231738
Which of the following does not show a resonance effect ?
1 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
2 It also shows resonance effect due to presence of resonating structure.
3 It also shows resonance effect.
4 It does not show resonance effect because it does not have planer structure because $\mathrm{NH}_{3}$ molecules have $\mathrm{sp}^{3}$ hybridisation with pyramidal shape.
Explanation:
To show resonance effect molecule must have conjugated system. The molecule which has alternate $\pi$-bond, positive I charge, negative charge and lone pair of electron show conjugation. It shows resonance, all of these are resonating structure. (b.) It also shows resonance effect due to presence of resonating structure. (c.) It also shows resonance effect. (d.) It does not show resonance effect because it does not have planer structure because $\mathrm{NH}_{3}$ molecules have $\mathrm{sp}^{3}$ hybridisation with pyramidal shape.
UPTU/ UPSEE-2018
GENERAL ORGANIC CHEMISTRY
231739
Among the given compound choose the two the yield same carbocation on ionisation
1 (ii),(iv)
2 (i),(ii)
3 (ii),(iii)
4 (i),(iii)
Explanation:
Compound I \& II will show same carbocation
UPTU/ UPSEE-2017
GENERAL ORGANIC CHEMISTRY
231741
Consider the following compounds Friedel-Crafts acylation can be used to obtain:
1 II, III, IV
2 I, II, IV
3 I, II, III
4 I, III, IV
Explanation:
Friedel Craft Acylation when benzene ring is treated with acid derivative in the presence of anhydrous $\mathrm{FeCl}_{3}$ then acylation is formed such reaction is called friedel craft acylation. (i) It has $+\mu$ group so it is ortho-para directing. (ii) $\mathrm{CH}_{3} \rightarrow$ hyper conjugation only at ortho and Para (iii) $\mathrm{NO}_{2} \rightarrow$ It is a deactivating group (iv) $\mathrm{N}-\mathrm{Me}_{2} \rightarrow$ It is Less basic So, only I \& II and IV is possible to give Ortho and para directing for Friedel Craft Acylation.
UPTU/ UPSEE-2017
GENERAL ORGANIC CHEMISTRY
231742
Transition state 2 (TS2) is structurally most likely as a/an
1 intermediate 2
2 product
3 intermediate 1
4 transition state 3 (TS3)
Explanation:
(A) : The transition state resembles that spaces which is energetically near to Hammond postulate. Hence, $\mathrm{TS}_{2}$ resemble intermediate 2 . Hommond postulate - According to this postulate the transition state will be more similar in structure to species that is closer to energetically. In exergonic reaction, transition state in nearly energetically closure to reactant than product.
231738
Which of the following does not show a resonance effect ?
1 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
2 It also shows resonance effect due to presence of resonating structure.
3 It also shows resonance effect.
4 It does not show resonance effect because it does not have planer structure because $\mathrm{NH}_{3}$ molecules have $\mathrm{sp}^{3}$ hybridisation with pyramidal shape.
Explanation:
To show resonance effect molecule must have conjugated system. The molecule which has alternate $\pi$-bond, positive I charge, negative charge and lone pair of electron show conjugation. It shows resonance, all of these are resonating structure. (b.) It also shows resonance effect due to presence of resonating structure. (c.) It also shows resonance effect. (d.) It does not show resonance effect because it does not have planer structure because $\mathrm{NH}_{3}$ molecules have $\mathrm{sp}^{3}$ hybridisation with pyramidal shape.
UPTU/ UPSEE-2018
GENERAL ORGANIC CHEMISTRY
231739
Among the given compound choose the two the yield same carbocation on ionisation
1 (ii),(iv)
2 (i),(ii)
3 (ii),(iii)
4 (i),(iii)
Explanation:
Compound I \& II will show same carbocation
UPTU/ UPSEE-2017
GENERAL ORGANIC CHEMISTRY
231741
Consider the following compounds Friedel-Crafts acylation can be used to obtain:
1 II, III, IV
2 I, II, IV
3 I, II, III
4 I, III, IV
Explanation:
Friedel Craft Acylation when benzene ring is treated with acid derivative in the presence of anhydrous $\mathrm{FeCl}_{3}$ then acylation is formed such reaction is called friedel craft acylation. (i) It has $+\mu$ group so it is ortho-para directing. (ii) $\mathrm{CH}_{3} \rightarrow$ hyper conjugation only at ortho and Para (iii) $\mathrm{NO}_{2} \rightarrow$ It is a deactivating group (iv) $\mathrm{N}-\mathrm{Me}_{2} \rightarrow$ It is Less basic So, only I \& II and IV is possible to give Ortho and para directing for Friedel Craft Acylation.
UPTU/ UPSEE-2017
GENERAL ORGANIC CHEMISTRY
231742
Transition state 2 (TS2) is structurally most likely as a/an
1 intermediate 2
2 product
3 intermediate 1
4 transition state 3 (TS3)
Explanation:
(A) : The transition state resembles that spaces which is energetically near to Hammond postulate. Hence, $\mathrm{TS}_{2}$ resemble intermediate 2 . Hommond postulate - According to this postulate the transition state will be more similar in structure to species that is closer to energetically. In exergonic reaction, transition state in nearly energetically closure to reactant than product.
231738
Which of the following does not show a resonance effect ?
1 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
2 It also shows resonance effect due to presence of resonating structure.
3 It also shows resonance effect.
4 It does not show resonance effect because it does not have planer structure because $\mathrm{NH}_{3}$ molecules have $\mathrm{sp}^{3}$ hybridisation with pyramidal shape.
Explanation:
To show resonance effect molecule must have conjugated system. The molecule which has alternate $\pi$-bond, positive I charge, negative charge and lone pair of electron show conjugation. It shows resonance, all of these are resonating structure. (b.) It also shows resonance effect due to presence of resonating structure. (c.) It also shows resonance effect. (d.) It does not show resonance effect because it does not have planer structure because $\mathrm{NH}_{3}$ molecules have $\mathrm{sp}^{3}$ hybridisation with pyramidal shape.
UPTU/ UPSEE-2018
GENERAL ORGANIC CHEMISTRY
231739
Among the given compound choose the two the yield same carbocation on ionisation
1 (ii),(iv)
2 (i),(ii)
3 (ii),(iii)
4 (i),(iii)
Explanation:
Compound I \& II will show same carbocation
UPTU/ UPSEE-2017
GENERAL ORGANIC CHEMISTRY
231741
Consider the following compounds Friedel-Crafts acylation can be used to obtain:
1 II, III, IV
2 I, II, IV
3 I, II, III
4 I, III, IV
Explanation:
Friedel Craft Acylation when benzene ring is treated with acid derivative in the presence of anhydrous $\mathrm{FeCl}_{3}$ then acylation is formed such reaction is called friedel craft acylation. (i) It has $+\mu$ group so it is ortho-para directing. (ii) $\mathrm{CH}_{3} \rightarrow$ hyper conjugation only at ortho and Para (iii) $\mathrm{NO}_{2} \rightarrow$ It is a deactivating group (iv) $\mathrm{N}-\mathrm{Me}_{2} \rightarrow$ It is Less basic So, only I \& II and IV is possible to give Ortho and para directing for Friedel Craft Acylation.
UPTU/ UPSEE-2017
GENERAL ORGANIC CHEMISTRY
231742
Transition state 2 (TS2) is structurally most likely as a/an
1 intermediate 2
2 product
3 intermediate 1
4 transition state 3 (TS3)
Explanation:
(A) : The transition state resembles that spaces which is energetically near to Hammond postulate. Hence, $\mathrm{TS}_{2}$ resemble intermediate 2 . Hommond postulate - According to this postulate the transition state will be more similar in structure to species that is closer to energetically. In exergonic reaction, transition state in nearly energetically closure to reactant than product.
231738
Which of the following does not show a resonance effect ?
1 $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
2 It also shows resonance effect due to presence of resonating structure.
3 It also shows resonance effect.
4 It does not show resonance effect because it does not have planer structure because $\mathrm{NH}_{3}$ molecules have $\mathrm{sp}^{3}$ hybridisation with pyramidal shape.
Explanation:
To show resonance effect molecule must have conjugated system. The molecule which has alternate $\pi$-bond, positive I charge, negative charge and lone pair of electron show conjugation. It shows resonance, all of these are resonating structure. (b.) It also shows resonance effect due to presence of resonating structure. (c.) It also shows resonance effect. (d.) It does not show resonance effect because it does not have planer structure because $\mathrm{NH}_{3}$ molecules have $\mathrm{sp}^{3}$ hybridisation with pyramidal shape.
UPTU/ UPSEE-2018
GENERAL ORGANIC CHEMISTRY
231739
Among the given compound choose the two the yield same carbocation on ionisation
1 (ii),(iv)
2 (i),(ii)
3 (ii),(iii)
4 (i),(iii)
Explanation:
Compound I \& II will show same carbocation
UPTU/ UPSEE-2017
GENERAL ORGANIC CHEMISTRY
231741
Consider the following compounds Friedel-Crafts acylation can be used to obtain:
1 II, III, IV
2 I, II, IV
3 I, II, III
4 I, III, IV
Explanation:
Friedel Craft Acylation when benzene ring is treated with acid derivative in the presence of anhydrous $\mathrm{FeCl}_{3}$ then acylation is formed such reaction is called friedel craft acylation. (i) It has $+\mu$ group so it is ortho-para directing. (ii) $\mathrm{CH}_{3} \rightarrow$ hyper conjugation only at ortho and Para (iii) $\mathrm{NO}_{2} \rightarrow$ It is a deactivating group (iv) $\mathrm{N}-\mathrm{Me}_{2} \rightarrow$ It is Less basic So, only I \& II and IV is possible to give Ortho and para directing for Friedel Craft Acylation.
UPTU/ UPSEE-2017
GENERAL ORGANIC CHEMISTRY
231742
Transition state 2 (TS2) is structurally most likely as a/an
1 intermediate 2
2 product
3 intermediate 1
4 transition state 3 (TS3)
Explanation:
(A) : The transition state resembles that spaces which is energetically near to Hammond postulate. Hence, $\mathrm{TS}_{2}$ resemble intermediate 2 . Hommond postulate - According to this postulate the transition state will be more similar in structure to species that is closer to energetically. In exergonic reaction, transition state in nearly energetically closure to reactant than product.