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JEE Advanced · Chemistry · 2014

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1
Different possible thermal\textbf{thermal} decomposition pathways for peroxyesters (RR-CO-O-O-R′R') are shown below. Match each pathway from List-I with the peroxyester structure from List-II whose R,R′R,R' groups are consistent with that pathway, and select the correct combination. Mechanism\textbf{Mechanism}: The generic peroxyester R-CO-O-O-R′R\text{-CO-O-O-}R' decomposes thermally by cleaving the weak O-O bond; depending on how stable the resulting R∙R^\bullet radical is (concerted vs. stepwise loss of CO2CO_2) and whether the alkoxy radical R′O∙R'O^\bullet is a tertiary/branched radical capable of β\beta-scission (fragmenting further into a smaller radical X∙X^\bullet plus a carbonyl compound) or a simple radical that does not fragment further, four distinct pathways result: Pathway P\textbf{Pathway P}: −CO2-CO_2 (concerted) →R∙+R′O∙\to R^\bullet + R'O^\bullet (no further fragmentation of either radical). Pathway Q\textbf{Pathway Q}: −CO2-CO_2 (concerted) →R∙+R′O∙→R∙+X∙+carbonyl compound\to R^\bullet + R'O^\bullet \to R^\bullet + X^\bullet + \text{carbonyl compound} (R′O∙R'O^\bullet undergoes further β\beta-scission). Pathway R\textbf{Pathway R}: stepwise, OO-OO homolysis first →RCO2∙+R′O∙→R∙+X′∙+carbonyl compound\to RCO_2^\bullet + R'O^\bullet \to R^\bullet + X'^\bullet + \text{carbonyl compound} (with loss of CO2CO_2 from RCO2∙RCO_2^\bullet occurring after R′O∙R'O^\bullet has already β\beta-scissioned). Pathway S\textbf{Pathway S}: stepwise, OO-OO homolysis first →RCO2∙+R′O∙→R∙+R′O∙\to RCO_2^\bullet + R'O^\bullet \to R^\bullet + R'O^\bullet (no further fragmentation of R′O∙R'O^\bullet). List-II (candidate peroxyester structures)\textbf{List-II (candidate peroxyester structures)}: 1. C6H5CH2-CO-O-O-CH31.\ C_6H_5CH_2\text{-CO-O-O-CH}_3 (R = benzyl, R' = methyl) 2. C6H5-CO-O-O-CH32.\ C_6H_5\text{-CO-O-O-CH}_3 (R = phenyl, R' = methyl) 3. C6H5CH2-CO-O-O-C(CH3)2CH2C6H53.\ C_6H_5CH_2\text{-CO-O-O-C(CH}_3)_2\text{CH}_2C_6H_5 (R = benzyl, R' = tertiary carbon bearing two methyls and a benzyl group) 4. C6H5-CO-O-O-C(CH3)2C6H54.\ C_6H_5\text{-CO-O-O-C(CH}_3)_2C_6H_5 (R = phenyl, R' = tertiary carbon bearing two methyls and a phenyl group)
Not attempted
A
P→\to1, Q→\to3, R→\to4, S→\to2
B
P→\to2, Q→\to4, R→\to3, S→\to1
C
P→\to4, Q→\to1, R→\to2, S→\to3
D
P→\to3, Q→\to2, R→\to1, S→\to4
2
Comprehension\textbf{Comprehension}: XX and YY are two volatile liquids with molar masses 10 g mol−110\ \text{g mol}^{-1} and 40 g mol−140\ \text{g mol}^{-1} respectively. Two cotton plugs, one soaked in XX and the other soaked in YY, are simultaneously placed at the two ends of a tube of length L=24 cmL = 24\ \text{cm} (shown below). The tube is filled with an inert gas at 1 atm and 300 K300\ K. Vapours of XX and YY diffuse from opposite ends and react to form a product first observed at a distance dd cm from the plug soaked in XX. XX and YY have equal molecular diameters; the inert gas and both vapours behave ideally. 35. \textbf{35. } The value of dd in cm (shown in the figure), as estimated from Graham's law, is:
Not attempted
A
8
B
12
C
16
D
20
3
Comprehension\textbf{Comprehension}: XX and YY are two volatile liquids with molar masses 10 g mol−110\ \text{g mol}^{-1} and 40 g mol−140\ \text{g mol}^{-1} respectively. Two cotton plugs, one soaked in XX and the other soaked in YY, are simultaneously placed at the two ends of a tube of length L=24 cmL = 24\ \text{cm} (shown below). The tube is filled with an inert gas at 1 atm and 300 K300\ K. Vapours of XX and YY diffuse from opposite ends and react to form a product first observed at a distance dd cm from the plug soaked in XX. XX and YY have equal molecular diameters; the inert gas and both vapours behave ideally. 36. \textbf{36. } The experimental value of dd is found to be smaller than the estimate obtained using Graham's law. This is due to:
Not attempted
A
Larger mean free path for XX as compared to that of YY.
