|
|
Niti KulkarniEduRev CAT |
|
Niti Kulkarni
EduRev CAT
|
Content, tests & courses saved by you for accessing later. (visible to you only)
Currently preparing for
 | IIT JAM |
 | Chemistry |
 | Government Jobs |
Top Scoring Tests by Niti
Test: Lanthanide- 1 | 38/40 |
Test: Reaction Mechanism Level - 1 | 55/60 |
Test: Pericyclic Level - 2 | 50/60 |
Test: Filling in the Blank - 1 | 91/120 |
Test: Gaseous State - 4 | 30/40 |
Test: Spellings- 1 | 34/45 |
Reactive Inter & Reaction Mechanism | 15/20 |
Test: Spectroscopy Level - 3 | 43/60 |
Test: Reaction Mechanism Level - 5 | 40/60 |
Test: Heterocyclic Level - 2 | 38/60 |
Discussed Questions
|
Parvathy Vs asked • Nov 25, 2020 |
Which of the following phenol would be the most acidic:
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Which of the following phenol would be the most acidic:
a)
b)
c)
d)
|
Uma Bharti answered |
Ewg enhances acudity cyanide is ewg
|
|
Parvathy Vs asked • Jun 21, 2020 |
The number of oxygen molecules that a molecule hemerythrin can transport is ___________
Correct answer is '8'. Can you explain this answer?
The number of oxygen molecules that a molecule hemerythrin can transport is ___________
|
Anushka Basak answered |
Answer:
Hemerythrin
Hemerythrin is a metalloprotein that is found in the muscles of some marine invertebrates. It is responsible for the transportation of oxygen in the blood.
Transportation of Oxygen
Hemerythrin can transport oxygen in two ways:
- By binding oxygen to iron ions in the protein
- By binding oxygen to the protein itself
Number of Oxygen Molecules Transported
Each hemerythrin molecule can transport up to 8 oxygen molecules. This is because there are 4 iron ions in each hemerythrin molecule, and each iron ion can bind to 2 oxygen molecules.
Conclusion
In summary, hemerythrin is a metalloprotein that is responsible for transporting oxygen in the blood. It can transport up to 8 oxygen molecules by binding them to iron ions in the protein or to the protein itself.
|
|
Parvathy Vs upvoted • May 25, 2020 |
The vibrational energy levels, v11 = 0 and v1 = 1 of a diatomic molecule are separated by 2143 cm-1. Its Anharmonicity (ωe Xe ) is 14 cm-1. The values of ωe (in cm-1) and first overtone (cm-1) of this molecules are respectively.- a)2171, 4300
- b)2157, 4286
- c)2157, 4314
- d)2171, 4258
Correct answer is option 'A'. Can you explain this answer?
The vibrational energy levels, v11 = 0 and v1 = 1 of a diatomic molecule are separated by 2143 cm-1. Its Anharmonicity (ωe Xe ) is 14 cm-1. The values of ωe (in cm-1) and first overtone (cm-1) of this molecules are respectively.
a)
2171, 4300
b)
2157, 4286
c)
2157, 4314
d)
2171, 4258
|
|
Pranavi Mishra answered |
|
|
Parvathy Vs asked • Mar 09, 2020 |
Products A and B are:
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Products A and B are:
a)
b)
c)
d)
|
|
Uma Bharti answered |
I think option 2 is correct
|
|
Parvathy Vs upvoted • Mar 03, 2020 |
The best procedure to improve resolution between two chromatographic peak is
- a)decreasing column length, decreasing band-width
- b)decreasing column length, increasing band-width
- c) increasing column length, increasing band-width
- d)increasing column length, decreasing band-width
Correct answer is option 'D'. Can you explain this answer?
The best procedure to improve resolution between two chromatographic peak is
a)
decreasing column length, decreasing band-width
b)
decreasing column length, increasing band-width
c)
increasing column length, increasing band-width
d)
increasing column length, decreasing band-width
|
Partho Gupta answered |
Here are the standard ways to improve resolution:
-
Increasing Column Length: This generally leads to better separation because it allows more time for the different components to separate.
-
Decreasing Band-Width (Peak Width): Narrower peaks mean less overlap and thus better resolution. This can be achieved by improving the efficiency of the column through factors such as packing material, particle size, and temperature control.
|
|
Parvathy Vs asked • Feb 28, 2020 |
Which statement is/are correct? - a)% of p-character in the orbitals forming P-P bonds in P4 is 75%
- b)The maximum number of atoms in B2H6 are 6 which are in same plane.
- c)PI3 upon hydrolysis gives monobasic acid and dibasic acid
- d)All of above
Correct answer is option 'D'. Can you explain this answer?
Which statement is/are correct?
a)
% of p-character in the orbitals forming P-P bonds in P4 is 75%
b)
The maximum number of atoms in B2H6 are 6 which are in same plane.
c)
PI3 upon hydrolysis gives monobasic acid and dibasic acid
d)
All of above
|
Jakir Ahmed answered |
Option A and C are correct.
P⁴, all P are SP³ hybrid, it is obvious P- character is 75%.
B2H6, Boron is sp³ hybrid, that means two bridging H are in the plane with two boron. among another four H, two H are out of the plane, and two are below the plane. therefore only 4 atoms are in plane
PI³ +H²O = H³PO³(di-basic) + HI(mono basic)
P⁴, all P are SP³ hybrid, it is obvious P- character is 75%.
B2H6, Boron is sp³ hybrid, that means two bridging H are in the plane with two boron. among another four H, two H are out of the plane, and two are below the plane. therefore only 4 atoms are in plane
PI³ +H²O = H³PO³(di-basic) + HI(mono basic)
|
|
Parvathy Vs upvoted • Feb 28, 2020 |
In fcc crystal lattice, edge length is 400 pm. Find the diameter of greatest sphere which can be fitted into the interstitial void without distortion of lattice: - a)158.54 pm
- b)117.08 pm
- c)129.54 pm
- d)414 pm
Correct answer is option 'B'. Can you explain this answer?
In fcc crystal lattice, edge length is 400 pm. Find the diameter of greatest sphere which can be fitted into the interstitial void without distortion of lattice:
a)
158.54 pm
b)
117.08 pm
c)
129.54 pm
d)
414 pm
|
|
Ameya Rane answered |
The species which by definition has zero standard molar enthalpy of formation at 298 K is: - a)Br2(g)
- b)Cl2(g)
- c)H2O(g)
- d)CH4(g)
Correct answer is option 'B'. Can you explain this answer?
The species which by definition has zero standard molar enthalpy of formation at 298 K is:
a)
Br2(g)
b)
Cl2(g)
c)
H2O(g)
d)
CH4(g)
|
|
Parvathy Vs answered • Feb 28, 2020 |
Standard molal enthalpy of formation is zero when an element is in its standard state ..hence its chlorine;
|
|
Parvathy Vs upvoted • Nov 23, 2019 |
Which of the following reaction lead to increase in ring size in the product:
- a)
- b)
- c)
- d)
Correct answer is option 'A,B,D'. Can you explain this answer?
Which of the following reaction lead to increase in ring size in the product:
a)
b)
c)
d)
|
Avishek Pan answered |
1st case mCPBA is electrophilic so here epoxidation does not for alpha-beta unsaturated ketone as electron density of the double bond is less so here BVO takes place and ring expansion occurs
2nd case Demjanov rearrangement occurs so ring expansion occurs
4th case CBr2 carbene generate so here ring expansion occurs
but for 3rd case the reagent should CF3COOOH so here epoxidation occ... more
2nd case Demjanov rearrangement occurs so ring expansion occurs
4th case CBr2 carbene generate so here ring expansion occurs
but for 3rd case the reagent should CF3COOOH so here epoxidation occ... more
|
|
Parvathy Vs asked • Nov 23, 2019 |
The major product formed in the following reaction is:
- a)CH3COCH3
- b)HOCH2CH2OH
- c)CH3CH2OH
- d)CH3OH
Correct answer is option 'B'. Can you explain this answer?
