Buddhadeb Chandra

Khera Pathi Road, Gwalior, India - 474002

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Buddhadeb Chandra

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Khera Pathi Road, Gwalior, India- 474002.

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5.0
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I am a qualified and experienced teacher and tutor with over 3 years experience in teaching Chemistry of CBSE and State Board level. Passionate to clear the basic concept of the students in chemistry especially to overcome the fear about different chemical reaction of Organic Chemistry. Now I am continuing my PhD in Chemical Science from DRDE, Gwalior.

Languages Spoken

Bengali Mother Tongue (Native)

English Proficient

Hindi Proficient

Education

Nagpur University 2014

Master of Science (M.Sc.)

Address

Khera Pathi Road, Gwalior, India - 474002

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BSc Tuition Overview

BSc Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in BSc Tuition

3

BSc Chemistry Subjects

Organic Functional Groups, Spectroscopy, Hydrocarbons And Stereochemistry, Polymer Chemistry, Organic Qualitative & Quantitative Analysis, Concepts In Inorganic Chemistry

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions, Group Classes

Taught in School or College

No

BSc Branch

BSc Chemistry

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Answers by Buddhadeb Chandra (4)

Answered on 05/11/2018 CBSE/Class 11/Science/Chemistry/Unit 4-Chemical Bonding and Molecular Structure CBSE/Class 11/Science/Chemistry

Is there any change in the hybridisation ofB and N atoms as a result of the following reaction ? BF3... ...more
Is there any change in the hybridisation ofB and N atoms as a result of the following reaction ?  BF3 + NH3 ——-> F3 B.NH3

In the given reaction the unshare pair (or loan pair) of electron present in sp3 hybridized non-bonding molecular orbital of NH3 donate to the vacant 2p orbital of BH3 to produce a co-ordination bond. During this bond formation hybridization of each of the participating atom remain unchange. i,e; N remain... ...more

In the given reaction the unshare pair (or loan pair) of electron present in sp3 hybridized non-bonding molecular orbital of NH3 donate to the vacant 2p orbital of BH3 to produce a co-ordination bond. During this bond formation hybridization of each of the participating atom remain unchange. i,e; N remain sp3 hybridized in NH3 and B remain sp2 hybrized in BF3.

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Answered on 28/10/2018 Hydrocarbons CBSE/Class 11/Science/Chemistry

Write the structures and names of products obtained in the reactions of sodium with a mixture of l-iodo-2-methylpropane... ...more
Write the structures and names of products obtained in the reactions of sodium with a mixture of l-iodo-2-methylpropane and 2-iodopropane.

The reaction of mixture of alkyl halide in presence sodium metal is known as Wurtz reaction. Since here the given alkyl iodides are unsymmetrical with each other, so there are probability of formation of total three types of higher alkene. 1) when they condence with each other formed 2,4-dimethyl pentane,... ...more

The reaction of mixture of alkyl halide in presence sodium metal is known as Wurtz reaction. Since here the given alkyl iodides are unsymmetrical with each other, so there are probability of formation of total three types of higher alkene. 1) when they condence with each other formed 2,4-dimethyl pentane, 2) when they react own self produce a) 2,5-dimethyl hexane (for former one) and b) 2,3-dimethyl butane (for later one).

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Answered on 26/10/2018 CBSE/Class 11/Science/Chemistry/Unit 14-Environmental Chemistry CBSE/Class 11/Science/Chemistry

Photochemical smog occurs in warm, dry and sunny climate. One of the following is not amongst the components... ...more
Photochemical smog occurs in warm, dry and sunny climate. One of the following is not amongst the components of photochemical smog, identify it.
(a) N02
(b) 03
(c) S02                                                       

(d) Unsaturated hydrocarbon

During warm, dry and sunny climate i,e; when the intensity of solar radiation is very high, oxygen presence in the air photo chemically react with NO2 and hydrocarbon evolved from automobile engines produces corresponding oxidized species, ozone, peroxyacyl nitrate, aldehyde, ketone etc. These gaseous... ...more

During warm, dry and sunny climate i,e; when the intensity of solar radiation is very high, oxygen presence in the air photo chemically react with NO2 and hydrocarbon evolved from automobile engines produces corresponding oxidized species, ozone, peroxyacyl nitrate, aldehyde, ketone etc. These gaseous molecule and the  particulates mix together in air to form photochemical smog. For SO2 given condition is not appropriate for photochemical reaction, it require humid weather. So the answer should be (C) SO2.