B
Larger mean free path for YY as compared to that of XX.
C
Increased collision frequency of YY with the inert gas as compared to that of XX with the inert gas.
D
Increased collision frequency of XX with the inert gas as compared to that of YY with the inert gas.
4
Comprehension\textbf{Comprehension}: An aqueous solution of metal ion M1M_1 reacts separately with reagents QQ and RR in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2M_2 always forms tetrahedral complexes with these same reagents (QQ and RR). Aqueous solution of M2M_2 on reaction with reagent SS (stoichiometric amount) gives a white precipitate, which dissolves in excess of SS. Scheme\text{Scheme}: Tetrahedral←excessQM1→excessRSquare planar\text{Tetrahedral} \xleftarrow[\text{excess}]{Q} M_1 \xrightarrow[\text{excess}]{R} \text{Square planar} Tetrahedral←excessQM2→excessRTetrahedral\text{Tetrahedral} \xleftarrow[\text{excess}]{Q} M_2 \xrightarrow[\text{excess}]{R} \text{Tetrahedral}, and separately M2→S, stoichiometricWhite precipitate→excessSprecipitate dissolvesM_2 \xrightarrow{S\text{, stoichiometric}} \text{White precipitate} \xrightarrow[\text{excess}]{S} \text{precipitate dissolves} 33. \textbf{33. } M1M_1, QQ and RR, respectively are:
Not attempted
A
Zn2+Zn^{2+}, KCNKCN and HClHCl
B
Ni2+Ni^{2+}, HClHCl and KCNKCN
C
Cd2+Cd^{2+}, KCNKCN and HClHCl
D
Co2+Co^{2+}, HClHCl, and KCNKCN
5
Comprehension\textbf{Comprehension}: An aqueous solution of metal ion M1M_1 reacts separately with reagents QQ and RR in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2M_2 always forms tetrahedral complexes with these same reagents (QQ and RR). Aqueous solution of M2M_2 on reaction with reagent SS (stoichiometric amount) gives a white precipitate, which dissolves in excess of SS. Scheme\text{Scheme}: Tetrahedral←excessQM1→excessRSquare planar\text{Tetrahedral} \xleftarrow[\text{excess}]{Q} M_1 \xrightarrow[\text{excess}]{R} \text{Square planar} Tetrahedral←excessQM2→excessRTetrahedral\text{Tetrahedral} \xleftarrow[\text{excess}]{Q} M_2 \xrightarrow[\text{excess}]{R} \text{Tetrahedral}, and separately M2→S, stoichiometricWhite precipitate→excessSprecipitate dissolvesM_2 \xrightarrow{S\text{, stoichiometric}} \text{White precipitate} \xrightarrow[\text{excess}]{S} \text{precipitate dissolves} 34. \textbf{34. } Reagent SS is:
Not attempted
A
K4[Fe(CN)6]K_4[Fe(CN)_6]
B
Na2HPO4Na_2HPO_4
C
K2CrO4K_2CrO_4
D
KOHKOH
6
Match each coordination compound in List-I with an appropriate pair of characteristics from List-II, and select the correct combination. (en=H2NCH2CH2NH2en = H_2NCH_2CH_2NH_2; atomic numbers: Ti=22Ti=22, Cr=24Cr=24, Co=27Co=27, Pt=78Pt=78.) List-I\textbf{List-I}: P. [Cr(NH3)4Cl2]Cl\textbf{P}.\ [Cr(NH_3)_4Cl_2]Cl. Q. [Ti(H2O)5Cl](NO3)2\quad \textbf{Q}.\ [Ti(H_2O)_5Cl](NO_3)_2. R. [Pt(en)(NH3)Cl]NO3\quad \textbf{R}.\ [Pt(en)(NH_3)Cl]NO_3. S. [Co(NH3)4(NO3)2]NO3\quad \textbf{S}.\ [Co(NH_3)_4(NO_3)_2]NO_3. List-II\textbf{List-II}: 1\textbf{1}. Paramagnetic and exhibits ionisation isomerism. 2\quad \textbf{2}. Diamagnetic and exhibits cis-trans isomerism. 3\quad \textbf{3}. Paramagnetic and exhibits cis-trans isomerism. 4\quad \textbf{4}. Diamagnetic and exhibits ionisation isomerism.
Not attempted
A
P→\to4, Q→\to2, R→\to3, S→\to1
B
P→\to3, Q→\to1, R→\to4, S→\to2
C
P→\to2, Q→\to1, R→\to3, S→\to4
D
P→\to1, Q→\to3, R→\to4, S→\to2
7
Assuming 2s2s-2p2p mixing is NOT\textbf{NOT} operative, the paramagnetic species among the following is:
Not attempted
A
Be2Be_2
B
B2B_2
C
C2C_2
D
N2N_2
8
Match the orbital overlap figures shown in List-I (labelled P, Q, R, S below — shaded lobes and unshaded/outline-only lobes represent opposite phases of the wavefunction) with the description given in List-II, and select the correct answer using the code given below the lists. List-II\textbf{List-II}: 11. pp-dd π\pi antibonding. 2\quad 2. dd-dd σ\sigma bonding. 3\quad 3. pp-dd π\pi bonding. 4\quad 4. dd-dd σ\sigma antibonding.
Not attempted
A
P→\to2, Q→\to1, R→\to3, S→\to4
B
P→\to4, Q→\to3, R→\to1, S→\to2
C
P→\to2, Q→\to3, R→\to1, S→\to4
D
P→\to4, Q→\to1, R→\to3, S→\to2
9
For the process H2O(l)→H2O(g)H_2O(l) \to H_2O(g) at T=100∘CT = 100^\circ C and 1 atmosphere pressure, the correct choice is:
Not attempted
A
ΔSsystem>0\Delta S_{system} > 0 and ΔSsurroundings>0\Delta S_{surroundings} > 0
B
ΔSsystem>0\Delta S_{system} > 0 and ΔSsurroundings<0\Delta S_{surroundings} < 0
C
ΔSsystem<0\Delta S_{system} < 0 and ΔSsurroundings>0\Delta S_{surroundings} > 0
D
ΔSsystem<0\Delta S_{system} < 0 and ΔSsurroundings<0\Delta S_{surroundings} < 0
10
For the elementary reaction M→NM \to N, the rate of disappearance of MM increases by a factor of 8 upon doubling the concentration of MM. The order of the reaction with respect to MM is:
Not attempted
A
4
B
3
C
2
D
1
11
For the identification of β\beta-naphthol using dye test, it is necessary to use:
Not attempted
A
Dichloromethane solution of β\beta-naphthol.
B
Acidic solution of β\beta-naphthol.
C
Neutral solution of β\beta-naphthol.
D
Alkaline solution of β\beta-naphthol.
12
Match the four starting materials (P, Q, R, S) given in List-I with the corresponding reaction schemes (I, II, III, IV) provided in List-II (each scheme is a reagent sequence ending in a stated final molecular formula), and select the correct combination. List-I (starting materials)\textbf{List-I (starting materials)}: P\textbf{P}: HC≡CHHC{\equiv}CH (acetylene). Q\textbf{Q}: resorcinol (1,3-dihydroxybenzene, C6H6O2C_6H_6O_2). R\textbf{R}: nitrobenzene (C6H5NO2C_6H_5NO_2). S\textbf{S}: pp-nitrotoluene (1-methyl-4-nitrobenzene, C7H7NO2C_7H_7NO_2). List-II (reaction schemes)\textbf{List-II (reaction schemes)}: 1. Scheme I\textbf{1. Scheme I}: (i) KMnO4KMnO_4, HO−HO^-, heat (ii) H+H^+, H2OH_2O (iii) SOCl2SOCl_2 (iv) NH3NH_3 ⟶\longrightarrow C7H6N2O3C_7H_6N_2O_3 2. Scheme II\textbf{2. Scheme II}: (i) Sn/HClSn/HCl (ii) CH3COClCH_3COCl (iii) conc. H2SO4H_2SO_4 (iv) HNO3HNO_3 (v) dil. H2SO4H_2SO_4, heat (vi) HO−HO^- ⟶\longrightarrow C6H6N2O2C_6H_6N_2O_2 3. Scheme III\textbf{3. Scheme III}: (i) red hot iron, 873 K (ii) fuming HNO3HNO_3, H2SO4H_2SO_4, heat (iii) H2S⋅NH3H_2S \cdot NH_3 (iv) NaNO2NaNO_2, H2SO4H_2SO_4 (v) hydrolysis ⟶\longrightarrow C6H5NO3C_6H_5NO_3 4. Scheme IV\textbf{4. Scheme IV}: (i) conc. H2SO4H_2SO_4, 60∘C60^\circ C (ii) conc. HNO3HNO_3, conc. H2SO4H_2SO_4 (iii) dil. H2SO4H_2SO_4, heat ⟶\longrightarrow C6H5NO4C_6H_5NO_4
Not attempted
A
P→\to1, Q→\to4, R→\to2, S→\to3
B
P→\to3, Q→\to1, R→\to4, S→\to2
C
P→\to3, Q→\to4, R→\to2, S→\to1
D
P→\to4, Q→\to1, R→\to3, S→\to2
13
The product formed in the reaction of SOCl2SOCl_2 with white phosphorous is:
Not attempted
A
PCl3PCl_3
B
SO2Cl2SO_2Cl_2
C
SCl2SCl_2
D
POCl3POCl_3
14
Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is: XeF6→Complete HydrolysisP+other productXeF_6 \xrightarrow{\text{Complete Hydrolysis}} P + \text{other product} P→OH−/H2OQP \xrightarrow{OH^-/H_2O} Q Q→slow disproportionation in OH−/H2OproductsQ \xrightarrow{\text{slow disproportionation in } OH^-/H_2O} \text{products}
Not attempted
A
0
B
1
C
2
D
3
15
Hydrogen peroxide in its reaction with KIO4KIO_4 and NH2OHNH_2OH respectively, is acting as a:
Not attempted
A
reducing agent, oxidising agent
B
reducing agent, reducing agent
C
oxidising agent, oxidising agent
D
oxidising agent, reducing agent
16
Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure below (Class I: the two dimethylbutane isomers; Class II: the two methylpentane isomers; Class III: nn-hexane). The correct order of their boiling points is:
Not attempted
A
I >> II >> III
B
III >> II >> I
C
II >> III >> I
D
III >> I >> II
17
