Class 11th
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New answer posted
5 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
52. Option (ii) II and III since in position isomerism, two or more compounds differ in the position of substituents, functional group, or multiple bonds but their molecular formula is same. In pentanone-2 and pentanone-3, position of ketonic group is different. CH3-CH2-CH2-CO-CH3 and CH3 - CH2-CO-CH2- CH3
New answer posted
5 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
51. Options (ii) and (iii) are correct since the electrophiles are positively charged or electron deficient species. They act as Lewis acids. AlCl3, SO3are Lewis acids while NO2+, CH3+, CH3-C+ = O are positively charged species. Therefore, the species involved in the above-mentioned options act as electrophiles.
New answer posted
5 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
50. Options (i), (iii) and (iv) are correct since the spatial arrangement of groups or atoms can be checked by doing two interchanges and bringing the H atom below the plane of the paper. Then find out the sequence of the left-out groups in a particular order whether clockwise or anticlockwise starting from the atom with the highest atomic number to the atom with lower atomic numbers. So, out of the given options option (ii) has the same spatial arrangement as (A) while in rest three spatial arrangements are different.
New answer posted
5 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
49. Options (i) and (iv) are correct since only in these two compounds all the carbon atoms are having same hybridization i.e., sp hybridized in option (i) and sp2 hybridized in option (iv).
New answer posted
5 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
48. Option (ii) is correct since the arrow shows the direction of electron transfer. In the given reaction, the electron flows from π bond towards the electron-deficient H+ group . This can be represented as-

New answer posted
5 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
47. Bromine is found to be more electronegative than Carbon which leads to heterolytic fission. Due to the higher electronegativity the electrons displace from carbon to Br. Therefore, CH3 gets positive charge and Br gets negative charge and hence option (ii) is found to be the correct one.
New answer posted
5 months agoContributor-Level 10
46. This is a Multiple Choice Questions as classified in NCERT Exemplar
Carbocation is correct since H+ attacks on propene and the delocalization of electrons can take place in two possible ways which can be shown as-

New answer posted
5 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
45. Option (iv) is found to be the correct one since the stabilisation of carboxylate ion depends upon the dispersal of the negative charge. And the dispersal depends upon two factors, i.e., +R effect of the carboxylate ion and Inductive effect of the halogens. In all the above structures, +R effect is common, but halogen atoms are uncommon. Therefore, dispersal of negative charge depends upon the halogen atoms. Fluorine is most electronegative, in structure (iv) two F atoms are present and hence more dispersal of negative charge is there.
New answer posted
5 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
44. Option (i)
is found to be the correct one since the order of electronegativity of the attached group is found to be:
Cl > Br > C > Mg.
Greater the electronegativity of the group attached to the C greater will be the positive charge. Therefore, in case (i) asterisk C will have the greatest positive charge.
New answer posted
6 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
43. Option (ii) 3,4- dimethyl hexane is correct since in selecting the longest continuous carbon chain both the ethyl groups are included while the third and fourth carbons have methyl groups as a substituent.
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