The major product formed in the following reaction is:
a)
CH3COCH3
b)
HOCH2CH2OH
c)
CH3CH2OH
d)
CH3OH
|
Pooja Choudhury answered |
HOCH2CH2OH, ethylene glycol will react with acid to give 5 membered ring with the carbonyl group.
|
|
Parvathy Vs asked • Nov 22, 2019 |
Salicylic acid is treated with bromine under two different conditions:
Predict the nature of [X] and [Y] in the following reactions
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Salicylic acid is treated with bromine under two different conditions:
Predict the nature of [X] and [Y] in the following reactions
a)
b)
c)
d)
|
Sagar Vairalkar answered |
D. The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. Given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth, calculate the number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate. Each D-glucose unit contributes
0.5 ... more
0.5 ... more
|
|
Parvathy Vs asked • Nov 22, 2019 |
What is the main product when 
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
What is the main product when
a)
b)
c)
d)
|
Saurabh Chavan answered |
Co2 and H2O liberate to form ketone.
Fehling’s solution can be used for distinguishing between:- a)CH3CHO and C6H5CHO
- b)CH3CHO and CH3COCH2OH
- c)Both (a) and (b)
- d)None of these.
Correct answer is option 'A'. Can you explain this answer?
Fehling’s solution can be used for distinguishing between:
a)
CH3CHO and C6H5CHO
b)
CH3CHO and CH3COCH2OH
c)
Both (a) and (b)
d)
None of these.
|
|
Parvathy Vs answered • Nov 22, 2019 |
Fehling can give reaction with aldehydes and only alpha hydroxy ketones..but they wont react with aromatic aldehydes and normal ketones
Compound A and B exhibit two singlets, each in their 1H NMR spectra. The expected chemical shifts are at 
:
- a)6.9 and 2.1 for A; 7.7 and 3.9 for B
- b)7.7 and 3.9 for A; 6.9 and 2.1 for B
- c)6.9 and 3.9 for A; 7.7 and 2.1 for B
- d)7.7 and 2.1 for A; 6.9 and 3.9 for B
Correct answer is option 'A'. Can you explain this answer?
Compound A and B exhibit two singlets, each in their 1H NMR spectra. The expected chemical shifts are at :
: a)
6.9 and 2.1 for A; 7.7 and 3.9 for B
b)
7.7 and 3.9 for A; 6.9 and 2.1 for B
c)
6.9 and 3.9 for A; 7.7 and 2.1 for B
d)
7.7 and 2.1 for A; 6.9 and 3.9 for B
|
|
Parvathy Vs answered • Nov 15, 2019 |
Option D is correct ..i guess bcz in a aromatic ring wen a EWG is present its value will be less than 7 and wen an ERG is present its value will be greater than 7. while for CH3 grp watched next to O will hav value around 3.9
The two fine–structure components of a nuclear magnet ic resonance transit ion are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant:- a)19.8 Hz
- b)0.066 Hz
- c)6.6 Hz
- d)date is insufficient
Correct answer is option 'A'. Can you explain this answer?
The two fine–structure components of a nuclear magnet ic resonance transit ion are observed at chemical shifts of 2.142 and 2.208 ppm in a 300 MHz NMR spectrometer. Calculate the coupling constant:
a)
19.8 Hz
b)
0.066 Hz
c)
6.6 Hz
d)
date is insufficient
|
|
Parvathy Vs answered • Nov 15, 2019 |
(2.208-2.142) ppm×300MHz= 19.8Hz
An intermediate in racemization of (R)-3-phenyl-2-butanone is:
... more
An intermediate in racemization of (R)-3-phenyl-2-butanone is:
|
|
Parvathy Vs answered • Nov 15, 2019 |
Is there any correction in b option ..is one of the OH grp CH3
The CORRECT order of stability for the following carbocations is
- a)I < III < IV < II
- b)III < II < IV < I
- c)II < IV < III < I
- d)IV < III < I < II
Correct answer is option 'C'. Can you explain this answer?
The CORRECT order of stability for the following carbocations is
a)
I < III < IV < II
b)
III < II < IV < I
c)
II < IV < III < I
d)
IV < III < I < II
|
|
Parvathy Vs answered • Nov 14, 2019 |
Carbocation at bridgehead position is less stable (brets rule)..so 2nd will be the last one..
now 1st one has stability both by phenyl and tertiary so it will be having first position..now amng the allylic carbocation and tertiary carbocation ..tertiary will be more stable
now 1st one has stability both by phenyl and tertiary so it will be having first position..now amng the allylic carbocation and tertiary carbocation ..tertiary will be more stable
Fetching relevant content for you
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup