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Organic Chemistry : Some Basic Principles & Techniques Test - 7

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Organic Chemistry : Some Basic Principles & Techniques Test - 7
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  • Question 1
    4 / -1

    Heterolysis of a carbon-chlorine bond produces

    Solution

    In C - Cl bond, the electron pair is carried away by Cl atom due to higher electronegativity.
    C - Cl → C+ + Cl-
    Hence, carbon cation and chlorine anion are formed.

  • Question 2
    4 / -1

    Which of the following is an electrophilic reagent?

    Solution

    Electron deficient species act as electrophiles i.e., NO+2.

  • Question 3
    4 / -1

    Which of the following sets of groups contains only electrophiles?

    Solution

    NO2+,AlCl3,SO3 and CH3C+ = O are electrophiles.

  • Question 4
    4 / -1

    Inductive effect involves

    Solution

    Due to displacement of a-electrons towards more electronegative atom the bond becomes polar. The polar bond induces polarityto the adjacent bonds, this is inductive effect.

  • Question 5
    4 / -1

    The increasing order of electron donating inductive effect of alkyl groups is

    Solution

    The order of +I effect of alkyl group is given as

  • Question 6
    4 / -1

    Inductive effect of which atom or group is taken as zero to compare inductive effect of other atoms?

    Solution

    Hydrogen does not exert I-effect. Its inductive effect is taken as zero. Electron releasing or electron withdrawing capability of other atoms are compared by hydrogen.

  • Question 7
    4 / -1

    Maximum -I effect is exerted by the group

    Solution

    -NO2 shows maximum electron with drawing or - I effect.

  • Question 8
    4 / -1

    Which one of the following acids would you expect to be the strongest?

    Solution

    Fluorine is most electronegative atom and exerts maximum -I effect Hence F - CH2COOH is the strongest acid.

  • Question 9
    4 / -1

     Few pairs of molecules are given below. Which bond of the molecule of the pairs is more polar?
    (i) H3C - H, H3C - Br
    (ii) H3C-NH2,H3C-0H
    (iii) H3C - OH, H3C - SH
    (iv) H3C - Cl, H3C - Br

    Solution

    (i) Br is more electronegative than H.
    (ii) O is more electronegative than N.
    (iii) O is more electronegative than S.
    (iv) C1 is more electronegative than Br.

  • Question 10
    4 / -1

    Which of the following is the correctorder of acidity of carboxylic acids?
    (i) CI3CCOOH > CI2CHCOOH > CICH2COOH
    (ii) CH3CH2COOH > (CH3)2CHCOOH > CCH3)3CCOOH
    (iii) F2CHCOOH > FCH2COOH > CICH2COOH

    Solution

    The strength of the acid increases with the inductive effect of the electronegative group attached to it (-I effect). The strength of the acid decreases with the presence of electron releasing group.

  • Question 11
    4 / -1

    Point out the incorrect statement about resonance?

    Solution

    The resonating structures should have same number of electron pairs.

  • Question 12
    4 / -1

    Which of the following ions is the most resonance stabilised?

    Solution

    Phenoxide ion shows maximum resonating structures.

  • Question 13
    4 / -1

    Hyperconjugation is

    Solution

    Hyperconjugation involves conjugation of a electrons of C - H bond and π electrons of multiple bond.
    It is noticed due to delocalisation of σ- and π-bonds. It is . also known as no bond resonance.

  • Question 14
    4 / -1

    The number of hyperconjugating structures shown by the carbocations are given below. Which one is not correctly matched?

    Solution

    shows 6 hyperconjuigatihg structures

  • Question 15
    4 / -1

    In which of the following species hyperconjugation is possible?

    Solution

    For hyperconjugation α-carbon with respect to sp2 hybridised carbon should have at least one hydrogen.

  • Question 16
    4 / -1

    Answer the following question based on the given paragraph.
    Stability of carbocation, alkylfree radical and alkene can be explained on the basis of hyperconjugation.
    In all these cases, there is presence ofhydrogen atom at theadjacent carbon atom ofsp2 hybridised carbon atom. Total number of hyperconjugating structures depends upon the number of hydrogen atoms present at adjacent carbon atom of sp2 C-atom.
    More the hyperconjugating structures, more is the stabihty of the ion.

    Decreasingorder of stabilityof following alkenes is
    (i) CH3 - CH = CH2
    (ii) CH3 - CH = CH - CH3
    (iii) 
    (iv) 

    Solution

  • Question 17
    4 / -1

    Which of the following alcohols on dehydration gives most stable carbocation? 

    Solution


    Tertiary carbocation is more stable than secondary and primary carbocatioris.

  • Question 18
    4 / -1

    Which of the following contains three pairs of electrons in valence shell?

    Solution

    In carbocations, the carbon atom with positive charge has only 6 electrons in its valence shell.

  • Question 19
    4 / -1

    Stability of alkyl carbocations can be explained by

    Solution

    Stability of alkyl carbocations can be explained by both inductive effect and hyperconjugation.

  • Question 20
    4 / -1

    In the given reaction two products are expected.

    The product (B) is formed as a major product because

    Solution



  • Question 21
    4 / -1

    The carbocation is less stable than

    Solution

    (CH3)3C+ is more stable than . Rest all are primary carbocations hence less stable.

  • Question 22
    4 / -1

    Complete the following reactions by filling most stable intermediate and the product.
    (i) 
    (ii) 

    Solution

  • Question 23
    4 / -1

    Which of the following statements is not true about the stability of carbanions?

    Solution

    Order of stability of carbanions is 1º > 2º > 3º;

  • Question 24
    4 / -1

    Which of the following carbanion expected to be most stable?

    Solution

    Carbanions are stabilised by electron withdrawing groups. - NO2 is stronger electron withdrawing group as compared to - CHO. At orthoposition, the effect is more pronounced.

  • Question 25
    4 / -1

    The order of decreasing stability of the following carbanions is
    (i) (CH3)3C-
    (ii) (CH3)2CH-
    (iii) CH3CH-2
    (iv) C6H5CH-2

    Solution

    Stability of carbanions decreases with increase in +l effect. + I effect is maximum in (CH3)3C-, followed by (CH3)2CH- and CH3CH-2. In C6H5CH-2, resonance effect increases the stability. Hence, the order of stability is (iv) > (iii) > (ii) > (i).

  • Question 26
    4 / -1

    Free radicals can undergo

    Solution

    Free radicals can undergo all given types of processes.

  • Question 27
    4 / -1

    The most stable free radical among the following Is

  • Question 28
    4 / -1

    Which of the following Is a characteristic feature of a free radical?

    Solution

    Free radical is formed as a result of homolytic fission hence it has an unpaired electron.

  • Question 29
    4 / -1

    The increasing order of stability of the following free radicals is

    Solution

    Radical is stabilized by α hydrogen. More the α-hydrogen more will be hyperconjugation more will be radical stabilization resonance. Resonance is also a good stabilizing factor and is given preference than hyperconjugation.

  • Question 30
    4 / -1

    Which of the following is a false statement?

    Solution

    The electronegative atom in the carbon chain produces -I effect.

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