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How would you distinguish between the components in each of the following pairs using chemical methods (preferably test-tube reactions)? a. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\) and \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}\) and \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C} \equiv \mathrm{CC}_{6} \mathrm{H}_{5}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C}_{6} \mathrm{H}_{5}\)

Short Answer

Expert verified
Use Tollens' test for part a, bromine water for part b, and bromine test for part c.

Step by step solution

01

Identifying Alkyne and Alkene

Part (a) involves distinguishing between \( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{C} \equiv \mathrm{CH} \) (1-butyne) and \( \mathrm{CH}_{3}\mathrm{C} \equiv \mathrm{CCH}_3 \) (2-butyne). Use the Tollens' test with silver nitrate in aqueous ammonia. 1-butyne, a terminal alkyne, reacts forming a silver acetylide precipitate, while 2-butyne does not.
02

Differentiating Alkynes from Dienes

For part (b), compare \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH} \) and \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} \) (1,3-butadiene). Use bromine water test; 1,3-butadiene decolorizes bromine water quickly due to multiple double bonds while 1-butyne only shows a slower reaction if it reacts at all.
03

Aromatic vs Non-aromatic Saturated Compound

In part (c), distinguish \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C} \equiv \mathrm{CC}_{6} \mathrm{H}_{5} \) (diphenylacetylene) from \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\mathrm{CH}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \) (diphenylethane) using a bromine test; diphenylacetylene does not react, while diphenylethane will decolorize bromine.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alkene and Alkyne Differentiation
Alkenes and alkynes are both unsaturated hydrocarbons, but they differ in types of bonds. Alkenes have at least one carbon-carbon double bond, whereas alkynes have at least one carbon-carbon triple bond. This difference is crucial in identifying them through chemical tests. One common method to differentiate them is using the Bromine Test.
  • Alkenes react readily with bromine water, resulting in a color change from brown to clear due to the addition reaction.
  • Alkynes also react with bromine but generally at a slower rate, depending on the specific structure of the alkyne. Terminal alkynes (with a free hydrogen) often react more slowly or not at all in simple bromine water tests.
Tollens' Test
Tollens' Test is a classic chemical test used to identify terminal alkynes. It involves using Tollens' reagent, which is silver nitrate in aqueous ammonia. This test capitalizes on the ability of terminal alkynes to form silver acetylide precipitates. This reaction occurs due to the acidic hydrogen at the terminal carbon, which is not available in internal alkynes like 2-butyne.
  • When Tollens' reagent is added to a terminal alkyne, look for the formation of a gray or black silver precipitate.
  • Internal alkynes, which lack the terminal hydrogen, will not produce a precipitate, giving a clear distinction between the two types.
Bromine Test
The Bromine Test is excellent for identifying unsaturation in hydrocarbons. This test detects alkenes and alkynes by their ability to decolorize bromine, a reddish-brown solution. Here's how it works:
  • Alkenes react with bromine water quickly, resulting in a clear solution. This occurs due to the addition of bromine across the double bond.
  • Alkynes react as well, but their rate can vary. Terminal alkynes might show no immediate color change, as in the case of 1-butyne.
  • Aromatic compounds usually do not react with bromine water under mild conditions, as their delocalized electrons provide stability, preventing the reaction.
Aromatic Compounds
Aromatic compounds are characterized by their ring-like structure and delocalized electrons, such as benzene rings. This stability often makes them less reactive than their non-aromatic counterparts in certain chemical tests.
- Many aromatic compounds will not decolorize bromine water, unlike other unsaturated compounds.
- For example, diphenylacetylene, despite being unsaturated, behaves like an aromatic compound due to its benzene rings. It does not easily react with bromine water.
Understanding this distinct reactivity helps distinguish aromatic compounds like diphenylacetylene from saturated or non-aromatic compounds like diphenylethane, which will react with bromine.

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Most popular questions from this chapter

a. Would you expect a carbon-nitrogen triple bond to be hydrogenated more, or less, easily than a carbon-carbon triple bond? b. Why is it difficult to hydrogenate a tetrasubstituted alkene such as 2,3 -dimethyl-2-butene?

Starting with cyclohexene, show how you could prepare each of the following compounds: a. the epoxide of cyclohexene b. cis-cyclohexane-1,2-diol c. trans-cyclohexane- 1,2 -diol

Draw structures for the products expected from the following reactions. Show configurations where significant. a. b. cis-2-butene \(\frac{\mathrm{D}_{2}, \mathrm{Pt}}{25^{\circ}}\) c. \(\mathrm{CH}_{2}=\mathrm{CHCOCH}_{3} \stackrel{\mathrm{H}_{2}, \mathrm{Pt}}{{ }_{25^{\circ}}}\) d. 1 -penten-3-yne \(\stackrel{\mathrm{H}_{2}, \mathrm{Pd}-\mathrm{Pb}}{\longrightarrow}\) e. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C} \equiv \mathrm{CC}_{6} \mathrm{H}_{5} \stackrel{\mathrm{H}_{2}, \mathrm{Pd}-\mathrm{Pb}}{\longrightarrow}\) f. 1,3 -dimethylcyclopentene \(\stackrel{\mathrm{H}_{2}, \mathrm{Pt}}{25^{\circ}}\)

A hydrocarbon of formula \(\mathrm{C}_{11} \mathrm{H}_{18}\) on reaction with ozone in dichloromethane gave, after the addition of water and finely divided zinc, three products in equimolar amounts that were identified as 2 -butanone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\right)\), methanal \(\left(\mathrm{CH}_{2} \mathrm{O}\right)\), and cyclohexane-1,4-dione \(\mathrm{Draw}\) the structure of the hydrocarbon \(\mathrm{C}_{11} \mathrm{H}_{18} .\)

Balance each of the following equations. You may need to add \(\mathrm{H}_{2} \mathrm{O}\) to one side or the other of the equations. a. \(\stackrel{\oplus}{\mathrm{K}}^{\ominus} \mathrm{M} \mathrm{O}_{4}+\mathrm{RCH}=\mathrm{CH}_{2} \rightarrow \mathrm{RCO}_{2} \stackrel{\oplus}{\mathrm{K}}+\mathrm{CH}_{2}=\mathrm{O}+\mathrm{MnO}_{2}\) b. \(\mathrm{CrO}_{3}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{3} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{Cr}^{3 \oplus}\)

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