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C5H11Br is least soluble in:
  • a)
    Ether
  • b)
    Alcohol
  • c)
    Water
  • d)
    None of these
Correct answer is 'C'. Can you explain this answer?

Rajeev Saxena answered
C5H11Br is only very slightly soluble in water. It is more soluble in organic solvents.

 Which of the following is not true about optical isomers?
  • a)
    They rotate the plane polarized light.
  • b)
    They are superimposable on their mirror image.
  • c)
    They posses at least one chiral carbon atom.
  • d)
    They are nonsuperimposable on their mirror image.
Correct answer is option 'B'. Can you explain this answer?

Nisha Pillai answered
Explanation:
Optical isomers are stereoisomers that exist in two mirror-image forms that are non-superimposable on each other. They are also known as enantiomers.

a) They rotate the plane polarized light: Optical isomers have the property of rotating the plane of polarized light in opposite directions. One isomer rotates the plane of polarized light in the clockwise direction and the other in the counterclockwise direction.

b) They are superimposable on their mirror image: This statement is false. Optical isomers are non-superimposable mirror images of each other. If we try to superimpose them, they will not match perfectly.

c) They possess at least one chiral carbon atom: Optical isomers possess chiral carbon atoms. Chiral carbon atoms are those carbon atoms that are attached to four different groups or atoms.

d) They are nonsuperimposable on their mirror image: As mentioned earlier, optical isomers are non-superimposable mirror images of each other. They have the same physical and chemical properties but differ in their biological activity, as they interact differently with other chiral molecules in living organisms.

Conclusion: Optical isomers are important in fields like medicinal chemistry, where knowing the activity of each isomer can help in designing drugs that are more effective and have fewer side effects. Therefore, it is crucial to understand the properties of optical isomers.

Which of the following is a tertiary halogenoalkanes?
  • a)
    2-Bromopentane
  • b)
    2-Bromo 3-methylpentane
  • c)
    Bromopentane
  • d)
    2-Bromo 2-methylpentane
Correct answer is option 'D'. Can you explain this answer?

Anaya Patel answered
√Br CH3 -( CH )- CH2 - CH2 - CH3 √CH3 The carbon with which the Br is bonded is bonded with another 3 carbon atoms. So haloalkane is 3.

Which one of the following is likely to give a precipitate with AgNO3 solution?
  • a)
    CHCl3
  • b)
    (CH3)3CCl
  • c)
    CH2=CH-Cl
  • d)
    CCl4
Correct answer is option 'B'. Can you explain this answer?

Gargi Ahuja answered
Precipitation Reaction

A precipitation reaction is a chemical reaction that produces a precipitate, which is an insoluble solid that forms from the reaction of two soluble compounds.

AgNO3 Solution

AgNO3 is a soluble salt that dissociates to give Ag+ and NO3- ions in an aqueous solution.

Likely Precipitate

A compound is likely to give a precipitate with AgNO3 solution if it contains Cl- ions, as AgCl is insoluble in water and will form a white precipitate.

Analysis of Options

a) CHCl3
CHCl3 does not contain Cl- ions and will not give a precipitate with AgNO3 solution.

b) (CH3)3CCl
(CH3)3CCl contains Cl- ions and will give a precipitate with AgNO3 solution.

c) CH2=CH-Cl
CH2=CH-Cl does not contain Cl- ions and will not give a precipitate with AgNO3 solution.

d) CCl4
CCl4 does not contain Cl- ions and will not give a precipitate with AgNO3 solution.

Conclusion

Out of the given options, only (CH3)3CCl contains Cl- ions and will give a precipitate with AgNO3 solution, making option 'B' the correct answer.

 Which of the following is not an aryl halide?
  • a)
    m-ClCH2C6H4CH2C(CH3)3
  • b)
    (CH3)3CCH2CH3
  • c)
    o-Br-C6H4CH(CH3)CH2CH3
  • d)
    p-ClC6H4CH2CH(CH3)2
Correct answer is 'B'. Can you explain this answer?

