Show how you would convert butan-1-ol to the following compounds using tosylate intermediates. You may use whatever additional reagents are needed.

(a)1-bromobutane (b)butan-1-amine, CH3CH2CH2CH2NH2

(c)butyl ethyl ether, CH3CH2CH2CH2OCH2CH3 (d)pentannitrile, CH3CH2CH2CH2CN

Short Answer

Expert verified

The hydroxyl group of a specific alcohol makes them polar. These groups can create hydrogen bonds with one another and with several other compounds.

Step by step solution

01

Alcohols

The hydroxyl group of a specific alcohol makes them polar. These groups can create hydrogen bonds with one another and with several other compounds.

02

Conversion of alcohols to other compounds

Alcohols can be transformed into tosylates using reagents like tosyl chloride and pyridine. The C-O bond of the particular alcohol is not broken during this transformation. This reaction happens with a retention of configuration.

03

Converting butan-1-ol to various compounds using tosylate intermediates

(a)The reaction of butan-1-ol with TsCl and pyridine leads to butyl tosylate. The butyl tosylate further reacts with NaBr to generate 1-bromobutane.The reaction can be given as:

Conversion of butan-1-ol to 1-bromobutane

(b) The reaction of butan-1-ol with TsCl and pyridine leads to butyl tosylate. The butyl tosylate further reacts with excess ammonia to generate butan-1-amine.The reaction can be given as:

Conversion of butan-1-ol to butan-1-amine

(c) The reaction of butan-1-ol with TsCl and pyridine leads to butyl tosylate. The butyl tosylate further reacts with NaOCH2CH3 generate butyl ethyl ether.The reaction can be given as:

Conversion of butan-1-ol to butyl ethyl ether

(d) The reaction of butan-1-ol with TsCl and pyridine leads to butyl tosylate. The butyl tosylate further reacts with KCN to generate pentannitrile. The reaction can be given as:

Conversion of butan-1-ol to pentannitrile

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Most popular questions from this chapter

Under normal circumstances, tertiary alcohols are not oxidized. However, when the tertiary alcohol is allylic, it can undergo a migration of the double bond (called an allylic shift) and subsequent oxidation of the alcohol. A particularly effective reagent for this reaction is Bobbitt’s reagent, similar to TEMPO used in many oxidations. (M. Shibuya et al., J. Org. Chem., 2008, 73, 4750.)

Show the expected product when each of these 3° allylic alcohols is oxidized by Bobbitt’s reagent.

(a)

(b)

(c)

(d)

Show how you would convert (S)-heptan-2-ol to

(a) (S)-2-chloroheptane.

(b) (R)-2-bromoheptane.

(c) (R)-heptan-2-ol.

Predict the products of the reactions of the following compounds with:

(1) chromic acid or excess sodium hypochlorite with acetic acid.

(2)PCC or NaOCl (1 equivalent) with TEMPO.

(a)cyclohexanol (b)1-methylcyclohexanol

(c)cyclopentylmethanol (d)cyclohexanone

(e)cyclohexane (f)1-phenylpropan-1-ol

(g)hexan-1-ol (h)acetaldehyde, CH3CHO

The compound shown below has three different types of OH groups, all with different acidities. Show the structure produced after this compound is treated with different amounts of NaH followed by a methylating reagent. Add a brief explanation.

(a)1 equivalent of NaH, followed by 1 equivalent of CH3l and heat

(b)2 equivalents of NaH, followed by 2 equivalents of CH3l and heat

(c) 3 equivalents of NaH, followed by 3 equivalents of CH3l and heat

Unknown Q is determined to have a molecular formula of C6H12O. Q is not optically active, and passing it through a chiral column does not separate it into enantiomers. Q does not react with Br2, nor with cold, dilute KMnO4 , nor does it take up H2under catalytic hydrogenation. Heating of Q with H2SO4 gives product R, of formula C6H10, which can be separated into enantiomers. Ozonolysis of a single enantiomer of R produces S, an acyclic, optically active ketoaldehyde of formula C6H10O2. Propose structures for compounds Q, R, and S, and show how your structures would react appropriately to give these results.

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