Answers 2 Comments
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Answered on 25/10/2018 Hydrocarbons CBSE/Class 11/Science/Chemistry

Arrange the following carbanions in order of their decreasing stability. (A) H3C-C ≡ C– ... ...more
Arrange the following carbanions in order of their decreasing stability.
(A) H3C-C ≡ C                                        
(B) H-C ≡ C
(C) H3C – CH2

(a) A>B>C
(b) B>A>C
(c) C>B>A
(d) C>A>B

Option (B) is the right answer. Explanation: Before solving such types of problem we should have the concept of conjugate acid-base theory. Here in question given three conjugate base of their acid 1) CH3-CCH, 2) HCCH, 3) H3C-CH2- respectively. We know that unstable conjugate base provides stable acids... ...more

Option (B) is the right answer.

Explanation: Before solving such types of problem we should have the concept of conjugate acid-base theory. Here in question given three conjugate base of their acid 1) CH3-CCH, 2) HCCH, 3) H3C-CH2- respectively. We know that unstable conjugate base provides stable acids and vice versa. If we order the  decreasing stability of 1), 2) and 3) then we get 1>2>3, one get one extra substituent than 2 and 3 is a charge entity. According to this 2 provides more stable conjugate base than 1 i,e; B>A. But for 3 it contain one negative charge even after absorption of  one proton. Therefor here conjugate acid base theory is not applicable. Hybridization play an important role in stabilization of particular entity. For A) and B) negative charge on sp hybridised molecular orbital (MO) whereas for C) it is on sp3 hybridized. Again sp MO provides extra stability to nagative charge because of its spherical size than dumble size sp3 orbital. Due to spherical size it remain more closely to nucleous and stabilized by delocalizing the charge around the nucleous. Due to dumble size sp3 MO remain far away from the nucleous. Thats why probability of delocalization is less. For this reason C) is highly unstable.

 

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Buddhadeb ChandraDirections

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BSc Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in BSc Tuition

3

BSc Chemistry Subjects

Organic Functional Groups, Spectroscopy, Hydrocarbons And Stereochemistry, Polymer Chemistry, Organic Qualitative & Quantitative Analysis, Concepts In Inorganic Chemistry

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions, Group Classes

Taught in School or College

No

BSc Branch

BSc Chemistry

Class 12 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 12 Tuition

5

Board

CBSE, ISC/ICSE, State

ISC/ICSE Subjects taught

Chemistry

CBSE Subjects taught

Chemistry

Taught in School or College

No

State Syllabus Subjects taught

Chemistry

Teaching Experience in detail in Class 12 Tuition

I have 5 years teaching experience on all branch of chemistry like physical, organic and inorganic chemistry. During taking classes I observed that students were fear about solving mathematical problem, especially in physical chemistry. Here my responsibility was to provide proper technique how they can easily understand the problem and save themselves from the fear of solving mathematical problems. Another students fears were on different types of organic reaction and their mechanism. In this case at first I clear their basic concept about reaction and how the reaction proceeds following particular pathway. During my B.Sc. I provide chemistry tuition for three years, state board level in West Bengal. During my M.Sc. from Nagpur university I provide chemistry tuition for two years, CBSE level. Now I pursuing my PhD from DRDE, Gwalior and looking some class 11 & 12 level students for providing sound chemistry tuition.

Class 11 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 11 Tuition

5

Board

CBSE, ISC/ICSE, State

ISC/ICSE Subjects taught

Chemistry

CBSE Subjects taught

Chemistry

Taught in School or College

No

State Syllabus Subjects taught

Chemistry

Teaching Experience in detail in Class 11 Tuition

The interesting experience about my class 11 Tuition was my first class on structure of atom. Here starting from Dalton's atomic theory via Bohr's atomic model to quantum mechanical model of atom. Most of the students are confused about all these things and started fearing the chapter. But I try to explain this chapter in front of the students just like a story. So that they can easily grab the whole matter. Another chapter is Chemical Thermodynamics, there are limitless fear in the students mind about this chapter. This problem was being solved with the help of diagramtical explanation and provide different techniques how to remember the related terminology.

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Answers by Buddhadeb Chandra (4)

Answered on 05/11/2018 CBSE/Class 11/Science/Chemistry/Unit 4-Chemical Bonding and Molecular Structure CBSE/Class 11/Science/Chemistry

Is there any change in the hybridisation ofB and N atoms as a result of the following reaction ? BF3... ...more
Is there any change in the hybridisation ofB and N atoms as a result of the following reaction ?  BF3 + NH3 ——-> F3 B.NH3

In the given reaction the unshare pair (or loan pair) of electron present in sp3 hybridized non-bonding molecular orbital of NH3 donate to the vacant 2p orbital of BH3 to produce a co-ordination bond. During this bond formation hybridization of each of the participating atom remain unchange. i,e; N remain... ...more

In the given reaction the unshare pair (or loan pair) of electron present in sp3 hybridized non-bonding molecular orbital of NH3 donate to the vacant 2p orbital of BH3 to produce a co-ordination bond. During this bond formation hybridization of each of the participating atom remain unchange. i,e; N remain sp3 hybridized in NH3 and B remain sp2 hybrized in BF3.

