Question

For compound of molecular formula \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br}\) if \(p, s\) and \(t\) represent possible structural isomers classiteev primary, secondary and tertiary bromides respectively, then correct relation is/are (A) \(p+s+t=8\) (B) \(p=s+t\) (C) \(t=\frac{-\mathrm{s}+\sqrt{\mathrm{s}^{2}+4 \mathrm{p}}}{2}\) (D) \(s^{2}=2 p+t\)

Short answer

The correct relation among primary, secondary, and tertiary bromides for the given compound is \(p = s + t\), where \(p = 5\), \(s = 4\), and \(t = 1\).

Step-by-step solution

Determine all possible structural isomers of the C5H11Br compound.

To generate all possible isomers for the given molecular formula, we must consider that a carbon atom will have 4 bonds and hydrogen atoms will have 1 bond, and that a single bromide atom will replace one of these hydrogens in each case. We start with a base C5H12 alkane structure and work our way through different combinations: 1. 1-Bromopentane (CH3CH2CH2CH2CH2Br) 2. 2-Bromopentane (CH3CH2CH2CH(Br)CH3) 3. 3-Bromopentane (CH3CH2CH(Br)CH2CH3) 4. 1,1-Dibromo-butane (CH3CH2CH(Br)CH2Br) 5. 1,2-Dibromo-butane (CH3CH(Br)CH(Br)CH3) 6. 1,3-Dibromo-butane (CH3CH(Br)CH2CH2Br) 7. 1-Bromo-2-methylbutane (CH3CH(Br)CH(CH3)CH3) 8. 2-Bromo-2-methylbutane (CH3C(Br)H2CH(CH3)CH3) Step 2: Classify the isomers

Classify the isomers into primary, secondary, and tertiary bromides.

We will classify each isomer by the category in which its bromide atom belongs: 1. 1-Bromopentane - Primary 2. 2-Bromopentane - Secondary 3. 3-Bromopentane - Secondary 4. 1,1-Dibromo-butane - Primary + Primary 5. 1,2-Dibromo-butane - Primary + Secondary 6. 1,3-Dibromo-butane - Primary + Primary 7. 1-Bromo-2-methylbutane - Secondary 8. 2-Bromo-2-methylbutane - Tertiary Step 3: Count the isomers and match the relation

Count the number of primary, secondary, and tertiary bromides and match the given relations.

After classifying, we have: - p (Primary bromides): 5 - s (Secondary bromides): 4 - t (Tertiary bromides): 1 Now let's test the given relations: (A) p + s + t = 8 --> 5 + 4 + 1 = 10 (incorrect) (B) p = s + t --> 5 = 4 + 1 (correct) (C) t = (-s + sqrt(s^2 + 4p))/2 --> 1 = (-4 + sqrt(4^2 + 4*5))/2 (incorrect) (D) s² = 2p + t --> 4² = 2*5 + 1 (incorrect) Only the relation (B) is correct.

Key concepts

Bromides Classification

Understanding the classification of bromides is essential for students studying organic chemistry, particularly when identifying structural isomers. Bromides are compounds containing a bromine atom, and in organic chemistry, they are categorized based on the carbon to which the bromine atom is attached.

Primary bromides have the bromine atom connected to a primary carbon, which is a carbon atom bonded to only one other carbon. Secondary bromides have the bromine atom attached to a secondary carbon, which is bonded to two other carbons. Finally, tertiary bromides are those in which the bromine atom is connected to a tertiary carbon, bonded to three other carbon atoms.

These classifications are crucial as they influence the reactivity and properties of the molecules, which is a topic of interest for examinations such as JEE Main and Advanced. In practical application, for a compound with the molecular formula C5H11Br, we explore all structural possibilities and attach the bromine atom to different carbons to deduce the primary, secondary, and tertiary bromides.

Isomers Counting

Isomers are compounds with the same molecular formula but different structural arrangements. Isomers counting is an important skill in organic chemistry, as it requires comprehending the complexity of carbon bonding and the possible structural variations within a molecule.

Approach to Counting Isomers

In the given example of C5H11Br, the counting of isomers begins with identifying the base alkane, pentane in this case, and then systematically replacing the hydrogen atoms with a bromine atom to generate different structures. Each unique structure is a different isomer. The number of primary, secondary, and tertiary bromides corresponds to the number of isomers categorized as p, s, and t respectively.

For competitive exams like JEE, students must be adept at counting isomers quickly and accurately, as it is a commonly tested concept. The attention to detail in recognizing mirror images and not double-counting enantiomers or mistakenly treating constitutional isomers as identical is crucial in mastering this topic.

JEE Main and Advanced Organic Chemistry

The Joint Entrance Examination (JEE) Main and Advanced tests are critical entry points for students seeking admission to premier engineering institutes in India. Organic chemistry is a significant part of the JEE syllabus and covers a vast array of topics including isomerism, reaction mechanisms, and the classification of organic compounds.

Questions on structural isomers, such as in the exercise involving C5H11Br, test students' understanding of not only the concept of isomerism but also their ability to apply this knowledge practically. Success in this area requires a strong foundation in organic chemistry concepts coupled with practiced problem-solving skills.

Students are encouraged to break down complex molecules into simpler structures and use logical reasoning to determine the number of potential isomers. This analytical approach is important both for the JEE and for a successful career in chemistry or related fields. Preparing for these competitive exams entails rigorous practice and a comprehensive grasp of organic chemistry principles, emphasizing the importance of exercises like the one discussed here.