Ankita Datta answered
 Because it is Open chain compound /Aliphatic Compound and not aryl halide as does not contain halogen atom. it's alkane simply.

Only One Option Correct Type
Direction (Q. Nos. 1-8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q. 
Which of the SN2 reaction is fastest?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Monika Devi answered
SN2 reaction means this will be applicable for primary or 1 degree carbon atom and attack by the reagent is forwarded so on doing mechanism we getted that CH3 carry + charge & Br get - charge so acetyl group that have negative charge goes on methyl group & Br react with electronic species

Elimination of bromine from 2-bromobutane results in the formation of
  • a)
    Predominantly 2-butene
  • b)
    Equimolar mixture of 1 and 2-butene
  • c)
     Predominantly 1-butene
  • d)
    Predominantly 2-butyne.
Correct answer is option 'A'. Can you explain this answer?

Palak Basak answered
Elimination of Bromine from 2-Bromobutane

The elimination reaction of bromine from 2-bromobutane is a type of E2 (bimolecular elimination) reaction. During this reaction, the bromine atom is removed from the carbon chain, resulting in the formation of a double bond.

E2 Reaction Mechanism
The E2 reaction occurs in a single step, involving the simultaneous removal of a proton and a leaving group (in this case, bromine) from adjacent carbon atoms. The leaving group takes a pair of electrons from the carbon-hydrogen bond, forming a double bond between the two carbon atoms.

The reaction mechanism can be summarized as follows:

1. The base (usually a strong base such as hydroxide ion, OH-) approaches the hydrogen atom on the carbon adjacent to the bromine atom.
2. The base abstracts the proton, creating a carbanion intermediate and leaving a bromide ion.
3. The carbanion intermediate then rearranges to form the most stable alkene through a transition state.
4. Finally, the bromide ion combines with a proton from the solvent to form HBr.

Stability of Alkenes
The stability of alkenes is determined by the degree of substitution at the double bond. The more substituted the double bond, the more stable the alkene. This stability is due to the increased electron density around the double bond, which is provided by the alkyl groups attached to the carbon atoms.

Explanation of the Correct Answer
In the case of 2-bromobutane, elimination of bromine results in the formation of 2-butene. This is because the double bond in 2-butene is more substituted than the double bond in 1-butene. Therefore, the stability of the alkene product, 2-butene, is higher compared to that of 1-butene.

The stability of alkenes follows the order: tertiary > secondary > primary. In 2-butene, the double bond is formed between two secondary carbons, making it more stable than 1-butene, which has a double bond between a primary and a secondary carbon.

Hence, the correct answer is option 'A' - Predominantly 2-butene.

It is important to note that while the formation of 2-butene is favored, a small amount of 1-butene may also be formed as a minor product due to the presence of a primary carbon adjacent to the double bond. However, the major product will be 2-butene.

One or More than One Options Correct Type
Direction (Q. Nos. 9-14) This section contains 6 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.
Q. 
 In a SN2 reaction rate of reaction depends on
  • a)
    concentration of substrate
  • b)
    concentration of nucleophile
  • c)
    nature of leaving group
  • d)
    number of lone pairs on donor atom
Correct answer is option 'A,B,C'. Can you explain this answer?

Arnab Chavan answered
In a SN2 reaction, Rate = k[substrate] [nucleophile]
Hence, rate depends both on concentration of substrate as well as nucleophile. Rate of reaction also depends on nature of leaving group. A better leaving group gives lower activation energy hence, greater rate of reaction. Number of lone pair has no role to play in nucieophilicity hence reactivity.

Which statement is true about SN2 mechanism?
  • a)
    The rate of reaction increases on increasing strength of the nucleophile
  • b)
    The reaction is faster in polar aprotic solvents
  • c)
    The rate of reaction increases as the leaving group ability increases
  • d)
    Ali'of the above
Correct answer is option 'D'. Can you explain this answer?

Arnab Chavan answered
Stronger the nucleophile, faster the SN2 reaction. Polar aprotic solvent solvate cations, makes anionic nucleophile more available for reaction, hence faster reaction. A better leaving group lowers the activation energy increasing rate of SN2 reaction.

Consider the reaction given below.
Q. 
The correct statement(s) applicable to the above reaction is/are
  • a)
    Cl- is substituted predominantly
  • b)
     Cl- is a better leaving group
  • c)
    substitution of I- in the above reaction required greater activation energy than for Cl-
  • d)
    addition of some Nal catalyse the substitution reaction
Correct answer is option 'A,C,D'. Can you explain this answer?

Steric hindrance plays the most important role in SN2 reaction. Hence, although Cl is poorer leaving group than I, Cl is substituted predominantly in the above reaction due to less steric hindrance at α-carbon. Addition of Nal replace Cl by I and substitution becomes easier.

What is the correct increasing order of reactivity of the  followings in SN2 reaction ?
I. CH2 = CH — Br
II. CH2 = CH— I
III. CH3CH2CH2 — I
IV. CH3OCH2CH2 — I 
  • a)
    I < II < III < IV
  • b)
    III < II < I < IV
  • c)
    II < III < IV < I
  • d)
    II < I < III < IV
Correct answer is option 'C'. Can you explain this answer?

Tejas Singh answered
Allyl bromide (I) is most reactive among the given halides as pi bonds from allylic position stabilises the transition state. Vinyl iodide (II) is least reactive due to partial double bond character. Electron withdrawing inductive effect of CH3O- increases reactivity of (IV) over (III)

The correct statement regarding a SN2 reaction is/are
  • a)
    reaction involving stronger bond formation to α-carbon occur at faster rate
  • b)
    reaction shows kinetic isotopic effect
  • c)
    substitution of   lower the activation energy of reaction
  • d)
    presence of electron withdrawing group in substrate increases reactivity
Correct answer is option 'B,C,D'. Can you explain this answer?

Anuj Iyer answered
Stronger bond formation at α-carbon affects the stability of product (thermodynamics) not the rate (kinetics) of reaction. Change of isotope at α-position or in halogens affect the rate, hence show kinetic isotopic effect. Presence of electron withdrawing group in substrate increases electrophilicity of α-carbon hence, increases its reactivity towards nucleophiles.

Consider the following SN2 reaction,
Q. 
Which of the following could increase the reactivity (rate) of reaction ?
  • a)
    Increasing NaOH concentration
  • b)
    Adding some Nal
  • c)
    Replacing CHsBr by 14CH3— Br
  • d)
     Using NaO18 H in place of NaOH
Correct answer is option 'A,B,D'. Can you explain this answer?

Anuj Iyer answered
Increasing concentration of nucleophile (NaOH) increases the reactivity. Adding some Nal also catalyse reaction because of following equilibrium.

Now reaction occur with CH3I which has better leaving group. Heavier isotope forms stronger covalent bond. Hence, replacing CH3Br by 14CH3Br stabilises substrate, decreases rate whereas replacing NaOH by NaO18H stabilises transition state increases rate of reaction.

What is the correct increasing order of reactivity of the following in the SN2 reaction?
  • a)
    I < II < III < IV
  • b)
    III < II < I < IV
  • c)
    II < III < IV < I
  • d)
    Ill < I < II < I V
Correct answer is option 'D'. Can you explain this answer?

 (IV) is most reactive as it is benzylic as well as electron withdrawing effect of — NO2 further increases the reactivity. (Ill) is least reactive due to resonance effect resulting in partial double bond character between carbon and chlorine. (I) is less reactive in SN2 reaction than II, due to greater steric hindrance in (I).

Consider the two lines shown in the diagram given below.
Q. 
In a SN2 reaction, these two lines compare the effect of the
  • a)
    substrate
  • b)
    nucleophile
  • c)
    leaving group
  • d)
    substrate or nucleophile
Correct answer is option 'D'. Can you explain this answer?

Substrates are same as they are at same potential energies. Nucleophiles are different as magnifested by different activation energies and different products with different potential energies,

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