Comprehension\textbf{Comprehension}: Schemes 1 and 2 describe sequential transformations of alkynes MM and NN (structures shown below). Consider only the major product formed in each step. Scheme-1\textbf{Scheme-1}: MM (HOCH2-C≡CH\text{HOCH}_2\text{-C}{\equiv}\text{CH}, prop-2-yn-1-ol) →1. NaNH2 (excess); 2. CH3CH2I (1 eq); 3. CH3I (1 eq); 4. H2, Lindlar’s catalystX\xrightarrow{\text{1. }NaNH_2\text{ (excess); 2. }CH_3CH_2I\text{ (1 eq); 3. }CH_3I\text{ (1 eq); 4. }H_2\text{, Lindlar's catalyst}} X Scheme-2\textbf{Scheme-2}: NN (CH3-C≡CHCH_3\text{-C}{\equiv}\text{CH}, propyne) →1. NaNH2 (2 eq); 2. 1-bromopropan-2-ol (shown below); 3. H3O+ (mild); 4. H2, Pd/C; 5. CrO3Y\xrightarrow{\text{1. }NaNH_2\text{ (2 eq); 2. 1-bromopropan-2-ol (shown below); 3. }H_3O^+\text{ (mild); 4. }H_2\text{, Pd/C; 5. }CrO_3} Y 31. \textbf{31. } The product XX is:
Not attempted
1-bromopropan-2-ol (Scheme-2 reagent)
A
Option (A): (Z)(Z)-1-methoxy-2-pentene, CH3CH2CH_3CH_2 and CH3OCH2CH_3OCH_2 groups cis
B
Option (B): same connectivity as (A) but trans\textit{trans} alkene
C
Option (C): ethoxy instead of methoxy, cis alkene
D
Option (D): ethoxy instead of methoxy, trans alkene
18
Comprehension\textbf{Comprehension}: Schemes 1 and 2 describe sequential transformations of alkynes MM and NN (structures shown below). Consider only the major product formed in each step. Scheme-1\textbf{Scheme-1}: MM (HOCH2-C≡CH\text{HOCH}_2\text{-C}{\equiv}\text{CH}, prop-2-yn-1-ol) →1. NaNH2 (excess); 2. CH3CH2I (1 eq); 3. CH3I (1 eq); 4. H2, Lindlar’s catalystX\xrightarrow{\text{1. }NaNH_2\text{ (excess); 2. }CH_3CH_2I\text{ (1 eq); 3. }CH_3I\text{ (1 eq); 4. }H_2\text{, Lindlar's catalyst}} X Scheme-2\textbf{Scheme-2}: NN (CH3-C≡CHCH_3\text{-C}{\equiv}\text{CH}, propyne) →1. NaNH2 (2 eq); 2. 1-bromopropan-2-ol (shown below); 3. H3O+ (mild); 4. H2, Pd/C; 5. CrO3Y\xrightarrow{\text{1. }NaNH_2\text{ (2 eq); 2. 1-bromopropan-2-ol (shown below); 3. }H_3O^+\text{ (mild); 4. }H_2\text{, Pd/C; 5. }CrO_3} Y 32. \textbf{32. } The correct statement with respect to product YY is:
Not attempted
A
It gives a positive Tollens test and is a functional isomer of XX.
B
It gives a positive Tollens test and is a geometrical isomer of XX.
C
It gives a positive iodoform test and is a functional isomer of XX.
D
It gives a positive iodoform test and is a geometrical isomer of XX.
19
The acidic hydrolysis of ether XX shown below is fastest when Ph3C-OR→acidPh3C-OH+ROH\text{Ph}_3C\text{-OR} \xrightarrow{\text{acid}} \text{Ph}_3C\text{-OH} + \text{ROH} (structure of XX shown below, where each Ph is a phenyl ring on the central carbon):
Not attempted
A
One phenyl group is replaced by a methyl group.
B
One phenyl group is replaced by a para\textit{para}-methoxyphenyl group.
C
Two phenyl groups are replaced by two para\textit{para}-methoxyphenyl groups.
D
No structural change is made to XX.
20
The major product in the following reaction is: Cl-CH2CH2CH2-CO-CH3→2. aq. acid1. CH3MgBr, dry ether, 0∘C\text{Cl-CH}_2\text{CH}_2\text{CH}_2\text{-CO-CH}_3 \xrightarrow[\text{2. aq. acid}]{\text{1. CH}_3\text{MgBr, dry ether, } 0^\circ C} ? (structure of the chloroketone starting material shown below):
Not attempted
A
Option (A)
B
Option (B)
C
Option (C)
D
Option (D)
21
A compound H2XH_2X with molar weight of 80 g is dissolved in a solvent having density of 0.4 g ml−10.4\ g\ ml^{-1}. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is
Not attempted
Your answer: —
Correct answer: 8
22
MX2MX_2 dissociates into M2+M^{2+} and X−X^- ions in an aqueous solution, with a degree of dissociation (α\alpha) of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is
Not attempted
Your answer: —
Correct answer: 2
23
If the value of Avogadro number is 6.023×1023 mol−16.023\times10^{23}\ mol^{-1} and the value of Boltzmann constant is 1.380×10−23 J K−11.380\times10^{-23}\ J\ K^{-1}, then the number of significant digits in the calculated value of the universal gas constant is
Not attempted
Your answer: —
Correct answer: 4
24
In an atom, the total number of electrons having quantum numbers n=4n=4, ∣ml∣=1|m_l|=1 and ms=−12m_s=-\frac{1}{2} is
Not attempted
Your answer: —
Correct answer: 6
25
The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of the peptide shown below is
Not attempted
Your answer: —
Correct answer: 1
26
Consider the following list of reagents: Acidified K2Cr2O7K_2Cr_2O_7, alkaline KMnO4KMnO_4, CuSO4CuSO_4, H2O2H_2O_2, Cl2Cl_2, O3O_3, FeCl3FeCl_3, HNO3HNO_3 and Na2S2O3Na_2S_2O_3. The total number of reagents that can oxidise aqueous iodide to iodine is
Not attempted
Your answer: —
Correct answer: 7
27
The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are)
Not attempted
Your answer: —
Correct answer: 3
28
Among PbSPbS, CuSCuS, HgSHgS, MnSMnS, Ag2SAg_2S, NiSNiS, CoSCoS, Bi2S3Bi_2S_3 and SnS2SnS_2, the total number of BLACK coloured sulfides is
Not attempted
Your answer: —
Correct answer: 7
29
A list of species having the formula XZ4XZ_4 is given below: XeF4XeF_4, SF4SF_4, SiF4SiF_4, BF4−BF_4^-, BrF4−BrF_4^-, [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+}, [FeCl4]2−[FeCl_4]^{2-}, [CoCl4]2−[CoCl_4]^{2-} and [PtCl4]2−[PtCl_4]^{2-}. Defining shape on the basis of the location of XX and ZZ atoms, the total number of species having a square planar shape is
Not attempted
Your answer: —
Correct answer: 4
30
Consider all possible isomeric ketones, including stereoisomers, of molecular weight 100. All these isomers are independently reacted with NaBH4NaBH_4 (stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is (are)
Not attempted
Your answer: —
Correct answer: 5
31
For the reaction I−+ClO3−+H2SO4→Cl−+HSO4−+I2I^- + ClO_3^- + H_2SO_4 \rightarrow Cl^- + HSO_4^- + I_2, the correct statement(s) in the balanced equation is/are:
Not attempted
A
Stoichiometric coefficient of HSO4−HSO_4^- is 6.
B
Iodide is oxidized.
C
Sulphur is reduced.
D
H2OH_2O is one of the products.
32
The pair(s) of reagents that yield paramagnetic species is/are
Not attempted
A
Na and excess of NH3NH_3
B
K and excess of O2O_2
C
Cu and dilute HNO3HNO_3
D
O2O_2 and 2-ethylanthraquinol
33
The correct statement(s) for orthoboric acid is/are
Not attempted
A
It behaves as a weak acid in water due to self ionization.
B
Acidity of its aqueous solution increases upon addition of ethylene glycol.
C
It has a three dimensional structure due to hydrogen bonding.
D
It is a weak electrolyte in water.
34
Upon heating with Cu2SCu_2S, the reagent(s) that give copper metal is/are
Not attempted
A
CuFeS2CuFeS_2
B
CuOCuO
C
Cu2OCu_2O
D
CuSO4CuSO_4
35
In a galvanic cell, the salt bridge
Not attempted
A
does not participate chemically in the cell reaction.
B
stops the diffusion of ions from one electrode to another.
C
is necessary for the occurrence of the cell reaction.
D
ensures mixing of the two electrolytic solutions.
36
Hydrogen bonding plays a central role in the following phenomena:
Not attempted
A
Ice floats in water.
B
Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions.
C
Formic acid is more acidic than acetic acid.
D
Dimerisation of acetic acid in benzene.
