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A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).
Correct answer is '6'. Can you explain this answer?
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Most Upvoted Answer
A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 ...
Molecular weight of B
Thus the increase in weight due to addition of Br2 is
= 374.22 − 54 = 320.22
Hence Br atom in the compound B is =320.22/80 = 4
Molecular weight of C is
So no. of H atom increased is 4 so a must be alkyne hence possible structure of the compound is:
Molecular weight of alkyl halide is 109 so it is ethyl bromide.
D. CH3−CH2−C ≡ C−CH2CH3
E.
So molecular weight of E is 114.
So x + y + z = 6
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Community Answer
A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 ...
Given:
- Molecular weight of hydrocarbon A = 54g
- Molecular weight of compound B = 539% more than A = 54 + (539/100)*54 = 330.06g
- Molecular weight of compound C after hydrogenation of A = 7.4% more than A = 54 + (7.4/100)*54 = 58.316g
- Molecular weight of alkyl bromide = 109g
- Molecular weight of hydrocarbon D formed after reaction of A with alkyl bromide = Molecular weight of A + Molecular weight of alkyl bromide = 54 + 109 = 163g
- Molecular weight of diketone E formed after reductive ozonolysis of D = xyz
Solution:
1. Finding the molecular formula of A:
- Since the molecular weight of A is given as 54g, we can assume it to be the empirical formula of A.
- The empirical formula of a hydrocarbon can be determined by dividing the molecular weight by the empirical formula weight.
- The empirical formula weight can be calculated using the molar mass of carbon and hydrogen (12g/mol and 1g/mol respectively).
- Empirical formula weight = (12 * number of carbon atoms) + (1 * number of hydrogen atoms)
- Let's assume the empirical formula of A as CxHy
- Empirical formula weight = (12x + 1y)g/mol
- Given that the molecular weight of A is 54g, we can set up the equation:
54 = (12x + 1y)g/mol
- Since we know that hydrocarbons only contain carbon and hydrogen, we can assume that the molar mass of A is mainly due to carbon.
- This implies that y = 0, as there are no hydrogen atoms present in the molecular formula of A.
- Substituting y = 0 in the equation, we get:
54 = 12x
- Solving for x, we get:
x = 54/12 = 4.5
- Since the number of carbon atoms cannot be a fraction, we round it off to the nearest whole number:
x = 5
- Therefore, the empirical formula of A is C5H0, which can be simplified to C5.
2. Finding the molecular formula of B:
- The molecular weight of B is given as 539% more than A.
- Molecular weight of B = Molecular weight of A + (539/100) * Molecular weight of A
- Substituting the values, we get:
Molecular weight of B = 54 + (539/100)*54 = 330.06g
- Since the empirical formula of A is C5, we can assume the molecular formula of B as C5n.
- The molecular formula weight can be calculated using the molar mass of carbon (12g/mol).
- Molecular formula weight = 12 * number of carbon atoms = 12 * 5n
- Setting up the equation:
330.06 = 12 * 5n
- Solving for n, we get:
n = 330.06 / (12 * 5) ≈ 5.5
- Since the number of carbon atoms cannot be a
- Molecular weight of hydrocarbon A = 54g
- Molecular weight of compound B = 539% more than A = 54 + (539/100)*54 = 330.06g
- Molecular weight of compound C after hydrogenation of A = 7.4% more than A = 54 + (7.4/100)*54 = 58.316g
- Molecular weight of alkyl bromide = 109g
- Molecular weight of hydrocarbon D formed after reaction of A with alkyl bromide = Molecular weight of A + Molecular weight of alkyl bromide = 54 + 109 = 163g
- Molecular weight of diketone E formed after reductive ozonolysis of D = xyz
Solution:
1. Finding the molecular formula of A:
- Since the molecular weight of A is given as 54g, we can assume it to be the empirical formula of A.
- The empirical formula of a hydrocarbon can be determined by dividing the molecular weight by the empirical formula weight.
- The empirical formula weight can be calculated using the molar mass of carbon and hydrogen (12g/mol and 1g/mol respectively).
- Empirical formula weight = (12 * number of carbon atoms) + (1 * number of hydrogen atoms)
- Let's assume the empirical formula of A as CxHy
- Empirical formula weight = (12x + 1y)g/mol
- Given that the molecular weight of A is 54g, we can set up the equation:
54 = (12x + 1y)g/mol
- Since we know that hydrocarbons only contain carbon and hydrogen, we can assume that the molar mass of A is mainly due to carbon.
- This implies that y = 0, as there are no hydrogen atoms present in the molecular formula of A.
- Substituting y = 0 in the equation, we get:
54 = 12x
- Solving for x, we get:
x = 54/12 = 4.5
- Since the number of carbon atoms cannot be a fraction, we round it off to the nearest whole number:
x = 5
- Therefore, the empirical formula of A is C5H0, which can be simplified to C5.
2. Finding the molecular formula of B:
- The molecular weight of B is given as 539% more than A.
- Molecular weight of B = Molecular weight of A + (539/100) * Molecular weight of A
- Substituting the values, we get:
Molecular weight of B = 54 + (539/100)*54 = 330.06g
- Since the empirical formula of A is C5, we can assume the molecular formula of B as C5n.
- The molecular formula weight can be calculated using the molar mass of carbon (12g/mol).
- Molecular formula weight = 12 * number of carbon atoms = 12 * 5n
- Setting up the equation:
330.06 = 12 * 5n
- Solving for n, we get:
n = 330.06 / (12 * 5) ≈ 5.5
- Since the number of carbon atoms cannot be a
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A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer?
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A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer?.
A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer?.
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A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer?, a detailed solution for A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer? has been provided alongside types of A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A hydrocarbon A of molecular weight 54g reacts with an excess of Br2 in CCl4 to give a compound B whose molecular weight is 539% more than that of A. however on catalytic hydrogenation with excess of H2 A forms C whose molecular weight is only 7.4% more than that of A. A reacts with an alkyl bromide of molecular weight 109g in the presence of NaNH2 to give another hydrocarbon D, which on reductive ozonolysis, yields diketone E, if the molecular weight of E is xyz then find the value of (x+y+z).Correct answer is '6'. Can you explain this answer? tests, examples and also practice JEE tests.
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