Answers 1 Comments
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Answered on 28/10/2018 Hydrocarbons CBSE/Class 11/Science/Chemistry

Write the structures and names of products obtained in the reactions of sodium with a mixture of l-iodo-2-methylpropane... ...more
Write the structures and names of products obtained in the reactions of sodium with a mixture of l-iodo-2-methylpropane and 2-iodopropane.

The reaction of mixture of alkyl halide in presence sodium metal is known as Wurtz reaction. Since here the given alkyl iodides are unsymmetrical with each other, so there are probability of formation of total three types of higher alkene. 1) when they condence with each other formed 2,4-dimethyl pentane,... ...more

The reaction of mixture of alkyl halide in presence sodium metal is known as Wurtz reaction. Since here the given alkyl iodides are unsymmetrical with each other, so there are probability of formation of total three types of higher alkene. 1) when they condence with each other formed 2,4-dimethyl pentane, 2) when they react own self produce a) 2,5-dimethyl hexane (for former one) and b) 2,3-dimethyl butane (for later one).

Answers 1 Comments
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Answered on 26/10/2018 CBSE/Class 11/Science/Chemistry/Unit 14-Environmental Chemistry CBSE/Class 11/Science/Chemistry

Photochemical smog occurs in warm, dry and sunny climate. One of the following is not amongst the components... ...more
Photochemical smog occurs in warm, dry and sunny climate. One of the following is not amongst the components of photochemical smog, identify it.
(a) N02
(b) 03
(c) S02                                                       

(d) Unsaturated hydrocarbon

During warm, dry and sunny climate i,e; when the intensity of solar radiation is very high, oxygen presence in the air photo chemically react with NO2 and hydrocarbon evolved from automobile engines produces corresponding oxidized species, ozone, peroxyacyl nitrate, aldehyde, ketone etc. These gaseous... ...more

During warm, dry and sunny climate i,e; when the intensity of solar radiation is very high, oxygen presence in the air photo chemically react with NO2 and hydrocarbon evolved from automobile engines produces corresponding oxidized species, ozone, peroxyacyl nitrate, aldehyde, ketone etc. These gaseous molecule and the  particulates mix together in air to form photochemical smog. For SO2 given condition is not appropriate for photochemical reaction, it require humid weather. So the answer should be (C) SO2.

Answers 2 Comments
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Answered on 25/10/2018 Hydrocarbons CBSE/Class 11/Science/Chemistry

Arrange the following carbanions in order of their decreasing stability. (A) H3C-C ≡ C– ... ...more
Arrange the following carbanions in order of their decreasing stability.
(A) H3C-C ≡ C                                        
(B) H-C ≡ C
(C) H3C – CH2

(a) A>B>C
(b) B>A>C
(c) C>B>A
(d) C>A>B

Option (B) is the right answer. Explanation: Before solving such types of problem we should have the concept of conjugate acid-base theory. Here in question given three conjugate base of their acid 1) CH3-CCH, 2) HCCH, 3) H3C-CH2- respectively. We know that unstable conjugate base provides stable acids... ...more

Option (B) is the right answer.

Explanation: Before solving such types of problem we should have the concept of conjugate acid-base theory. Here in question given three conjugate base of their acid 1) CH3-CCH, 2) HCCH, 3) H3C-CH2- respectively. We know that unstable conjugate base provides stable acids and vice versa. If we order the  decreasing stability of 1), 2) and 3) then we get 1>2>3, one get one extra substituent than 2 and 3 is a charge entity. According to this 2 provides more stable conjugate base than 1 i,e; B>A. But for 3 it contain one negative charge even after absorption of  one proton. Therefor here conjugate acid base theory is not applicable. Hybridization play an important role in stabilization of particular entity. For A) and B) negative charge on sp hybridised molecular orbital (MO) whereas for C) it is on sp3 hybridized. Again sp MO provides extra stability to nagative charge because of its spherical size than dumble size sp3 orbital. Due to spherical size it remain more closely to nucleous and stabilized by delocalizing the charge around the nucleous. Due to dumble size sp3 MO remain far away from the nucleous. Thats why probability of delocalization is less. For this reason C) is highly unstable.

 

Answers 2 Comments
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Buddhadeb Chandra describes himself as Tutor. He conducts classes in BSc Tuition, Class 11 Tuition and Class 12 Tuition. Buddhadeb is located in Khera Pathi Road, Gwalior. Buddhadeb takes at students Home. He has 5 years of teaching experience . Buddhadeb has completed Master of Science (M.Sc.) from Nagpur University in 2014. HeĀ is well versed in English, Bengali and Hindi.

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