InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
The shape of `(CH_3)^(oplus)` is………. |
| Answer» Correct Answer - Planar | |
| 102. |
Explain why `(CH_3)_3C^(oplus)` is more stable than `CH_3 overset (oplus) C H_2` and `overset (oplus) C H_3` is less stable. |
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Answer» Hyperconjugation interaction in `(CH_3)_3 C^(oplus) (I)` is greater than that of `C` in `CH_3 overset (oplus) CH_2 (II)`, as `(I)` has nine `(C - H)` bonds and `(II)` has three `(C - H)` bonds. In `overset (oplus) CH_3`, the vacant `p` orbital is perpendicular to the plane in which `(C- H)` bonds lies , hence cannot overlap with it. Thus, `overset (oplus) CH_3` lacks hyperconjugative stability. |
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| 103. |
Number of chiral carbon atoms is isA. 2B. 3C. 4D. 1 |
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Answer» Correct Answer - A The given structure has two chiral centers. |
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| 104. |
How many optically active isomers are possible for A. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - A Tartric acid has two optically active isomers and one meso form |
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| 105. |
Categorise the following molecules/ions ad nucleophile or electrophile. (1) `HS^(Ө)` (2) `BF_3` (3) `C_2H_5O^(Ө)` (4) `(CH_3)_3 ddot N` (5) `C1^(Ө)` (6) `CH_3 - overset (oplus) C = O` (7) `:overset (Ө) NH_2` (8) `overset (oplus) N O_2`. |
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Answer» Nucleophiles : `HS^(Ө), C_2 H_5 O^(Ө),(CH_3) _3 ddot N, overset (ddot (Ө)) N H_2` These species have unshareped pair of `overline e^, s` which can be donated and shared with an electrophile. Electrophilies : `BF_3, C1^(oplus), CH_3 - C^(oplus) = O, overset (oplus) NO_2` Relative sites have only six valence `overline e^, s` and can accept `overline e` pair from a nucleophile. |
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| 106. |
Observe these compound and give answer of following questions. Q. In the given Aldose which can form same osazone (I). D-glucose (II). D-glucose (III). D-mannose (IV). D-galactoseA. I & IIIB. I,III,IVC. I,IID. III,IV |
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Answer» Correct Answer - A All C-2 epimers give same osazone. |
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| 107. |
Statement-I: per molecule of Glucose consumes two molecule of `Ph-NH-NH_(2)` during formation of osazone. Statement-II: Structure of Glucose osazone is: A. If both Statement-I & Statement-II are True & the Statement-II is not a correct explanantion of the statement I.B. If both statement-I & statement-II are true but statement-II is not a correct explanation of the statement-IC. If statement-I is true but the statement-II is falseD. If statement-I is false but the statement-II is true. |
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Answer» Correct Answer - D 3 Moles of `Ph-NHNH_(2)` are consumed. |
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| 108. |
Give the correct IUPAC name of `{:(" "CH_(3)),(" |"),(CH_(3).CH_(2)OCH.CH_(2).CH_(2)CH_(2)Cl):}`A. 2-ethoxy-5-chloropentaneB. 1-chloro-4-ethoxy-4-methylbutaneC. 1-chloro-4-ethoxypentaneD. Ethyl-1-chloropentylether |
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Answer» Correct Answer - A According to IUPAC system, ether are named as alkoxy alkanes. The larger alkyl groups forms the parent chain while lower alkyl group is taken ethernal oxygen and forms a part of alkoxy group. `{:(" ".^(1)CH_(3)),(" |"),(CH_(3)-CH_(2)-O-underset(2)(C)H-underset(3)(C)H_(2)-underset(4)(C)H_(2)-underset(5)(C)H_(2)Cl),(" 2-ethoxy-5-chloropentane"):}` |
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| 109. |
The term anomer of glucose refers to:A. Isomers of glucose that differ in configurations at carbon one and four (C-1 and C-4)B. A mixture of (D)-glucose and (L)-glucoseC. Enantiomers of glucoseD. Isomers of glucose that differ in configuration at carbon one (C-1) |
| Answer» Correct Answer - D | |
| 110. |
Which of the following has a bond formed by overlap of `sp^(3)-sp`, hybrid orbitals?A. `CH_(3) - C -= C - H`B. `CH_(3)-CH = CH - CH_(3)`C. `CH_(2) = CH - CH = CH_(2)`D. `HC -= CH` |
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Answer» Correct Answer - A `underset(sp^(3))(CH_(3))-underset(sp)(C)-=underset(sp)(C)-H` |
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| 111. |
Carbene intermediates are produced by the photolysis of diazomethane `(CH_2N_2)` or kenene `(CH_2 = C = O)`. They are also produced by the reaction of `CHX_3` with base or by Simmons-Smith reaction. There are two types of carbenes, singlet and triplet. They are so called due to their spin state. In which reaction, the insertion of methylene increases potential energy ?A. `CH_(2) = CH_(2) +:CH_(2) rarr CH_(3) - CH =CH_(2)`B. C. `Me-Ch=CH_(2)+ :CH_(2) rarr Me-CH=CH-CH_(3)`D. |
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Answer» Correct Answer - B (b) Due to ring strain in cyclopropane |
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| 112. |
The compounds in which `C` uses its `sp^(3)`- hybrid orbitals for bond formation are:A. HCOOHB. `(H_(2)N)_(2)CO`C. `(CH_(3))_(3)COH`D. `CH_(3)CHO` |
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Answer» Correct Answer - C::D Carbon forms four `sigma`-bonds is `sp^(3)`-hybridized. |
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| 113. |
Among the following compounds, one that will not shown keto-enol tautomerism isA. B. C. D. |
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Answer» Correct Answer - B HCN and HCN are tautomers |
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| 114. |
Among the following how many of them will show keto-enol tautomerism? |
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Answer» Correct Answer - 5 2,3,6,5,7,8 compunds contains `alpha`-Hydrogen |
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| 115. |
Keto-enol tautomerism is observed in :A. `C_(6)H_(5)-overset(O) overset(||)C-H`B. `C_(6)H_(5)-overset(O) overset(||)C-CH_(3)`C. `C_(6)H_(5)-overset(O) overset(||)C-C_(6)H_(5)`D. `C_(6)H_(5)-overset(O) overset(||)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-C_(6)H_(5)` |
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Answer» Correct Answer - B For a compound to exhibit keto-enol isomerism, presence of `alpha`-hydrogen is necessary. This is satisfied only by acetophenone |
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| 116. |
Keto-enol tautomerism is observed in :A. `H_(5)C_(6)-overset(O)overset(||)(C)-H`B. `H_(5)C_(6)-overset(O)overset(||)(O)-C_(6)H_(5)`C. `H_(5)C_(6)-overset(O)overset(||)(C)-CH_(3)`D. `H_(5)C_(6)-overset(O)overset(||)(C)-CH_(2)-overset(O)overset(||)(C)-CH_(3)` |
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Answer» Correct Answer - C::D `underset("(Keto form)")(C_(6)H_(5)-overset(O)overset(||)(C)-CH_(3))` and `underset("(enol form)")(C_(6)H_(5)-overset(OH)overset(|)(C)=CH_(2))` `underset("(keto form)")(C_(6)H_(5)=overset(O)overset(||)(C)-CH_(2)-overset(O)overset(||)(C)-CH_(3))` and `{:(C_(6)H_(5)-C-CH=C-CH_(3)),(" ||"" |"),(" "O" "OH),(" (enol form)"):}` |
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| 117. |
The number of optical isomers of `CH_(3)CH(OH)CH(OH)CHO` isA. ZeroB. 2C. 3D. 4 |
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Answer» Correct Answer - D `CH_(3)-underset(OH)underset(|)overset(H)overset(|)(C^(**))-overset(H)overset(|)underset(OH)underset(|)(C^(**))-CHO` It has 2 optically active carbon atoms and it has no plane of symmetry. So, number of optical isomers `= 2^(2) = 4`. |
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| 118. |
In the reaction `CH_(3)CHO+HCNtoCH_(3)CH(OH)CN` a chiral centre is produced. This product would beA. LaevorotatoryB. Meso compoundC. DextrorotatoryD. Racemic mixture |
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Answer» Correct Answer - D `CH_(3)CHO+HCN rarr CH_(3)-underset(OH)underset(|)overset(**)(CH)-CN` (Both d and l forms are obtained) Hence, product will be a racemic mixture. |
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| 119. |
In which of the following lone-pair indicated is involved in resonance:A. B. C. D. |
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Answer» Correct Answer - B::C::D See the conjugate system. |
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| 120. |
IUPAC name of crotonaldehyde isA. Prop-2-ene-1-alB. PropenalC. But-2-ene-1-alD. Butenal |
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Answer» Correct Answer - C `underset("But-2-en-1-al")(underset(4)(C)H_(3)-underset(3)(C)H=underset(2)(C)H-underset(1)(C)HO)` |
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| 121. |
Arrange the following compounds in increasing order of basicity : A. `SgtRgtQgtP`B. `SgtRgtPgtQ`C. `PgtQgtRgtS`D. `RgtQgtPgtS` |
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Answer» Correct Answer - B S is most basic becaues of stearic inhibition of resonance. |
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| 122. |
The predominant form of histamine present in human blood is (`pK_(a)` , Histidine=6.0)A. B. C. D. |
| Answer» Correct Answer - B | |
| 123. |
The increasing order of basicity of the following compounds is : A. `(b) lt(a) lt (d) lt (c)`B. `(d) lt (b) lt (a) lt (c)`C. `(a) lt (b) lt (c) lt (d)`D. `(b) lt (a) lt (c) lt (d)` |
| Answer» Correct Answer - A | |
| 124. |
`{:("COlumn(I)","Column(II)"),("Acid",pK_(a)),((a)CH_(3)-CO_(2)H,(p)5.69),((b)(CH_(3))_(3)overset(o+)NCH_(2)CO_(2)H,(q)4.27),((c)(CH_(3))_(3)overset(o+)N(CH_(2))_(4)CO_(2)H,(r)1.83),((d)overset(-)O_(2)C-CH_(2)-CO_(2)H,(s)4.80):}` |
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Answer» Correct Answer - a-s; b-r; c-q; d-p `pHK_(a) prop (1)/("acidic strength") prop -I` effect of attached groups |
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| 125. |
Match the names of carboxylic acids in column I with `pk_(a)` value in column II. `{:("Column I","Column II"),((A)." Benzoic acid",(P)." 4.17"),((B)." Ethanoic acid",(Q)." 4.14"),((C)." o-methyl benzoic acid",(R)." 4.74"),((D)." p-florobenzoic acid",(S)." 3.91"):}` |
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Answer» Correct Answer - (A). P, (B)R, (C)S,(D)Q With increase in electron withdrawing groups acidity `(K_(a))` of carboxylic acid increases decreasing the value of `pK_(a)` |
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| 126. |
Write increasing order of basic strength of following: (i). (a) `F^(ɵ)`, (b). `Cl^(ɵ)`, (c). `Br^(ɵ)`, (d). `I^(ɵ)` (A). `a gt b gt c gt d` (B). `d lt a lt c lt b` (C). `d gt a gt b gt c` (D). `a gt d gt c gt b` (ii). (a). `CH_(3)^(ɵ)`, (b). `Cl^(ɵ)`, (c) `OH^(ɵ)`, (d). `F^(ɵ)` (A). `a gt b gt c gt d` (B). `d lt a lt c lt b` (C). `d gt a gt b gt c` (D). `a gt d gt c gt b` (iii). (a). `R-NH_(2)`, (b). `Ph-NH_(2)`, (c). `R-underset(O)underset(||)(C)-NH_(2)` (A). `a gt b gt c` (B). `a lt c lt b` (C). `a lt b lt c` (D). `a gt c gt b` (iv). (a). `NH_(3)`, (b). `MeNH_(2)`, (c). `Me_(2)NH` , (d). `Me_(2)N` (A). `a lt b lt c lt d` (B). `d lt a lt c lt b` (C). `d gt a gt b gt c` (D). `a gt d gt c gt b` |
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Answer» Correct Answer - (i)A, (ii)A, (iii)A, (iv)A Basic strength : The anion which is less stable take more `H^(+)`. (iii). In option (B) & (C) 1. pair involve in resonance. (iv). In gas phase order `-3^(@)gt2^(@)gt1^(@)gtNH_(3)` |
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| 127. |
The order of basicity among the following compounds is A. `II gt I gt IV gt III`B. `I gt IV gt III gt II`C. `IV gt II gt III gt I`D. `IV gt I gt II gt III` |
| Answer» Correct Answer - D | |
| 128. |
Arrange the following in increasing order of their `pK_(a)` values. (x) `CH_(3)-underset(O)underset(||)overset(O)overset(||)S-O-H` (y) `CH_(3)-overset(O)overset(||)S-O-H` (z) `CH_(3)-OH`A. `yltxltz`B. `xltyltz`C. `yltzltx`D. `xgtzgty` |
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Answer» Correct Answer - B Stability of reaction hybrid `prop` no. of equivalent R.S So stability order is `Me-underset(3-eq.R.S)underset(O)underset(||)overset(O)overset(||)S-O^(Ө)gtMeunderset(2-eq.R.S)overset(O)overset(||)C-O^(Ө)gtMe-O^(Ө)` So Acidic, strenght order is `xgttgtz` so `pK_(a)` is `zgtygtx` |
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| 129. |
Record the following sets of compounds according to increasing `pK_(a)(=-logKa)` (a). cyclohexane carboxylic acid (A). `3 lt 2 lt 1` (B). `3 gt 2 gt 1` (C). `1 lt 2 lt 3` (D). `3 lt 1 lt 2` (b). 1-butyne, 1-butene, butane (A). `3 lt 2 lt 1` (B). `3 lt 2 lt 1` (C). `1 lt 3 lt 2` (D). `3 lt 2 gt 1` (c). Propanoic acid, 3-bromopropanoic acid, 2-nitropropanoic acid (A). 3lt2lt1 (B). 3gt2gt1 (C). 1lt2lt3 (D). 3lt1lt2 |
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Answer» Correct Answer - (a)A, (b)A, (c)A (c). Acidic strength `alpha.Kaprop(1)/(pKa)` |
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| 130. |
The IUPAC name of the following compound is `{:((CH_(3))_(2)CH-CH_(2)CH=CH-CH=CH-CHCH_(3)),(" |"),(" "C_(2)H_(5)):}`A. 1,1,7,7-tetramethyl-2,5-octadieneB. 2,8-dimethyl-3, 6-decadieneC. 1,5-di-iso-propyl-1, 4-hexadieneD. 2,8-dimethyl-4,6-decadiene |
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Answer» Correct Answer - D `{:(" "CH_(3)),(" |"),(underset("2,8-Dimethyl-4,6-decadiene")(overset(1)(C)H_(3)overset(2)(C)Hoverset(3)(C)H_(2)overset(4)(C)H)=overset(5)(C)Hoverset(6)(C)H=overset(7)(C)Hoverset(8)(C)HCH_(3)),(" |"),(" "overset(9)(C)H_(2)overset(10)(C)H_(3)):}` |
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| 131. |
The compounds `CH_(3)CH = CHCH_(3)` and `CH_(3)CH_(2)CH = CH_(2)`A. Are tautomersB. Are position isomersC. Contain same number of `sp^(3) - sp^(3), sp^(3) - sp^(2)` and `sp^(2) - sp^(2)` carbon-carbon bondsD. Exist together in dynamic equilibrium |
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Answer» Correct Answer - B The two isomers differ in the position of the double bond so they are called position isomers. |
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| 132. |
The correct order of increasing basic nature of the following bases is A. `(ii) lt (v) lt (i) lt (iii) lt (iv)`B. `(v) lt (ii) lt (i) lt (iii) lt (iv)`C. `(ii) lt (v) lt (i) lt (iv) lt (iii)`D. `(v) lt (ii) lt (i) lt (iv) lt (iii)` |
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Answer» Correct Answer - A Presence of electron-withdrawing (-I or -M group) like `-NO_(2)` at p-position willdecrease the basicity, so (ii) will be the least basic, whereas presence of electron-donating (+I or +M) group like `-OCH_(3)` at p-position in (iv) will increase the basicity so (iv) will be the most basic. Hence correct order of increasing basic character is `(ii) lt (v) lt (i) lt (iii) lt (iv)`. |
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| 133. |
The correct order of increasing basic nature for the bases `NH_(3) CH_(3)NH_(2) and (CH_(3))_(2)NH` is :A. `CH_(3)NH_(2) lt NH_(3) lt (CH_(3))_(2)NH`B. `(CH_(3))_(2)NH lt NH_(3) lt CH_(3)NH_(2)`C. `NH_(3) lt CH_(3)NH_(2) lt (CH_(3))_(2)NH`D. `CH_(3)NH_(2) lt (CH_(3))_(2)NH lt NH_(3)` |
| Answer» Correct Answer - C | |
| 134. |
Which is most stable carbocationA. `CH_(3)CH_(2)^(+)`B. `CH_(3)^(+)`C. `CH_(3)-overset(+)(C)H-CH_(3)`D. `CH_(3)CH_(2)-overset(+)(C)H_(2)` |
| Answer» Correct Answer - C | |
| 135. |
As the base changes from `RNH_(2)` to `R_(2)NH_(1)` to `R_(3)N` the basicityA. `R_(2)NH gt R_(3)N gt RNH_(2)`B. `RNH_(2) gt R_(3)H gt R_(2)NH`C. `RNH_(2) gt R_(2)NH gt R_(3)N`D. `R_(3)N gt RNH_(2) gt R_(2)NH.` |
| Answer» Correct Answer - A | |
| 136. |
The increasing order of `pK_(a)` values of the following compounds is : A. `C lt B lt A lt D`B. `B lt C lt D lt A`C. `B lt C lt A lt D`D. `D lt A lt C lt B` |
| Answer» Correct Answer - C | |
| 137. |
Arrange the following amines in the decreasing order of basicity : A. `I gt III gt II`B. `III gt I gt II`C. `III gt II gt I`D. `I gt II gt III` |
| Answer» Correct Answer - B | |
| 138. |
Arrange the following in decreasing order of the basicity. `CH_(2)=CHCH_(2)NH_(2),CH_(3)CH_(2)CH_(2)NH_(2),CH-=C CH_(2)NH_(2)`A. `IgtIIgtIII`B. `IIgtIgtIII`C. `IIIgtIIgtI`D. `IIgtIIIgtI` |
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Answer» Correct Answer - B `-I` power of triple bonded carbon is greater than double bonded and single bonded carbon. |
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| 139. |
Arrange the following compounds in decreasing order of basicity : A. `PgtQgtRgtS`B. `QgtPgtRgtS`C. `RgtSgtQgtP`D. `QgtPgtSgtR` |
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Answer» Correct Answer - A P is more basic as it is `2^(@)` amine. R is more bacis than S, as size increases tendency to donate `lp e^(-)` decreases to avoid stearic repuision. |
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| 140. |
Which amongst the following is the strongest acid ?A. `CHBr_(3)`B. `CHCl_(3)`C. `CHl_(3)`D. `CH(CN)_(3)` |
| Answer» Correct Answer - D | |
| 141. |
Basic nature of `H_(3)O^(+),H_(2)O` and `OH^(-)` is in orderA. `H_(3)O^(+) lt H_(2)O lt OH^(-)`B. `H_(2)O lt OH^(-) lt H_(3)O^(-)`C. `OH^(-) lt H_(2)O lt H_(3)O^(+)`D. `OH^(-) =H_(3)O^(+)=H_(2)O` |
| Answer» Correct Answer - A | |
| 142. |
The strongest acid amongst the following compounds is :A. `CH_(3)COOH`B. `HCOOH`C. `CH_(3)CH_(2)CH(Cl)CO_(2)H`D. `ClCH_(2)CH_(2)CH_(2)COOH` |
| Answer» Correct Answer - C | |
| 143. |
Find the order of basic strength .(If R=Me) ? `(I) R_(4)N^(+)OH^(-) " "(II) R_(3)N" "(III) R_(2)NH" "(IV)RNH_(2)`A. `I gt III gt IV gt II`B. `IV gt III gt I gt II`C. `II gt IV gt III gt I`D. `II gt IV gt I gt III` |
| Answer» Correct Answer - A | |
| 144. |
Arrang the anions (p) `overset(-)CH_(3),(q)overset(-)NH_(2),(r)OH^(-), (s)F^(-),` in decreasing order of their basic strength .A. `pgtqgtrgts`B. `qgtpgtrgts`C. `rgtqgtpgts`D. `rgtpgtqgts` |
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Answer» Correct Answer - A Basic strength `prop(1)/("Acidic strength of conjugated acid")` |
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| 145. |
Which carbocation is the most stable-A. B. C. D. |
| Answer» Correct Answer - B | |
| 146. |
The most stable carbocation is:A. B. C. D. |
| Answer» Correct Answer - B | |
| 147. |
The groups which when present in para position tend to decrease the acidity of phenol are-A. `-NO_(2)`B. `-CN`C. `-OCH_(3)`D. `-F` |
| Answer» Correct Answer - C | |
| 148. |
Out of the four `pK_(a)` values 3.75, 9.88, 15.54 and 19.30 , the highest `pK_(a)` value corresponds toA. acetoneB. formic acidC. phenolD. methanol |
| Answer» Correct Answer - A | |
| 149. |
Select the acid with the highest `K_(a)` (i.e., lowest `pK_(a))`.A. B. C. D. |
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Answer» Correct Answer - B Acidic strength `prop=I,-H-Mprop(1)/(+I,H,R)` |
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| 150. |
Which is the following phenol has lowest `pK_(a)` ?A. B. C. D. |
| Answer» Correct Answer - D | |