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Chapter 2: Problem 3

Write the structural formula of a hydrocarbon of molecular formula (a) \(\mathrm{C}_{6} \mathrm{H}_{12}\) having only secondary hydrogens (b) \(\mathrm{C}_{5} \mathrm{H}_{12}\) having only primary hydrogens (c) \(\mathrm{C}_{6} \mathrm{H}_{14}\) having only primary and tertiary hydrogens

Short Answer

Expert verified
Question: Draw a hydrocarbon structural formula that satisfies the following conditions: A structure for C6H12 having only secondary hydrogens, A structure for C5H12 having only primary hydrogens, and A structure for C6H14 having only primary and tertiary hydrogens. Answer: The hydrocarbon structures that satisfy the given conditions are as follows: 1. C6H12 with only secondary hydrogens: H H H | | | H-C-C-C-H | | | H H H 2. C5H12 with only primary hydrogens: H H H H H | | | | | H-C-C-C-C-H 3. C6H14 with only primary and tertiary hydrogens: H H H | | | H-C-C-C-H | | H-C-H

Step by step solution

01

Understanding primary, secondary, and tertiary hydrogens

Primary hydrogens are attached to a carbon atom that is bonded to only one other carbon atom. Secondary hydrogens are attached to a carbon atom that is bonded to two other carbon atoms. Tertiary hydrogens are attached to a carbon atom that is bonded to three other carbon atoms.
02

Drawing a structure for C6H12 having only secondary hydrogens

To satisfy the condition of having only secondary hydrogens, we can construct a 6-carbon cyclic structure. In a cyclic structure, every carbon is bonded to two other carbons. A hexagonal cyclic structure will have the formula C6H12, as shown below: ``` H H H | | | H-C-C-C-H | | | H H H ```
03

Drawing a structure for C5H12 having only primary hydrogens

To ensure that C5H12 has only primary hydrogens, we should have a straight-chain structure with no branching. In this structure, there are two end carbon atoms, each with two primary hydrogens; and three other carbon atoms in the middle with one primary hydrogen each. ``` H H H H H | | | | | H-C-C-C-C-H ```
04

Drawing a structure for C6H14 having only primary and tertiary hydrogens

To construct a structure for C6H14 with only primary and tertiary hydrogens, we need a core carbon atom with three other carbons attached to it. Three primary hydrogens are then attached to each of the three outer carbon atoms, and the central carbon is connected to a hydrogen. This gives us a total of 6 carbon atoms and 14 hydrogen atoms, meeting the molecular formula requirement: ``` H H H | | | H-C-C-C-H | | H-C-H

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Most popular questions from this chapter

Directions: Each question contains Statement-1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement-1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True Statement 1 Oxymercuration reduction of an alkene gives an alcohol in which the -OH group is added to a more branched carbon of the double bond, while hydroboration- oxidation of an alkene gives an alcohol in which the -OH group is added to a less branched carbon of the alkene. and Statement 2 Hydration of an alkene in the presence of an acid is one of the standard methods of preparation of an alcohol form alkenes.

Alkynes are less reactive than alkenes towards electrophilic reagents. But the alkyne chemistry is useful for organic synthesis, in the form of hydration of the triple bond, formation of metalacetylides, selective reduction of alkynes and few other related reactions. Acetylene and HCHO react together in the presence of \(\mathrm{NaOCH}_{3}\) to give (a) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{OH}\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{HOH}_{2} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3}-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CH}_{3}\)

Which of the following reactions will result in 2,2-dibromopropane? (a) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}+2 \mathrm{HBr} \rightarrow\) (b) \(\mathrm{CH}_{3}-\mathrm{CBr}=\mathrm{CH}_{2}+\mathrm{HBr} \rightarrow\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{Br}_{2} \rightarrow\) (d) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHBr}+\mathrm{HBr} \rightarrow\)

The best method of preparation of an alkane with odd number of carbons from two alkylhalides is (a) Wurtz reaction (b) Kolbe's electrolysis (c) Corey-House synthesis (d) Ullmann reaction

Alkynes are less reactive than alkenes towards electrophilic reagents. But the alkyne chemistry is useful for organic synthesis, in the form of hydration of the triple bond, formation of metalacetylides, selective reduction of alkynes and few other related reactions.
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