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Chapter 4: Problem 143

Five alcohols (A) \(-(\mathrm{E})\) are given below (A) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{2} \mathrm{OH}\) (B) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{C}_{3} \mathrm{H}_{7}\) (C) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{C}_{3} \mathrm{H}_{7}\) (D) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (E) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) The ease of dehydration in the decreasing order will be (a) \(\mathrm{D}>\mathrm{C}>\mathrm{E}>\mathrm{B}>\mathrm{A}\) (b) \(C>D>B>A>E\) (c) \(\mathrm{D}>\mathrm{C}>\mathrm{A}>\mathrm{B}>\mathrm{E}\) (d) \(\mathrm{D}>\mathrm{E}>\mathrm{B}>\mathrm{A}>\mathrm{C}\)

Short Answer

Expert verified
Question: Arrange the given alcohols in the decreasing order of ease of dehydration: (A) CH3(CH2)4CH2OH, (B) C2H5CH(OH)C3H7, (C) (CH3)2C(OH)C3H7, (D) (CH3)2C(OH)CH(CH3)2, and (E) C2H5CH(OH)CH(CH3)2. Answer: The correct order of ease of dehydration is (a): D>C>E>B>A.

Step by step solution

01

Find the structure of alcohol (A)

The structure of alcohol A is given by \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{2}\mathrm{OH}\)
02

Find the structure of alcohol (B)

The structure of alcohol B is given by \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{C}_{3}\mathrm{H}_{7}\)
03

Find the structure of alcohol (C)

The structure of alcohol C is given by \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{C}_{3}\mathrm{H}_{7}\)
04

Find the structure of alcohol (D)

The structure of alcohol D is given by \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH})\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\)
05

Find the structure of alcohol (E)

The structure of alcohol E is given by \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH})\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\)
06

Determine the carbocation stability for each alcohol

Based on the structures, we can find the carbocation formed during the dehydration as: (A): Primary carbocation (B): Secondary carbocation (C): Tertiary carbocation (D): Tertiary carbocation (with 3 alkyl groups) (E): Secondary carbocation (with 2 alkyl groups)
07

Order of carbocation stability

The order of carbocation stability in general is: Tertiary > Secondary > Primary. Thus, we can give the order as: (D) > (C) > (E) > (B) > (A)
08

Determine the correct order of ease of dehydration

Based on our analysis of carbocation stability, the ease of dehydration in decreasing order is: \(\mathrm{D}>\mathrm{C}>\mathrm{E}>\mathrm{B}>\mathrm{A}\) Therefore, the correct answer is (a): \(\mathrm{D}>\mathrm{C}>\mathrm{E}>\mathrm{B}>\mathrm{A}\).

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Most popular questions from this chapter

IUPAC name of allyl chloride is (a) 2 -chloropropene (b) 3 -chloropropene (c) 1 -chloropropene (d) 1 -(chloromethyl)ethane

Reaction of an alkyl halide (A) with alcoholic KOH forms a product which on ozonolysis gives a mixture of propanone and methanal. The structure of the alkyl halide (A) is (a) 2 -Bromobutane (b) 1,4 -Dibromobutane (c) 2 -Bromo-2-methyl propane (d) 2,3 -Dibromobutane

The compound with zero dipole moment is (a) 1,2 -dichlorobenzene (b) 1,3 -dichlorobenzene (c) 1,4 -dichlorobenzene (d) 1,4 -dihydroxybenzene

The correct decreasing order of reactivity of the halides for \(\mathrm{S}_{\mathrm{N}} 1\) reaction is (A) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\) (B) \(\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}\) (d) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{AgF} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~F}+\mathrm{AgCl}\). This reaction is called (a) Finkelstein reaction (b) Balz-Schiemann reaction (c) Swarts reaction (d) Gattermann reaction
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