37
An ideal gas in a thermally insulated vessel at internal pressure P1P_1, volume V1V_1 and absolute temperature T1T_1 expands irreversibly against zero external pressure (shown below) to a final state P2P_2, V2V_2, T2T_2. For this expansion,
Not attempted
A
q=0q = 0
B
T2=T1T_2 = T_1
C
P2V2=P1V1P_2V_2 = P_1V_1
D
P2V2γ=P1V1γP_2V_2^{\gamma} = P_1V_1^{\gamma}
38
3-(aminomethyl)benzamide is treated with acetic anhydride in CH2Cl2CH_2Cl_2 (shown below). The major product(s) formed is/are
Not attempted
A
Acetamide at the benzylic −CH2NH2-CH_2NH_2 nitrogen; the aromatic −CONH2-CONH_2 is unchanged; by-product CH3COOHCH_3COOH (structure shown)
B
Acetamide at the aromatic amide nitrogen; the benzylic −CH2NH2-CH_2NH_2 is unchanged; by-product CH3COOHCH_3COOH (structure shown)
C
Both nitrogens acetylated (imide formed at the amide N); by-product H2OH_2O (structure shown)
D
Ammonium-acetate salt at the benzylic nitrogen with the amide nitrogen acetylated (structure shown)
39
The correct combination of names for isomeric alcohols with molecular formula C4H10OC_4H_{10}O is/are
Not attempted
A
tert-butanol and 2-methylpropan-2-ol
B
tert-butanol and 1, 1-dimethylethan-1-ol
C
n-butanol and butan-1-ol
D
isobutyl alcohol and 2-methylpropan-1-ol
40
The reactivity of compound ZZ (3-tert-butylphenol) with different halogens under appropriate conditions is shown below: with X2=I2X_2 = I_2 a mono-halo substituted derivative forms; with X2=Br2X_2 = Br_2 a di-halo substituted derivative forms; with X2=Cl2X_2 = Cl_2 a tri-halo substituted derivative forms. The observed pattern of electrophilic substitution can be explained by
Not attempted
A
the steric effect of the halogen
B
the steric effect of the tert-butyl group
C
the electronic effect of the phenolic group
D
the electronic effect of the tert-butyl group
Hint\textbf{Hint}: The −OH-OH group activates the ortho/para positions strongly; think about which of those activated positions are sterically blocked by the bulky tert-butyl group, and how the size/reactivity of the halogen itself changes whether it can still attack a hindered position.
Deconstruction\textbf{Deconstruction}: Compound Z has OHOH at C1 and −C(CH3)3-C(CH_3)_3 at C3. The activated (ortho/para to OHOH) positions are C2, C4 and C6. C2 is flanked by both OHOH and the bulky tert-butyl group (most hindered); C4 is ortho to tert-butyl (moderately hindered); C6 is only ortho to OHOH, para to tert-butyl (least hindered). Strategy\textbf{Strategy}: Correlate the number of positions substituted with (i) the intrinsic reactivity/size of the halogen electrophile and (ii) the steric blocking by the bulky tert-butyl group, while noting that −OH-OH's strong activation is what makes all three positions reactive at all. Execution\textbf{Execution}: I2I_2 is a bulky, weakly electrophilic halogen — it can only attack the least hindered activated position (C6), giving mono-substitution. Br2Br_2, more reactive/smaller, can additionally reach C4, giving di-substitution. Cl2Cl_2, the smallest and most reactive halogen, can overcome even the severe steric hindrance at C2 next to the tert-butyl group, giving tri-substitution. This pattern is explained by: (A) the steric effect of the halogen itself (size decreases I2→Br2→Cl2I_2 \to Br_2 \to Cl_2, so more positions become accessible); (B) the steric effect of the bulky tert-butyl group (it is what makes C2 and, to a lesser extent, C4 hard to reach); (C) the electronic (activating, ortho/para-directing) effect of the phenolic −OH-OH group (without it none of these positions would be reactive at all towards electrophilic halogenation). Option (D), the electronic effect of tert-butyl, is a weak inductive/hyperconjugative activation and is not what differentiates the three positions — the differentiation here is steric, not electronic, on the tert-butyl group's part. Verification\textbf{Verification}: If sterics of the halogen and of the tert-butyl group were irrelevant, all three halogens would give the same substitution pattern since all three positions are similarly activated electronically by OHOH; the fact that the degree of substitution tracks exactly with halogen size/reactivity confirms sterics (A, B) are controlling, on top of (C)'s baseline activation. Final Answer\textbf{Final Answer}: (A), (B) and (C).