistent in the ortho- position of phenol, since the (-OH) group  is smaller in size in comparision to (-COOH) group.

    i. In table EWG (-I or -R effect) are represented by downward vector sign (↓) and EDG (+I or +R or H.C. effect) by upward vector sign (↑).

    Acidic character of substituted aromatic acid, phenol, and basic character of substituted aromatic amines















    Other Concepts :

    Concept 1 :
    Structure of carbocationStructure and properties

    The charged carbon atom in a carbocation is a "sextet", i.e. it has only six electrons in its outer valence shell instead of the eight valence electrons that ensures maximum stability (octet rule). Therefore, carbocations are often reactive, seeking to fill the octet of valence electrons as well as regain a neutral charge. One could reasonably assume a carbocation to have sp3 hybridization with an empty sp3 orbital giving positive charge. However, the reactivity of a carbocation more closely resembles sp2 hybridization with a trigonal planar molecular geometry. An example is the methyl cation, 

    Order of stability of examples of tertiary (III), secondary (II), and primary (I) alkylcarbenium ions, as well as the methyl cation (far right). Carbocations are often the target of nucleophilic attack by nucleophiles like hydroxide (OH-) ions or halogen ions.

    Carbocations typically undergo rearrangement reactions from less stable structures to equally stable or more stable ones with rate constants in excess of 109/sec. This fact complicates synthetic pathways to many compounds. For example, when 3-pentanol is heated with aqueous HCl, the initially formed 3-pentyl carbocation rearranges to a statistical mixture of the 3-pentyl and 2-pentyl. These cations react with chloride ion to produce about 1/3 3-chloropentane and 2/3 2-chloropentane.

    A carbocation may be stabilized by resonance by a carbon-carbon double bond next to the ionized carbon. Such cations as allyl cation 

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