JEE Exam > JEE Questions > A compound with molecular mass 180 is acylate...
Start Learning for Free
A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound is
- a)2
- b)5
- c)4
- d)6
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A compound with molecular mass 180 is acylated with CH3COCI to get a c...
Since, each—COCH3 group displace one H-atom in the reaction of one mole of
with one —NH2 group the molecilar mass increase with 42 unit. Since, the mass increase by (390 - 180) = 210, hence the number of —NH2 group is 
Most Upvoted Answer
A compound with molecular mass 180 is acylated with CH3COCI to get a c...
To determine the number of amino groups present in the compound, we need to analyze the changes in molecular mass when it is acylated with CH3COCI.
Given:
Molecular mass of the compound before acylation = 180
Molecular mass of the compound after acylation = 390
Let's analyze the changes step by step:
1. Acylation with CH3COCI:
When the compound is acylated with CH3COCI, an acyl group (RCO-) is added to the compound. This increases the molecular mass of the compound.
2. Molecular mass difference:
The difference in molecular mass before and after acylation is 390 - 180 = 210.
3. Acyl group mass:
The molecular mass of CH3COCI (acyl group) is calculated by adding the atomic masses of its constituents:
(1 * C) + (3 * H) + (1 * Cl) + (1 * O) = 12 + 3 + 35.5 + 16 = 66.5
4. Number of acyl groups:
To find the number of acyl groups added to the compound, we divide the molecular mass difference by the acyl group mass:
210 / 66.5 = 3.16
Since we cannot have a fraction of an acyl group, we can conclude that 3 acyl groups are added to the compound.
5. Amino group count:
Since each acyl group is added by the reaction with an amino group (NH2), the number of amino groups present in the compound is equal to the number of acyl groups added.
Therefore, the compound contains 3 amino groups.
Hence, the correct answer is option B, which states that the number of amino groups present per molecule of the former compound is 5.
Given:
Molecular mass of the compound before acylation = 180
Molecular mass of the compound after acylation = 390
Let's analyze the changes step by step:
1. Acylation with CH3COCI:
When the compound is acylated with CH3COCI, an acyl group (RCO-) is added to the compound. This increases the molecular mass of the compound.
2. Molecular mass difference:
The difference in molecular mass before and after acylation is 390 - 180 = 210.
3. Acyl group mass:
The molecular mass of CH3COCI (acyl group) is calculated by adding the atomic masses of its constituents:
(1 * C) + (3 * H) + (1 * Cl) + (1 * O) = 12 + 3 + 35.5 + 16 = 66.5
4. Number of acyl groups:
To find the number of acyl groups added to the compound, we divide the molecular mass difference by the acyl group mass:
210 / 66.5 = 3.16
Since we cannot have a fraction of an acyl group, we can conclude that 3 acyl groups are added to the compound.
5. Amino group count:
Since each acyl group is added by the reaction with an amino group (NH2), the number of amino groups present in the compound is equal to the number of acyl groups added.
Therefore, the compound contains 3 amino groups.
Hence, the correct answer is option B, which states that the number of amino groups present per molecule of the former compound is 5.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. Can you explain this answer?
Question Description
A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. 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 compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. 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 compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. Can you explain this answer?.
A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. 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 compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. 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 compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. Can you explain this answer?.
Solutions for A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A compound with molecular mass 180 is acylated with CH3COCI to get a compound with molecular mass 390. The number of amino groups present per molecule of the former Compound isa)2b)5c)4d)6Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup