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All questions of Reaction Intermediates for Chemistry Exam

 Arrange stability of the given carbocations in decreasing order:
  • a)
    I < II < III < IV 
  • b)
     IV < III < II < I 
  • c)
    IV < II < III < I
  • d)
    II < IV < III < I
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered
Correct Answer :- c
Explanation : The carbocation here is stabilized when there is an electron-donating group present.
More the activating effect of the group, more stable is the cation. Out of given options, I is most stable due to ability of oxygen to activate the ring with its lone pair of electrons.
Out of III and I,III is more stable as the lone pair of nitrogen in II is less available due to Ac group. IV is least stable due to electron- withdrawing effect of Cl atom.
The correct answer is : IV < II < III < I
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 Identify correct C—O bond length order:
  • a)
    I > II > III 
  • b)
    II > III > I
  • c)
     I > III > II   
  • d)
     III > I > II
     
Correct answer is option 'B'. Can you explain this answer?

Vedika Singh answered
I is the shortest because of double bond.
III is next because of resonance.
II is last as there is no kind of bond strengthening.
Hence, B is correct.

 Identify correct acidic strength order in the following compounds:

a)II > III > I
b) III > II > I 
c)I > II > III
d)III > I > II
Correct answer is option 'C'. Can you explain this answer?

Chirag Verma answered
Correct Answer :- C
Explanation : The correct option is A. Since −OE+ which increases the acidity of the compound and is distance dependent. In I −OE+ group is nearer to the substituents. So, its more acidic than the other two. SO the correct order will be i>ii>iii.

 The relative stability of the following carbocations is:
  • a)
     I > II > III 
  • b)
     II > III > I 
  • c)
     I > III > II 
  • d)
    III > II > I
Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered
Correct Answer :- d
Explanation : The dispersal of the charge stabilizes the carbocation. More the number of alkyl groups, the greater the dispersal of positive charge and therefore, more the stability of carbocation,is an electron donating group, thus it will increases the stability of carbocation, hence the expected order is, (iii)>(ii)>(i).

 The acidity for the following compounds increases in the order:
CH3​CH2​CH(Br)COOH, CH3CH(Br)CH2​COOH, (CH3​)2CHCOOH
  • a)
     I < II < III 
  • b)
     II < III < I 
  • c)
     III < II < I 
  • d)
     II < I < III
     
Correct answer is option 'C'. Can you explain this answer?

Edurev.iitjam answered
a. EWG ( e−− withdrawing groups) increases the acidic strength, whereas EDG ( e−− donating groups) decrease the acidic strength.
b. Nearer is the EWG to the source [(−COOH)group], stronger is the acid, i.e., α− substituted halo acid stronger than β−orγ− substituted halo acid.
Increasing order of acidic strength:
(III)<(II)<(I).

 The relative acidity of the indicated H in each of the following is:
  
 
  • a)
     I > II > III 
  • b)
    II > III > I 
  • c)
     I > III > II 
  • d)
     III > I > II
     
Correct answer is option 'D'. Can you explain this answer?

Subha Som answered
In 3 the h is attached with acidic carboxyl gr
in 1 the -ve charge which generate after removing h will get stability by carbonyl gr of aldehyde
but in 2 the carbonyl gr is less available in ester gr

 The relative nucleophilicity in polar, protic, solvents of the following is:
 
  • a)
    I > II > III 
  • b)
    II > III > I 
  • c)
     I > III > II 
  • d)
     III > I > II
     
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered
Correct Answer :- a
Explanation : In a polar protic solvent (CH3OH), nucleophilicity increases down a column of the periodic table. So S- is more nucleophilic than O-.
For two species with the same attacking atom, the more basic is the more nucleophilic, so CH3CH2O- is more nucleophilic than CH3CO2-
CO2- give identical str resonance hybrids more solvated, very less nucleophilic.

 Which of the following carbonium ion is most stable?
 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Pooja Choudhury answered
A is the correct option as 9- alpha Hydrogens are present. More alpha hydrogen, more hyperconjugation, more stability.

A compound  shows a large dipole moment. Which of the following resonance structures
can be used to adequately explain this observation:
  • a)
    I only 
  • b)
     III and IV 
  • c)
     II and III 
  • d)
     IV only
Correct answer is option 'B'. Can you explain this answer?

Pooja Choudhury answered
Both (IV) and (III) member ring are aromatic due to charge separation
in case of 5 member ring total 6 electron and 3 member ring 2 electron are present
according to 4n+2 rule they are aromatic
and all the carbon are sp2 hybridize

 Rate of reaction of CH3COCl/AlCl3 with each of the following is:

 
  • a)
     III > II > I 
  • b)
    II > III > I
  • c)
     I > III > II 
  • d)
     III > I > II
     
Correct answer is option 'A'. Can you explain this answer?

Subha Som answered
In iii the ome gr has strong +R effect
in ii the br gr has weak +R effect
in i the co gr has -R effect
hence we see the relative order iii>ii>i

The increasing order of stability of the following free radicals is .
  • a)
    (CH3​)2​CH˙<(CH3​)3​C˙<(C6​H5​)2​CH˙<(C6​H5​)3​C˙
  • b)
    (C6​H5​)3​C˙<(C6​H5​)2​CH˙<(CH3​)3​C˙<(CH3​)2​CH˙
  • c)
    (C6​H5​)2​CH˙<(C6​H5​)3​C˙<(CH3​)3​C˙<(CH3​)2​CH˙
  • d)
    (CH3​)2​CH˙<(CH3​)3​C˙<(C6​H5​)3​C˙<(C6​H5​)2​CH˙
Correct answer is option 'A'. Can you explain this answer?

Akshat Saini answered
The increasing order of stability of the given free radicals is:

a) (CH3)2CH

In this radical, there are two methyl groups attached to the central carbon atom. The presence of these methyl groups increases the electron density around the central carbon atom, making it more stable. This is due to the inductive effect, where the methyl groups donate electron density through the sigma bonds to the central carbon atom. As a result, the central carbon atom is better able to stabilize the unpaired electron in the radical.

Therefore, (CH3)2CH is more stable than other radicals without any electron-donating groups.

The acidity of the protons H in each of the following is:
  
  • a)
     III > II > I 
  • b)
     II > III > I 
  • c)
     I > III > II 
  • d)
    III > I > II
     
Correct answer is option 'A'. Can you explain this answer?

Akash Unplugged answered
Ans is (a)
As we dont compare the acidity of acids with carbonyls so it is quite obvious that acetic acid will have more acidity than carbonyls.
and in ll option it is active methylene compound, which is more acidic than acetone.because in ll option there is two conjugation.
thanks,

The relative nucleophilicity in polar, protic, solvents of the following is:
  (I) CH3OH                  (II) CH3SH                (III) CH3NH2
 
  • a)
    I > II > III 
  • b)
     II > III > I 
  • c)
    I > III > II 
  • d)
     III > I > II
Correct answer is option 'B'. Can you explain this answer?

Aashna Shah answered
< iii="" />< />
b)III < ii="" />< />
c)II < i="" />< />
d)I < ii="" />< />

Correct answer: d) I < ii="" />< />

Explanation:
In polar, protic solvents, nucleophilicity is generally directly proportional to basicity. The stronger the basicity of a species, the stronger its ability to donate a pair of electrons and act as a nucleophile.

Among the given species, CH3NH2 has the highest basicity due to the presence of a lone pair of electrons on the nitrogen atom. It can donate this lone pair easily and act as a strong nucleophile. Hence, it is the most nucleophilic species among the given options.

Next comes CH3SH, which is less basic than CH3NH2 but still more basic than CH3OH. Sulfur is less electronegative than oxygen, which makes the lone pair of electrons on the sulfur more available for donation. Hence, CH3SH is a stronger nucleophile than CH3OH.

Finally, CH3OH has the lowest basicity among the given options. Its lone pair of electrons is involved in hydrogen bonding with the solvent molecules, which reduces its availability for donation. Hence, it is the least nucleophilic species among the given options.

Which of the following is most reactive as a nucleophile?
  • a)
    PhO− 
  • b)
    PhCH2O−
  • c)
    None of these
  • d)
    PhS−
Correct answer is option 'D'. Can you explain this answer?

Pooja Choudhury answered
The most nucleophile is dependent on the compound which has a tendency to give an electron. In the compound there is oxygen and one has sulphur. The sulphur has big in size that why it has more tendency to donate an electron. The correct answer is PhS-

 Cyclopentadiene has a pKa = 15, whereas cyclopentane has a pKa > 50. This is because:
 
  • a)
     Cyclopentadiene is particularly unstable.
     
  • b)
    Cyclopentadiene contains no lone pairs.
  • c)
    Cyclopentadiene is a 4π anti-aromatic compound 
  • d)
    Cyclopentadiene is a 4π anti-aromatic compound and after deprotonation it is aromatic.
Correct answer is option 'D'. Can you explain this answer?

Raksha Pillai answered
Since after deprotonation cyclopentadiene becomes stable by aromaticity, So, it will definitely have a higher acidic strength than that of cyclopentane and consequently lower pKa value.

Further, no such stability factor has a role in the case of cyclopentane so it has lower acidic strength comparatively.

The reaction of (+) 2-iodobutane and Nal* in acetone was studied by measuring the rate of incorporation of I* (ki) and the rate of racemisation(kr)
(+) CH3CH(I)CH2CH3 + Nal* → CH3CH(I*)CH2CH3 + Nal
For this reaction, the relationship between kr and ki is
  • a)
    ki = 2kr
  • b)
    ki = (1/2)kr
  • c)
    ki = kr
  • d)
    ki = (1/3)kr
Correct answer is option 'B'. Can you explain this answer?

Pooja Choudhury answered
The given reaction is : (+)CH3CH(I)CH2CH3 + Nal* → CH3CH(I*)CH2CH3 + Nal

This is an example of an SN2 type reaction.
  • SN2 reaction involves the product formation with inversion of configuration. When one molecule of the original substances undergoes inversion, one molecule of product is formed with inversion of configuration.
  • This trier that when one original optically active molecule undergoes inversion, the inverted product and another origins, optically active molecules lose their optical activity due to the occurrence of racemisation i.e. when one molecule is inverted, actually two molecules are racemised.
  • Hence, it can be concluded that the rate of racemisation is double to the rate of inversion. Hence, mathematically, 2ki = kr

Which of the following statements about resonance structures is false?
  • a)
    Individual resonance structure are imaginary, not real.
  • b)
    Resonance forms differ only in the placement of their -or non-bonding electron or unpairedelectron.
  • c)
    In valid resonance structures, all atoms from the second row of the periodic table must have an octet of electrons. 
  • d)
    Different resonance structures of a substance do not have to be equivalent.
Correct answer is option 'C'. Can you explain this answer?

Vandana Gupta answered

Explanation:

Resonance Structures:

- Resonance structures are a way of representing the delocalization of electrons in a molecule or ion.
- They are used when a single Lewis structure cannot accurately represent the bonding in a molecule.
- Resonance structures are not individual, real structures but rather theoretical representations of the possible electron distributions in a molecule.

Differences in Resonance Structures:

- Resonance forms differ only in the placement of their π or non-bonding electrons.
- The actual structure of the molecule is a combination or hybrid of all the resonance structures.

Equivalence of Resonance Structures:

- Different resonance structures of a substance do not have to be equivalent.
- Some resonance structures may contribute more to the overall structure of the molecule than others.
- The more stable resonance structures are typically those where formal charges are minimized and octet rules are obeyed.

Validity of Resonance Structures:

- In valid resonance structures, all atoms from the second row of the periodic table must have an octet of electrons.
- This means that elements like carbon, nitrogen, oxygen, and fluorine follow the octet rule in resonance structures.

Therefore, the false statement among the options provided is that different resonance structures of a substance do not have to be equivalent. In reality, some resonance structures may contribute more significantly to the overall structure of the molecule based on stability considerations.

 Identify correct acidic strength order in the following compounds:
  • a)
    I > II > III 
  • b)
    II > III > I 
  • c)
    I > III > II 
  • d)
     III > I > II
     
Correct answer is option 'B'. Can you explain this answer?

Uma Bharti answered
AFTER REMOVING H+NEGATIVE CHARGE IS DELOCALISED BOTH SIDE IN 2ND AND 3RD SO 1ST IS LEAST ACIDIC..in 3rd..NH GROUP PRESNT ITS GIVES ELCTRON SL IT DESTABLUSHES NEGATIVE CHARGE

 List the following carbocations in order of decreasing stability (starting with the most stable):
  • a)
     II, III, I, IV 
  • b)
     III, IV, II, I 
  • c)
     III, IV, I, II 
  • d)
     I, II, IV, III
Correct answer is option 'B'. Can you explain this answer?

Isha Bose answered
Correct Answer :- b
Explanation : I have an sp2 hybridized carbocation which is highly unstable. Hence, I will be the least stable of all. III will be the most stable as it is tertiary carbocation.
Among II and IV, IV has 4 alpha− hydrogens whereas II has only, α− hydrogens. Also IV is a secondary carbocation while II is a primary one. Hence IV will be more stable than II.
Hence, the decreasing order of stability or stabilization energy is III>IV>II>I

Two of the great Mughals wrote their own memories. There were
  • a)
    Babar and Humayun
  • b)
    Humayun and Jahangir
  • c)
    Babar and Jahangir
  • d)
    Jahangir and Shahjahan
Correct answer is option 'C'. Can you explain this answer?

Arun Khatri answered

The two great Mughals who wrote their own memories are Babar and Jahangir. Below is a detailed explanation of the answer:
1. Introduction:
- The question asks about two Mughals who wrote their own memories.
- We need to identify the correct pair of Mughals.
2. Possible Options:
The given options are:
A) Babar and Humayun
B) Humayun and Jahangir
C) Babar and Jahangir
D) Jahangir and Shahjahan
3. Analyzing the Options:
- We need to determine which pair of Mughals wrote their own memories.
- We will analyze each option to find the correct answer.
4. Option A) Babar and Humayun:
- It is known that Babar wrote his own memoir called "Baburnama."
- However, there is no record of Humayun writing his own memoirs.
- Therefore, this option is incorrect.
5. Option B) Humayun and Jahangir:
- While Humayun did not write his own memoirs, Jahangir did.
- The memoirs written by Jahangir are known as "Tuzk-e-Jahangiri."
- Therefore, this option is incorrect.
6. Option C) Babar and Jahangir:
- As mentioned earlier, Babar wrote his own memoir called "Baburnama."
- Jahangir also wrote his own memoirs known as "Tuzk-e-Jahangiri."
- Both Babar and Jahangir wrote their own memories, making this option correct.
7. Option D) Jahangir and Shahjahan:
- While Jahangir did write his own memoirs, there is no record of Shahjahan writing his own memoirs.
- Therefore, this option is incorrect.
8. Conclusion:
- After analyzing all the options, we can conclude that the correct answer is option C) Babar and Jahangir.
- Both Babar and Jahangir wrote their own memoirs, namely "Baburnama" and "Tuzk-e-Jahangiri," respectively.

 Rank, from the most stabilized to the least stabilized, the following free radicals according to their stabilization energies:
(I) CH3CH2 (II) CH2CH3 (III) (CH3)2CH          (IV) (CH2=CH—CH2) 
  • a)
     IV > III > II > I 
  • b)
     I > IV > III > II 
  • c)
     III > IV > I > II 
  • d)
     III > IV > II > I 
Correct answer is option 'A'. Can you explain this answer?

Jaya Sen answered
To rank the free radicals according to their stabilization energies, we need to consider the factors that contribute to stabilization.

The factors that contribute to stabilization of free radicals include:
1. Hyperconjugation: The presence of adjacent sigma bonds can provide stabilization through the overlap of electron density from the sigma bond into the empty p orbital of the free radical.
2. Inductive effect: The presence of electron-donating alkyl groups can stabilize the free radical.
3. Resonance: If the free radical can form resonance structures, it will be more stabilized.

Now, let's analyze each free radical:

(I) CH3CH2: This free radical has one adjacent sigma bond and one alkyl group, providing some stabilization through hyperconjugation and the inductive effect.

(II) CH2CH3: This free radical also has one adjacent sigma bond and one alkyl group, providing the same level of stabilization as (I).

(III) (CH3)2CH: This free radical has two adjacent sigma bonds and one alkyl group, providing more stabilization through hyperconjugation and the inductive effect compared to (I) and (II).

(IV) (CH2=CH: This free radical has one adjacent sigma bond and one pi bond. The presence of the pi bond allows for resonance stabilization, which is a stronger stabilizing factor compared to hyperconjugation and the inductive effect.

Therefore, the ranking of the free radicals from most stabilized to least stabilized is as follows:
1. (CH2=CH
2. (CH3)2CH
3. CH3CH2
4. CH2CH3

Which one of the following carbocation would you expect a H migration?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Amar Chawla answered
In option B and C, after carbocation rearrangement, it will remain 2°, therefore, rearrangement will not take place.
In option D, after carbocation rearrangement 3° will become 2° or 1°, therefore, rearrangement will not take place.
In option A, after rearrangement, carbocation will become 3° from 2°, therefore, this rearrangment will take place.

 Alkyne hydrogens are more acidic than alkene or alkane hydrogens because:
  • a)
    The alkyne carbon has higher’s character.
  • b)
    The anion formed is more stable
  • c)
    The electrons in the sp orbital are closer to the nucleus.
  • d)
    All of the above. 
Correct answer is option 'D'. Can you explain this answer?

Aditi Basak answered
Electronegativity than the alkene or alkane carbon, resulting in a more stable conjugate base.
b)The alkyne carbon has a greater electron density, making it easier for the hydrogen to be abstracted.
c)The alkyne carbon has a higher degree of s-character in its hybrid orbital, making the C-H bond more acidic.
d)All of the above

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